Instructor's Solutions Manual to Calculus: One and Several Variables [10 ed.] 9780470140628
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SECTION 1.2
1
CHAPTER 1 SECTION 1.2 1.
rational
2.
rational
3.
rational
4.
irrational
5.
rational
6.
irrational
7.
rational
8.
rational
9.
rational
10.
rational
11.
3 = 0.75 4
12.
0.33
1.414 22 7
√ 16
16.
π
−2
5x < −1
Ans:
Ans:
(−2, ∞)
x
0
[−1, ∞)
− 15
9. x2 − x − 6 ≥ 0 (x − 3)(x + 2) ≥ 0
(−5, −4)
11. 2x2 + x − 1 ≤ 0
Ans: 12.
(∞, −2] ∪ [3, ∞)
3x2 + 4x − 4 ≥ 0
(x − 5)(x + 1) > 0
(2x − 1)(x + 1) ≤ 0
(3x − 2)(x + 2) ≥ 0
Ans: (−∞, −1) ∪ (5, ∞)
Ans:
Ans:
13. x(x − 1)(x − 2) > 0
[−1, 1/2]
14. x(2x − 1)(3x − 5) ≤ 0
(−∞, −2] ∪ [2/3, ∞)
15. x3 − 2x2 + x ≥ 0 x(x − 1)2 ≥ 0
Ans:
(0, 1) ∪ (2, ∞)
Ans:
(−∞, 0] ∪ [ 12 , 53 ]
Ans:
[0, ∞)
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SECTION 1.3
16. x2 − 4x + 4 ≤ 0
17. x3 (x − 2)(x + 3)2 < 0
18. x2 (x − 3)(x + 4)2 > 0
(x − 2)2 ≤ 0 {2}
Ans:
Ans:
(0, 2)
19. x2 (x − 2)(x + 6) > 0 Ans:
Ans:
(3, ∞)
20. 7x(x − 4)2 < 0
(−∞, −6) ∪ (2, ∞)
Ans:
(−∞, 0)
21. (−2, 2)
22. (−∞, −1] ∪ [1, ∞)
23. (−∞, −3) ∪ (3, ∞)
24. (0, 2)
25. ( 32 , 52 )
26. (− 32 , 52 )
27. (−1, 0) ∪ (0, 1)
28. (− 12 , 0) ∪ (0, 12 )
29. ( 32 , 2) ∪ (2, 52 )
30. (− 32 , 12 ) ∪ ( 12 , 52 )
31. (−5, 3) ∪ (3, 11)
32. ( 23 , 83 )
33. (− 58 , − 38 )
7 34. ( 12 , 10 )
35. (−∞, −4) ∪ (−1, ∞)
36. (−∞, −2) ∪ ( 43 , ∞)
37. |x − 0| < 3 or |x| < 3
38. |x − 0| < 2 or |x| < 2
39. |x − 2| < 5
40. |x − 2| < 2 b + a b − a 42. x − < 2 2
41. |x − (−2)| < 5 or |x + 2| < 5
43. |x − 2| < 1
=⇒
|2x − 4| = 2|x − 2| < 2, so |2x − 4| < A true for A ≥ 2.
44. |x − 2| < A
=⇒
2|x − 2| = |2x − 4| < 2A provided that 0 < A ≤
45. |x + 1| < A
=⇒
|3x + 3| = 3|x + 1| < 3A provided that 0 < A ≤
46. |x + 1| < 2
=⇒
3|x + 1| = |3x + 3| < 6
=⇒
|2x − 4| < 3
=⇒
|3x + 3| < 4
3 2
4 3
=⇒
|3x + 3| < A
provided that A ≥ 6 47. (a) 48.
√ 1 1 < √ θ1 . Let m1 = tan θ1 , m2 = tan θ2 . The angle α between l1 and l2 is the smaller of θ2 − θ1 and 180◦ − [θ2 − θ1 ]. In the first case tan α = tan[θ2 − θ1 ] =
tan θ2 − tan θ1 m 2 − m1 = >0 1 + tan θ2 tan θ1 1 + m2 m1
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SECTION 1.6 m2 − m1 In the second case, tan α = tan[180◦ − (θ2 − θ1 )] = − tan(θ2 − θ1 ) = − fk (x) on (1, ∞)
SECTION 1.7 9 2
15 2
1.
(f + g)(2) = f (2) + g(2) = 3 +
3.
(f · g)(−2) = f (−2)g(−2) = 15 ·
5.
(2f − 3g)( 12 ) = 2f ( 12 ) − 3g( 12 ) = 2 · 0 − 3 ·
= 7 2
=
105 2 9 4
2.
(f − g)(−1) = 6
4.
f (1) = 0 g
= − 27 4
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SECTION 1.7 6. 7.
f + 2g f
(−1) = 1
(f ◦ g)(1) = f [g(1)] = f (2) = 3
9. (f + g)(x) = f (x) + g(x) = x − 1; (f − g)(x) = f (x) − g(x) = 3x − 5;
(g ◦ f )(1) = g(0), undefined.
8.
dom (f + g) = (−∞, ∞) dom (f − g) = (−∞, ∞)
(f · g)(x) = f (x)g(x) = −2x + 7x − 6; dom (f · g) = (−∞, ∞) 2x − 3 (f /g)(x) = ; dom (f /g) = {x : x = 2} 2−x 2
1 ; dom (f + g) = (−∞, 0) ∪ (0, ∞) x 1 (f − g)(x) = f (x) − g(x) = x2 − x − 1 − ; dom (f − g) = (−∞, 0) ∪ (0, ∞) x x4 − 1 (f · g)(x) = f (x)g(x) = ; dom (f · g) = (−∞, 0) ∪ (0, ∞) x 3 x −x ; dom (f /g) = (−∞, 0) ∪ (0, ∞) [g(0) is undefined.] (f /g)(x) = 2 x +1 √ √ 11. (f + g)(x) = x + x − 1 − x + 1; dom (f + g) = [1, ∞) √ √ (f − g)(x) = x − 1 + x + 1 − x; dom (f − g) = [1, ∞) √ √ √ √ (f · g)(x) = x − 1 x − x + 1 = x x − 1 − x2 − 1; dom (f · g) = [1, ∞) √ √ x−1 √ (f /g)(x) = ; dom (f /g) = {x : x ≥ 1 and x = 12 (1 + 5)} x− x+1 10. (f + g)(x) = f (x) + g(x) = x2 + x − 1 +
12. (f + g)(x) = sin2 x + cos 2x; dom (f + g) = (−∞, ∞) (f − g)(x) = sin2 x − cos 2x; (f · g)(x) = sin x cos 2x; 2
2
sin x ; cos 2x
dom (f − g) = (−∞, ∞)
dom (f · g) = (−∞, ∞)
2n + 1 π, n = 0, ±1, ±2, · · ·} 4 √ √ √ √ 13. (a) (6f + 3g)(x) = 6(x + 1/ x) + 3( x − 2/ x) = 6x + 3 x; x > 0 √ √ √ √ √ (b) (f − g)(x) = x + 1/ x − ( x − 2/ x) = x + 3/ x − x; x > 0 √ x x+1 ; x > 0, x = 2 (c) (f /g)(x) = x−2 (f /g)(x) =
dom (f /g) = {x : x =
14.
⎧ ⎪ ⎪ ⎨ 1 − x, x ≤ 1 (f + g)(x) =
2x − 1, 1 < x < 2 ⎪ ⎪ ⎩ 2x − 2, x ≥ 2
⎧ ⎪ ⎪ ⎨ 1 − x, x ≤ 1 (f − g)(x) = 2x − 1, 1 < x < 2 ⎪ ⎪ ⎩ 2x, x ≥ 2 (f · g)(x) =
0, x < 2 1 − 2x, x ≥ 2
21
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SECTION 1.7
15.
16.
17.
18.
19.
20.
21.
22.
23.
(f ◦ g)(x) = 2x2 + 5; dom (f ◦ g) = (−∞, ∞)
25.
(f ◦ g)(x) =
√
24.
x2 + 5; dom (f ◦ g) = (−∞, ∞) 26.
y = 0 on (0, c]
(f ◦ g)(x) = (2x + 5)2 ; dom (f ◦ g) = (−∞, ∞) (f ◦ g)(x) = x +
√
x; dom (f ◦ g) = [0, ∞)
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SECTION 1.7 27. (f ◦ g)(x) = 28. (f ◦ g)(x) = 29. (f ◦ g)(x) = 30. (f ◦ g)(x) =
23
x ; dom (f ◦ g) = {x : x = 0, 2} x−2 x2 √ √
1 ; dom (f ◦ g) = {x = ±1} −1
1 − cos2 2x = | sin 2x|; dom (f ◦ g) = (−∞, ∞) 1 − 2 cos x; dom (f ◦ g) = [0, π/3] ∪ [5π/3, 2π]
31. (f ◦ g ◦ h) = 4 [g(h(x))] = 4 [ h(x) − 1 ] = 4(x2 − 1);
dom (f ◦ g ◦ h) = (−∞, ∞)
32. (f ◦ g ◦ h)(x) = f (g(h(x))) = f (g(x2 )) = f (4x2 ) = 4x2 − 1; dom (f ◦ g ◦ h) = (−∞, ∞) 33. (f ◦ g ◦ h)
1 1 = = 2h(x) + 1 = 2x2 + 1; g(h(x)) 1/[2h(x) + 1]
34. (f ◦ g ◦ h)(x) = f (g(h(x))) = f (g(x )) = f 2
35. Take
f (x) =
1 x
since
1 2x2 + 1
=
1 + x4 = F (x) = f (g(x)) = f 1 + x2
dom (f ◦ g ◦ h) = (−∞, ∞)
1/(2x2 + 1) + 1 = 1 + (2x2 + 1) = 2x2 + 2 1/(2x2 + 1)
1 + x2 1 + x4
.
36. Take f (x) = ax + b since f (g(x)) = f (x2 ) = ax2 + b = F (x) 37. Take
f (x) = 2 sin x
38. Take f (x) = 39. Take
√
g(x) =
since
a2 − x,
since
2/3
1 1− 4 x
2 sin 3x = F (x) = f (g(x)) = f (3x). f (g(x)) = f (−x2 ) = since
40. Take g(x) = a2 x2 (x = 0), since a2 x2 + 41. Take
g(x) = 2x3 − 1 (or −(2x3 − 1))
1 1− 4 x
a2 − (−x2 ) =
√
a2 + x2 = F (x).
2 = F (x) = f (g(x)) = [ g(x)]3 .
1 1 . = F (x) = f (g(x)) = g(x) + a2 x2 g(x) since
(2x3 − 1)2 + 1 = F (x) = f (g(x)) = [ g(x)]2 + 1.
1 = F (x) = f (g(x)) = sin(g(x)). x √ 43. (f ◦ g)(x) = f (g(x)) = g(x) = x2 = |x|; √ (g ◦ f )(x) = g(f (x)) = [f (x)]2 = [ x ]2 = x, x ≥ 0
42. Take g(x) =
1 x
since sin
44. (f ◦ g)(x) = f (g(x)) = 3g(x) + 1 = 3x2 + 1, 45. (f ◦ g)(x) = f (g(x)) = 1 − sin2 x = cos2 x;
(g ◦ f )(x) = g(f (x)) = (f (x))2 = (3x + 1)2 (g ◦ f )(x) = g(f (x)) = sin f (x) = sin(1 − x2 ).
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SECTION 1.7
46. (f ◦ g)(x) = f (g(x)) = (x − 1) + 1 = x;
(g ◦ f )(x) = g(f (x)) =
3 (x3 + 1) − 1 = x.
47. (f + g)(x) = f (x) + g(x) = f (x) + c; quadg(x) = c. 48. (f ◦ g)(x) = f (g(x)) = g(x) + c implies f (x) = x + c. 49. (f g)(x) = f (x)g(x) = c f (x) implies g(x) = c. 50. (f ◦ g)(x) = f (g(x)) = c g(x) implies f (x) = cx. 51. (a) The graph of g is the graph of f shifted 3 units to the right. dom (g) = [3, a + 3], range (g) = [0, b]. (b) The graph of g is the graph of f shifted 4 units to the left and scaled vertically by a factor of 3. dom (g) = [−4, a − 4], range (g) = [0, 3b]. (c) The graph of g is the graph of f
scaled horizontally by a factor of 2. dom (g) = [0, a/2],
range (g) = [0, b]. (d) The graph of g is the graph of f
scaled horizontally by a factor of
range (g) = [0, b]. 52. even: (f g)(−x) = f (−x)g(−x) = (−f (x))(−g(x)) = f (x)g(x) = (f g)(x). 53.
f g is even since (f g)(−x) = f (−x)g(−x) = f (x)g(x) = (f g)(x).
54. odd: (f g)(−x) = f (−x)g(−x) = (f (x))(−g(x)) = −f (x)g(x) = −(f g)(x). 55. (a) If f is even, then
f (x) =
(b) If f is odd, then
f (x) =
56. (a) f (x) = x2 + x,
(b)
−x, −1 ≤ x < 0 1,
x < −1.
x,
−1 ≤ x < 0
−1,
x < −1.
f (x) = −x2 − x
57.
g(−x) = f (−x) + f [−(−x)] = f (−x) + f (x) = g(x)
58.
h(−x) = f (−x) − f [−(−x)] = f (−x) − f (x) = −[f (x) − f (−x)] = −h(x)
59. f (x) =
1 1 [f (x) + f (−x)] + [f (x) − f (−x)] 2 2 even
odd
1 2.
dom (g) = [0, 2a],
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SECTION 1.8
25
60.
61. (a) (f ◦ g)(x) =
5x2 + 16x − 16 (2 − x)2
62. (a) (g ◦ f )(x) =
3(x2 − 4) 6 − x2
(b) (g ◦ k)(x) = x
(b) (k ◦ g)(x) = x
(c) (f ◦ k ◦ g)(x) = x2 − 4
(c) (g ◦ f ◦ k)(x) = −
18(2x + 3) + 18x + 27
x2
63. (a) For fixed a, varying b varies the y-coordinate of the vertex of the parabola. (b) For fixed b, varying a varies the x-coordinate of the vertex of the parabola (c) The graph of −F is the reflection of the graph of F in the x-axis. 64. a =
1 , 4
b=−
49 16
65. (a) For c > 0, the graph of cf is the graph of f scaled vertically by the factor c; for c < 0, the graph of cf is the graph of f scaled vertically by the factor |c| and then reflected in the x-axis. (b) For c > 1, the graph of f (cx) is the graph of f compressed horizontally; for 0 < c < 1, the graph of f (cx) is the graph of f stretched horizontally; for −1 < c < 0, the graph of f (cx) is the graph of f stretched horizontally and reflected in the y-axis; for c < −1, the graph of f (cx) is the graph of f compressed horizontally and reflected in the y-axis. 66. (a) The graph of f (x − c) is the graph of f (x) shifted c units to the right if c > 0 and |c| units to the left if c < 0. (b) a changes the amplitude, b changes the period, c shifts the graph right or left |c/b| units.
SECTION 1.8 1. Let S be the set of integers for which the statement is true. Since 2(1) ≤ 21 , S contains 1. Assume now that k ∈ S. This tells us that 2k ≤ 2k , and thus 2(k + 1) = 2k + 2 ≤ 2k + 2 ≤ 2k + 2k = 2(2k ) = 2k+1 . ∧ (k ≥ 1) This places k + 1 in S.
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SECTION 1.8 We have shown that 1∈S
and that k ∈ S
implies
k + 1 ∈ S.
It follows that S contains all the positive integers. 2. Use
1 + 2(n + 1) = 1 + 2n + 2 ≤ 3n + 2 < 3n + 3n = 2 · 3n < 3n+1 .
3. Let S be the set of integers for which the statement is true. Since (1)(2) = 2 is divisible by 2, 1 ∈ S. Assume now that k ∈ S. This tells us that k(k + 1) is divisible by 2 and therefore (k + 1)(k + 2) = k(k + 1) + 2(k + 1) is also divisible by 2. This places k + 1 ∈ S. We have shown that 1 ∈ S and that k ∈ S implies
k + 1 ∈ S.
It follows that S contains all the positive integers. 4. Use 5. Use
1 + 3 + 5 + · · · + (2(n + 1) − 1) = n2 + 2n + 1 = (n + 1)2 12 + 22 + · · · + k 2 + (k + 1)2 = 16 k(k + 1)(2k + 1) + (k + 1)2 = 16 (k + 1)[k(2k + 1) + 6(k + 1)] = 16 (k + 1)(2k 2 + 7k + 6) = 16 (k + 1)(k + 2)(2k + 3) = 16 (k + 1)[(k + 1) + 1][2(k + 1) + 1].
6. Use 13 + 23 + · · · + n3 + (n + 1)3 = (1 + 2 + · · · + n)2 + (n + 1)3 2 n(n + 1) = + (n + 1)3 (by example 1) 2 n4 + 6n3 + 13n2 + 12n + 4 4 2 (n + 1)(n + 2) = 2
=
= [1 + 2 + · · · + n + (n + 1)]2 7. By Exercise 6 and Example 1
13 + 23 + · · · + (n − 1)3 = [ 12 (n − 1)n]2 = 14 (n − 1)2 n2 < 14 n4 and 13 + 23 + · · · + n3 = [ 12 n(n + 1)]2 = 14 n2 (n + 1)2 > 14 n4 .
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SECTION 1.8
27
8. By Exercise 5, 12 + 22 + · · · + (n − 1)2 = 16 (n − 1)n(2n − 1) < 13 n3 12 + 22 + · · · + n2 = 16 n(n + 1)(2n + 1) > 13 n3 9. Use 1 1 1 1 1 √ + √ + √ + ··· + √ + √ n n+1 1 2 3 √ > n+ √
10. Use
√ √ √ 1 n+1− n √ = n + 1. √ √ n+1+ n n+1− n
1 1 1 n 1 n(n + 2) + 1 n+1 1 + + + ··· + = + = = 1·2 2·3 3·4 (n + 1)(n + 2) n + 1 (n + 1)(n + 2) (n + 1)(n + 2) n+2
11. Let S be the set of integers for which the statement is true. Since 32(1)+1 + 21+2 = 27 + 8 = 35 is divisible by 7, we see that 1 ∈ S. Assume now that k ∈ S. This tells us that 32k+1 + 2k+2 is divisible by 7. It follows that 32(k+1)+1 + 2(k+1)+2 = 32 · 32k+1 + 2 · 2k+2 = 9 · 32k+1 + 2 · 2k+2 = 7 · 32k+1 + 2(32k+1 + 2k+2 ) is also divisible by 7. This places k + 1 ∈ S. We have shown that 1∈S
k∈S
and that
implies
k + 1 ∈ S.
It follows that S contains all the positive integers. 12. n ≥ 1 :
True for n = 1.
For the induction step, use
9n+1 − 8(n + 1) − 1 = 9 · 9n − 8n − 9 − 64n + 64n = 9(9n − 8n − 1) + 64n 13. For all positive integers n ≥ 2,
1 1− 2
1 1− 3
1 ··· 1 − n
=
1 . n
To see this, let S be the set of integers n for which the formula holds. Since 1 − now that k ∈ S. This tells us that 1−
1 2
1−
1 3
1 1 ··· 1 − = k k
1 2
= 12 ,
2 ∈ S. Suppose
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REVIEW EXERCISES and therefore that k 1 1 1 1 1 1 1 1 1− 1− ··· 1 − 1− = 1− = = . 2 3 k k+1 k k+1 k k+1 k+1 This places k + 1 ∈ S and verifies the formula for n ≥ 2.
14. The product is
n+1 ; 2n
use
n+1 2n
1−
1 (n + 1)2
n+1 = 2n
n2 + 2n (n + 1)2
=
n+2 2(n + 1)
15. From the figure, observe that adding a vertex VN +1 to an N -sided polygon increases the number of diagonals by (N − 2) + 1 = N − 1. Then use the identity 1 2 N (N
− 3) + (N − 1) = 12 (N + 1)(N + 1 − 3).
16. From the figure for Exercise 15, observe that adding a vertex (VN +1 ) to an N -sided polygon increases the angle sum by 180◦ . 17. To go from k to k + 1, take A = {a1 , . . . , ak+1 } and B = {a1 , . . . , ak } . Assume that B has 2k subsets: B1 , B2 , . . . B2k . The subsets of A are then B1 , B2 , . . . , B2k together with B1 ∪ {ak+1 } , B2 ∪ {ak+1 } , . . . , B2k ∪ {ak+1 } . This gives 2(2k ) = 2k+1 subsets for A. √ 18. Assuming that we can construct a line segment of length k, construct a right triangle with side √ √ lengths 1 and k. Then the hypotenuse is a line segment of length k + 1. 19. n = 41
CHAPTER 1. REVIEW EXERCISES 1. rational
2. rational
3. irrational
4. rational
5. bounded below by 1
6. bounded above by 1
7. bounded; lower bound −5, upper bound 1 9. 2x2 + x − 1 = (2x − 1)(x + 1); x = 12 , −1 10. no real roots
8. bounded; lower bound −1, upper bound
1 4
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REVIEW EXERCISES
29
11. x2 − 10x + 25 = (x − 5)2 ; x = 5 12. 9x3 − x = x(3x + 1)(3x − 1); x = 0,
1 3,
− 13
13. 5x − 2 < 0
14. 3x + 5 < 12 (4 − x) 3x + 5 < 2 −
5x < 2 x
0 (x + 2)(x − 2)
18.
Ans: (−2, −1) ∪ (2, ∞) 19. |x − 2| < 1
x2 − 4x + 4 ≤0 x2 − 2x − 3 (x − 2)2 ≤0 (x − 3)(x + 1) Ans: (−1, 3)
20. |3x − 2| ≥ 4
−1 < x − 2 < 1
3x − 2 ≥ 4 or 3x − 2 ≤ −4
Ans: (1, 3)
Ans: (−∞, − 23 ) ∪ [2, ∞)
2 >2 x+4 2 2 > 2 or < −2 x+4 x+4 2 If >2 x+4 x + 4 > 0 and 2 > 2x + 8
5 −1 x+1 x < −6
21.
22.
−4 < x < −3 2 If < −2 x+4 x + 4 < 0 and 2 > −2x − 8 −5 < x < −4
Ans: (−∞, −6) ∪ (4, ∞)
Ans: (−5, −4) ∪ (−4, −3) 23. d(P, Q) =
24. d(P, Q) = 25. x = 2
(1 −
2)2
+ (4 −
(−3))2
√ = 5 2;
midpoint:
2+1 4−3 , 2 2
√ (−1 − (−3)2 + (6 − (−4))2 = 2 26; midpoint:
=
3 , 2
−3 − 1 −4 + 6 , 2 2
1 2
= (−2, 1)
26. y = −3
27. The line l : 2x − 3y = 6 has slope m = 2/3. Therefore, an equation for the line through (2, −3) perpendicular to l is: y + 3 = − 32 (x − 2) or 3x + 2y = 0
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REVIEW EXERCISES
28. The line l : 3x + 4y = 12 has slope m = −3/4. Therefore, an equation for the line through (2, −3) parallel to l is: y + 3 = − 34 (x − 2) or 3x + 4y = −6 29. x − 2y = −4 3x + 4y = 3
⇒ 5x = −5
⇒ x = −1;
(−1,
3 2 ).
30. 4x − y = −2 3x + 2y = 0
⇒ 11x = −4
⇒ x=
−4 11 ;
6 ( −4 11 , 11 ).
31. Solve the equations simultaneously: 2x2 = 8x − 6 2x2 − 8x + 6 = 0 2(x − 1)(x − 3) = 0 the line and the parabola intersect at (1, 2) and (3, 18). 32. The line tangent to the circle at the point (2, 1) is perpendicular to the radius at that point. The center of the circle is at (−1, 3). The slope of the line through (−1, 3) and (2, 1) is −2/3. Therefore an equation for the line tangent to the circle at (2, 1) is y − 1 = 32 (x − 2) or
3x − 2y = 4.
33. domain: (−∞, ∞); range: (−∞, 4]
34. domain: (−∞, ∞); range: (−∞, ∞)
35. domain: [4, ∞); range: [0, ∞)
36. domain: [− 12 , 12 ]; range; [0, 12 ]
37. domain: (−∞, ∞); range: [1, ∞)
38. domain: (−∞, ∞); range: [0, ∞)
39. domain: (−∞, ∞); range: [0, ∞)
40. domain: (−∞, ∞); range: (−∞, ∞) y
y 5
4
4 3
3
2 1 2 1
2
3
4
x 1
-2
-1
1 -1
2
x
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REVIEW EXERCISES 41. x = 76 π, 43. x =
11 6 π
42. x = 13 π,
3π 2
44. x = 0,
45.
31
2 4π 5π 3 π, 3 , 3
1 2 3 π, 3 π,
π,
4 5 3 π, 3 π,
46.
2π
y
y 1 1 -6
-3
3
6
-6
x
-3
3
6
-1
-1
47.
48. y 3
y 1
1 -6
-6
-3
3
6
-3
3 -1
x
-1
-3
49. (f + g)(x) = (3x + 2) + (x2 − 1) = x2 + 3x + 1,
dom (f + g) = (−∞, ∞).
(f − g)(x) = (3x + 2) − (x − 1) = 3 + 3x − x ,
dom (f − g) = (−∞, ∞).
2
2
(f · g)(x) = (3x + 2)(x − 1) = 3x + 2x − 3x − 2, dom (f · g) = (−∞, ∞). f 3x + 2 (x) = 2 , dom (f /g) = (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). g x −1 2
3
2
50. (f + g)(x) = x2 − 4 + x + 1/x = x2 + x + 1/x − 4, (f − g)(x) = x − x − 1/x − 4, 2
dom (f + g) = (−∞, 0) ∪ (0, ∞).
dom (f − g) = (−∞, 0) ∪ (0, ∞).
(f · g)(x) = (x − 4)(x + 1/x) = x3 − 3x − 4/x, dom (f · g) = (−∞, 0) ∪ (0, ∞). f x2 − 4 x3 − 4x (x) = = 2 , dom (f /g) = (−∞, 0) ∪ (0, ∞). g x + 1/x x +1 2
51. (f + g)(x) = cos2 x + sin 2x, dom (f + g) = [0, 2π]. (f − g)(x) = cos2 x − sin 2x, dom (f − g) = [0, 2π]. (f · g)(x) = cos2 x(sin 2x) = 2 cos3 x sin x, dom (f · g) = [0, 2π]. f cos2 x (x) = = 12 cot x, dom(f /g) : x ∈ (0, 2π), x = 12 π, π, 32 π. g sin 2x 52. (f ◦ g)(x) = (x + 1)2 − 2(x + 1) = x2 − 1, dom (f ◦ g) = (−∞, ∞). (g ◦ f )(x) = x2 − 2x + 1 = (x − 1)2 , dom (f ◦ g) = (−∞, ∞).
√ (x2 − 5) + 1 = x2 − 4, dom (f ◦ g) = (−∞, −2] ∪ [2, ∞). √ 2 (g ◦ f )(x) = x + 1 − 5 = x − 4, dom (g ◦ f ) = [−1, ∞).
53. (f ◦ g)(x) =
54. (f ◦ g)(x) =
1 − sin2 2x = | cos 2x|, dom (f ◦ g) = (−∞, ∞). √ (g ◦ f )(x) = sin 2 1 − x2 , dom (g ◦ f ) = [−1, 1].
6
x
x
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REVIEW EXERCISES
55. (a) y = kx. (b) If b = ka, then αb = αka = k(αb). Hence, Q is a point on l. (c) If α > 0, P, Q are on the same side of the origin; if α < 0, P, Q are on opposite sides of the origin. 56. (a) Set α = − a2 . Then a = −2α and the quadratic equation can be written as x2 − 2αx + b = 0
or
(x − α)2 − (α2 − b) = 0.
if α2 − b > 0, set β 2 = (α2 − b) and we have (x − α)2 − β 2 = 0; if α2 − b = 0, we have (x − α)2 = 0; if α2 − b < 0, set −β 2 = (α2 − b) and we have (x − α)2 + β 2 = 0. (b) x = α + β or x = α − β. (c) x = α. (d) (x − α)2 + β 2 > 0 for all x. 57. Since |a| = |a − b + b| ≤ |a − b| + |b| by the given inequality, we have |a| − |b| ≤ |a − b|. 58. (a) P = 12 πD (b) A = 18 πD2
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SECTION 2.1 CHAPTER 2 SECTION 2.1 1. (a) 2
(b) −1
(c) does not exist
(d) −3
2. (a) −4
(b) −4
(c) −4
(d) 2
3. (a) does not exist
(b) −3
(c) does not exist
(d) −3
4. (a) 1
(b) does not exist
(c) does not exist
(d) 1
5. (a) does not exist
(b) does not exist
(c) does not exist
(d) 1
6. (a) 1
(b) −2
(c) does not exist
(d) −2
7. (a) 2
(b) 2
(c) 2
(d) −1
8. (a) 2
(b) 2
(c) 2
(d) 2
9. (a) 0
(b) 0
(c) 0
(d) 0
(b) does not exist
(c) does not exist
(d) 1
10. (a) does not exist 11.
c = 0, 6
12.
c=3
13.
−1
14.
−3
15.
12
16.
5
17.
1
18.
does not exist
19.
3 2
20.
does not exist
21.
does not exist
22.
does not exist
23. 24. 25.
26. 27. 28. 30.
lim
x→3
2x − 6 = lim 2 = 2 x→3 x−3
x2 − 6x + 9 (x − 3)2 = lim = lim (x − 3) = 0 x→3 x→3 x − 3 x→3 x−3 lim
lim
x→3
x−3 x−3 1 = lim ; does not exist = lim x→3 x − 3 x2 − 6x + 9 x→3 (x − 3)2
x2 − 3x + 2 (x − 1)(x − 2) = lim = lim (x − 1) = 1 x→2 x→2 x→2 x−2 x−2 lim
lim
x→2
x−2 x−2 1 = lim = lim =1 x2 − 3x + 2 x→2 (x − 1)(x − 2) x→2 x − 1
does not exist 1 lim x + =2 x→1 x
29.
does not exist
31.
2x − 5x2 = lim (2 − 5x) = 2 x→0 x→0 x lim
33
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SECTION 2.1
x−3 x−3 1 1 = lim = lim − =− 32. lim x→3 6 − 2x x→3 2(3 − x) x→3 2 2 33.
x2 − 1 (x − 1)(x + 1) = lim = lim (x + 1) = 2 x→1 x − 1 x→1 x→1 x−1
34.
x3 − 1 (x − 1)(x2 + x + 1) = lim = lim x2 + x + 1 = 3 x→1 x − 1 x→1 x→1 (x − 1)
35.
0
36.
does not exist
37.
1
38.
3
39.
16
40.
0
41.
does not exist
42.
2
43.
4
44.
0
45.
does not exist
46.
2
lim
lim
47.
√ lim
x→1
x2 + 1 − x−1
√
2
√ √ √ √ ( x2 + 1 − 2 )( x2 + 1 + 2 ) √ √ x→1 (x − 1)( x2 + 1 + 2 )
= lim
x2 − 1 x+1 1 2 √ √ = lim √ √ = √ =√ x→1 (x − 1)( x2 + 1 + 2 2 2 2 ) x→1 x2 + 1 + 2
= lim 48.
√ lim
x→5
x2 + 5 − x−5
√
30
√ √ √ √ ( x2 + 5 − 30 )( x2 + 5 + 30 ) √ √ x→5 (x − 5)( x2 + 5 + 30 )
= lim
x2 − 25 x+5 5 10 √ √ √ = √ =√ = lim √ 2 x→5 (x − 5)( x2 + 5 + x→5 2 30 30 30 ) x + 5 + 30
= lim 49. lim √
x→1
√ x2 − 1 x2 − 1 2x + 2 + 2 √ = lim √ 2x + 2 − 2 x→1 2x + 2 − 2 2x + 2 + 2 √ √ (x − 1)(x + 1) 2x + 2 + 2 (x − 1)(x + 1) 2x + 2 + 2 = lim = lim x→1 x→1 2x + 2 − 4 2(x − 1) √ (x + 1) 2x + 2 + 2 2·4 = =4 = lim x→1 2 2
50.
0
51.
2
52.
0.167
53.
3 2
54.
1 12
55.
(i) 5
(ii) does not exist
56.
(i) 0 (ii)
57.
(i) −
58.
(i) −
17 32
59.
c = −1
60.
c=
61.
c = −2
2 3
5 2
5 4
(ii) 0
(ii) −
1 2
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SECTION 2.2 62.
c=3
63.
lim f (x) = 1;
64.
x→0
lim g(x) = 0
lim f (x) = 1;
x→0
lim g(x) = 0
x→0
x→0
SECTION 2.2 2.
x2 (1 + x) = lim x→0 x→0 2x
4 3
5.
x4 − 1 = lim (x3 + x2 + x + 1) = 4 x→1 x − 1 x→1
6.
does not exist
7.
does not exist
8.
x2 − 1 (x − 1)(x + 1) x+1 = lim ; = lim x→1 x2 − 2x + 1 x→1 x→1 x − 1 (x − 1)2
9.
−1
10.
does not exist
11.
does not exist
12.
−1
13.
0
14.
0
1.
1 2
3.
x(1 + x) 1+x ; = lim 2 x→0 x→0 2x 2x
4.
15.
lim
1 2
x(1 + x) = 0
does not exist
lim
lim f (x) = lim (x2 − x) = 2
x→2+
lim
lim
does not exist
16.
x→2+
lim f (x) = lim− 1 = 1
x→1−
x→1
17.
1
18.
9
19.
1
20.
10
21.
δ1
22.
2 and 3
23.
δ = 12 = .05
24.
δ=
25.
2 = .02
26.
δ = 5 = 0.5
27.
δ = 1.75
28.
δ = .22474
29. for = 0.5 take δ = 0.24;
for = 0.25 take δ = 0.1
30. for = 0.5 take δ = 0.06;
for = 0.25 take δ = 0.03
31. for = 0.5 take δ = 0.75;
for = 0.25 take δ = 0.43
32. for = 0.5 take δ = 0.125;
for = 0.1 take δ = 0.028
33. for = 0.25 take δ = 0.23;
for = 0.1 take δ = 0.14
34. for = 0.5 take δ = 0.25;
for = 0.1 take δ = 0.05
35. Since |(2x − 5) − 3| = |2x − 8| = 2|x − 4|, we can take δ = 12 : if
0 < |x − 4| < 12
then, |(2x − 5) − 3| = 2|x − 4| < .
1 5
= 0.1
35
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SECTION 2.2
36. Since |(3x − 1) − 5| = |3x − 6| = 3|x − 2|, we can take δ = 13 : if
0 < |x − 2| < 13
then, |(3x − 1) − 5| = 3|x − 2| < .
37. Since |(6x − 7) − 11| = |6x − 18| = 6|x − 3|, we can take δ = 16 : if
0 < |x − 3| < 16
then |(6x − 7) − 11| = 6|x − 3| < .
38. Since |(2 − 5x) − 2| = | − 5x| = 5|x|, we can take δ = 15 : if
0 < |x| < 15
then, |(2 − 5x) − 2| = 5|x| < .
39. Since |1 − 3x| − 5 = |3x − 1| − 5 ≤ |3x − 6| = 3|x − 2|, we can take δ = 13 : if 40. Take δ = :
if
0 < |x − 2| < 13
0 < |x − 2| < then
then
|1 − 3x| − 5 ≤ 3|x − 2| < .
||x − 2| − 0| = |x − 2| <
41. Statements (b), (e), (g), and (i) are necessarily true. 42. Suppose A = B and take = 43.
44.
1 2
|A − B|. Then |A − B| < =
1 1 = x−1 2 1 1 (iii) lim − =0 x→3 x − 1 2 (i) lim
x→3
x 1 = x2 + 2 3 x 1 − =0 (iii) lim x→1 x2 + 2 3 (i) lim
x→1
45. By (2.2.6) parts (i) and (iv) with L = 0
1 2
|A − B| which is impossible.
1 1 = (3 + h) − 1 2 1 1 (iv) lim − = 0 x→3 x − 1 2
(ii) lim
h→0
1+h 1 lim = (1 + h)2 + 2 3 x 1 (iv) lim 2 − = 0 x→1 x + 2 3 (ii)
h→0
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SECTION 2.2 46. (a) Suppose lim f (x) = L, and let > 0. Then there exists δ > 0 such that x→c
if
0 < |x − c| < δ,
then
|f (x) − L| < .
But then we also have if and therefore, (b)
0 < |x − c| < δ,
then
||f (x)| − |L|| ≤ |f (x) − L| < ,
lim |f (x)| = |L|
x→c
(i) Set f (x) = −2 for all x, and let c = 0. Then lim |f (x)| = 2 = |2| but
lim f (x) = −2 = 2.
x→0
(ii) Set
f (x) =
lim |f (x)| = 1
x→0
x→0
−1
if x < 0
1
if x > 0,
but
c = 0. Then
lim f (x)
x→0
does not exist.
47. Let > 0. If lim f (x) = L,
x→c
then there must exist δ > 0 such that (∗)
if
0 < |x − c| < δ
then
|f (x) − L | < .
Suppose now that 0 < |h| < δ. Then 0 < |(c + h) − c| < δ and thus by (∗) |f (c + h) − L | < . This proves that lim f (x) = L
if
x→c
then
lim f (c + h) = L.
h→0
If, on the other hand, lim f (c + h) = L,
h→0
then there must exist δ > 0 such that (∗∗)
if
0 < |h| < δ
then
|f (c + h) − L | < .
Suppose now that 0 < |x − c| < δ.
37
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SECTION 2.2 Then by (∗∗) |f (c + (x − c)) − L | < . More simply stated, |f (x) − L | < . This proves that if
lim f (c + h) = L
h→0
then
lim f (x) = L.
x→c
48. See comment after (2.2.9). 49. (a)
√ Set δ = c. By the hint, if
(b)
√ 0 < |x − c| < c
Set δ = 2 . If 0 < x < 2 ,
then
√ √ 1 then | x − c | < √ |x − c| < . c √ √ | x − 0| = x < .
50. Take δ = minimum of 1 and /5. If 0 < |x − 2| < δ, Therefore x2 − 4 = |x − 2| |x + 2| < (/5)(5) = .
then
|x − 2| < /5,
|x + 2| < 5 and
|x3 − 1| = |x2 + x + 1| |x − 1| < 7|x − 1| < 7(/7) = . 51. Take δ = minimum of 1 and /7.
If 0 < |x − 1| < δ,
then
0 12 . From the second, we conclude that L < 12 . Clearly no such number L exists. 56. We begin by assuming that lim− f (x) = L and showing that x→c
lim f (c − |h|) = L.
h→0
Let > 0. Since lim− f (x) = L, there exists δ > 0 such that x→c
if c − δ < x < c then
(∗)
|f (x) − L | < .
Suppose now that 0 < |h| < δ. Then c − δ < c − |h| < c and, by (∗), |f (c − |h|) − L | < . Thus
lim f (c − |h|) = L.
h→0
Conversely we now assume that lim f (c − |h|) = L. Then for > 0 there exists δ > 0 such that h→0
(∗∗)
if
0 < |h| < δ
then
|f (c − |h|) − L | < .
Suppose now that c − δ < x < c. Then 0 < c − x < δ so that, by (∗∗), |f (c − (c − x)) − L | = |f (x) − L | < . Thus
lim f (x) = L.
x→c−
57. We begin by assuming that lim f (x) = L and showing that + x→c
lim f (c + |h|) = L.
h→0
Let > 0. Since lim f (x) = L, there exists δ > 0 such that + x→c
(∗)
if c < x < c + δ
then
|f (x) − L | < .
Suppose now that 0 < |h| < δ. Then c < c + |h| < c + δ and, by (∗), |f (c + |h|) − L | < . Thus
lim f (c + |h|) = L.
h→0
Conversely we now assume that lim f (c + |h|) = L. Then for > 0 there exists δ > 0 such that h→0
(∗∗)
if
0 < |h| < δ
then
|f (c + |h|) − L | < .
Suppose now that c < x < c + δ. Then 0 < x − c < δ so that, by (∗∗), |f (c + (x − c)) − L | = |f (x) − L | < . Thus
lim f (x) = L.
x→c+
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SECTION 2.2
58. Suppose that lim f (x) = L and let > 0. Then there exists δ > 0 such that x→c
0 < |x − c| < δ,
if
|f (x) − L| <
then
=⇒
lim [f (x) − L] = 0.
x→c
Now suppose that lim [f (x) − L] = 0 and let > 0. Then there exists δ > 0 such that x→c
if
0 < |x − c| < δ,
|f (x) − L − 0| <
then
=⇒
lim |f (x) − L| <
x→c
which implies lim f (x) = L x→c
59. (a) Let = L. Since lim f (x) = L, there exists δ > 0 such that if 0 < |x − c| < δ then x→c
L − f (x) ≤ |L − f (x)| = |f (x) − L| < L Therefore, f (x) > L − L = 0 for all x ∈ (c − δ, c + δ); take γ = δ. (b) Let = −L and repeat the argument in part (a). 60. Counterexample:
Set
f (x) =
Then
f (−1) = 1 > 0
and
x if x = −1 if x = −1.
1
lim f (x) = lim x = −1. By Exercise 51 (b), f (x) < 0 for all x = −1
x→−1
x→−1
in an interval of the form (−1 − γ, −1 + γ), γ > 0. 61. (a) Let lim f (x) = L and x→c
lim g(x) = M,
x→c
and let > 0. There exist positive numbers δ1 and δ2
such that |f (x) − L| < /2
if
0 < |x − c| < δ1
|g(x) − M | < /2
if
0 < |x − c| < δ2
and
Let δ = min (δ1 , δ2 ). Then M − L = M − g(x) + g(x) − f (x) + f (x) − L ≥ [M − g(x)] + [f (x) − L] ≥ − /2 − /2 = − for all x such that 0 < |x − c| < δ. Since is arbitrary, it follows that M ≥ L. (b) No. For example, if f (x) = x2 and g(x) = |x| on (−1, 1), then f (x) < g(x) on (−1, 1) except at x = 0, but lim x2 = lim |x| = 0. x→0
x→0
62. Let = 1. Since lim f (x) = L, there exists a δ > 0 such that if 0 < |x − c| < δ, then |f (x) − L| < x→c
= 1. Thus, −1 < f (x) − L < 1
or
L − 1 < f (x) < L + 1
Take B = max {|L − 1|, |L + 1| }. Then |f (x)| < B for all x such that 0 < |x − c| < δ.
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SECTION 2.3
41
SECTION 2.3 1. (a)
3
(b)
2. (a)
5
(b) −8
3.
lim
x→4
1 1 − x 4
since lim
x→4
4
1 x−4
(c) −2 (c)
(d)
0
does not exist
= lim
x→4
4−x 4x
1 does not exist. x−4
1 x−4
(e)
does not exist
(d)
0
= lim
x→4
(e)
(f)
4/5
−1 1 =− ; 4x 16
1 3
(f)
9
Theorem 2.3.2 does not apply
4. Theorem 2.3.10 does not apply since lim (x2 + x − 12) = 0. x→3
x + x − 12 (x + 4)(x − 3) = lim = lim (x + 4) = 7 x→3 x→3 x−3 x−3 2
lim
x→3
−3
5.
3
6.
49
7.
8.
9
9.
5
10.
2
− 23 20
13.
−1
11.
does not exist
14.
0
15.
16.
1 = lim (h − 1) = −1 lim h 1 − h→0 h→0 h
17.
18.
lim
x→2
x−2 1 = lim = x2 − 4 x→2 x + 2
12.
1 4
19.
does not exist 1 = lim (h + 1) = 1 lim h 1 + h→0 h→0 h
lim
x→2
x2 − 4 x+2 = lim =4 x→2 x−2 1
(x2 − x − 6)2 (x − 3)2 (x + 2)2 = lim = lim (x + 3)2 (x + 2) = 0 x→−2 x→−2 x→−2 x+2 x+2 √ √ √ 1 x−2 x−2 x+2 x−4 √ = lim ·√ 21. lim = lim = x→4 x − 4 x→4 x − 4 4 x + 2 x→4 (x − 4)( x + 2)
20.
lim
√ √ √ x−1 ( x − 1)( x + 1) √ 22. lim √ = lim = lim ( x + 1) = 2 x→1 x→1 x − 1 x→1 x−1 23.
lim
x→1
x2 − x − 6 (x + 2)(x − 3) x−3 2 = lim = lim =− 2 2 x→1 x→1 (x + 2) (x + 2) x+2 3
24.
(x2 − x − 6) (x − 3)(x + 2) x−3 ; = lim = lim x→−2 x→−2 x→−2 x + 2 (x + 2)2 (x + 2)2
25.
lim
1 − 1/h2 h2 − 1 (h + 1)(h − 1) h+1 = lim 2 = lim = lim ; h→0 h − h h→0 h→0 1 − 1/h h(h − 1) h
lim
1 − 1/h2 h2 − 1 = −1 = lim h→0 h2 + 1 1 + 1/h2
26.
lim
h→0
h→0
does not exist
does not exist
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28.
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SECTION 2.3 lim
1 − 1/h h−1 = lim = −1 1 + 1/h h→0 h + 1
lim
1 + 1/h h2 + h =0 = lim h→0 h2 + 1 1 + 1/h2
h→0
h→0
t2 + 6t + 5 (t + 1)(t + 5) t+5 = lim = lim =4 2 t + 3t + 2 t→−1 (t + 1)(t + 2) t→−1 t + 2
lim
t→−1
√ 30.
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x2 − 4 = lim x−2 x→2+
lim
x→2+
√
√ √ x+2 x−2 x+2 ; = lim √ x−2 x→2+ x−2
does not exist
31. lim
t + a/t t2 + a a = lim 2 = t→0 t + b/t t +b b
32.
lim
x2 − 1 (x − 1)(x + 1) x+1 2 = lim = lim 2 = 3 2 x − 1 x→1 (x − 1)(x + x + 1) x→1 x + x + 1 3
lim
x5 − 1 (x − 1)(x4 + x3 + x2 + x + 1) x4 + x3 + x2 + x + 1 5 = lim = lim = 4 3 2 x→1 x − 1 x→1 (x − 1)(x + x + x + 1) x3 + x2 + x + 1 4
t→0
33.
x→1
x→1
34.
1 1+ h
lim h
2
h→0
1 35. lim h 1 + 2 h→0 h 36.
lim
x→−4
37.
lim
x→−4
38.
lim
x→−4
= lim (h2 + h) = 0 h→0
= lim
h→0
3x 8 + x+4 x+4 2x 8 + x+4 x+4 2x 8 − x+4 x+4
h2 + 1 ; h
does not exist
= lim
3x + 8 ; x+4
= lim
2x + 8 = lim 2 = 2 x→−4 x+4
= lim
2x − 8 ; x+4
x→−4
x→−4
x→−4
does not exist
does not exist
39. (a) lim
x→4
(b) lim
x→4
(c) lim
x→4
(d) lim
x→4
40. (a) lim
x→3
(b) lim
x→3
1 1 4−x − = lim =0 x→4 x 4 4x 1 1 1 1 4−x 1 1 − = lim = lim − =− x→4 x→4 x 4 x−4 4x x−4 4x 16 1 1 (4 − x)(x − 2) − (x − 2) = lim =0 x→4 x 4 4x
2 1 1 1 4−x 1 − ; does not exist = lim = lim x→4 4x(x − 4)2 x→4 4x(4 − x) x 4 x−4 x2 + x + 12 ; x−3
does not exist
x2 + x − 12 (x + 4)(x − 3) = lim = lim (x + 4) = 7 x→3 x→3 x−3 x−3
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SECTION 2.3 (c) lim
(x2 + x − 12)2 (x + 4)2 (x − 3)2 = lim = lim (x + 4)2 (x − 3) = 0 x→3 x→3 x−3 x−3
(d) lim
x2 + x − 12 (x + 4)(x − 3) x+4 ; = lim = lim 2 2 x→3 x→3 x − 3 (x − 3) (x − 3)
41. (a) lim
f (x) − f (4) (x2 − 4x) − (0) = lim = lim x = 4 x→4 x→4 x−4 x−4
x→3
x→3
x→4
does not exist
(b) lim
f (x) − f (1) x2 − 4x + 3 (x − 1)(x − 3) = lim = lim = lim (x − 3) = −2 x→1 x→1 x→1 x−1 x−1 x−1
(c) lim
f (x) − f (1) x2 − 4x + 3 (x − 1)(x − 3) = lim = lim = lim (x − 1) = 2 x→3 x→3 x→3 x−3 x−3 x−3
(d) lim
f (x) − f (2) x2 − 4x + 4 = lim ; does not exist x→3 x−3 x−3
x→1
x→3
x→3
42. (a) lim
x→3
f (x) − f (3) x3 − 27 = lim = lim (x2 + 3x + 9) = 27 x→3 x − 3 x→3 x−3
(b) lim
f (x) − f (2) x3 − 8 (x − 2)(x2 + 2x + 4) = lim = lim ; x→3 x − 3 x→3 x−3 x−3
(c) lim
f (x) − f (3) x3 − 27 = lim =0 x→3 x − 2 x−2
(d) lim
f (x) − f (1) x3 − 1 = lim = lim (x2 + x + 1) = 3 x→1 x − 1 x→1 x−1
x→3
x→3
x→1
43
does not exist
43. f (x) = 1/x, g(x) = −1/x with c = 0 44. Set, for instance,
f (x) =
0, x < c
g(x) =
1, x > c,
1,
xc
45. True. Let lim [f (x) + g(x)] = L. If lim g(x) = M exists, then lim f (x) = lim [f (x) + g(x) − g(x)] = x→c
x→c
x→c
L − M also exists. This contradicts the fact that lim f (x) does not exist.
x→c
x→c
46. False, because lim g(x) = lim [f (x) + g(x) − f (x)] = lim [f (x) + g(x)] − lim f (x). exists. x→c
47. True. If lim
x→c
x→c
f (x) = L exists, then
x→c
lim
x→c
f (x)
x→c
f (x) = L2 also exists.
48. False. Set f (x) = −1 for all x, c = 0. 49. False;
for example set f (x) = x and
c=0
50. False; for example, neither limit need exist: set f (x) = 51. False;
1 (x − 3)2
for example, set f (x) = 1 − x2 , g(x) = 1 + x2 , and c = 0.
and g(x) =
2 . (x − 3)2
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SECTION 2.3
52. (a) If f (x) ≥ g(x) then |f (x) − g(x)| = f (x) − g(x) and 1 2
{[f (x) + g(x)] + |f (x) − g(x)|} =
1 2
{f (x) + g(x) + f (x) − g(x)}
=
1 2
· 2f (x) = f (x) = max {f (x), g(x)}.
If f (x) ≤ g(x) then |f (x) − g(x)| = −[f (x) − g(x)] = g(x) − f (x) and 1 2
{[f (x) + g(x)] + |f (x) − g(x)|} =
1 2
{f (x) + g(x) + g(x) − f (x)}
=
1 2
· 2g(x) = g(x) = max {f (x), g(x)}.
(b) min {f (x), g(x)} = 53. If
lim f (x) = L
x→c
1 2
{[f (x) + g(x)] − |f (x) − g(x)|}
and
lim g(x) = L,
1 {[f (x) x→c 2
lim h(x) = lim
x→c
then
x→c
1 [f (x) x→c 2
= lim
+ g(x)] − |f (x) − g(x)|} | (x) x→c f
+ g(x)] − lim
− g(x)|
= 12 (L + L) − 12 (L − L) = L. A similar argument works for H. 54. (a) Let > 0. Since lim f (x) = L, there exists δ1 > 0 such that x→c
if
0 < |x − c| < δ1
then
(∗)
|f (x) − L| <
Let δ2 = min {|xi − c| : 1 ≤ i ≤ n, xi = c}. Then (∗∗)
f (x) = g(x)
Now, choose δ = min {δ1 , δ2 }. Then, if
if
0 < |x − c| < δ
|f (x) − L| < f (x) = g(x)
0 < |x − c| < δ2
by (*), and by (**).
Therefore, if 0 < |x − c| < δ, then |g(x) − L| <
=⇒
lim g(x) = L.
x→c
(b) If lim g(x) exists, then lim f (x) must exist by part (a). x→c
x→c
55. (a) Suppose on the contrary that lim g(x) does exist. Let L = lim g(x). Then x→c
x→c
lim f (x)g(x) = lim f (x) · lim g(x) = 0 · L = 0.
x→c
x→c
x→c
This contradicts the fact that lim f (x)g(x) = 1 x→c
(b) lim g(x) exists since lim g(x) = lim x→c
x→c
x→c
f (x)g(x) 1 = . f (x) L
56. Suppose lim f (x) does not exist. Let g(x) = −f (x). Then lim g(x) does not exist. x→c
x→c
Now, lim [f (x) + g(x)] = lim [f (x) − f (x)] = lim 0 = 0 exists. This contradicts the hypothesis. x→c
x→c
x→c
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SECTION 2.4 57. f (x) = 2x2 − 3x,
a = 2, lim
x→a
f (2) = 2 f (x) − f (a) 2x2 − 3x − 2 = lim x→2 x−a x−2 = lim
x→2
58. f (x) = x3 + 1, lim
x→a
a = −1,
(x − 2)(2x + 1) = lim (2x + 1) = 5 x→2 x−2
f (−1) = 0
f (x) − f (a) x3 + 1 = lim x→−1 x + 1 x−a = lim
x→−1
59. f (x) =
√
x,
(x + 1)(x2 − x + 1) = lim (x2 − x + 1) = 3 x→−1 x+1
a = 4,
f (4) = 2 √ √ √ f (x) − f (a) x−2 x−2 x+2 = lim = lim ·√ lim x→a x→4 x − 4 x→4 x − 2 x−a x+2 = lim
x→4
60. f (x) = 1/(x + 1),
x−4 1 1 √ = lim √ = 4 (x − 4)( x + 2) x→4 x + 2
a = 1,
f (1) = 1/2
1 1 − f (x) − f (a) lim = lim x + 1 2 x→a x→1 x−a x−1 −(x − 1) −1 1 2(x + 1) = lim =− = lim x→1 x→1 2(x + 1) x−1 4 f (x + h) − f (x) x+h−x = lim =1 h→0 h h f (x + h) − f (x) (x + h)2 − x2 x2 + 2xh + h2 − x2 (b) lim = lim = lim h→0 h→0 h→0 h h h = lim (2x + h) = 2x
61. (a) lim
h→0
h→0
f (x + h) − f (x) (x + h)3 − x3 x3 + 3x2 h + 3xh2 + h3 − x3 (c) lim = lim = lim h→0 h→0 h→0 h h h 2 2 = lim (3x + 3xh + h ) = 3x2 h→0
f (x + h) − f (x) (x + h)4 − x4 x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 − x4 = lim = lim (d) lim h→0 h→0 h→0 h h h 3 2 2 3 3 = lim (4x + 6x h + 4xh + h ) = 4x h→0
f (x + h) − f (x) (x + h)n − xn (e) lim = lim = nxn−1 h→0 h→0 h h
for any positive integer n.
SECTION 2.4 1. (a) f is discontinuous at x = −3, 0, 2, 6 (b)
at −3, neither; f is continuous from the right at 0; at 2 and 6, neither
45
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SECTION 2.4
2. g is continuous on (−4, −1), (−1, 3], (3, 5), (5, 8] 3.
continuous
4.
continuous
5.
continuous
6.
continuous
7.
continuous
8.
jump discontinuity
9.
removable discontinuity
10.
removable discontinuity
11.
jump discontinuity
12.
removable discontinuity
13.
continuous
14.
indefinite discontinuity
15.
infinite discontinuity
16.
17.
infinite discontinuity
18. y
y 3 2
1
1
-1
x
1
-1
no discontinuities
x
1
no discontinuities
19.
20. y
y 10
4 2
-3 -2
-2 -1
1
2
3
x
2
-10
infinite discontinuity at −3
no discontinuities 21.
22. y 4 2 -2
2
4
x
-2 -4
continuous at −2 infinite discontinuity at 3
removable discontinuity at 1
x
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SECTION 2.4 23.
24.
jump discontinuity at −2
no discontinuities 25.
26.
jump discontinuity at −1
jump discontinuities at 0 and 2 27.
28.
removable discontinuity at −2;
jump discontinuity at −3
jump discontinuity at 3
removable discontinuity at −1
47
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SECTION 2.4
29.
30.
(One possibility)
(One possibility)
31.
f (1) = 2
32.
impossible
33.
impossible;
34.
f(1)=0
lim f (x) = −1;
x→1−
lim f (x) = 1
x→1+
35. Since lim− f (x) = 1 x→1
and
lim f (x) = A − 3 = f (1), take A = 4.
x→1+
36. Since lim− f (x) = 4A2 = f (2) x→2
1 = 0.
This gives
lim f (x) = 2(1 − A), we need 4A2 = 2(1 − A) or 2A2 + A −
and
x→2+
A = 12 , −1.
37. The function f is continuous at x = 1 iff f (1) = lim− f (x) = A − B
and
x→1
are equal;
that is, A − B = 3. The function f is discontinuous at x = 2 lim f (x) = 6
and
x→2−
A − B = 3, 4B − A = 6
lim f (x) = f (2) = 4B − A
38. Discontinuous at x = 1 : Continuous at x = 2 :
A − B = 3, 3B − 3 = 6
=⇒
lim f (x) = lim f (x)
x→1−
4B − A = 6, and A − B = 3
=⇒
39.
c = −3
40.
c = 2, d = −3
42.
f (5) = 0
43.
f (5) =
45.
nowhere;
see Figure 2.1.8
x = 2,
1 3
46.
and all non-integral values of x
=⇒
B = 3 :
with
A − B = 3, B = 3.
A − B = 3. 4B − A = 6.
=⇒
x→2
Now,
=⇒
x→1+
lim f (x) = lim f (x) +
x→2−
iff
x→2+
that is, iff 4B − A = 6. More simply we have A − B = 3
are unequal;
47. x = 0,
lim f (x) = 3
x→1+
4B − A = 6 and B = 3. 41.
f (5) =
1 6
44.
f (5) =
1 4
continuous only at x = 0.
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SECTION 2.4
49
48. (a)
(b) Hc is continuous on (−∞, c) ∪ [c, ∞). P,c is continuous on (−∞, c) ∪ [c, c + ) ∪ [c + , ∞). δc is continuous on (−∞, c) ∪ (c, ∞). 49. Refer to (2.2.6). Use the equivalence of (i) and (ii) setting L = f (c). 50. (a) Let = f (c) > 0. By the continuity of f at c, there exists δ > 0 such that |f (x) − f (c)| < f (c)
for all x ∈ (c − δ, c + δ).
This implies that f (x) > 0 for all x ∈ (c − δ, c + δ). (b) Apply the result in part (a) to h(x) = −f (x). (c) Apply the result in part (a) to h(x) = g(x) − f (x). 51. Suppose that g does not have a non-removable discontinuity at c. Then either g is continuous at c or it has a removable discontinuity at c. In either case, lim g(x) as x → c exists. Since g(x) = f (x) except at a finite set of points x1 , x2 , . . . , xn ,
lim f (x) exists as x → c by Exercise 54, Section 2.3.
52. (a) Choose any point c and let > 0. Since f is continuous at c, there exists δ > 0 such that if |x − c| < δ
then |f (x) − f (c)| < .
Now, since | |f (x)| − |f (c)| | ≤ |f (x) − f (c)|, it follows that | |f (x)| − |f (c)| | < if |x − c| < δ. Thus, |f | is continuous at c. (b)
Let
f (x) =
1,
x = 1
−1,
x = 1.
Then f is not continuous at x = 1.
However, |f (x)| = 1 for all x is continuous everywhere.
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SECTION 2.5 (c)
Let
f (x) =
1,
x rational
Then f is nowhere continuous.
−1, x irratiional.
However, |f (x)| = 1 for all x is continuous everywhere. 53. By implication, f is defined on (c − p, c + p). The given inequality implies that B ≥ 0. If B = 0, then f ≡ f (c) is a constant function and hence is continuous. Now assume that B > 0. Let > 0 and let δ = min{/B, p}. If |x − c| < δ then x ∈ (c − p, c + p) and |f (x) − f (c)| ≤ B|x − c| < B · δ ≤ B ·
= B
Thus, f is continuous at c. 54. Choose any point c ∈ (a, b), and let > 0. Now choosing δ = , we have if |x − c| < δ, x ∈ (a, b),
then |f (x) − f (c)| ≤ |x − c| < δ = .
Thus, f is continuous at c, and it follows that f is continuous on (a, b.) f (c + h) − f (c) f (c + h) − f (c) · h = lim · lim h = L · 0 = 0. h→0 h→0 h→0 h→0 h h Therefore f is continuous at c by Exercise 47.
55.
lim [f (c + h) − f (c)] = lim
56. Let fe (x) =
1 2
[f (x) + f (−x)]
and fo (x) =
1 2
[f (x) − f (−x)]. Then fe is an even function, fo is
an odd function, each function is continuous on (−∞, ∞), and f = fe + fo . 57. 59.
5 2
58.
lim f (x) = −1;
x→0−
lim f (x) = 1;
x→0+
1 2
f is discontinuous for all k.
60. 2
SECTION 2.5
sin 3x 3x
1.
sin 3x = lim 3 x→0 x→0 x
3.
3x 3 = lim x→0 sin 5x x→0 5
5.
lim
sin 4x 4x sin 4x 2x = lim · · = 2(1)(1) = 2 sin 2x x→0 2x 4x sin 2x
lim
sin 3x sin 3x 3 3 = lim = 5x 5 x→0 3x 5
6.
lim
lim
x→0
x→0
5x sin 5x
= 3(1) = 3 =
3 3 (1) = 5 5
2.
4.
lim
x→0
2x sin x = 2 lim = 2(1) = 2 x→0 x sin x
sin 3x sin 3x 3 3 = lim = x→0 2x 2 x→0 3x 2 lim
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SECTION 2.5 sin x2 7. lim = lim x x→0 x→0 x 8.
lim
x→0
sin x2 x2
sin2 3x 9 = lim 11. lim x→0 5x2 x→0 5 lim
x→0
sin x2 = 0(1) = 0 x→0 x2
= lim x · lim x→0
sin x2 =1 x2
sin 3x 3x
sin x2 x2
2 =
lim
does not exist
= 0(1) = 0
9 9 (1) = 5 5
tan2 3x 9 9 1 sin2 3x 9 = lim = (1)(1) = · · 2 2 2 x→0 4 cos 3x 4x (3x) 4 4
13.
2x 2x cos 3x 2 = lim = lim x→0 tan 3x x→0 sin 3x x→0 3
14.
lim
lim
x→0
sin x (sin x)/x = lim ; 2 x→0 x x→0 x
9.
sin2 x2 10. lim = lim sin x2 2 x→0 x→0 x
12.
3x sin 3x
cos 3x =
4x 4x sin 3x = lim =0 cot 3x x→0 cos 3x
15.
2 2 (1)(1) = 3 3 x =1 x→0 sin x
lim x csc x = lim
x→0
1 cos x − 1 1 − cos x = − lim =0 x→0 2x 2 x→0 x 1 + cos 2x x2 x2 x2 (1 + cos 2x) 17. lim = lim · = lim x→0 1 − cos 2x x→0 1 − cos 2x x→0 1 + cos 2x sin2 2x 2 2x 1 1 1 = lim (1 + cos 2x) = (1)(2) = x→0 4 sin 2x 4 2
16.
lim
x2 − 2x = lim 18. lim x→0 sin 3x x→0
x−2 3
3x sin 3x
=
2 − 3
(1) = −
2 3
2 sin 2x 1 − sec2 2x − tan2 2x − sin2 2x 1 19. lim = lim −4 = −4 = lim = lim 2 x→0 x→0 x→0 x cos2 2x x→0 x2 x2 2x cos2 2x 20.
lim
x→0
1 sin x 1 1 = lim = 2x csc x 2 x→0 x 2
1 − cos 4x 22. lim = lim x→0 x→0 9x2 = lim
1 − cos 4x 9x2
2x2 + x x = lim (2x + 1) =1 x→0 sin x x→0 sin x
21.
1 + cos 4x 1 + cos 4x
lim
= lim
x→0
1 − cos2 4x 9x2 (1 + cos 4x)
sin2 4x sin2 4x 16 16 1 8 1 = lim = ·1· = · + cos 4x) 9 x→0 (4x)2 1 + cos 4x 9 2 9
x→0 9x2 (1
tan 3x 1 3 sin 3x 23. lim 2 = lim = lim x→0 2x + 5x x→0 x(2x + 5) cos 3x x→0 2x + 5 24.
2
2
lim x (1 + cot 3x) = lim
x→0
x→0
x2 cos2 3x x + sin2 3x 2
sin 3x 3x
1 3 3 = (1)(1) = cos 3x 5 5
(3x)2 cos2 3x = lim x + · x→0 9 sin2 3x 2
= 0 + (1)
1 1 = 9 9
51
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SECTION 2.5
1 −1 sec x − 1 1 − cos x cos = lim x = lim =0 25. lim x→0 x sec x x→0 x→0 1 x x cos x √ 1 − cos x 1 − cos(π/4) 4−2 2 = = 26. lim x π/4 π x→π/4
28.
29.
lim
x→0
sin2 x x sin2 x ; does not exist = lim · x(1 − cos x) x→0 1 − cos x x2
cos x cos (h + π/2) − sin h = lim = lim = −1 h→0 h→0 h h x→π/2 x − π/2 lim
h = x − π/2 30.
31.
32.
27.
√ 2 2 π
lim
x→π
cos (h + π/2) = cos h cos π/2 − sin h sin π/2
sin x − sin(x − π) = lim = −1 x − π x→π x−π
lim
x→π/4
lim
x→π/6
sin (x + π/4) − 1 sin (h + π/2) − 1 cos h − 1 = lim = lim =0 h→0 h→0 x − π/4 h h ∧ h = x − π/4 sin[x + (π/3)] − 1 sin[x − (π/6) + (π/2)] − 1 cos[x − (π/6)] − 1 = lim = lim =0 x − (π/6) x − (π/6) x − (π/6) x→π/6 x→π/6
33. Equivalently we will show that lim cos (c + h) = cos c. The identity h→0
cos (c + h) = cos c cos h − sin c sin h gives
lim cos (c + h) = cos c lim cos h − sin c
h→0
34.
h→0
a b
35.
0
lim sin h
h→0
36.
= (cos c)(1) − (sin c)(0) = cos c.
a b
37.
38. Let > 0. There exists δ1 > 0 such that if
0 < |x| < δ1 ,
then |f (x) − L| < .
Take δ = δ1 /|a| : if
0 < |x| < δ,
then 0 < |ax| = |a| |x| < δ1
and |f (ax) − L| < .
1
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SECTION 2.5 39. f (x) = sin x;
a = π/4
40. f (x) = cos x;
a = π/3
53
sin π4 + h − sin π4 f (a + h) − f (a) sin (π/4) cos h + cos (π/4) sin h − sin (π/4) lim = lim = lim h→0 h→0 h→0 h h h − sin (π/4) (1 − cos h) + cos (π/4) sin h = lim h→0 h √ 1 − cos h sin h 2 + cos (π/4) lim = cos (π/4) = = − sin (π/4) lim h→0 h→0 h h 2 √ √ 2 2 π = x− tangent line: y − 2 2 4
cos f (a + h) − f (a) lim = lim h→0 h→0 h
π 3
+ h − cos π3 cos (π/3) cos h − sin (π/3) sin h − cos (π/3) = lim h→0 h h
− cos (π/3) (1 − cos h) − sin (π/3) sin h h→0 h √ 1 − cos h sin h 3 − sin (π/3) lim = − sin (π/3) = − = − cos (π/3) lim h→0 h→0 h h 2 √ 3 π 1 x− y− =− 2 2 3 = lim
tangent line: 41. f (x) = cos 2x;
a = π/6
cos 2 f (a + h) − f (a) lim = lim h→0 h→0 h
π 6
+ h − cos 2 π6 cos (2h + π/3) − cos (π/3) = lim h→0 h h
cos (π/3) cos 2h − sin (π/3) sin 2h − cos (π/3) h 1 − cos 2h sin 2h − sin (π/3) lim 2 = − cos (π/3) lim 2 h→0 h→0 h h √ = − cos (π/3) · 2 · 0 − sin (π/3) · 2 · 1 = − 3 √ 1 π tangent line: y − = − 3 x − 2 6 = lim
h→0
42. f (x) = sin 3x;
a = π/2
sin 3 f (a + h) − f (a) lim = lim h→0 h→0 h
π
+ h − sin 3 π2 sin (3h + 3π/2) − sin (3π/2) = lim h→0 h h
= lim
sin (3π/2) cos 3h + cos (3π/2) cos 3h − sin (3π/2) h
= lim
1 − cos 3h 1 − cos 3h = 3 lim =3·0=0 h→0 h 3h
h→0
h→0
tangent line: y − (−1) = 0 [x − (3π/2)] 43. For x = 0,
2
or y = −1
|x sin(1/x)| = |x| | sin(1/x)| ≤ |x|. Thus, −|x| ≤ |x sin(1/x)| ≤ |x|
Since lim (−|x|) = lim |x| = 0, x→0
x→0
the result follows by the pinching theorem.
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SECTION 2.5
44. Since 0 ≤ cos2 [1/(x − π)] ≤ 1 for all x = π, we have |(x − π) cos2 [1/(x − π)]| ≤ |x − π|. Thus, −|x − π| ≤ |(x − π) cos[1/(x − π)]| ≤ |x − π| Since lim (−|x − π|) = lim |x − π| = 0, the result follows by the pinching theorem. x→π
x→π
45. For x close to 1(radian), 0 < sin x ≤ 1. Thus, 0 < |x − 1| sin x ≤ |x − 1| and the result follows by the pinching theorem. 46. Clearly, |f (x)| ≤ 1 for all x. Therefore, |x f (x)| ≤ |x|
which implies
− |x| ≤ xf (x) ≤ |x|
for all x.
The result follows by the pinching theorem. 47. Suppose that there is a number B such that |f (x)| ≤ B for all x = 0. Then |x f (x)| ≤ B |x| and −B |x| ≤ x f (x) ≤ B |x| The result follows by the pinching theorem. 48. We have f (x) = x ·
f (x) f (x) ≤ B, x = 0. and x x
Therefore, the result follows from Exercise 47. f (x) − L ≤ B for x = c. Then 49. Suppose that there is a number B such that x−c f (x) − L 0 ≤ |f (x) − L| = (x − c) ≤ B|x − c| x−c By the pinching theorem, lim |f (x) − L| = 0 x→c
which implies
lim f (x) = L.
x→c
Let δ > 0 be such that for |δ − c| < p, if |x − c| < δ then |f (x)| < . B In other words, B|f (x)| < .
50. Let > 0.
But |g(x)| ≤ B, so |f (x)g(x)| ≤ B|f (x)|. Thus |f (x)g(x)| < . 51. 53.
lim
x→0 1 3
20x − 15x2 = 10 sin 2x
52. 54.
lim
x→0 1 2
tan x does not exist x2
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SECTION 2.6
55
SECTION 2.6 1. Set f (x) = 2x3 − 4x2 + 5x − 4. Then f is continuous on [1, 2] and f (1) = −1 < 0.
f (2) = 6 > 0.
By the intermediate-value theorem there is a c in [1, 2] such that f (c) = 0. 2. Set f (x) = x4 − x − 1. Then f is continuous on [−1, 1] and f (−1) = 1 > 0,
f (1) = −1 < 0.
By the intermediate-value theorem there is a c in [−1, 1] such that f (c) = 0. π 3. Set f (x) = sin x + 2 cos x − x2 Then f is continuous on [0, ] and f (0) = 2 > 0, 2 π By the intermediate-value theorem there is a c in [0, ] such that f (c) = 0. 2
π π2 f( ) = 1 − < 0. 2 4
π 4. Set f (x) = 2 tan x − x. Then f is continuous on [0, ] and f (0) = 0 < 1, 4 π By the intermediate-value theorem there is a c in [0, ] such that f (c) = 1. 4
π π f ( ) = 2 − > 1. 4 4
1 1 1 1 . Then f is continuous on [ , 1] and f ( ) = > 0, 2x 4 4 16 1 By the intermediate-value theorem there is a c in [ , 1] such that f (c) = 0. 4
5. Set f (x) = x2 − 2 +
5
f (1) = −
1
6. Set f (x) = x 3 + x 3 . Then f is continuous on [−1, 1] and f (−1) = −2 < 1,
1 < 0. 2
f (1) = 2 > 1.
By the intermediate-value theorem there is a c in [−1, 1] such that f (c) = 1. 7. Set f (x) = x3 −
√
x + 2. Then f is continuous on [1, 2] and f (1) = 1 −
√
3 < 0,
f (2) = 6 > 0.
By the intermediate-value theorem there is a c in [1, 2] such that f (c) = 0. √ i.e. c3 = c + 2. 8. Set f (x) =
√
x2 − 3x − 2. Then f is continuous on [3, 5] and f (3) = −2 < 0,
f (5) =
√
10 + 2 > 0.
By the intermediate-value theorem there is a c in [3, 5] such that f (c) = 0. 9. f (x) is continuous on [0, 1];
f (0) = 0 < 1 and f (1) = 4 > 1.
By the intermediate value theorem there is a c in (0, 1) such that f (c) = 1. 1 1 > 0 and f (3) = − < 0, 2 2 By the intermediate value theorem there is a c in (2, 3) (hence in (1, 4)) such that f (c) = 0.
10. f (x) is continuous on [2, 3];
11. Set f (x) = x3 − 4x + 2.
f (2) =
Then f (x) is continuous on [−3, 3].
Checking the integer values on this interval, f (−3) = −13 < 0,
f (−2) = 2 > 0,
f (0) = 2 > 0,
f (1) = −1 < 0, and f (2) = 2 > 0.
By the intermediate value theorem there are roots in (−3, −2), (0, 1) and (1, 2). 12. Set f (x) = x2 .
Then f (x) is continuous on [1, 2],
f (1) = 1 < 2 and f (2) = 4 > 2.
By the intermediate value theorem there is a c in (1, 2) such that f (c) = 2.
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SECTION 2.6
13.
14.
15.
16.
17.
18.
19.
20.
22.
23.
24.
21.
Impossible
Impossible
Impossible
25. If f (0) = 0 or if f (1) = 1, we have a fixed point. If f (0) = 0 and f (1) = 1, then set g(x) = x − f (x). g is continuous on [ 0, 1 ] and g(0) < 0 < g(1). Therefore there exists a number c ∈ (0, 1) such that g(c) = c − f (c) = 0. 26. Set h(x) = f (x) − g(x). Then h is continuous on [a, b], and h(a) = f (a) − g(a) < 0,
h(b) = f (b) −
g(b) > 0. By the intermediate value theorem there exists a number c ∈ (a, b) such that h(c) = 0. Thus, f (c) = g(c). 27. (a) If f (0) = 1 or if f (1) = 0, we’re done. Suppose f (0) = 1 and f (1) = 0. The diagonal from (0, 1) to (1, 0) has equation y = 1 − x, 0 ≤ x ≤ 1. Set g(x) = (1 − x) − f (x). g is continuous on [0, 1] and g(0) = 1 − f (0) > 0; g(1) = −f (1) < 0. Therefore there exists a number c ∈ (0, 1) such that g(c) = (c − 1) − f (c) = 0. (b) Suppose g(0) = 0 and g(1) = 1. Set h(x) = g(x) − f (x); h is continuous on [0, 1]. If h(0) = 0 or if h(1) = 0, we’re done. If not, then h(0) < 0 and h(1) > 0. Therefore there exists a number c ∈ (0, 1) such that h(c) = 0. The same argument works for the other case. 28. Let f (x) = x3 and repeat the argument given in Exercise 37. 29. The cubic polynomial P (x) = x3 + ax2 + bx + c is continuous on (−∞, ∞).. Writing P as b a c 3 P (x) = x 1 + + 2 + 3 x = 0 x x x
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SECTION 2.6
57
it follows that P (x) < 0 for large negative values of x and P (x) > 0 for large positive values of x. Thus there exists a negative number N such that P (x) < 0 for
x < N, and a positive number M
such that P (x) > 0 for x > M . By the intermediate-value theorem, P has a zero in [N, M ]. 30. (a) Let S be the set of positive integers for which the statement is true. Then 1 ∈ S by hypothesis. Now assume that k ∈ S. Then ak < bk and ak+1 = (a)ak < (a)bk < (b)bk = bk+1 . Therefore, k + 1 ∈ S and S is the set of positive integers. (b) Clearly 0 is the unique nth root of 0. Choose any positive number x and let f (t) = tn − x. Since f (0) = −x < 0 and f (t) → ∞ as t → ∞, there exists a number c > 0 such that f (c) = 0. The number c is an nth root of x. The uniqueness follows from part (a). and T (4000) ∼ = 98.0995, T (4500) ∼ = 97.9478. Thus,
31. The function T is continuous on [4000, 4500]];
by the intermediate-value theorem, there is an elevation h between 4000 and 4500 meters such that T (h) = 98. 32. Think of the equator as being a circle and choose a reference point P and a positive direction. For example, choose P to be 0◦ longitude and let “eastward” be the positive direction. Using radian measure, let x, 0 ≤ x ≤ 2π denote the coordinate of a point x radians from P. Then, x and x + π are diametrically opposite points on the equator. Let T (x) be the temperature at the point x, and let f (x) = T (x) − T (x + π). If f (0) = 0, then the temperatures at the points 0 and π are equal. If f (x) = 0, then f (0) = T (0) − T (π) and f (π) = T (π) − T (2π) = t(π) − T (0) have opposite sign. Thus, there exists a point c ∈ (0, π) at which f (c) = 0, and T (c) = T (c + π). 33. Let A(r) denote the area of a circle with radius r, r ∈ [0, 10]. Then A(r) = πr 2 is continuous on ∼ 314. Since 0 < 250 < 314 it follows from the intermediate [0, 10], and A(0) = 0 and A(10) = 100π = value theorem that there exists a number c ∈ (0, 10) such that A(c) = 250. 34. Let x and y be the dimensions of a rectangle in R. Then, 2x + 2y = P and y = function
A(x) = xy = x
P −x 2
=
P x − x2 , 2
P − x. The area 2
x ∈ [0, P/2]
is continuous. Therefore, A has a maximum value on [0, P/2]. Since 2 P2 P P 2 − x− A(x) = x − x = 2 16 4 it is clear that the rectangle with maximum area has dimensions x = y =
P . 4
35. Inscribe a rectangle in a circle of radius R. Introduce a coordinate system with the origin at the center of the circle and the sides of the rectangle parallel to the coordinate axes. Then the area of the rectangle
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SECTION 2.6 is given by A(x) = 4x R2 − x2 ,
x ∈ [0, R].
Since A is continuous on [0, R], A has a maximum value. 36. f (0) = −4, f (1) = 2. 37. f (−3) = −9,
Thus, f has a zero in (0, 1) at r = 0.771.
f (−2) = 5;
f (0) = 3,
f (1) = −1;
f (1) = −1,
f (2) = 1
Thus, f
has a zero in
(−3, −2, ) in (0, 1) and in (1, 2). r1 = −2.4909,
r2 = 0.6566,
38. f (−2) = −25, f (−1) = 1;
and r3 = 1.8343 f (0) = 1, f (1) = −1;
f (1) = −1, f (2) = 27
Thus, f has a zero in
(−2, −1, ) in (0, 1) and in (1, 2). r1 = −1.3888,
r2 = 0.3345,
and r3 = 1.2146
39. f (−2) = −5.6814, f (−1) = 1.1829;
f (0) = 0.5, f (1) = −0.1829;
f (1) = −0.1829, f (2) = 6.681
Thus, f has a zero in (−2, −1), in (0, 1) and in (1, 2). r1 = −1.3482,
r2 = 0.2620,
and r3 = 1.0816
40. f satisfies the hypothesis of the intermediate-value theorem:
c∼ = −0.83
41. f is not continuous at x = 1. Therefore f does not satisfy the hypothesis of the intermediate-value theorem. However, 1 f (−3) + f (2) = = f (c) for c ∼ = 0.163 2 2 42. f is not continuous at x = −π/2, π/2, 3π/2;
1 2
[f (−π) + f (2π)] = 0;
f (x) = 0 for all x ∈ [−π, 2π].
43. f satisfies the hypothesis of the intermediate-value theorem: f (π/2) + f (2π) 1 = = f (c) for c ∼ = 2.38, 4.16, 5.25 2 2 44. f is bounded. max (f ) = 6 [f (0) = 6] min (f ) ∼ = −0.376 [f (1.46) ∼ = −0.376]
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PROJECT 2.6 45. f is bounded. max (f ) = 1
[f (1) = 1]
min (f ) = −1
[f (−1) = −1]
46. f is unbounded
47. f is bounded. no max f is not defined at x = 0 min (f ) ∼ = 0.3540
PROJECT 2.6 2−1 < 0.001 =⇒ 2n > 1000 =⇒ n > 9. 2n The minimum number of iterations required is n = 10. √ 1449 2 1.415034. 1024 2−1 < 0.0001 =⇒ 2n > 10, 000 =⇒ n > 13. (b) 2n The minimum number of iterations required is n = 14.
1. (a)
2−1 < 0.00001 2n
=⇒
2n > 100, 000
=⇒
n > 16.
The minimum number of iterations required is n = 17. 2. f (x) = x3 + x − 9; f (1) = −7 and f (2) = 1.
Therefore, f (c) = 0 for some c ∈ (1, 2).
(a) A minimum of 7 iterations are required; c
245 1.9140625. 128
Accurate to 7 decimal places, the root is c 1.9201751. (b)
2−1 < 0.001 2n
=⇒
2n > 1, 000
=⇒
n > 9.
59
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REVIEW EXERCISES The minimum number of iterations required is n = 10. 2−1 < 0.0001 2n
2n > 10, 000
=⇒
=⇒
n > 13.
The minimum number of iterations required is n = 14. 3. f (x) = sin x + x + 3; f (−3) ≈ −0.142 and f (−2) ≈ 0.091.
Therefore, f (c) = 0 for some c ∈ (−3, −2).
−279 (a) A minimum of 7 iterations are required; c −2.17969. 128 Accurate to 6 decimal places, the root is c −2.179727. (b)
2−1 < 0.00001 2n
=⇒
2n > 100, 000
=⇒
n > 16.
The minimum number of iterations required is n = 17. 2−1 < 0.000001 2n
=⇒
2n > 1, 000, 000
=⇒
The minimum number of iterations required is n = 20. 4. For f (x) = x2 − 2, the first three iterations are: c1 = 43 ,
c2 =
5 3
≈ 1.6667,
c3 =
3 2
= 1.5
For f (x) = x + x − 9, the first three iterations are: 3
15 8
c1 =
= 1.875,
c2 =
31 16
= 1.9375,
c3 =
61 32
= 1.90625.
For f (x) = sin x + x + 3, the first three iterations are: c1 ≈ −2.39056,
c2 ≈ −2.19528,
c3 ≈ −2.09764.
CHAPTER 2. REVIEW EXERCISES 1.
x2 − 3 32 − 3 = = 1. x→3 x + 3 3+3
2.
lim
lim
x→2 x2
x2 + 4 22 + 4 8 = 2 = . + 2x + 1 2 + 2(2) + 1 9
3.
(x − 3)2 0 = = 0. x→3 x + 3 6
4.
lim
5.
6. 7.
lim
x→3 x2
x2 − 9 (x − 3)(x + 3) x+3 = lim = lim = 6. − 5x + 6 x→3 (x − 2)(x − 3) x→3 x − 2
lim
x−2 x−2 = lim = 1. |x − 2| x→2+ x − 2
lim
|x| −x 1 = lim =− . x − 2 x→−2 x − 2 2
x→2+
x→−2
lim
x→0
1 1 − x x − = lim = 1. x→0 x x x
n > 19.
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REVIEW EXERCISES √
√ x−3 x−3 1 = lim = lim √ 8. lim+ ; does not exist. + + |x − 3| x − 3 x→3 x→3 x→3 x−3 9.
10.
lim+
|x − 1| x−1 |x − 1| −x + 1 |x − 1| = lim = 0 and limx→1− = limx→1− = 0. limx→1 = 0. + x x x x x x→1
lim+
x3 − 1 x3 − 1 x3 − 1 x3 − 1 − − = lim = 1 and lim = lim = −1; does not exist. x→1 x→1 |x3 − 1| x→1+ x3 − 1 |x3 − 1| −x3 + 1
x→1
x→1
√ 11.
12.
lim
√ x−1 x−1 1 1 √ = lim √ = lim √ = . x→1 ( x − 1)( x + 1) x→1 x−1 2 x+1
lim+
√ x2 − 2x − 3 (x − 3)(x + 1) √ √ = lim+ = lim+ x − 3(x + 1) = 0. x→3 x→3 x−3 x−3
x→1
x→3
√ 13.
lim
x→3+
√ 14.
15.
16.
lim
x→4
√ x2 − 2x − 3 x+1 = lim √ , does not exist. + x−3 x→3 x−3
√ √ x+5−3 ( x + 5 − 3)( x + 5 + 3) x−4 1 √ √ = lim = . = lim x→4 x→4 (x − 4)( x + 5 + 3) x−4 6 (x − 4)( x + 5 + 3)
x3 − 8 (x − 2)(x2 + 2x + 4) x2 + 2x + 4 3 = lim = lim = . x→2 x4 − 3x2 − 4 x→2 (x − 2)(x + 2)(x2 + 1) x→2 (x + 2)(x2 + 1) 5 lim
lim
x→0
5x 5 2x 5 = lim = . sin 2x x→0 2 sin 2x 2
tan2 2x 4 = lim 17. lim x→0 x→0 3 cos2 2x 3x2 18.
lim x csc 4x = lim
x→0
x→0
sin 2x 2x
2 =
4 . 3
4x 1 = . 4 sin 4x 4
19.
x2 − 3x x(x − 3) cos x = lim = −3 x→0 tan x x→0 sin x
20.
cos x sin(π/2 − x) 1 = lim =− . 2 x→π/2 2x − π x→π/2 −2(π/2 − x)
21.
22. 23.
24.
lim
lim
lim
x→0
sin 3x = lim 5x2 − 4x x→0
sin 3x 3 =− . 5 4 4 3x( x − ) 3 3
5x2 5x2 5 = . = lim x→0 1 − cos 2x x→0 2 sin2 x 2 lim
lim
x→−π
x+π x+π = lim = −1. x→−π sin x − sin(x + π)
x2 − 4 (x − 2)(x + 2) x+2 1 = lim = lim 2 = . 3 2 x→2 x − 8 x→2 (x − 2)(x + 2x + 4) x→2 x + 2x + 4 3 lim
61
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26.
27.
28. 29. 30.
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REVIEW EXERCISES lim
x→2−
x−2 1 x−2 1 = lim − = lim − =− . |x2 − 4| x→2− (x − 2)(x + 2) x→2− x + 2 4
1 − 2/x x2 − 2x x 1 = lim = . = lim 2 2 x→2 1 − 4/x x→2 x − 4 x→2 x + 2 2 lim
lim+
x→1
√ x2 − 3x + 2 (x − 2)(x − 1) √ √ = lim+ = lim+ (x − 2) x − 1 = 0. x→1 x→1 x−1 x−1
1 − 9/x2 1 − 9/9 = = 0. x→3 1 + 3/x 1 + 3/3 lim
lim f (x) = 3(2) − 22 = 2.
x→2
lim f (x) = (−2)2 − 3 = 1 and lim − f (x) = 3 − 2 = 1. Hence lim f (x) = 1.
x→−2+
x→−2
x→−2
31. (a) False
(b) False
(c) True
(d) False
(e) True
32. Since lim x2 − 2x − 3 = 0, we must have lim (2x2 − 3ax + x − a − 1) = 0. x→3
x→3
lim (2x2 − 3ax + x − a − 1) = 18 − 9a + 3 − a − 1 = 0;
x→3
−10a + 20 = 0;
a = 2.
2x2 − 5x − 3 (2x + 1)(x − 3) 7 = lim = . x→3 x2 − 2x − 3 x→3 (x + 1)(x − 3) 4 lim
33. (a) y 6 4 2
-3
-2
-1
1
2
3
x
-2 -4 -6
(b) (i) 1
(ii) 0
(iii) does not exist
(iv) −6
(v) 4
(vi) does not exist
(c) (i) f is continuous from the left at −1; f is not continuous from the right at −1. (ii) f is not continuous from the left at 2; f is continuous from the right at 2.
1 − cos x 1 − cos x 34. (a) lim cos = cos lim = cos 0 = 1 x→0 x→0 2x 2x π sin x π sin x (b) lim cos = cos lim = cos 12 π = 0 x→0 x→0 2x 2x
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REVIEW EXERCISES
63
35. For f to be continuous at 2, we must have lim (2x2 − 1) = 7 = A = lim (x3 − 2Bx) = 8 − 4B.
x→2−
x→2+
These equations imply that A = 7 and B =
1 . 4
36. For f to continuous at x = −1, we must have lim (Ax + B) = B − A = −2
x→−1−
or
A = B + 2.
For f to be discontinuous at x = 2, we must have lim (2Bx − A) = 4B − A = f (2) = 4.
x→2+
A = B + 2 and 4B − A = 4 imply that B = 2. Necessary and sufficient conditions for f to be continuous at x = −1 and discontinuous at x = 2 are: A = B + 2 with B = 2. x2 − 2x − 15 (x − 5)(x + 3) = lim = −8; set f (−3) = −8. x→−3 x→−3 x+3 x+3 √ √ √ x+1−2 ( x + 1 − 2)( x + 1 + 2) x−3 1 √ √ 38. lim = lim = lim = ; set f (3) = 14 . x→3 x→3 x→3 (x − 3)( x + 1 + 2) x−3 4 (x − 3)( x + 1 + 2) 37.
39.
40.
lim
lim
x→0
sin πx π sin πx = lim = π; set f (0) = π. x→0 x πx
1 − cos2 x sin2 x 1 − cos x 1 − cos x 1 + cos x 1 1 = = . = · · · 2 2 2 x x 1 + cos x x 1 + cos x x2 1 + cos x 1 − cos x sin2 x 1 1 = ; set f (0) = 12 . = lim · x→0 x→0 x2 x2 1 + cos x 2
Therefore, lim
41. (a) False; need f continuous (c) False; need f continuous on [a, b]
(b) True; intermediate-value theorem (d) False; consider f (x) = x2 on [−1.1].
42. Set f (x) = x3 − 3x − 4. Then f (2) = −2 and f (3) = 14. By the intermediate-value theorem, f (x) has zero in [2, 3]. 43. Set f (x) = 2 cos x − x + 1. Then f (1) = 2 cos 1 > 0 and f (2) = 2 cos 2 − 1 < 0. By the intermediatevalue theorem, f (x) as zero in [1, 2]. 44. Let > 0. |(5x − 4) − 6| = |5x − 10| = 5|x − 2|. If 0 < |x − 2| < δ = /5, then |(5x − 4) − 6| < . Therefore, lim (5x − 4) = 6. x→2
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REVIEW EXERCISES
45. If 0 < |x − (−4)| = |x + 4| < 1, then 2x + 5 < 0 and |2x + 5| = −2x − 5. Let > 0. |2x + 5| − 3 = | − 2x − 5 − 3| = | − 2x − 8| = 2|x + 4|. If δ = min {1, /2}, then |2x + 5| − 3 < . Therefore,
lim |2x + 5| = 3.
x→−4
46. Let > 0.
√ √ √ x − 5 − 2 = x − 5 − 2 · √x − 5 + 2 = √ x − 9 < |x − 9|. x−5+2 x − 5 + 2 √ √ If 0 < |x − 9| < δ = , then x − 5 − 2 < . Therefore, lim x − 5 = 2. x→9
f (x) = L exists. Then x→0 x f (x) x f (x) lim f (x) = lim = lim x = L lim x = 0. x→0 x→0 x→0 x x→0 x
47. Suppose that lim
48. Suppose that lim g(x) = l and that f is continuous at l. Let > 0. By the continuity of f at x→c
l, there is a number δ1 > 0 such that |f (y) − f (l)| < whenever |y − l| < δ1 . Since lim g(x) = l, x→c
there is a positive number δ such that |g(x) − l| < δ1 whenever 0 < |x − c| < δ. Therefore, if 0 < |x − c| < δ, then |g(x) − l| < δ1 and |f (g(x) − f (l)| < . Therefore, lim f (g(x) = l. x→c
⎧ ⎨ sin πx , x 49. (a) Yes. For example, set f (x) = ⎩ π,
x>0 x=0
(b) No. By the continuity of f (from the right) at 0, f (1/n) = 0 for each postive integer implies that f (0) = 0. 50. (a) No. Suppose that f and g are everywhere continuous. If f (c) = g(c), then there is an interval (c − p, c + p) on which f (c) = g(c). (b) No. Suppose that f and g are everywhere continuous. If f (x) = g(x) on [a, b], then f (a) = g(a) and f (b) = g(b). Therefore, by the continuity of f and g, there exist postive numbers p and q such that f (x) = g(x) on (a − p, b + q).
x(1 − x),
0≤x≤1
0, f and g are everywhere continuous and f (x) = g(x) only on (0, 1).
otherwise.
(c) Yes. For example, set f (x) = 0 for all x and g(x) =
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SECTION 3.1 CHAPTER 3 SECTION 3.1 f (x + h) − f (x) [2 − 3(x + h)] − [2 − 3x] −3h = lim = lim = lim −3 = −3 h→0 h→0 h→0 h h→0 h h
1. f (x) = lim 2. f (x) = lim
h→0
f (x + h) − f (x) k−k = lim = lim 0 = 0 h→0 h→0 h h
f (x + h) − f (x) [5(x + h) − (x + h)2 ] − (5x − x2 ) = lim h→0 h→0 h h
3. f (x) = lim
5h − 2xh − h2 = lim (5 − 2x − h) = 5 − 2x h→0 h→0 h
= lim
f (x + h) − f (x) [2(x + h)3 + 1] − [2x3 + 1] = lim h→0 h→0 h h
4. f (x) = lim
2(x3 + 3x2 h + 3xh2 + h3 ) − 2x3 6x2 h + 6xh2 + 2h3 = lim h→0 h→0 h h
= lim
= lim (6x2 + 6xh + 2h2 ) = 6x2 h→0
f (x + h) − f (x) (x + h)4 − x4 = lim h→0 h→0 h h
5. f (x) = lim
(x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 ) − x4 h→0 h
= lim
= lim (4x3 + 6x2 h + 4xh2 + h3 ) = 4x3 h→0
1 1 − f (x + h) − f (x) x + h + 3 x + 3 6. f (x) = lim = lim h→0 h→0 h h
lim
(x + 3) − (x + h + 3) −h = lim h→0 h(x + h + 3)(x + 3) h(x + h + 3)(x + 3)
lim
−1 −1 = (x + h + 3)(x + 3) (x + 3)2
h→0
h→0
f (x + h) − f (x) = lim h→0 h→0 h
7. f (x) = lim = lim
h→0
√
x+h−1− h
√
x−1
(x + h − 1) − (x − 1) 1 1 √ √ √ = lim √ = √ 2 x−1 h( x + h − 1 + x − 1 ) h→0 x + h − 1 + x − 1
f (x + h) − f (x) [(x + h)3 − 4(x + h)] − [x3 − 4x] = lim h→0 h→0 h h
8. f (x) = lim
3x2 h + 3xh2 + h3 − 4h = lim (3x2 + 3xh + h2 − 4) = 3x2 − 4 h→0 h→0 h
= lim
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SECTION 3.1
1 1 − 2 2 f (x + h) − f (x) (x + h) x 9. f (x) = lim = lim h→0 h→0 h h x2 − (x2 + 2hx + h2 ) −2x − h 2 = lim 2 =− 3 2 2 2 h→0 h→0 hx (x + h) x (x + h) x
= lim
1 1 √ −√ f (x + h) − f (x) x x+h = lim 10. f (x) = lim h→0 h→0 h h √ √ √ √ √ √ x− x+h x+ x+h x− x+h x − (x + h) √ √ = lim √ √ √ √ lim √ √ = lim √ √ h→0 h x x + h h→0 h x x + h h→0 x+ x+h h x x+h x+ x+h lim
h→0
−h −1 √ √ = √ √ √ 2x x h x x+h x+ x+h
11. f (x) = x2 − 4x;
c = 3:
difference quotient: f (3 + h) − f (3) (3 + h)2 − 4(3 + h) − (−3) = h h 2h + h2 9 + 6h + h2 − 12 − 4h + 3 = =2+h h h f (3 + h) − f (3) f (3) = lim = lim (2 + h) = 2 h→0 h→0 h =
Therefore,
12. f (x) = 7x − x2 ;
c = 2:
difference quotient: f (2 + h) − f (2) 7(2 + h) − (2 + h)2 − (10) = h h 3h − h2 14 + 7h − 4 − 4h − h2 − 10 = =3−h h h f (2 + h) − f (2) = lim (3 − h) = 3 f (2) = lim h→0 h→0 h =
Therefore,
13. f (x) = 2x3 + 1;
c = 1:
difference quotient: f (−1 + h) − f (−1) 2(−1 + h)3 + 1 − (−1) = h h 2 −1 + 3h − 3h2 + h3 + 2 6h − 6h2 + 2h3 = = = 6 − 6h + 22 h h f (−1 + h) − f (−1) = lim (6 − 6h + 2h2 ) = 6 Therefore, f (−1) = lim h→0 h→0 h 14. f (x) = 5 − x4 ;
c = −1:
difference quotient: f (1 + h) − f (1) 5 − (1 + h)4 − (4) = h h
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SECTION 3.1 =
67
5 − 1 − 4h − 6h2 − 4h3 − h4 − 4 h
−4h − 6h2 − 4h3 − h4 = −4 − 6h − 4h2 − h3 h f (1 + h) − f (1) = lim −4 − 6h − 4h2 − h3 = −4 f (3) = lim h→0 h→0 h =
Therefore,
8 ; c = −2: x+4 difference quotient:
15. f (x) =
8 8 −4 −4 f (−2 + h) − f (−2) (−2 + h) + 4 = = h+2 h h h −4 8 − 4h − 8 = = h(h + 2) h+2 Therefore, 16. f (x) =
√
f (−2) = lim
h→0
6 − x;
f (−2 + h) − f (−2) −4 = lim = −2 h→0 h + 2 h
c = 2:
difference quotient: f (2 + h) − f (2) = h
√
= √
6 − (2 + h) − (2) h
4−h−2 h
√ 4−h−2 4−h+2 = ·√ h 4−h+2 −1 = √ 4−h+2 Therefore,
f (2) = lim
h→0
f (2 + h) − f (2) −1 1 = lim √ =− h→0 h 4 4−h+2
17. f (4) = 4. Slope of tangent at (4, 4) is f (4) = −3.
Tangent y − 4 = −3(x − 4).
1 1 18. f (4) = 2. Slope of tangent at (4, 2) is f (4) = √ = . 4 2 4 19. f (−2) =
Tangent y − 2 =
1 (x − 4). 4
1 1 1 1 1 . Slope of tangent at (−2, ) is . Tangent: y − = (x + 2). 4 4 4 4 4
20. f (2) = −3. Slope of tangent at (2, −3) is f (2) = −3(2)2 = −12.
Tangent: y + 3 = −12(x − 2) ;
21. (a) f is not continuous at c = −1 and c = 1; f has a removable discontinuity at c = −1 and a jump discontinuity at c = 1. (b) f is continuous but not differentiable at c = 0 and c = 3. 22. (a) f is not continuous at c = 2; f has a jump discontinuity at 2 (b) f is continuous but not differentiable at c = −2 and c = 3.
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SECTION 3.1
23.
at x = −1
24.
at x =
5 2
25.
at x = 0
26.
at x = −2, 2
27.
at x = 1
28.
at x = 2
29. f (1) = 4 lim
f (1 + h) − f (1) 4(1 + h) − 4 = lim− =4 h→0 h h
lim
f (1 + h) − f (1) 2(1 + h)2 + 2 − 4 = lim =4 h h h→0+
h→0−
h→0+
30. f (1) = 6 lim−
f (1 + h) − f (1) 3(1 + h)2 − 3 = lim− =6 h→0 h h
lim
f (1 + h) − f (1) [2(1 + h)3 + 1] − 3 = lim =6 h h h→0+
h→0
h→0+
31. f (−1) does not exist f (−1 + h) − f (−1) h−0 = lim− =1 h→0 h h f (−1 + h) − f (−1) h2 − 0 lim = lim =0 h h h→0+ h→0+ lim
h→0−
32. f (3) does not exist because f is not continuous at x = 3. 33.
34.
35.
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SECTION 3.1 36.
37.
69
38.
39. Since f (1) = 1 and lim f (x) = 2, f is not continuous at 1. Therefore, by (3.1.3), f is not differentiable x→1+
at 1.
40. (a) f (x) =
2(x + 1),
x 0.
f (0 + h) − f (0) (h + 1)2 − 1 = lim− = lim− (h + 2) = 2, h→0 h→0 h→0 h h 2 f (0 + h) − f (0) (h − 1) − 1 lim = lim = lim (h − 2) = −2. h h h→0+ h→0+ h→0+
(b) lim−
41. Continuity at x = 1 : lim− f (x) = 1 = lim f (x) = A + B. Thus A + B = 1. + x→1
x→1
Differentiability at x = 1 : lim−
h→0
f (1 + h) − f (1) (1 + h)3 − 1 f (1 + h) − f (1) = lim− = 3 = lim =A h→0 h h h h→0+
Therefore, A = 3,
=⇒
42. Continuity at x = 2 A = −2, B =
=⇒
B = −2.
4B + 2A = 2;
differentiability at x = 2
43.
=⇒
4B + A = 4;
3 2
f (x) = c, c any constant.
45. f (x) = |x + 1|;
or
f (x) =
0,
x = −1
1,
x = −1.
1,
x = 0
0,
x = 0.
44.
f (x) =
48.
f (x) = − |x|
46. f (x) = |x2 − 1| 47.
f (x) = 2x + 5
49. (a) lim f (x) = lim− f (x) = f (2) = 2 x→2+
(b) lim− h→0
lim
h→0+
x→2
Thus, f is continuous at x = 2.
f (2 + h) − f (2) (2 + h)2 − (2 + h) − 2 = lim− =3 h→0 h h f (2 + h) − f (2) 2(2 + h) − 2 − 2 = lim =2 h h h→0+
f is not differentiable at 2.
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SECTION 3.1
50.
√ √ (x + h) x + h − x x h→0 h √ √ √ x x+h− x +h x+h = lim h→0 h √ xh = lim √ √ + lim x + h h→0 h h→0 x+h+ x √ √ √ = 12 x + x = 32 x
f (x) = lim
51. (a) f is not continuous at 0 since lim− f (x) = 1 and lim f (x) = 0. Therefore f is not differentiable x→0
at 0.
x→0+
(b) y
1
1
x
⎧ h ⎪ ⎪ lim = 1, ⎨ f (h) − 0 h→0 h = 52. (a) lim ⎪ h→0 0 h ⎪ ⎩ lim = 0, h→0 h
h rational h irrational.
f (0 + h) − f (0) does not exist. h ⎧ h2 ⎪ ⎪ lim = 0, h rational ⎨ g(h) − 0 h→0 h = (b) lim ⎪ h→0 h 0 ⎪ ⎩ lim = 0, h irrational h→0 h Therefore, lim
h→0
Therefore, g is differentiable at 0 and g (0) = 0. 53. (a) If the tangent line is horizontal, then the normal line is vertical: x = c. (b) If the tangent line has slope f (c) = 0, then the normal line has slope − y − f (c) = −
1 f (c)
1 f (c)
(x − c).
(c) If the tangent line is vertical, then the normal line is horizontal: y = f (c). 54. The center of the circle. 55. The normal line has slope −4 : y − 2 = −4(x − 4).
:
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SECTION 3.1 56.
71
y 4
2
2
x
4
57. f (x) = x2 , f (x) = 2x
3 . 2 The normal line at (3, 9) has equation y − 9 = − 16 (x − 3) and x-intercept x = 57. 111 3 . s = 57 − = 2 2
The tangent line at (3, 9) has equation y − 9 = 6(x − 3) and x-intercept x =
58. (a) The slope of the normal at P is y/x. (b) The slope of the tangent at P is −x/y. √ √ √ 1 − (x + h)2 − 1 − x2 1 − (x + h)2 − 1 − x2 1 − (x + h)2 + 1 − x2 √ = · h h 1 − (x + h)2 + 1 − x2
(c)
=
1 − (x + h)2 − (1 − x2 )
√ h 1 − (x + h)2 + 1 − x2
= h lim
h→0
−2xh − h2 1 − (x + h)2 +
1 − x2
=
−2x − h 1 − (x + h)2 +
√
| sin(1/x)| ≤ 1
it follows that
lim f (x) = f (0) = 0
x→0
and
and
− x2 ≤ g(x) ≤ x2
lim g(x) = g(0) = 0,
x→0
which implies that f and g
are continuous at 0. (b) lim
h sin(1/h) − 0 = lim sin(1/h) h→0 h
(c) lim
h2 sin(1/h) − 0 = lim h sin(1/h) = 0. Thus g is differentiable at 0 and g (0) = 0. h→0 h
h→0
h→0
1 − x2
√ 1 − x2 .
−x ≤ f (x) ≤ x Thus
√
f (x + h) − f (x) −2x − h 2x x √ = lim =− √ =− h→0 h y 2 1 − x2 1 − (x + h)2 + 1 − x2
where y = 59. (a) Since
does not exist.
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SECTION 3.1 f (2 + h) − f (2) [(2 + h)3 + 1] − (23 + 1) 12h + 6h2 + h3 = lim = lim = 12 h→0 h→0 h→0 h h h f (x) − f (2) (x3 + 1) − 9 (x − 2)(x2 + 2x + 4) f (2) = lim = lim = lim = lim (x2 + 2x + 4) = 12 x→2 x→2 x→2 x→2 x−2 x−2 x−2
60. f (2) = lim
f (1 + h) − f (1) [(1 + h)2 − 3(1 + h)] − (−2)] −h + h2 = lim = lim = −1 h→0 h→0 h→0 h h h
61. f (1) = lim
f (x) − f (1) (x2 − 3x) − (−2) (x − 2)(x − 1) = lim = lim = lim (x − 2) = −1 x→1 x→1 x→1 x→1 x−1 x−1 x−1 √ 1 + (3 + h) − 2 f (3 + h) − f (3) 4+h−2 h 1 62. f (3) = lim = lim = lim = lim √ = h→0 h→0 h→0 h→0 h 4 + h + 2 h h h 4 √ f (x) − f (3) 1+x−2 x−3 1 1 √ = lim √ = f (3) = lim = lim = lim x→3 x→1 x→3 (x − 3)( 1 + x + 2) x→3 x−3 x−3 4 1+x+2 f (1) = lim
f (−1 + h) − f (−1) (−1 + h)1/3 + 1 = lim h→0 h→0 h h
63. f (−1) = lim
(−1 + h)1/3 + 1 (−1 + h)2/3 − (−1 + h)1/3 + 1 · h→0 h (−1 + h)2/3 − (−1 + h)1/3 + 1
= lim
= lim
h→0
f (−1) = lim
x→−1
h 1 = 3 h [−1 + h)2/3 − (−1 + h)1/3 + 1]
f (x) − f (−1) x1/3 + 1 x1/3 + 1 x2/3 − x1/3 + 1 = lim = lim · 2/3 x→−1 x→−1 x − (−1) x+1 x+1 x − x1/3 + 1 = lim
x→−1
(x +
1 x+1 = 1/3 3 − x + 1)
1)(x2/3
1 1 − f (0 + h) − f (0) −h −1 1 64. f (0) = lim = lim h + 2 2 = lim = lim =− h→0 h→0 h→0 2h(h + 2) h→0 2(h + 2) h h 4 1 1 − f (x) − f (0) −x −1 1 x + 2 2 = lim f (0) = lim = lim = lim =− x→0 x→0 x→0 2x(x + 2) x→0 2(x + 2) x x 4
65. (a) D =
(2 + h)5/2 − 25/2 h
−1≤h≤1
(b) f (2) ∼ = 7.071
66. (a) D =
(2 + h)2/3 − 22/3 h
−1≤h≤1
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SECTION 3.1
73
(b) f (2) ∼ = 0.529
5 5 67. (a) f (x) = √ ; f (3) = √ 2 5x − 4 2 11 3(3 − 2x) 2 ; f (−1) = 13 − (c) f (x) = − 2 + 3x)2 2 + 3x 68. (a) f (1) does not exist
(b) f (x) = −2x + 16x3 − 6x5 ; f (−2) = 68
(b) f (−2) = 0
(c) f (3) does not exist
69.
(c) f (x) = 10x − 21x2
(d) f (x) = 0 at x = 0,
70.
(c) f (x) = 3x2 + 2x − 4
(d) f (x) = 0 at x ∼ = −1.5352, 0.8685
71. (a) f ( 32 ) = − 11 4 ;
tangent line: y −
21 8
3 = − 11 4 x− 2 ,
10 21
normal line: y −
21 8
=
4 11
x−
3 2
(c) (1.453, 1.547) 72. (a) Set f (x) = x3 . Then f (x + h) − f (x) (x + h)3 − x3 = h h =
x3 + 3x2 h + 3xh2 + h3 − x3 h
=
3x2 h + 3xh2 + h3 = 3x2 + 3xh + h2 h
f (x + h) − f (x) = lim (3x2 + 3xh + h2 ) = 3x2 . h→0 h→0 h (b) Let S be the set of integers for which the statement is true. We have shown that 1, 2, 3 ∈ S. (x + h)k − xk Assume that k ∈ S. This tells us that if f (x) = xk , then f (x) = lim = kxk−1 . h→0 h Set f (x) = xk+1 . Then, by the hint, lim
(x + h)k+1 − xk+1 x(x + h)k − x · xk + h(x + h)k = lim h→0 h→0 h h k (x + h) − xk = lim x + lim (x + h)k h→0 h→0 h
f (x) = lim
= x · kxk−1 + xk = (k + 1)xk . Therefore, k + 1 ∈ S. Thus S contains the set of positive integers.
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SECTION 3.2
SECTION 3.2 1.
F (x) = −1
2.
F (x) = 2
3.
F (x) = 55x4 − 18x2
4.
F (x) =
−6 x3
5.
F (x) = 2ax + b
6.
F (x) = x3 − x2 + x − 1
7.
F (x) = 2x−3
8.
F (x) =
x3 (2x) − (x2 + 2)3x2 x2 + 6 = − x6 x4
9. G (x) = (x2 − 1)(1) + (x − 3)(2x) = 3x2 − 6x − 1
10. F (x) = 1 +
11. G (x) =
(1 − x)(3x2 ) − x3 (−1) 3x2 − 2x3 = (1 − x)2 (1 − x)2
12. F (x) =
(cx − d)a − (ax − b)c bc − ad = 2 (cx − d) (cx − d)2
13. G (x) =
(2x + 3)(2x) − (x2 − 1)(2) 2(x2 + 3x + 1) = 2 (2x + 3) (2x + 3)2
14. G (x) =
(x + 1)(28x3 ) − (7x4 + 11)(1) 21x4 + 28x3 − 11 = 2 (x + 1) (x + 1)2
1 x2
15. G (x) = (x3 − 2x)(2) + (2x + 5)(3x2 − 2) = 8x3 + 15x2 − 8x − 10 16. G (x) =
(x2 − 1)(3x2 + 3) − (x3 + 3x)2x x4 − 6x2 − 3 = (x2 − 1)2 (x2 − 1)2
17. G (x) =
(x − 2)(1/x2 ) − (6 − 1/x)(1) −2(3x2 − x + 1) = (x − 2)2 x2 (x − 2)2
x2 (4x3 ) − (1 + x4 )2x 2(x4 − 1) = 4 x x3 1 1 8 9 (72x7 − 72x8 ) = −80x9 + 81x8 − 64x7 + 63x6 19. G (x) = (9x − 8x ) 1 − 2 + x + x x 18. G (x) =
20. G (x) =
−1 x2
1 1+ 2 x
21. f (x) = −(x − 2)−2 ; 22. f (x) = 3x3 + 2x;
−2 2 3 1 1 + 1+ =− 2 − 3 − 4 3 x x x x x
f (0) = − 14 , f (0) = 0,
f (1) = −1
f (1) = 5
23. f (x) =
(1 + x2 )(−2x) − (1 − x2 )(2x) −4x = ; (1 + x2 )2 (1 + x2 )2
24. f (x) =
(x2 + 2x + 1)(4x + 1) − (2x2 + x + 1)(2x + 2) (x + 1)(4x + 1) − 2(2x2 + x + 1) = ; 2 2 (x + 2x + 1) (x + 1)3
f (0) = −1,
f (1) =
1 4
f (0) = 0,
f (1) = −1
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SECTION 3.2 25. f (x) =
(cx + d)a − (ax + b)c ad − bc = , (cx + d)2 (cx + d)2
26. f (x) =
(cx2 + bx + a)(2ax + b) − (ax2 + bx + c)(2cx + b) ; (cx2 + bx + a)2
27. f (x) = xh (x) + h(x);
ad − bc , d2
f (1) =
f (0) =
ad − bc (c + d)2
b(a − c) 2(a − c) , f (1) = a2 a+b+c
f (0) = 0h (0) + h(0) = 0(2) + 3 = 3
28. f (x) = 6xh(x) + 3x2 h (x) − 5; 29. f (x) = h (x) +
h (x) , [h(x)]2
30. f (x) = h (x) +
h(x) − xh (x) ; h2 (x)
31. f (x) =
f (0) =
75
f (0) = −5
f (0) = h (0) +
h (0) 2 20 =2+ 2 = [h(0)]2 3 9
f (0) = h (0) +
h(0) 1 7 =2+ = h2 (0) 3 3
(x + 2)(1) − x(1) 2 = , 2 (x + 2) (x + 2)2
slope of tangent at (−4, 2) : f (−4) = 12 ;
equation for tangent: y − 2 = 12 (x + 4)
32. f (x) = (x3 − 2x + 1)(4) + (4x − 5)(3x2 − 2) = 16x3 − 15x2 − 16x + 14 slope of tangent at (2, 15) : f (2) = 50;
equation for tangent: y − 15 = 50(x − 2)
33. f (x) = (x2 − 3)(5 − 3x2 ) + (5x − x3 )(2x); slope of tangent at (1, −8) : f (1) = (−2)(2) + (4)(2) = 4; 10 ; x2 slope of tangent at (−2, 9) : f (−2) = − 32 ;
equation for tangent: y + 8 = 4(x − 1)
34. f (x) = 2x +
equation for tangent: y − 9 = − 32 (x + 2)
35. f (x) = (x − 2)(2x − 1) + (x2 − x − 11)(1) = 3(x − 3)(x + 1); f (x) = 0 at x = −1, 3; (−1, 27), (3, −5) 36. f (x) = 2x +
37. f (x) =
16 2(x3 + 8) = ; 2 x x2
f (x) = 0 at x = −2; (−2, 12)
(x2 + 1)(5) − 5x(2x) 5(1 − x2 ) = , f (x) = 0 at x = ±1; (−1, −5/2), (1, 5/2) (x2 + 1)2 (x2 + 1)2
38. f (x) = (x + 2)(2x − 2) + (x2 − 2x − 8)(1) = 3x2 − 12 = 3(x2 − 4); f (x) = 0 at x = ±2; (−2, 0), (2, −32) 39. f (x) = x4 − 8x2 + 3;
f (x) = 4x3 − 16x = 4x(x2 − 4) = 4x(x − 2)(x + 2)
(a) f (x) = 0 at x = 0, ±2 (b) f (x) > 0 on (−2, 0) ∪ (2, ∞) (c) f (x) < 0 on (−∞, −2) ∪ (0, 2)
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SECTION 3.2
40. f (x) = 3x4 − 4x3 − 2;
f (x) = 12x3 − 12x2 = 12x2 (x − 1)
(a) f (x) = 0 at x = 0, 1. (b) f (x) > 0 on (1, ∞). (c) f (x) < 0 on (−∞, 0) ∪ (0, 1) 8 x3 − 8 (x − 2)(x2 + 2x + 4) 4 ; f (x) = 1 − = = x2 x3 x3 x3 (a) f (x) = 0 at x = 2.
41. f (x) = x +
(b) f (x) > 0 on (−∞, 0) ∪ (2, ∞) (c) f (x) < 0 on (0, 2) (x2 + 4)(2x − 2) − (x2 − 2x + 4)(2x) 2(x2 − 4) x2 − 2x + 4 ; f (x) = = x2 + 4 (x2 + 4)2 (x2 + 4)2 (a) f (x) = 0 at x = 2, −2.
42. f (x) =
(b) f (x) > 0 on (−∞, −2) ∪ (2, ∞) (c) f (x) < 0 on (−2, 2) 43. slope of line 4; slope of tangent f (x) = −2x; −2x = 4 at x = −2; (−2, −10) 44. slope of line 3/5; slope of tangent f (x) = 3x2 − 3; 2 46 perpendicular when 3x2 − 3 = − 53 ; x = ± 23 ; − 23 , 46 27 , 3 , − 27 45.
f (x) = x3 + x2 + x + C
47.
f (x) =
3x2 1 2x3 − + +C 3 2 x
46.
f (x) = x4 − x2 + 4x + C
48.
f (x) =
x5 x4 √ + + x+C 5 2
49. We want f to be continuous at x = 2. That is, we want lim f (x) = f (2) = lim f (x).
x→2−
x→2+
This gives 8A + 2B + 2 = 4B − A.
(1) We also want
lim f (x) = lim f (x).
x→2−
x→2+
This gives (2)
12A + B = 4B.
Equations (1) and (2) together imply that A = −2 and B = −8. 50. First we need, Next we need,
lim f (x) = lim f (x), +
x→−1−
x→−1
lim f (x) = lim f (x) +
x→−1−
x→−1
=⇒ =⇒
Solving these equations gives A = 5, B = −3.
A + B = −B − A + 4 or A + B = 2. −2A = 5B + A or 3A + 5B = 0.
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SECTION 3.2 slope of tangent at (5, 5) is f (5) = −4
51.
tangent y − 5 = −4(x − 5) intersects x-axis at ( 25 4 , 0) normal y − 5 = 14 (x − 5) intersects x-axis at (−15, 0) area of triangle is 1 2 (5)(15
+
25 4 )
425 8
=
slope of tangent at (2, 5) is f (2) = −4
52.
tangent y − 5 = −4(x − 2) intersects x-axis at ( 13 4 , 0) normal y − 5 = 14 (x − 2) intersects x-axis at (−18, 0) area of triangle is 1 2 (5)(18
+
13 4 )
=
425 8
53. If the point (1, 3) lies on the graph, we have f (1) = 3 and thus (∗)
A + B + C = 3.
If the line 4x + y = 8 (slope −4) is tangent to the graph at (2, 0), then f (2) = 0 and f (2) = −4. Thus, (∗∗)
4A + 2B + C = 0
and
4A + B = −4.
Solving the equations in (∗) and (∗∗), we find that A = −1, 54. First, f (1) = 0 and f (1) = 3
Next, f (2) = 9 and f (2) = 18
=⇒ =⇒
B = 0,
C = 4.
A + B + C + D = 0 and 3A + 2B + C = 3 8A + 4B + 2C + D = 9 and 12A + 4B + C = 18
Solving these equations gives A = 3, B = −6, C = 6, D = −3. 55. Let f (x) = ax2 + bx + c.
Then f (x) = 2ax + b and f (x) = 0 at x = −b/2a.
56. The derivative of p is the quadratic p (x) = 3ax2 + 2bx + c. Its discriminant is D = (2b)2 − 4(3a)(c) = 4b2 − 12ac (a) p has two horizontal tangents iff p has two real roots iff D > 0. (b) p has exactly one horizontal tangent iff p has only one real root iff D = 0. (c) p has no horizontal tangent iff p has no real roots iff D < 0.
77
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SECTION 3.2
57. Let f (x) = x3 − x. The secant line through (−1, f (−1)) = (−1, 0) and (2, f (2)) = (2, 6) has slope 6−0 m= = 2. Now, f (x) = 3x2 − 1 and 3c2 − 1 = 2 implies c = −1, 1. 2 − (−1) 58. msec = f (x) =
f (3) − f (1) = 3−1
3 4
− 2
1 2
=
1 8
(x + 1)(1) − x(1) 1 = ; 2 (x + 1) (x + 1)2
f (c) =
1 8
=⇒
√ c = −1 ± 2 2
59. Let f (x) = 1/x, x > 0. Then f (x) = −1/x2 . An equation for the tangent line to the graph of f at the point (a, f (a)), a > 0, is
y = (−1/a2 )x + 2/a. The y-intercept is 2/a and the x-intercept is 2a.
The area of the triangle formed by this line and the coordinate axes is: A =
1 2
(2/a)(2a) = 2 square
units. 60. Let (x, y) be the point on the graph that the tangent line passes through. 3x2 (x − 2).
Thus x = 2 or x = −1.
The lines are y − 8 = 12(x − 2) and y + 1 = 3(x + 1).
61. Let (x, y) be the point on the graph that the tangent line passes through. x − 2 = (3x2 − 1)(x + 2). 62. (a) f (c) = c3 ;
f (x) = 3x2 , so x3 − 8 =
Thus x = 0 or x = −3.
f (x) = 3x2 − 1, so x3 −
The lines are y = −x and y + 24 = 26(x + 3).
f (x) = 3x2 and f (c) = 3c2 . Tangent line: y − c3 = 3c2 (x − c) or y = 3c2 x − 2c3 .
(b) We solve the equation 3c2 x − 2c3 = x3 : x3 − 3c2 x + 2c3 = 0
=⇒
(x − c)(x2 + cx − 2c2 ) = 0
=⇒
(x − c)2 (x + 2c) = 0
Thus, the tangent line at x = c, c = 0 intersects the graph at x = −2c. 63. Since f and f + g are differentiable, g = (f + g) − f is differentiable. The functions f (x) = |x| and g(x) = −|x| are not differentiable at x = 0 yet their sum f (x) + g(x) ≡ 0 is differentiable for all x. 64. No. If f and f g are differentiable, then g = 65. Since
fg will be differentiable where f (x) = 0. f
f f (x) 1 (x) = = f (x) · , g g(x) g(x)
it follows from the product and reciprocal rules that g(x)f (x) − f (x)g (x) 1 g (x) 1 f = + f (x) · (x) = f · (x) = f (x) − . 2 g g g(x) [g(x)] [g(x)]2 66.
(f gh) (x) = [(f g)(x) · h(x)] = (f g)(x)h (x) + h(x)[(f g)(x)] = f (x)g(x)h (x) + h(x)[f (x)g (x) + g(x)f (x)] = f (x)g(x)h(x) + f (x)g (x)h(x) + f (x)g(x)h (x)
−1 1 1 3 2 67. F (x) = 2x 1 + (2x (2x3 − x + 1) + (x2 + 1) (6x2 − 1) − x + 1) + (x + 1) 1 + x x2 x
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SECTION 3.2 1 68. G (x) = √ 2 x
1 1 + 2x
(x + x − 1) + 2
√
x
−2 (1 + 2x)2
(x + x − 1) + 2
√
x
1 1 + 2x
(2x + 1)
69. g(x) = [f (x)]2 = f (x) · f (x) g (x) = f (x)f (x) + f (x)f (x) = 2f (x)f (x) 70. g (x) = 2(x3 − 2x2 + x + 2)(3x2 − 4x + 1) 71. (a) f (x) = 0 at x = 0, −2
(b) f (x) > 0 on (−∞, −2) ∪ (0, ∞)
(c) f (x) < 0 on (−2, −1) ∪ (−1, 0) 72. (a) f (x) = 0 at x = 0, 1,
(b) f (x) > 0 on (−∞, 0) ∪ 1, 52 ∪ (5/2, ∞)
5 2
(c) f (x) < 0 on (0, 1) 73. (a) f (x) = 0 for all x = 0
(b) f (x) > 0 on (0, ∞)
(c) f (x) < 0 on (−∞, 0) √ 74. (a) f (x) = 0 at x = − 3 4 ∼ = −1.587 √ (c) f (x) < 0 on −∞, − 3 4 ∪ (0, ∞) 75. (a)
sin(0 + 0.001) − sin 0 ∼ = 0.99999 0.001
√ (b) f (x) > 0 on − 3 4, 0
sin(0 − 0.001) − sin 0 ∼ = 0.99999 −0.001
sin[(π/6) + 0.001] − sin(π/6) ∼ = 0.86578 0.001
sin[(π/6) − 0.001] − sin(π/6) ∼ = 0.86628 −0.001
sin[(π/4) + 0.001] − sin(π/4) ∼ = 0.70675 0.001
sin[(π/4) − 0.001] − sin(π/4) ∼ = 0.70746 −0.001
sin[(π/3) + 0.001] − sin(π/3) ∼ = 0.49957 0.001
sin[(π/3) − 0.001] − sin(π/3) ∼ = 0.50043 −0.001
sin[(π/2) + 0.001] − sin(π/2) ∼ sin[(π/2) − 0.001] − sin(π/2) ∼ = −0.0005 = 0.0005 0.001 −0.001 (b) cos 0 = 1, cos(π/6) ∼ = 0.866025, cos(π/4) ∼ = 0.707107, cos(π/3) = 0.5, cos(π/2) = 0 (c) If f (x) = sin x then f (x) = cos x. 76. (a)
(b)
x = −2, 0,
5 4
x1 = −2.732, x3 = 0.732,
x2 = −0.618, x4 = 1.618
79
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SECTION 3.3
SECTION 3.3 1.
dy = 12x3 − 2x dx
2.
dy = 2x − 8x−5 dx
3.
1 dy = 1+ 2 dx x
4.
(1 − x)2 − 2x(−1) dy 2 = = dx (1 − x)2 (1 − x)2
5.
dy 1 − x2 (1 + x2 )(1) − x(2x) = = dx (1 + x2 )2 (1 + x2 )2
6.
y = x3 − x2 − 2x;
7.
(1 − x)2x − x2 (−1) x(2 − x) dy = = dx (1 − x)2 (1 − x)2
8. y =
9.
10.
11. 13. 14. 15.
dy = 3x2 − 2x − 2 dx
(3 + 3x)(2 − 2x) − (2x − x2 )(3) 2x − x2 dy 2 − 2x − x2 ; = = 3 + 3x dx (3 + 3x)2 3(1 + x)2
(x3 − 1)3x2 − (x3 + 1)3x2 dy −6x2 = = dx (x3 − 1)2 (x3 − 1)2 (1 + x)2x − x2 (1) dy x2 + 2x = = 2 dx (1 + x) (1 + x)2 d (2x − 5) = 2 dx
12.
d (5x + 2) = 5 dx
d [(3x2 − x−1 )(2x + 5)] = (3x2 − x−1 )2 + (2x + 5)(6x + x−2 ) = 18x2 + 30x + 5x−2 dx d 2 (2x + 3x−1 )(2x − 3x−2 ) = (2x2 + 3x−1 )(2 + 6x−3 ) + (2x − 3x−2 )(4x − 3x−2 ) = 12x2 + 27x−4 dx t4 d 2t3 (t3 − 2) (2t3 − 1)4t3 − t4 (6t2 ) = = dt 2t3 − 1 (2t3 − 1)2 (2t3 − 1)2
2t3 + 1 t4
2 4 2(t3 + 2) − 5 =− 2 t t t5
=
d dt
=
(1 − 2u)2 − 2u(−2) 2 = (1 − 2u)2 (1 − 2u)2
=
u(2 − u3 ) (u3 + 1)(2u) − u2 (3u2 ) = (u3 + 1)2 (u3 + 1)2
2 1 + 4 t t
16.
d dt
17.
d du
18.
d du
19.
d du
20.
d 2 d 2 u (1 − u2 )(1 − u3 ) = [u − u4 − u5 + u7 ] = 2u − 4u3 − 5u4 + 7u6 du du
2u 1 − 2u u2 u3 + 1
u u − u−1 u+1
=−
(u − 1)(1) − u (u + 1)(1) − u − (u − 1)2 (u + 1)2 1 1 2(1 + u2 ) =− − = − (u − 1)2 (u + 1)2 (u2 − 1)2 =
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SECTION 3.3 21.
d dx
22.
d dx
23.
24.
25.
x3 + x2 + x + 1 x3 − x2 + x − 1
x3 + x2 + x − 1 x3 − x2 + x + 1
81
(x3 − x2 + x − 1)(3x2 + 2x + 1) − (x3 + x2 + x + 1)(3x2 − 2x + 1) (x3 − x2 + x − 1)2 −2(x4 + 2x2 + 1) −2 = 2 = (x + 1)2 (x − 1)2 (x − 1)2 =
(x3 − x2 + x + 1)(3x2 + 2x + 1) − (x3 + x2 + x − 1)(3x2 − 2x + 1) (x3 − x2 + x + 1)2 4 2 −2x + 8x + 2 = 3 (x − x2 + x + 12 ) =
d d dy = (x + 1) [(x + 2)(x + 3)] + (x + 2)(x + 3) (x + 1) = (x + 1)(2x + 5) + (x + 2)(x + 3) dx dx dx dy = (3)(9) + (4)(5) = 47. At x = 2, dx dy = (x + 1)(x2 + 2)(3x2 ) + (x + 1)(x3 + 3)(2x) + (x2 + 2)(x3 + 3)(1) dx dy = 3(6)(12) + 3(11)(4) + (6)(11) = 414. At x = 2, dx d (x + 2) [(x − 1)(x − 2)] − (x − 1)(x − 2)(1) dy dx = dx (x + 2)2 =
(x + 2)(2x − 3) − (x − 1)(x − 2) (x + 2)2
At x = 2, 26. y =
dy 4(1) − 1(0) 1 = = . dx 16 4
x4 − x2 − 2 ; x2 + 2
At x = 2,
dy (x2 + 2)(4x3 − 2x) − (x4 − x2 − 2)(2x) = dx (x2 + 2)2
dy 6(28) − (10)4 32 = = dx 36 9
27.
f (x) = 21x2 − 30x4 , f (x) = 42x − 120x3
28.
f (x) = 10x4 − 24x3 + 2, f (x) = 40x3 − 72x2
29.
f (x) = 1 + 3x−2 , f (x) = −6x−3
30.
f (x) = 2x + 2x−3 , f (x) = 2 − 6x−4
31. f (x) = 2x2 − 2x−2 − 3, 32. f (x) = 4x − 9x−1 , 33.
dy = x2 + x + 1 dx d2 y = 2x + 1 dx2 d3 y =2 dx3
f (x) = 4x + 4x−3 ,
f (x) = 4 + 9x−2 , 34.
f (x) = 4 − 12x−4
f (x) = − 18x−3
dy = 10 + 50x dx
35.
dy = 8x − 20 dx
d2 y = 50 dx2
d2 y =8 dx2
d3 y =0 dx3
d3 y =0 dx3
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40.
T1: PBU November 25, 2006
SECTION 3.3 dy = dx
1 2
x2 −
d2 y =x− dx2
39.
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1 2
x+1
37.
dy = 3x2 + 3x−4 dx
38.
d2 y = 6x − 12x−5 dx2
1 2
dy = 3x2 − 2x−2 dx d2 y = 6x + 4x−3 dx2
d3 y d3 y d3 y −6 = 1 = 6 + 60x = 6 − 12x−4 dx3 dx3 dx3 d d d d x (x − x2 ) = [x(1 − 2x)] = [x − 2x2 ] = 1 − 4x dx dx dx dx d2 d d2 2 2 −1 (x − 3x) ) = (x − 3x)(1 − x−2 ) (x + x 2 2 dx dx dx d2 2 [x − 3x − 1 + 3x−1 ] dx2 d (2x − 3 − 3x−2 ) = 2 + 6x−3 = dx =
41.
d4 d3 d2 d 4 3 [−24x] = −24 [3x − x ] = [3 − 4x ] = [−12x2 ] = 4 3 2 dx dx dx dx d5 d4 4 3 2 [ax + bx + cx + dx + e] = [4ax3 + 3bx2 + 2cx + d] dx5 dx4
42.
43. 44.
d2 dx2
=
d3 [12ax2 + 6bx + 2c] dx3
=
d d2 [24a] = 0 [24ax + 6b] = 2 dx dx
d2 d2 d2 3 (1 + 2x) 2 (5 − x ) = 2 [(1 + 2x)(−6x)] = 2 −6x − 12x2 = −24 dx dx dx d3 1 d2 4 d3 1 2 2 [x − 5x ] = − 10) (12x dx3 x dx2 dx3 x
d2 d3 [12x − 10x−1 ] = 2 [12 + 10x−2 ] 3 dx dx d [−20x−3 ] = 60x−4 = dx
=
45.
y = x4 −
x3 + 2x2 + C 3
46.
y=
x2 1 + 2 + 3x + C 2 x
47.
y = x5 −
1 +C x4
48.
y=
5 2x6 + 3 − 2x + C 3 3x
49. Let p(x) = ax2 + bx + c. Then p (x) = 2ax + b p (1) = 2a = 4 =⇒ a = 2;
and
p (x) = 2a. Now
p (1) = 2(2)(1) + b = −2 =⇒ b = −6;
p(1) = 2(1)2 − 6(1) + c = 3 =⇒ c = 7 Thus p(x) = 2x2 − 6x + 7.
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SECTION 3.3 50. p(x) = ax3 + bx2 + cx + d
p (−1) = 6
=⇒
a=1
p (x) = 3ax2 + 2bx + c
p (−1) = −2
p (x) = 6ax + 2b
p (−1) = 3
=⇒
c=4
p (x) = 6a
p(−1) = 0
=⇒
d=3
=⇒
b=2
Therefore, p(x) = x3 + 2x2 + 4x + 3. f (n) (x) = n!
51. (a) If k = n,
(b) If k > n,
f (k) (x) = n(n − 1)(n − 2) · · · (n − k + 1)xn−k .
(c) If k < n, dn = n! an dxn x2 53. Let f (x) = 0
(b)
52. (a)
(0) = lim+ (a) f+ h→0
f− (0) = lim− h→0
dk = 0 if k > n. dxk
x≥0 x≤0 f (0 + h) − f (0) h2 − 0 = lim = 0, + h h h→0 f (0 + h) − f (0) 0 = lim− = 0 h→0 h h
Therefore, f is differentiable at 0 and f (0) = 0. 2x x ≥ 0 (b) f (x) = (d) 0 x≤0 f (0 + h) − f (0) 2h − 0 = lim = 2, h h h→0+ h→0+ f (0 + h) − f (0) 0 f− (0) = lim− = lim− = 0 h→0 h→0 h h (0) = f− (0), f (0) does not exist. Since f+
(c) f+ (0) = lim
54. Let g(x) =
f (k) (x) = 0.
x3
x≥0
0
x 0.
f’ f
10 -2
-1
1
2
3
x
-10 -20
71. (a) f (x) =
1 2
x3 − 3x2 + 3x + 3;
(b)
f (x) =
3 2
x2 − 6x + 3 (c) The line tangent to the graph is horizontal; the graph turns from rising to falling, or from falling to rising.
(d) x1 ∼ = 0.586, x2 ∼ = 3.414
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SECTION 3.4 1 2
72. (a) f (x) =
f (x) =
x3 − 3x2 + 4x + 1;
3 2
x2 − 6x + 4.
87
(b)
(c) The line tangent to the graph is horizontal; the graph turns from rising to falling, or from falling to rising. (d) x1 ∼ = 0.845, x2 ∼ = 3.155
SECTION 3.4 1.
A = πr2 ,
3.
A=
5. y =
6.
dA dA = 2πr. When r = 2, = 4π. dr dr
dA 1 2 dA z , = z. When z = 4, = 4. 2 dz dz
dy −(2x + 1) 1 , = 2 . x(1 + x) dx x (1 + x)2
At x = 2,
dy = 3x2 − 24x + 45 = 3(x − 3)(x − 5); dx
7. V =
dV = 3s2 . ds
2.
V = s3 ,
4.
dy = −x−2 . dx
When s = 4,
When x = −1,
5 dy =− . dx 36
dy = 0 at x = 3, 5. dx
dS dS = 8πr and = 8πr0 at r = r0 . dr dr
9. y = 2x2 + x − 1,
10. (a) A = π
dy = 4x + 1; dx
dS =1 dr
dy = 4 at x = 34 . dx
2 d π π = d2 ; A = d 2 4 2
=⇒
r0 =
1 . 8π
Therefore x0 = 34 . (b) A = π
C 2π
2 =
√
C2 ; 4π
√ 2 3 dV 3 2 2 = w , = w . 4 dw 4 √ √ 3 √ z 3 3 dV 3 2 2 2 2 2 3 = z , = z . (b) z = s + w = 3s , z = s 3. V = s = √ 9 dz 3 3
√ 11. (a) w = s 2, V = s3 =
12. A = bh = (constant) 13. (a)
w √ 2
=⇒
3
h=
A ; b
1 dA = r2 dθ 2
2A (c) θ = 2 r 14. (a)
dy = −1. dx
4 3 dV πr , = 4πr2 = the surface area of the ball. 3 dr
8. S = 4πr2 ;
so
dA = 2πr dh
(c) h =
A − r; 2πr
dh A bh h =− 2 =− 2 =− db b b b (b)
dθ −4 −4A = 3 = 3 dr r r
1 2 r θ 2
=
dA = rθ dr
−2θ r (b)
dA = 2π(2r + h) dr
dh A −2πr(r + h) − 2πr2 2r + h =− −1= =− 2 dr 2πr 2πr2 r
dV = 48. ds
dA C = dC 2π
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SECTION 3.5
15. y = ax2 + bx + c, z = bx2 + ax + c. dy dz = 2ax + b, = 2bx + a. dx dx 1 dy dz = iff 2ax + b = 2bx + a. With a = b, this occurs only at x = . dx dx 2 16. Set k(x) = f (x)g(x)h(x).
Then k (x) = f (x)g(x)h (x) + f (x)g (x)h(x) + f (x)g(x)h(x) and k (1) = (0)(2)(0) + (0)(−1)(−2) + (1)(2)(−2) = −4
SECTION 3.5 y = 4x3 + 4x = 4x(x2 + 1)
1. y = x4 + 2x2 + 1, y = (x2 + 1)2 ,
y = 2(x2 + 1)(2x) = 4x(x2 + 1) y = 6x5 − 6x2 = 6x2 (x3 − 1)
2. y = x6 − 2x3 + 1, y = (x3 − 1)2 ,
y = 2(x3 − 1)(3x2 ) = 6x2 (x3 − 1)
3. y = 8x3 + 12x2 + 6x + 1, y = (2x + 1)3 ,
y = 3(2x + 1)2 (2) = 6(2x + 1)2
4. y = x6 + 3x4 + 3x2 + 1, y = (x2 + 1)3 ,
2
f (x) = 6x5 + 12x3 + 6x = 6x(x2 + 1)2
y = 3(x2 + 1)2 (2x) = 6x(x2 + 1)2
5. y = x2 + 2 + x−2 , y = (x + x−1 ) ,
y = 24x2 + 24x + 6 = 6(2x + 1)2
y = 2x − 2x−3 = 2x(1 − x−4 ) y = 2(x + x−1 )(1 − x−2 ) = 2x(1 + x−2 )(1 − x−2 ) = 2x(1 − x−4 ) y = 36x3 − 36x2 + 8x = 4x(3x − 2)(3x − 1)
6. y = 9x4 − 12x3 + 4x2 ,
y = 2(3x2 − 2x)(6x − 2) = 4x(3x − 2)(3x − 1)
y = (3x2 − 2x)2 ,
7. f (x) = −1(1 − 2x)−2 8. f (x) = 5(1 + 2x)4
d (1 − 2x) = 2(1 − 2x)−2 dx
d (1 + 2x) = 10(1 + 2x)4 dx
9. f (x) = 20(x5 − x10 )19
d (x5 − x10 ) = 20(x5 − x10 )19 (5x4 − 10x9 ) dx
d (x2 + x−2 ) = 6(x2 + x−2 )2 (x − x−3 ) dx 3 3 d 1 1 1 1 x− =4 x− 1+ 2 11. f (x) = 4 x − x dx x x x
10. f (x) = 3(x2 + x−2 )2
12. f (t) = (1 + t)−4 ;
f (t) = −4(1 + t)−5
13. f (x) = 4(x − x3 + x5 )3 14. f (t)= 3(t − t2 )2
d (1 + t) = −4(1 + t)−5 dt
d 3 (x − x3 + x5 ) = 4(x − x3 + x5 ) (1 − 3x2 + 5x4 ) dx
d (t − t2 ) = 3(t − t2 )2 (1 − 2t) dt
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SECTION 3.5 15. f (t)= 4(t−1 + t−2 )3 16. f (x) = 3 17. f (x) = 4
4x + 3 5x − 2 3x 2 x +1
d −1 (t + t−2 ) = 4(t−1 + t−2 )3 (−t−2 − 2t−3 ) dt
2
3
d dx d dx
4x + 3 5x − 2 3x 2 x +1
=3
=4
4x + 3 5x − 2 3x 2 x +1
2
3
(5x − 2)4 − (4x + 3)5 69(4x + 3)2 =− 2 (5x − 2) (5x − 2)4
(x2 + 1)3 − 3x(2x) 324x3 (1 − x2 ) = (x2 + 1)2 (x2 + 1)5
2 d (2x + 1)2 + (x + 1)2 18. f (x) = 3 (2x + 1)2 + (x + 1)2 dx 2 = 3 (2x + 1)2 + (x + 1)2 [2(2x + 1)(2) + 2(x + 1)(1)] 2 = 6 (2x + 1)2 + (x + 1)2 (5x + 3) 3 −2 3 −2 x x d x3 x2 x x2 x x2 19. f (x) = − (x2 + x + 1) + + + + =− + +x 3 2 1 dx 3 2 1 3 2
20. f (x) = 2[(6x + x5 )−1 + x] 21.
d [(6x + x5 )−1 + x] = 2[(6x + x5 )−1 + x][1 − (6x + x5 )−2 (6 + 5x4 )] dx
dy du −2u dy = = · (2) dx du dx (1 + u2 )2 −4 dy = = −1. dx 4
At x = 0, we have u = 1 and thus 22.
23.
dy du dy = = (1 − u−2 ) · 4(3x + 1)3 (3) dx du dx dy = 0. At x = 0, we have u = 1 and thus dx dy du (1 − 4u)2 − 2u(−4) 2 dy = = · 4(5x2 + 1)3 (10x) = · 40x(5x2 + 1)3 dx du dx (1 − 4u)2 (1 − 4u)2 At x = 0, we have u = 1 and thus
24.
dy dy du = = (3u2 − 1) · dx du dx
−2 (1 + x)2
At x = 0, we have u = 1 and thus 25.
dy = 2(−2) = −4. dx
dy dy du dx (1 + u2 )(−7) − (1 − 7u)(2u) = = (2x)(2) dt du dx dt (1 + u2 )2 =
26.
2 dy = (0) = 0. dx 9
4x(7x4 + 12x2 − 2) 4(2t − 5)[7(2t − 5)4 + 12(2t − 5)2 − 2] 7u2 − 2u − 7 (4x) = = 2 2 4 2 2 (1 + u ) (x + 2x + 2) [(2t − 5)4 + 2(2t − 5)2 + 2]2
dy du dx dy = = 2u dt du dx dt = 10u ·
(1 + x2 )(−7) − (1 − 7x)(2x) (1 + x2 )2
(5)
10[1 − 7(5t + 2)][7(5t + 2)2 − 2(5t + 2) − 7] 7x2 − 2x − 7 = (1 + x2 )2 [1 + (5t + 2)2 ]2
89
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SECTION 3.5 dy ds dt 1 dy = = 2(s + 3) · √ · (2x) dx ds dt dx 2 t−3 At x = 2, we have t = 4 so that s = 1 and thus
28.
dy dy ds dt 2 = = dx ds dt dx (1 − s)2 At x = 2, we have t =
√
1+
1 t2
2 and s =
√
1 dy = 2(4) (4) = 16. dx 2·1
1 √ 2 x
2/2. Thus
2 dy √ = dx (1 − 2/2)2
1+
1 2
1 3 √ = √ . 2 2 3 2−4
29. (f ◦ g) (0) = f (g(0))g (0) = f (2)g (0) = (1)(1) = 1 30. (f ◦ g) (1) = f (g(1))g (1) = f (1)g (1) = (1)(0) = 0 31. (f ◦ g) (2) = f (g(2))g (2) = f (2)g (2) = (1)(1) = 1 32. (g ◦ f ) (0) = g (f (0))f (0) = g (1)f (0) = (0)(2) = 0 33. (g ◦ f ) (1) = g (f (1))f (1) = g (0)f (1) = (1)(1) = 1 34. (g ◦ f ) (2) = g (f (2))f (2) = g (1)f (2) = (0)(1) = 0 35. (f ◦ h) (0) = f (h(0))h (0) = f (1)h (0) = (1)(2) = 2 36. (f ◦ h ◦ g) (1) = f (h(g(1))) h (g(1))g (1) = f (2)h (1)g (1) = (1)(1)(0) = 0 37. (g ◦ f ◦ h) (2) = g (f (h(2))) f (h(2))h (2) = g (1)f (0)h (2) = (0)(2)(2) = 0 38. (g ◦ h ◦ f ) (0) = g (h(f (0))) h (f (0))f (0) = g (2)h (1)f (0) = (1)(1)(2) = 2 39. f (x) = 4(x3 + x)3 (3x2 + 1) f (x) = 3(4)(x3 + x)2 (3x2 + 1)2 + 4(x3 + x)3 (6x) = 12(x3 + x)2 [(3x2 + 1)2 + 2x(x3 + x)] 40. f (x) = 10(x2 − 5x + 2)9 (2x − 5) f (x) = 9(10)(x2 − 5x + 2)8 (2x − 5)2 + 10(x2 − 5x + 2)9 (2) = (10)(x2 − 5x + 2)8 9(2x − 5)2 + 2(x2 − 5x + 2) 41. f (x) = 3 f (x) =
x 1−x
2 ·
1 3x2 = 2 (1 − x) (1 − x)4
6x(1 − x)4 − 3x2 (4)(1 − x)3 (−1) 6x(1 + x) = 8 (1 − x) (1 − x)5
1 x 42. f (x) = √ (2x) = √ 2 2 2 x +1 x +1
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SECTION 3.5 √ f (x) =
x x2 + 1(1) − x √ 2 1 x +1 = √ 2 2 (x + 1)3/2 x2 + 1
43. 2xf (x2 + 1)
44. f
45. 2f (x)f (x)
46.
47. f (x) = −4x(1 + x2 )−3 ;
(a)
x=0
d dx
x−1 x+1
=
2 f (x + 1)2
x−1 x+1
[f (x) + 1]f (x) − [f (x) − 1]f (x) 2f (x) = 2 [f (x) + 1] [f (x) + 1]2
(b) x < 0
48. f (x) = 2(1 − x2 )(−2x) = −4x(1 − x2 ); (c)
x−1 x+1
(c)
x>0
(a) x = −1, 0, 1
(b)
− 1 < x < 0, x > 1
x < −1, 0 < x < 1
49. f (x) =
1 − x2 ; (1 + x2 )2
(a)
x = ±1
(b) −1 < x < 1
(c)
x < −1,
x>1
50. f (x) = (1 − x2 )3 + x(3)(1 − x2 )2 (−2x) = (1 − x2 )2 (1 − 7x2 ); √ √ √ (a) x = ±1, x = ± 17 7 (b) − 17 7 < x < 17 7 √ √ (c) x < −1, −1 < x < − 17 7, 17 7 < x < 1, x > 1 51.
n! (1 − x)n+1
52.
(−1)n+1 n! (1 + x)n+1
53.
n! bn
54.
(−1)n n! abn (bx + c)n+1
55.
y = (x2 + 1)3 + C
56.
y=
(x2 − 1)2 +C 2
57.
y = (x3 − 2)2 + C
58.
y=
(x3 + 2)3 +C 3
59. L (x) =
x2
2x 1 · 2x = 2 +1 x +1
60. H (x) = 2f (x)f (x) − 2g(x)g (x) = 2f (x)g(x) − 2g(x)f (x) = 0 61. T (x) = 2f (x) · f (x) + 2g(x) · g (x) = 2f (x) · g(x) − 2g(x) · f (x) = 0 62. (a) Suppose f is even: [f (x)] = [f (−x)] = f (−x)(−1) = −f (−x); (b) Suppose f is odd:
[f (x)] = −[f (−x)] = −f (−x)(−1) = f (−x);
thus f (−x) = −f (x). thus f (−x) = f (x).
63. Suppose P (x) = (x − a)2 q(x), where q(a) = 0. Then P (x) = 2(x − a)q(x) + (x − a)2 q (x)
and P (x) = 2q(x) + 4(x − a)q (x) + (x − a)2 q (x),
and it follows that P (a) = P (a) = 0, and P (a) = 0.
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SECTION 3.5 Now suppose that P (a) = P (a) = 0 and P (a) = 0. P (a) = 0
=⇒
P (x) = (x − a)g(x)
for some polynomial g.
Then P (x) = g(x) + (x − a)g (x) and P (a) = 0
=⇒
g(a) = 0 and so g(x) = (x − a)q(x) for some polynomial q.
Therefore, P (x) = (x − a)2 q(x). Finally, P (a) = 0 implies q(a) = 0. 64. Suppose P (x) = (x − a)3 q(x), where q(a) = 0. Then P (x) = 3(x − a)2 q(x) + (x − a)3 q (x) P (x) = 6(x − a)q(x) + 6(x − a)2 q (x) + (x − a)3 q (x) P (x) = 6q(x) + 18(x − a)q (x) + 9(x − a)2 q (x) + (x − a)3 q (x) and it follows that P (a) = P (a) = P (a) = 0, P (a) = 0. Now suppose that P (a) = P (a) = P (a) = 0 and P (a) = 0. P (a) = 0
=⇒
P (x) = (x − a)g(x)
for some polynomial g.
Then P (x) = g(x) + (x − a)g (x) and P (a) = 0
=⇒
g(a) = 0 and so g(x) = (x − a)h(x) for some polynomial h.
Therefore, P (x) = (x − a)2 h(x). Now P (x) = 2h(x) + 4(x − a)h (x) + (x − a)2 h (x) and P (a) = 0
=⇒
h(a) = 0 and so h(x) = (x − a)q(x) for some polynomial q.
Therefore, P (x) = (x − a)3 q(x). Finally, P (a) = 0 implies q(a) = 0. 65. Let P be a polynomial function of degree n. The number a is a root of P of multiplicity k, (k < n) if and only if P (a) = P (a) = · · · = P (k−1) (a) = 0 and P (k) (a) = 0. √
3 2 x , 66. A = 4
67.
√ 2 3 h. Now where x = 3 √ √ √ dA 3 dA dx 2 3 2 3 = = x· = h; dh dx dh 2 3 3
√ dA = 4 when h = 2 3 dh
dr dr dV dV dr = 2 cm/sec. By the chain rule, = = 4πr2 = 8πr2 . dt dt dr dt dt At the instant the radius is 10 centimeters, the volume is increasing at the rate V =
4 3
πr3
and
dV = 8π(10)2 = 800π cm3 /sec. dt 68. V =
4 3
πr3 , S = 4πr2 , and
dV = 200. dt dS dS dr dV = · · dt dr dV dt 1 · 200 = 8πr · 4πr2
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SECTION 3.5 =
93
400 r
= 80 when r = 5 The surface area is increasing 80 cm2 /sec. at the instant the radius is 5 centimeters. 2k dF dr 2k dF (49 − 9.8t), 0 ≤ t ≤ 10 = · = 3 · (49 − 9.8t) = dt dr dt r (49t − 4.9t2 )3 dF 2k 2k dF (19.6); (19.6). (3) = (7) = − (b) dt (102.9)3 dt (102.9)3
69. (a)
1 1 ; tangent T : y = − 12 x − 54 . 70. (a) f (9) = −2, f (9) = − 12
y
(b)
(c) (7.4, 10.8)
f
2
4
6
8
10
x
-1 -2
T
71. (a) f (1) = 12 , f (1) = − 12 ; tangent T : y = − 12 x + 1. y
(b)
(c) (0.8, 1.2)
1
f 1
x
2 T
72.
4 d4 d = 288 35x5 + 20x3 + x x2 4 x2 + 1 dx dx
1 f (1/x) d f =− 73. (a) dx x x2 (c)
74. (a)
(c) 75.
d dx
2 4x f (x2 − 1)/(x2 + 1) x −1 d f = (b) dx x2 + 1 (1 + x2 )2
f (x) f (x) = 1 + f (x) [1 + f (x)]2
d [u1 (u2 (x))] = u1 (u2 (x)) u2 (x) dx
(b)
d [u1 (u2 (u3 (x)))] = u1 [u2 (u3 (x)] u2 (u3 (x)) dx u3 (x)
d [u1 (u2 [u3 (u4 (x))])] = u1 [u2 (u3 [u4 (x)])] u2 (u3 [u4 (x)]) u3 [u4 (x)]u4 (x) dx
d2 2 [f (g(x))] = [g (x)] f [g(x)] + f [g(x)]g (x) dx2
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SECTION 3.6
SECTION 3.6 1.
dy = −3 sin x − 4 sec x tan x dx
2.
dy = 2x sec x + x2 sec x tan x dx
3.
dy = 3x2 csc x − x3 csc x cot x dx
4.
dy = 2 sin x cos x dx
5.
dy = −2 cos t sin t dt
6.
dy = 6t tan t + 3t2 sec2 t dt
7. 8. 10.
12. 13.
15.
√ d √ √ √ √ d √ √ dy = 4 sin3 u (sin u ) = 4 sin3 u cos u ( u ) = 2u−1/2 sin3 u cos u du du du dy = csc u2 − 2u2 csc u2 cot u2 du √ dy 1 = − √ sin x dx 2 x
9.
d dy = sec2 x2 (x2 ) = 2x sec2 x2 dx dx
11.
dy = 4[x + cot πx]3 [1 − π csc2 πx] dx
dy = 3(x2 − sec 2x)2 (2x − 2 sec 2x tan 2x) = 6(x2 − sec 2x)2 (x − sec 2x tan 2x) dx dy d2 y = − sin x = cos x, dx dx2
14.
dy d2 y = − cos x = − sin x, dx dx2
dy (1 + sin x)(− sin x) − cos x (cos x) − sin x − (sin2 x + cos2 x) = = = −(1 + sin x)−1 2 dx (1 + sin x) (1 + sin x)2 d2 y d (1 + sin x) = cos x (1 + sin x)−2 = (1 + sin x)−2 dx2 dx
16.
dy = 3 tan2 (2πx) sec2 (2πx)(2π) = 6π tan2 (2πx) sec2 (2πx) dx d2 y = 6π(2) tan(2πx) sec2 (2πx) sec2 (2πx)(2π) + 6π tan2 (2πx)(2) sec(2πx)[sec(2πx) tan(2πx)](2π) dx2 = 24π 2 tan(2πx) sec2 (2πx)[sec2 (2πx) + tan2 (2πx)]
17.
dy d = 3 cos2 2u (cos 2u) = −6 cos2 2u sin 2u du du d2 y d d = −6[cos2 2u (sin 2u) + sin 2u (cos2 2u)] du2 du du = −6[2 cos3 2u + sin 2u (−4 cos 2u sin 2u)] = 12 cos 2u [2 sin2 2u − cos2 2u]
18.
dy = 5 sin4 (3t) cos(3t)(3) = 15 sin4 (3t) cos(3t) dt d2 y = 15(4) sin3 (3t) 3 cos2 (3t) + 15 sin4 (3t)[−3 sin(3t)] = 45 sin3 (3t)[4 cos2 (3t) − sin2 (3t)] dt2
19.
dy = 2 sec2 2t, dt
d2 y d = 4 sec 2t (sec 2t) = 8 sec2 2t tan 2t dt2 dt
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SECTION 3.6 d2 y = −4(2) csc(4u)[− csc(4u) cot(4u)(4)] = 32 csc2 (4u) cot(4u) du2
20.
dy = −4 csc2 4u; du
21.
dy = x2 (3 cos 3x) + 2x sin 3x dx d2 y = [x2 (−9 sin 3x) + 2x(3 cos 3x)] + [2x(3 cos 3x) + 2(sin 3x)] dx2 = (2 − 9x2 ) sin 3x + 12x cos 3x
22.
(1 − cos x) cos x − sin x (−[− sin x]) cos x − 1 −1 dy = = = 2 2 dx (1 − cos x) (1 − cos x) 1 − cos x d2 y = sin x(1 − cos x)−2 dx2
23. y = sin2 x + cos2 x = 1 so
d2 y dy = 2 =0 dx dx
24. y = sec2 x − tan2 x = 1 so
d2 y dy = 2 =0 dx dx
25. 26. 27.
d4 d3 d2 d (− cos x) = sin x (sin x) = (cos x) = (− sin x) = 4 3 2 dx dx dx dx d3 d2 d d4 (sin x) = cos x (cos x) = 3 (− sin x) = 2 (− cos x) = 4 dx dx dx dx d 2d d 2 d2 t 2 (t cos 3t) = t (cos 3t − 3t sin 3t) dt dt dt dt d 2 [t (−3 sin 3t − 3 sin 3t − 9t cos 3t)] dt d = [−6t2 sin 3t − 9t3 cos 3t] dt
=
= (−18t2 cos 3t − 12t sin 3t) + (27t3 sin 3t − 27t2 cos 3t) = (27t3 − 12t) sin 3t − 45t2 cos 3t 28.
d d d d 2 t (cos t ) = −t sin t2 (2t) = −2t2 sin t2 dt dt dt dt = −4t sin t2 − 2t2 cos t2 (2t) = −4t(sin t2 + t2 cos t2 )
29.
d d [f (sin 3x)] = f (sin 3x) (sin 3x) = 3 cos 3xf (sin 3x) dx dx
30.
d [sin f (3x)] = cos[f (3x)]f (3x)(3) = 3f (3x) cos[f (3x)] dx
31.
dy = cos x; dx
slope of tangent at (0, 0) is 1; tangent: y = x.
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33.
34. 35.
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SECTION 3.6 √ dy = sec2 x; slope of tangent at (π/6, 3/3) is sec2 (π/6) = 4/3; dx √ tangent: y − 13 3 = 43 x − 16 π π √ dy = − csc2 x; slope of tangent at ( , 3) is −4, an equation for dx 6 √ π tangent: y − 3 = −4 x − . 6 dy = − sin x; dx
slope of tangent at (0, 1) is 0;
√ π √ dy = sec x tan x, slope of tangent at ( , 2) is 2, an equation for dx 4 √ √ π tangent is y − 2 = 2 x − . 4 √ dy = − csc x cot x, slope of tangent at (π/3, 2 3/3) is − 2/3; dx √ tangent: y − 23 3 = − 23 x − 13 π .
37.
dy = − sin x; x = π dx
39.
√ dy = cos x − 3 sin x; dx
40.
√ dy = − sin x − 3 cos x; dx
41.
π 3π dy = 2 sin x cos x = sin 2x; x = , π, dx 2 2
42.
dy = −2 sin x cos x = − sin 2x; dx
43.
dy = sec2 x − 2; dx
44.
dy = −3 csc2 x + 4; dx
45.
46.
tangent: y = 1.
dy = cos x; dx
38. dy =0 dx
dy =0 dx
dy =0 dx
gives
x=
gives
dy =0 dx
1 tan x = √ ; 3
gives
√ tan x = − 3;
π 7π , 6 6 x=
x=
2 csc x = ± √ ; 3
π 3π 5π 7π , , , 4 4 4 4 x=
π 2π 4π 5π , , , 3 3 3 3
dy dy = 2 sec x tan x + sec2 x; since sec x is never zero, =0 dx dx 7π 11π , 2 tan x + sec x = 0 so that sin x = −1/2; x = 6 6 dy = − csc2 x + 2 csc x cot x; dx 2 cot x − csc x = 0
so that
2π 5π , 3 3
π 3π , π, 2 2
√ sec x = ± 2;
gives
x=
x = 12 π, x = 32 π
since csc x is never zero, cos x = 1/2;
x=
π 5π , 3 3
gives
dy =0 dx
gives
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SECTION 3.6 f (x) = 1 − 2 sin x.
47. f (x) = x + 2 cos x, (a) xf (x) = 0 : f :
+ 0 π 6
1 − 2 sin x = 0,
sin x = 1/2,
-
+ 5π 6
x = π/6, 5π/6.
2π
(b) f (x) > 0 on (0, π/6) ∪ (5π/6, 2π), 48. f (x) = x −
√
f (x) = 1 − √ 1 − 2 cos x = 0,
2 sin x,
(a) f (x) = 0 : f :
− 0
√
(c) f (x) < 0 on (π/6, 5π/6).
2 cos x. cos x =
√
2/2,
+
π 4
7 π 2π 4
(b) f (x) > 0 on (π/4, 7π/4),
(a) f (x) = 0 :
cos x − sin x = 0,
+ 0
(c) f (x) < 0 on (0, π/4) ∪ (7π/4, 2π).
f (x) = cos x − sin x.
49. f (x) = sin x + cos x,
f :
-
cos x = sin x,
5π 4
50. f (x) = sin x − cos x, (a) f (x) = 0 :
2π
(c) f (x) < 0 on (π/4, 5π/4).
f (x) = cos x + sin x.
cos x + sin x = 0, +
0
x = π/4, 5π/4.
+
π 4
(b) f (x) > 0 on (0, π/4) ∪ (5π/4, 2π),
f :
x = π/4, 7π/4.
cos x = − sin x,
3π 4
x = 3π/4, 7π/4.
+ 7π 2π 4
(b) f (x) > 0 on (0, 3π/4) ∪ (7π/4, 2π),
(c) f (x) < 0 on (3π/4, 7π/4).
dy du dx dy = = (2u)(sec x tan x)π = 2π sec2 πt tan πt dt du dx dt dy = 2 sec πt (sec πt tan πt)π = 2π sec2 πt tan πt (b) y = sec2 πt − 1, dt
51. (a)
2 2 1 1 1 dy du dx dy = = 3 (1 + u) (− sin x)(2) = 3 (1 + cos 2t) (− sin 2t) 52. (a) dt du dx dt 2 2 2 = 3(cos4 t)(−2 sin t cos t) = −6 cos5 t sin t 3 1 dy (1 + cos 2t) = cos6 t; = 6 cos5 t(− sin t) = −6 cos5 t sin t (b) y = 2 dt
53. (a)
3 3 1 1 1 dy du dx dy − = = 4 (1 − u) (− sin x)(2) = 4 (1 − cos 2t) sin 2t dt du dx dt 2 2 2 = 4 sin6 t (2 sin t cos t) = 8 sin7 t cos t
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SECTION 3.6 4
1 (1 − cos 2t) (b) y = 2 54. (a)
= sin8 t,
dy = 8 sin7 t cos t dt
dy du dx dy = = (−2u)(− csc x cot x)(3) = (−2 csc(3t)(− csc(3t) cot(3t)3 dt du dx dt
= 6 csc2 (3t) cot(3t) dy = −2 csc(3t)[− csc(3t) cot(3t)](3) = 6 csc2 (3t) cot(3t) (b) y = 1 − csc2 (3t); dt n odd (−1)(n+1)/2 sin x, dn (cos x) = 55. n n/2 dx (−1) cos x, n even −1 d cos x sin x(− sin x) − cos x(cos x) d = = − csc2 x (cot x) = = dx dx sin x sin2 x sin2 x 1 d −1 1 sin x d (sec x) = = (− sin x) = · = sec x tan x (b) 2 dx dx cos x cos x cos x cos x 1 −1 cos x d d −1 (cos x) = (c) (csc x) = = · = − csc x cot x 2 dx dx sin x sin x sin x sin x
56. (a)
57.
π d d π (cos x) = sin − x = − cos − x = − sin x dx dx 2 2
58. Differentiating both sides, 2 cos 2x = 2(cos2 x − sin2 x).
Thus cos 2x = cos2 x − sin2 x.
59. f (0) = lim
sin(0 + h) − sin 0 sin h sin x = lim = lim x→0 x h→0 h h
60. f (0) = lim
cos(0 + h) − cos 0 cos h − 1 cos x − 1 = lim = lim x→0 h→0 h h x
h→0
h→0
61.
f (x) = 2 sin x + 3 cos x + C
62.
f (x) = tan x + cot x + C
63.
f (x) = sin 2x + sec x + C
64.
f (x) =
− cos 3x csc 2x + +C 3 2
65.
f (x) = sin(x2 ) + cos 2x + C
66.
f (x) =
tan(x3 ) + sec 2x + C 3
67. (a) f (x) = sin(1/x) + x cos(1/x)(−1/x2 ) = sin(1/x) − (1/x) cos(1/x) g (x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2 ) = 2x sin(1/x) − cos(1/x)
(b) lim g (x) = lim [2x sin(1/x) − cos(1/x)] = − lim cos(1/x) does not exist x→0
x→0
x→0
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SECTION 3.6
99
68. (a) f must be continuous at 0: lim f (x) = lim cos x = 1,
x→0+
x→0+
lim f (x) = lim− (ax + b) = b; thus b = 1
x→0−
x→0
Differentiable at 0 : lim
f (h) − f (0) cos h − 1 = lim+ = 0, h h h→0
lim
f (h) − f (0) ah + 1 − 1 = lim+ =a h h h→0
h→0+
h→0−
Therefore, f is differentiable at 0 if a = 0 and b = 1. (b) y 1 f
x
1
69. (a) Continuity:
√
√ 3 2πa 2πa lim + (ax + b) = + b; thus +b= 3 3 2 x→2π/3
3 , lim − sin x = 2 x→2π/3 Differentiability: lim
x→2π/3−
1 cos x = − , 2
lim
x→2π/3+
(a) = a; thus a = −
Therefore, f is differentiable at 2π/3 if a = − 12 and b = (b)
1 2
√
1 2
3 + 13 π
y 1
g
2π 3
x
70. (a) Continuity: lim (1 + a cos x) = 1 +
x→π/3−
1 2
a,
lim [b + sin (x/2)] = b +
x→π/3+
1 2
which implies b =
1 2
+
Differentiability: √ lim − (−a sin x) = − 12 3 a,
x→π/3
lim + [ 12 cos (x/2)] =
x→π/3
1 4
√
Therefore, f is differentiable at π/3 if a = − 12 and b = 14 .
3 which implies a = − 12 .
1 2
a.
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SECTION 3.6 y
(b)
1
f
x
π 3
71. Let y(t) = A sin ωt + B cos ωt. Then y (t) = ω A cos ωt − ω B sin ωt
and y (t) = −ω 2 A sin ωt − ω 2 B cos ωt
Thus, d2 y + ω 2 y = 0. dt2 72. (a) θ = a sin(ωt + φ);
θ = aω cos(ωt + φ);
θ = −aω 2 sin(ωt + φ)
Thus, θ satisfies the equation. (b) θ = a sin(ωt + φ0 ) = a sin(ωt) cos φ0 − a cos(ωt) sin φ0 = A sin(ωt) + B cos(ωt) where A = −a sin φ0 , B = a cos φ0 73. A = 74. c =
1 2
√
c2 sin x;
dA 1 = c2 cos x dx 2 dc 1 ab sin x = √ (2ab sin x) = √ 2 2 2 dx 2 a + b − 2ab cos x a + b2 − 2ab cos x
a2 + b2 − 2ab cos x;
75. (a) f (4p) (x) = k 4p cos kx,
f (4p+1) (x) = −k 4p+1 sin kx,
f (4p+2) (x) = −k 4p+2 cos kx,
f (4p+3) (x) = k 4p+3 sin kx, p = 0, 1, 2, . . . (b) m = k 2 , k = 1, 2, 3, . . . 76. f (x) = A cos f (0) = 2
√
=⇒
2 x + B sin
√ 2 x;
A = 2;
√ √ √ √ f (x) = −A 2 sin 2 x + B 2 cos 2 x √ 3 3 2 f (0) = −3 =⇒ B = − √ = − 2 2
77. f has horizontal at the points with x-coordinate 78. f (x) = sin x − sin2 x has horizontal tangents at: 79. f (x) = sin x, f (x) = cos x; y = x.
1 3 2 π, 2 π,
π
1 6, 4
∼ = 3.39, ∼ = 6.03
π 5π 1 3π , 2 , 0 , 6 , 4 , 2 , −2
f (0) = 0, f (0) = 1. Therefore an equation for the tangent line T is
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SECTION 3.7
101
The graphs of f and T are: y T f
1
-
π 2
x
π 2
-1
| sin x − x| < 0.01 on (−0.39, 0.39). 80. f (x) = tan x, f (x) = sec2 x;
f (π/4) = 1, f (π/4) = 2. Therefore an equation for the tangent line
T is y = 2x + 1 − 12 π. The graphs of f and T are: y f T 1 π 4
-1
π 2
x
| tan x − (2x + 1 − 12 π)| < 0.01 on (0.712, 0.852).
SECTION 3.7 1.
x2 + y 2 = 4 2x + 2y
3.
dy =0 dx dy −x = dx y 4.
dy =0 dx dy −4x = dx 9y
√ √ x+ y = 4 1 dy 1 √ + √ =0 2 x 2 y dx √ y dy =−√ dx x
x4 + 4x3 y + y 4 = 1 4x3 + 12x2 y + 4x3
x3 + y 3 − 3xy = 0 dy 2 2 dy 3x + 3y −3 y+x =0 dx dx dy y − x2 = 2 dx y −x
4x2 + 9y 2 = 36 8x + 18y
5.
2.
dy dy + 4y 3 =0 dx dx dy x3 + 3x2 y =− 3 dx x + y3
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SECTION 3.7 x2 − x2 y + xy 2 + y 2 = 1
6. 2x − 2xy − x2
dy dy dy + y 2 + 2xy + 2y =0 dx dx dx dy 2xy − 2x − y 2 = dx 2xy + 2y − x2
(x − y)2 − y = 0 dy dy − =0 2(x − y) 1 − dx dx
7.
dy 2(x − y) = dx 2(x − y) + 1 (y + 3x)2 − 4x = 0 dy 2(y + 3x) +3 −4 = 0 dx
8.
2 dy = −3 + dx y + 3x 9.
sin (x + y) = xy dy dy cos (x + y) 1 + =x +y dx dx dy y − cos (x + y) = dx cos (x + y) − x
10. tan xy = xy;
dy sec (xy) y + x dx
2
=y+x
dy ; dx
dy y =− dx x
y 2 + 2xy = 16
11. 2y
dy dy + 2x + 2y = 0 dx dx dy + y = 0. (x + y) dx
Differentiating a second time, we have d2 y dy (x + y) 2 + dx dx Substituting
−y dy = , dx x+y
x2 − 2xy + 4y 2 = 3 dy dy + 8y =0 dx dx dy = 0. x − y + (4y − x) dx 2x − 2y − 2x
dy 2+ dx
= 0.
we have
d2 y y (x + y) 2 − dx (x + y) 12.
2x + y x+y
= 0,
d2 y 2xy + y 2 16 = = . dx2 (x + y)3 (x + y)3
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SECTION 3.7 Differentiating a second time, we have dy dy d2 y dy + 4 −1 + (4y − x) 2 = 0. 1− dx dx dx dx Substituting
y−x dy = , dx 4y − x
we have d2 y 3(x2 − 2xy + 4y 2 ) −9 = − = . 2 3 dx (4y − x) (4y − x)3
y 2 + xy − x2 = 9
13. 2y
dy dy +x + y − 2x = 0. dx dx
Differentiating a second time, we have 2 dy dy d2 y d2 y dy 2 + −2 = 0 + 2y 2 + x 2 + dx dx dx dx dx 2 dy d2 y dy (2y + x) 2 + 2 + − 1 = 0. dx dx dx Substituting
2x − y dy = , dx 2y + x
we have
(2x − y)2 + (2x − y)(2y + x) − (2y + x)2 d2 y (2y + x) 2 + 2 =0 dx (2y + x)2 d2 y 10(y 2 + xy − x2 ) 90 = = . 2 3 dx (2y + x) (2y + x)3
14.
x2 − 3xy = 18 2x − 3y − 3x
dy = 0. dx
Differentiating a second time, we have 2−3 Substituting
2x − 3y dy = , dx 3x
dy d2 y dy −3 − 3x 2 = 0 dx dx dx
we have
d2 y 6(x − 3y) 6(x2 − 3xy) 6(18) 12 = − = − =− =− 3 dx2 9x2 9x3 9x3 x 15.
4 tan y = x3 4 sec2 y
dy = 3x2 dx dy 3 = x2 cos2 y dx 4
d2 y dy 3 3 2 2 x cos x 2 cos y(− sin y) = y + dx2 2 4 dx =
3 9 x cos2 y − x4 sin y cos3 y 2 8
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SECTION 3.7
16.
sin2 x + cos2 y = 1 dy =0 dx dy =0 sin 2x − sin 2y dx dy sin 2x = dx sin 2y
2 sin x cos x − 2 cos y sin y
dy sin 2y(cos 2x)2 − sin 2x(cos 2y)2 d2 y dx = dx2 sin2 2y Substituting
dy sin 2x = and using the double angle formulas, we find that dx sin 2y 8 cos2 x cos2 y(sin2 y − sin2 x) − sin2 x sin2 y(cos2 y − cos2 x) d2 y =0 = dx2 sin3 2y
since sin2 x = sin2 y and cos2 x = cos2 y from the original equation. dy = 0. dx 5 dy = . Then, dx 8
17. x2 − 4y 2 = 9,
2x − 8y
At (5, 2), we get
d2 y 2−8 y 2 + dx At (5, 2) we get
2 25 d y =0 2−8 2 2 + dx 64
At (2, −1) we get 19. cos (x + 2y) = 0
dy dx
2 = 0.
so that
9 d2 y =− . dx2 128
dy dy + 3y 2 = 0. dx dx dy dy dy +3 = 0 so = 0. Differentiating again, 4−4+8 dx dx dx 2 dy d2 y dy d2 y dy +4 + 4x 2 + 6y + 3y 2 2 = 0 2+4 dx dx dx dx dx
18. x2 + 4xy + y 3 + 5 = 0, At (2, −1), we get
2x + 4y + 4x
d2 y 2 d2 y = 0 so =− 2 dx dx2 11 dy − sin (x + 2y) 1 + 2 = 0. dx
2 + 11
dy = −1/2. Then, dx 2 2 d y dy − sin (x + 2y) 2 2 = 0. − cos (x + 2y) 1 + 2 dx dx
At (π/6, π/6),
we get
At (π/6, π/6),
we get − cos
π π 2 (0) − sin 2 2
2
d2 y dx2
=0
so that
d2 y = 0. dx2
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SECTION 3.7 20. x = sin2 y
1 = 2 sin y cos y
At (1/2, π/4),
we get
At (1/2, π/4),
we get
21.
dy = 1. Differentiating again, dx 2 dy d2 y 0 = 2 cos 2y + sin 2y 2 . dx dx d2 y = 0. dx2 22.
2x + 3y = 5 2+3
9x2 + 4y 2 = 72
dy =0 dx
slope of tangent at (−2, 3) :
18x + 8y −2/3
dy =0 dx
slope of tangent at (2, 3) :
− 32
tangent: y − 3 = − 23 (x + 2)
tangent: y − 3 = − 32 (x − 2)
normal: y − 3 = 32 (x + 2)
normal: y − 3 = 23 (x − 2)
23.
x2 + xy + 2y 2 = 28 2x + x
x3 − axy + 3ay 2 = 3a3
24.
dy dy + y + 4y =0 dx dx
slope of tangent at (−2, −3) :
25.
dy dy = sin 2y . dx dx
3x2 − ay − ax
−1/2
dy dy + 6ay =0 dx dx
slope of tangent at (a, a) :
− 25
tangent: y + 3 = − 12 (x + 2)
tangent: y − a = − 25 (x − a)
normal: y + 3 = 2(x + 2)
normal: y − a = 52 (x − a)
dy dx dy 1 = − sin y dx 1 π −2 slope of tangent at , : √ 2 3 3 2 π 1 tangent: y − = − √ x− 3 2 3 √ 3 1 π x− normal: y − = 3 2 2 x = cos y
26.
tan xy = x dy 2 sec (xy) y + x =1 dx
slope of tangent at (1, π/4) :
2−π 4
2−π π = (x − 1) 4 4 4 π (x − 1) y− =− 4 2−π
tangent: y − normal:
27.
dy d 1 3 −1/2 = (x3 + 1)−1/2 (x3 + 1) = x2 (x3 + 1) dx 2 dx 2
29.
d 1 dy = (2x2 + 1)−3/4 (2x2 + 1) = x(2x2 + 1)−3/4 dx 4 dx
30.
dy 3x + 4 = 13 (x + 1)−2/3 (x + 2)2/3 + (x + 1)1/3 23 (x + 2)−1/3 = dx 3(x + 1)2/3 (x + 2)1/3
31.
−x −x x(2x2 − 5) dy √ = 2 − x2 √ + 3 − x2 √ =√ dx 3 − x2 2 − x2 2 − x2 3 − x2
32.
√ dy = 32 (x4 − x + 1)1/2 (4x3 − 1) = 32 x4 − x + 1 (4x3 − 1) dx
28.
dy = 13 (x + 1)−2/3 dx
105
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SECTION 3.7
33.
d dx
34.
d dx
35.
d dx
36.
QC: PBU/OVY
JWDD027-Salas-v1
d dx
√
1 x+ √ x
√
3x + 1 2x + 5
x
x2 + 1 x
1 1 1 d (x1/2 + x−1/2 ) = x−1/2 − x−3/2 = x−3/2 (x − 1) dx 2 2 2
1 = 2
3x + 1 2x + 5
−1/2
(2x + 5)3 − (3x + 1)2 (2x + 5)2
13 = 2(2x + 5)2
2x + 5 3x + 1
d x(x2 + 1)−1/2 dx 1 2 −3/2 = x − (x + 1) (2x) + (x2 + 1)−1/2 = (x2 + 1)−3/2 2 =
x2 + 1
√
=
=
37. (a)
x
1 2
√
(x2 + 1)−1/2 (2x) − (x2 + 1)1/2 = x2
x2 − x2 + 1 2 −1 x +1 = √ x2 x2 x2 + 1
(b)
(c)
1 dy = (a2 + x2 )−1/2 (2x) = x(a2 + x2 )−1/2 ; dx 2
38. y = (a2 + x2 )1/2 ;
d2 y 1 a2 2 2 −1/2 2 2 −3/2 x(a = (a + x ) − + x ) (2x) = dx2 2 (a2 + x2 )3/2 b dy = (a + bx)−2/3 ; dx 3
39. y = (a + bx)1/3 ; 40. y = x(a2 − x2 )1/2 ;
dy = (a2 − x2 )1/2 + dx
d2 y −2b2 (a + bx)−5/3 = 2 dx 9 1 2
x(a2 − x2 )−1/2 (−2x) = (a2 − x2 )1/2 − x2 (a2 − x2 )−1/2
1 2 d2 y x(2x2 − 3a2 ) 1 2 2 −1/2 2 2 −1/2 2 2 −3/2 = (a − x ) (−2x) − 2x(a − x ) − x − x ) (−2x) = − (a 2 2 dx2 (a2 − x2 )3/2 √ √ √ √ √ 1 1 dy 1 1 2 √ = √ tan x + x sec x 41. y = x tan x; = √ tan x + sec2 x dx 2 2 x 2 x 2 x √ √ √ √ √ √ √ √ √ d2 y 2 x sec2 x (1/2 x) − tan x(1/ x) = + sec x sec x tan x(1/2 x) 2 dx 4x √ √ √ √ √ x sec2 x − tan x + 2x sec2 x tan x √ = 4x x √
42. y =
√
√ √ x sin x;
√ √ √ 1 dy = √ sin x + x cos x dx 2 x
1 √
2 x
√ √ 1 1 = √ sin x + cos x 2 2 x
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SECTION 3.7
107
√ √ √ √ √ √ √ d2 y 2 x cos x(1/2 x) − sin x(1/ x) 1 = − sin x(1/2 x) dx2 4x 2 √ √ √ √ x cos x − sin x sin x − √ = 4 x 4x3/2 dy = 0 so that the slope of the normal line is dx −1 y = (x = 0). dy/dx x
43. Differentiation of x2 + y 2 = r2 gives 2x + 2y
Let (x0 , y0 ) be a point on the circle. Clearly, if x0 = 0, the normal line, x = 0, passes through the origin.
If x0 = 0, the normal line is y0 y − y0 = (x − x0 ), x0
which simplifies to y =
y0 x, x0
a line through the origin. dy = 1; dx
dy 1 = dx 2y √ When x = a, y = ± a.
44. y 2 = x;
2y
√
1 √ 2 a √ 1 Equation of tangent line: y − a = √ (x − a) 2 a
Slope of tangent at (a,
x-intercept:
a) :
√ 1 − a = √ (x − a) 2 a
√ Slope of tangent at (a, − a) : Equation of tangent line: y + x-intercept:
=⇒
x = −a
1 − √ 2 a √
1 a = − √ (x − a) 2 a
√ 1 a = − √ (x − a) 2 a
=⇒
x = −a
dy = 2p and the slope of a tangent is given by m1 = p/y. dx 2 2 For the parabola y = p − 2px, we obtain m2 = −p/y as the slope of a tangent. The parabolas
45. For the parabola y 2 = 2px + p2 , we have 2y
intersect at the points (0, ±p). At each of these points m1 m2 = −1; the parabolas intersect at right angles.
46. For y = 2x, the slope is m1 = 2. For x2 − xy + 2y 2 = 28, we have 2x − y − x
dy dy + 4y =0 dx dx
so
dy y − 2x = m2 = dx 4y − x
At a point of intersection of the line and the curve, we have m2 = 0 since y = 2x. Thus tan α = | − m1 | = 2
=⇒
α∼ = 1.107(radians) ∼ = 63.4◦
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SECTION 3.7 dy dy dy = 2x; for x = y 3 we have 3y 2 = 1 or m2 = = 1/3y 2 . dx dx dx At (1, 1), m1 = 2, m2 = 1/3 and m1 − m2 2 − (1/3) =1 ⇒ α= π = tan α = 1 − m1 m2 1 + 2(1/3) 4
47. For y = x2 we have m1 =
At (0, 0), m1 = 0 and m2 is undefined. Thus α = π/2. 48. (x − 1)2 + y 2 = 10; x2 + (y − 2)2 = 5;
2(x − 1) + 2y 2x + 2(y − 2)
dy = 0, dx
dy = 0, dx
m1 = m2 =
dy x−1 =− dx y
x dy =− dx y−2
The circles intersect at the points (−2, 1) and (2, 3). 3+2 = 1. Thus α = π . At (−2, 1) : m1 = 3, m2 = −2 and tan α = 1 + (3)(−2) 4 (−1/3) + 2 = 1. Thus α = π . At (2, 3) : m1 = −1/3, m2 = −2 and tan α = 1 + (−1/3)(−2) 4 49. The hyperbola and the ellipse intersect at the points (±3, ±2). For the hyperbola, the ellipse
x dy = and for dx y
4x 4x2 dy = − . The product of these slopes is − 2 . This product is −1 at each of the points dx 9y 9y
of intersection. Therefore the hyperbola and ellipse are orthogonal.
50. The curves intersect at the points (±1, 1). For the ellipse,
dy 3x =− dx 2y
and for y 3 = x2 we have
dy 2x 6x2 x2 = 2 . The product of these slopes is − 3 = − 3 . This product is −1 at each of the points of dx 3y 6y y intersection. Therefore the curves are orthogonal.
51. For the circles,
dy x dy y = − , y = 0, and for the straight lines, = m = , x = 0. Since the product dx y dx x
of the slopes is −1, it follows that the two families are orthogonal trajectories. 52. For the parabolas, m1 =
dy dy 1 2x = , y = 0, and for the ellipses, m2 = = − , y = 0. Let (x0 , y0 ) dx 2ay dx y
be a point of intersection of a parabola and an ellipse. Then m 1 · m2 =
x0 1 2x0 = − 2 = −1 since x0 = ay02 . · − 2ay0 y0 ay0
53. The line x + 2y + 3 = 0 has slope m = −1/2. Thus, a line perpendicular to this line will have slope 2. A tangent line to the ellipse 4x2 + y 2 = 72 has slope m =
dy 4x 4x = − . Setting − = 2 gives y = −2x. dx y y
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SECTION 3.7
109
Substituting into the equation for the ellipse, we have 4x2 + 4x2 = 72
⇒
8x2 = 72
⇒
x = ±3
It now follows that y = ∓6 and the equations of the tangents are: at (3, −6) :
y + 6 = 2(x − 3)
or
y = 2x − 12;
at (−3, 6) :
y − 6 = 2(x + 3)
or y = 2x + 12.
54. The line 2x + 5y − 4 = 0 has slope m = −2/5. A tangent line to the hyperbola 4x2 − y 2 = 36 has slope 4x y y 2 dy = . Therefore a normal line to the hyperbola will have slope m = − . Setting − =− dx y 4x 4x 5 gives y = 8x/5. Substituting this into the equation for the hyperbola, we have 4x2 −
64 2 x = 36 25
=⇒
x = ±5
It now follows that y = 8 when x = 5 and y = −8 when x = −5. The equations of the normals are: at (5, 8) :
y − 8 = − 25 (x − 5)
at (−5, −8) :
or
y + 8 = − 25 (x + 5)
y = − 25 x + 10; or y = − 25 x − 10.
55. Differentiate the equation (x2 + y 2 )2 = x2 − y 2 implicitly with respect to x : dy dy 2 2 = 2x − 2y 2(x + y ) 2x + 2y dx dx Now set dy/dx = 0. This gives 2x(x2 + y 2 ) = x x2 + y 2 =
1 2
(x = 0)
Substituting this result into the original equation, we get x2 − y 2 = Now
√
√
x2 + y 2 = 1/2
6 , ⇒ x=± 4 x2 − y 2 = 1/4
1 4
y=±
2 4
Thus, the points on the curve at which the tangent line is horizontal are: √ √ √ √ √ √ √ √ ( 6/4, 2/4), ( 6/4, − 2/4), (− 6/4, 2/4), (− 6/4, − 2/4). y 1/3 dy dy = 0, and =− dx dx x 1/3 y1 Thus, the slope at (x1 , y1 ), x1 = 0 is: m = − x1 1/3 y1 (b) m = 0 : − = 0 =⇒ y1 = 0 and x1 = ± a; (a, 0), (−a, 0) x1
56. (a) x2/3 + y 2/3 = a2/3 ;
2 −1/3 3x
+ 23 y −1/3
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SECTION 3.7 m=1:
−
m = −1 :
−
y1 x1
y1 x1
1/3 =1
√ y1 = −x1 and x1 = ± 14 a 2;
=⇒
1 √ 1 √ 1 √ 1 √ a 2, − a 2 , − a 2, a 2 4 4 4 4
1/3 = −1
√ y1 = x1 and x1 = ± 14 a 2;
=⇒
1 √ 1 √ 1 √ 1 √ a 2, a 2 , − a 2, − a 2 4 4 4 4
57. Differentiate the equation x1/2 + y 1/2 = c1/2 implicitly with respect to x : y 1/2 1 −1/2 1 −1/2 dy dy + y x = 0 which implies =− 2 2 dx dx x An equation for the tangent line to the graph at the point (x0 , y0 ) is 1/2 y0 (x − x0 ) y − y0 = − x0 The x- and y-intercepts of this line are a = (x0 y0 )1/2 + x0 Now
and b = (x0 y0 )1/2 + y0
respectively.
2 1/2 1/2 a + b = 2(x0 y0 )1/2 + x0 + y0 = x0 + y0 = c.
58. The circle has equation x2 + (y − a)2 = 1 and 2x + 2(y − a) A tangent line to the parabola has slope 4x = −
x y−a
=⇒
dy = 4x. Now dx
x(4y − 4a + 1) = 0
It follows that 1 y =a− 4 √ 15 15 Points of intersection: ± , 4 8 √ 3 33h = 2/3 59. (a) d(h) = h h (b) d(h) → ∞ as h → 0−
=⇒
dy x dy = 0. Thus =− . dx dx y−a
4y − 4a + 1 = 0 since x = 0
=⇒
√ 15 x=± 4
=⇒
and as h → 0+
(c) The graph of f has a vertical tangent at (0, 0). √ 3 3 3 h2 = 1/3 h h (b) d(h) → −∞ as h → 0− ;
60. (a) d(h) =
and d(h) → ∞
(c) The graph is said to have a “cusp” at (0, 0). 61. f (x) > 0 on (−∞, ∞)
as
h → 0+
y=
15 8
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SECTION 3.7 √
3 ; 3
62. (a) f (x) = 0 at x = 63.
64.
67.
x = t,
(b) f (x) > 0 on
√ 4 − t2
y=
√
x = t,
dy 3 =− dx (2,1) 2
dy =3 dx (3,4)
65.
3/3, ∞ ;
(c) f (x) < 0 on
√ 0, 3/3
√ y = − 4 − t2
66.
dy =0 dx (0,π/6)
√ dy =− 3 √ dx (1,3 3) dy = −1; dx (3,3)
68. (b) x = 3 implies y = 3; (c)
tangent line: y − 3 = −(x − 3) or x + y = 6.
y
4 3 2
-2
2
3
4
x
-2
69. (a) The graph of x4 = x2 − y 2 is:
(b) Differentiate the equation x4 = x2 − y 2 implicitly with respect to x :
y
4x3 = 2x − 2y 0.05
dy dx
Now set dy/dx = 0. This gives 4x3 = 2x -0.5 -0.5
0.5
x
which implies x = ±
√
2 2 .
111
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REVIEW EXERCISES
70. The graph of (2 − x)y 2 = x3 is: y 15 10 5 5
10
15
x
-5 -10 -15
REVIEW EXERCISES f (x + h) − f (x) [(x + h)3 − 4(x + h) + 3] − [x3 − 4x + 3] = lim h→0 h→0 h h
f (x) = lim
1.
3x2 h + 3xh2 + h3 − 4h = lim (3x2 + 3xh + h2 − 4) = 3x2 − 4. h→0 h→0 h
= lim
√ 1 + 2(x + h) − 1 + 2x f (x + h) − f (x) f (x) = lim = lim h→0 h→0 h h √ √ 1 + 2(x + h) − 1 + 2x 1 + 2(x + h) + 1 + 2x · = lim √ h→0 h 1 + 2(x + h) + 1 + 2x
2.
= lim
h→0
h
2h 1 + 2(x + h) +
√
1
=√ 1 + 2x 1 + 2x
1 1 − f (x + h) − f (x) g (x) = lim = lim x + h − 2 x − 2 h→0 h→0 h h
3.
−1 (x − 2) − (x + h − 2) −1 = lim = . h→0 h(x + h − 2)(x − 2) h→0 (x + h − 2)(x − 2) (x − 2)2
= lim
4.
F (x + h) − F (x) (x + h) sin (x + h) − x sin x = lim h→0 h→0 h h
F (x) = lim
(x + h)(sin x cos h + cos x sin h) − x sin x h→0 h
= lim
x sin x(cos h − 1) + h sin x cos h + x cos x sin h + h cos x sin h h cos h − 1 sin h + lim sin x cos h + x cos x lim + lim cos x sin h = x sin x lim h→0 h→0 h→0 h→0 h h = lim
h→0
= sin x + x cos x. 5. y = 23 x−1/3
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REVIEW EXERCISES 6. y = 2( 34 )x−1/4 − 4(− 14 )x−5/4 = 7. y =
3 2
x−1/4 + x−5/4 .
(x3 − 1)(2x + 2) − (1 + 2x + x2 )3x2 x4 + 4x3 + 3x2 + 2x + 2 = − . (x3 − 1)2 (x3 − 1)2
8. f (t) = 3(2 − 3t2 )2 (−6t) = −18t(2 − 3t2 )2 . 1 x f (x) = − (a2 − x2 )−3/2 (−2x) = 2 . 2 (a − x2 )3/2
9. f (x) = (a2 − x2 )−1/2 ;
b 2b b b . = 2 a− 10. y = 2 a − x x2 x x
2 2 2b 6b b b − 3 =− 3 a+ 2 . 11. y = 3 a + 2 x x x x 12. y =
√ √ x 3x . 2 + 3x + (2 + 3x)−1/2 (3) = 2 + 3x + √ 2 2 2 + 3x
13. y = sec
2
√
2x + 1
1
2 (2x
−1/2
+ 1)
√ sec2 2x + 1 (2) = √ . 2x + 1
14. g (x) = x2 [− sin (2x − 1)(2)] + 2x cos (2x − 1) = 2x cos (2x − 1) − 2x2 sin (2x − 1). 15. F (x) = (x + 2)2 12 (x2 + 2)−1/2 (2x) + 16. y =
√
√ x(x + 2)2 x2 + 2 (2)(x + 2) = 2(x + 2) x2 + 2 + √ . x2 + 2
(a2 − x2 )2x − (a2 + x2 )(−2x) 4a2 x = 2 . 2 2 2 (a − x ) (a − x2 )2
17. h (t) = sec t2 + t(2t) sec t2 tan t2 + 6t2 = sec t2 + 2t2 tan t2 sec t2 + 6t2 . 18. y =
(1 + cos x)2 cos 2x − sin 2x(− sin x) 2 cos 2x sin 2x sin x = . + 2 (1 + cos x) 1 + cos x (1 + cos x)2
1 19. s = 3
2 − 3t 2 + 3t
−2/3
·
(2 + 3t)(−3) − (2 − 3t)(3) 4 =− . 2 4/3 (2 + 3t) (2 + 3t) (2 − 3t)2/3
√ √ 2θ2 20. r = θ2 ( 12 )(3 − 4θ)−1/2 (−4) + 2θ 3 − 4θ = 2θ 3 − 4θ − √ . 3 − 4θ 21. f (θ) = −3 csc2 (3θ + π). 22. y =
(1 + x2 )(sin 2x + 2x cos 2x) − x sin 2x(2x) sin 2x + 2x cos 2x + 2x3 cos 2x − x2 sin 2x = . 2 2 (1 + x ) (1 + x2 )2
23. f (x) = 13 x−2/3 + 12 x−1/2 = 24. f (x) =
1 1 + 1/2 ; 3x2/3 2x
f (64) =
1 1 1 + = . 48 16 12
x x2 1 √ (−2x) + 8 − x2 = 8 − x2 − √ ; 2 8 − x2 8 − x2
f (2) = 0.
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REVIEW EXERCISES
2
2
2
2
25. f (x) = x (2)(π) sin πx cos πx + 2x sin πx = 2x sin πx + 2πx sin πx cos πx; 26. f = −3 csc2 3x;
√ 1 π 3 f (1/6) = + . 12 72
f (π/9) = −4.
27. f (x) = 2x3 − x2 + 3, f (x) = 6x2 − 2x;
f (1) = 4.
Tangent line: y − 4 = 4(x − 1) or y = 4x; normal line: y − 4 = − 14 (x − 1) or y = − 14 x + 28. f (x) =
2x − 3 (3x + 4)2 − (2x − 3)3 17 , f (x) = = ; 3x + 4 (3x + 4)2 (3x + 4)2
f (−1) = 17.
1 (x + 1). normal line: y + 5 = − 17
Tangent line: y + 5 = 17(x + 1);
29. f (x) = (x + 1) sin 2x, f (x) = 2(x + 1) cos 2x + sin 2x; Tangent line: y = 2x;
17 4 .
f (0) = 2.
normal line: y = − 12 x.
√ √ √ x2 30. f (x) = x 1 + x2 , f (x) = 1 + x2 + √ ; f (1) = 32 2. 1 + x2 √ √ √ √ 3 Tangent line: y − 2 = 2 2 (x − 1); normal line: y − 2 = − 13 2 (x − 1) 31. f (x) = − sin(2 − x)(−1) = sin(2 − x), f (x) = cos(2 − x)(−1) = − cos(2 − x). 32. f (x) = 32 (x2 + 4)1/2 (2x) = 3x(x2 + 4)1/2 , f = 3x( 12 )(x2 + 4)−1/2 (2x) + 3(x2 + 4)1/2 =
(x2
3x2 6x2 + 12 + 3(x2 + 4)1/2 = √ . 1/2 + 4) x2 + 4
33. y = x cos x + sin x, y = 2 cos x − x sin x. 34. g (u) = 2 tan u sec2 u, g (u) = 2 tan u · 2 sec u · sec u tan u + 2 sec4 u = 4 sec2 u tan2 u + 2 sec4 u. 35.
(−1)n n!bn
36.
37. 3x2 y + x3 y + y 3 + 3xy 2 y = 0,
y = −
38. (1 + 2y ) sec2 (x + 2y) = 2xy + x2 y ,
1 xy 3xy √ 40. 2x + 3 y + √ = − 2 , 2 y y y
y =
41. 2x + 2y + 2xy − 6yy = 0, y =
y 3 + 3x2 y . x3 + 3xy 2
y =
39. 6x2 + 3 cos y − 3xy sin y = 2y + 2xy ,
(−1)n n!abn (bx + c)n+1
2xy − sec2 (x + 2y) . 2 sec2 (x + 2y) − x2
y =
6x2 + 3 cos y − 2y . 2x + 3x sin y
2y 2 − 4xy 3 − 6y 7/2 . 2xy + 3xy 5/2
x+y ; 3y − x
at (3, 2) y =
5 . 3
Tangent line: y − 2 = 53 (x − 3) or 5x − 3y = 9; normal line: y − 2 = − 35 (x − 3) or 3x + 5y = 19.
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REVIEW EXERCISES 42. 2y cos 2x + y sin 2x − xy cos y − sin y = 0, y =
sin y − 2t cos 2x ; sin 2x − x cos y
115
at ( 14 π, 12 π) y = 1.
Tangent line: y − 12 π = x − 14 π or y = x + 14 π; normal line: y − 12 π = −(x − 14 π) or y = −x + 34 π. 43. f (x) = 3(x − 4)(x − 2). f :
+
2
+ 4
(a) f (x) = 0 at x = 2, 4,
(b) f (x) > 0 on (−∞, 2) ∪ (4, ∞),
(c) f (x) < 0 on (2, 4). 44. f (x) =
2 − 4x2 . (1 + 2x2 )2
f :
-
+ -
1 2
1 2
√ (a) f (x) = 0 at x = ± 2/2, √ √ (c) f (x) < 0 on (−∞, − 2/2) ∪ ( 2/2, ∞).
√ √ (b) f (x) > 0 on (− 2/2, 2/2),
45. f (x) = 1 + 2 cos 2x. f :
+
π 3
-
+ 2π 3
4π 3
+ 5π 3
(a) f (x) = 0 at x = π/3, 2π/3, 4π/3, 5π/3, (b) f (x) > 0 on (0, π/3) ∪ (2π/3, 4π/3) ∪ (5π/3, 2π), (c) f (x) < 0 on (π/3, 2π/3) ∪ (4π/3, 5π/3). 46. f (x) = f :
√
3 + 2 sin x. +
4π 3
0
5π 3
2π
(a) f (x) = 0 at x = 4π/3, 5π/3,
(b) f (x) > 0 on (0, 4π/3) ∪ (5π/3, 2π),
(c) f (x) < 0 on (4π/3, 5π/3). 47. y =
√ x.
(a) 1,
(b) 3,
(c)
1 3.
48. y = 3x2 . The tangent line to the curve y = x3 at the point (a, a3 ) has equation y − a3 = 3a2 (x − a)
or
y = 3a2 x − 2a3 .
Since this line must pass through (0, 2), we have 2 = −2a3
which implies a = −1.
There is one tangent line that passes through (0, 2), namely y = 3x + 2.
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REVIEW EXERCISES
49. y = 3x2 − 1. The tangent line to the curve y = x3 − x at the point (a, a3 − a) has equation y − (a3 − a) = (3a2 − 1)(x − a)
or
y = (3a2 − 1)x − 2a3 .
Since this line must past through (−2, 2), we have 2 = −6a2 + 2 − 2a3
which gives
6a2 + 2a3 = 0
and
2a2 (3 + a) = 0.
Thus, a = 0, −3. There are two tangent lines that pass through (−2, 2) : y = −x, y = 26x + 54. 50. The curve y = Ax2 + Bx + C passes through the points (1, 3) and (2, 3). This implies that A + B + C = 3 and 4A + 2B + C = 3. y = 2Ax + B. The line x − y + 1 = 0 has slope 1. At x = 2, y = 4A + B. Therefore we have 4A + B = 1. The solution set of the system of equations A+B+C = 3 4A + 2B + C = 3 4A + B = 1 is: A = 1, B = −3, C = 5;
y = x2 − 3x + 5.
51. The curve y = Ax3 + Bx2 + Cx + D passes through the points (1, 1) and (−1, −9). This implies that A + B + C + D = 1 and −A + B − C + D = −9. y = 3Ax2 + 2Bx + C. The line y = 5x − 4 has slope 5; the line y = 9x has slope 9. At x = 1, y = 3A + 2B + C; at x = −1, y = 3A − 2B + C. Therefore we have 3A + 2B + C = 5 and 3A − 2B + C = 9. The solution set of the system of equations A+B+C +D = 1 −A + B − C + D = −9 3A + 2B + C = 5 3A − 2B + C = 9 is: A = 1, B = −1, C = 4, D = −3; 52.
y = x3 − x2 + 4x − 3.
1 1 1 xn − (x + h)n 1 − n = h (x + h)n x h (x + h)n xn 1 xn − xn − nxn−1 h − (terms of the form Cxk hm , m ≥ 2) = h (x + h)n xn =
−nxn−1 − (terms of the form Cxk hm , m ≥ 1) . (x + h)n xn
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117
Therefore,
1 −nxn−1 − (terms of the form Cxk hm , m ≥ 1) 1 1 = lim lim − h→0 h (x + h)n h→0 xn (x + h)n xn =
−nxn−1 −n = n+1 . x2n x
f (x) = 2x − 2.
53. Let f (x) = x2 − 2x + 1;
(1 + h)2 − 2(1 + h) + 1 (1 + h)2 − 2(1 + h) + 1 − 0 = lim = f (1) = 0 h→0 h→0 h h √ √ 54. Let f (x) = x; f (x) = 1/2 x. √ 1 9+h−3 lim = f (9) = h→0 h 6 lim
55. Let f (x) = sin x;
f (x) = cos x. sin( 16 π + h) − lim h→0 h
1 2
√ 3 1 = f ( π) = 6 2
f (x) = 5x4 .
56. Let f (x) = x5 ;
x5 − 32 = f (2) = 80 x→2 x − 2 lim
57. Let f (x) = sin x;
f (x) = cos x. lim
x→π
sin x = f (π) = −1 x−π
58. (a) From Figure A, M increases on [a, b], is constant on [b, e], and increases on [e, ∞); M is differentiable at b (f (b) = 0), M is not differentiable at e (f (e) = 0). (b) From Figure B, m is constant on [a, c], decreases on [c, d], and is constant on [d, ∞); m is not differentiable at c (f (c) = 0), m is differentiable at d (f (d) = 0).
a
b
e
Figure A
a
b
e
Figure B
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SECTION 4.1
CHAPTER 4 SECTION 4.1 1. f is differentiable on (0, 1), continuous on [0, 1]; and f (0) = f (1) = 0. √ √ 3 3 2 2 f (c) = 3c − 1; 3c − 1 = 0 =⇒ c = − ∈ / (0, 1) 3 3 2. f is differentiable on (−2, 2), continuous on [−2, 2]; and f (−2) = f (2) = 0. f (c) = 4c3 − 4c;
4c(c2 − 1) = 0 =⇒ c = 0, ±1
3. f is differentiable on (0, 2π), continuous on [0, 2π]; and f (0) = f (2π) = 0. f (c) = 2 cos 2c; Thus, c =
2 cos 2c = 0 =⇒ 2c =
π 3π 5π 7π , , , 4 4 4 4
π π nπ + nπ, and c = + , n = 0, ±1, ±2 . . . 2 4 2
4. f is differentiable on (0, 8), continuous on [0, 8]; and f (0) = f (8) = 0. f (c) =
2 3
c−1/3 −
3 6. f (c) = √ − 4, 2 c 7. f (c) = 3c ,
2 3
c−1/3 ,
9. f (c) = √ (−
2 c1/3 − 1 3 c2/3
f (c) = 0
=⇒
2c = 3
=⇒
−10 − (−1) f (b) − f (a) = = −3; b−a 4−1
−c , 1 − c2
4−1 3 f (b) − f (a) = = ; b−a 8−1 7
2 3
c = 3/2 3 √ − 4 = −3 2 c
c−1/3 =
0−1 f (b) − f (a) = = −1; b−a 1−0
√
1√ 2 is not in [a, b]) 2
c = 1.
10. f (c) = 3c − 3, 2
f (b) − f (a) −2 − 2 = = −2; b−a 1 − (−1)
3 7
=⇒
3c − 3 = −2 2
(14)3 93
=⇒
=⇒
c=
−x(5 − x2 ) √ , (3 + x2 )2 1 − x2
12. (a) f (x) =
2 3
x−1/3 =
f (c) = 0 for c in (−1, 1) implies c = 0.
2 = 0 for all x ∈ (−1, 1). 3x1/3
(b) f (0) does not exist. Therefore, f is not differentiable on (−1, 1).
1√ 2 2
√ 3 c=± 3
11. f is continuous on [−1, 1], differentiable on (−1, 1) and f (−1) = f (1) = 0. f (x) =
c = 9/4
1√ − 39 is not in [a, b] 3
c=
−c = −1 1 − c2
=⇒
f (b) − f (a) 1√ 27 − 1 39 = = 13; 3c2 = 13 =⇒ c = b−a 3−1 3
2
8. f (c) =
c−2/3 =
4−1 f (b) − f (a) = = 3; b−a 2−1
5. f (c) = 2c,
2 3
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119
13. No. By the mean-value theorem there exists at least one number c ∈ (0, 2) such that f (c) =
f (2) − f (0) 3 = > 1. 2−0 2
14. No, by Rolle’s theorem: f (2) = f (3) = 1 but there is no value c ∈ (2, 3) such that f (c) = 0. 15. By the mean-value theorem there is a number c ∈ (2, 6) such that f (6) − f (2) = f (c)(6 − 2) = f (c)4. Since 1 ≤ f (x) ≤ 3 for all x ∈ (2, 6), it follows that 4 ≤ f (6) − f (2) ≤ 12. 16. f (x) = x2 + x + 3, f (x) = 2x + 1. The slope of the line through (−1, 3) and (2, 9) is 2. Setting 2x + 1 = 2, we get x = 12 . The point on the graph of f where the tangent line is parallel to the line through (−1, 3) and (2, 9) is: (1/2, 15/4). 17. f is everywhere continuous and everywhere differentiable except possibly at x = −1. f is continuous at x = −1: as you can check, lim f (x) = 0,
x→−1−
lim f (x) = 0,
and f (−1) = 0.
x→−1+
f is differentiable at x = −1 and f (−1) = 2: as you can check, lim
h→0−
f (−1 + h) − f (−1) =2 h
and
f (−1 + h) − f (−1) = 2. h
lim
h→0+
Thus f satisfies the conditions of the mean-value theorem on every closed interval [ a, b ].
2,
x ≤ −1
3x − 1,
x > −1
f (x) =
2
f (2) − f (−3) 6 − (−4) = = 2. 2 − (−3) 2 − (−3)
f (c) = 2 with c ∈ (−3, 2) 18. f is continuous and differentiable everywhere; 2 3x , x ≤ 1 f (x) = 3, x > 1 f (2) − f (−1) 6−1 5 = = 2 − (−1) 3 3
For c ≤ 1, f (c) = 3c = 2
For c > 1, f (c) = 3 =
5 3
5 3
√ =⇒
c=±
5 3
iff
c = 1 or −3 < c ≤ −1.
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SECTION 4.1
19. Let f (x) = Ax2 + Bx + C. Then f (x) = 2Ax + B. By the mean-value theorem f (c) =
f (b) − f (a) (Ab2 + Bb + C) − (Aa2 + Ba + C) = b−a b−a =
A(b2 − a2 ) + B(b − a) = A(b + a) + B b−a
Therefore, we have 2Ac + B = A(b + a) + B =⇒ c = 20.
f (1) − f (−1) = 1 and f (x) = −1/x2 < 0; 1 − (−1)
a+b 2
f is not continuous at 0.
f (1) − f (−1) = 0 and f (x) is never zero. This result does not violate the mean-value theorem 1 − (−1) since f is not differentiable at 0; the theorem does not apply. 2x − 4, x ≥ 1/2 22. f (x) = −2x − 2, x < 1/2 2, x > 1/2 f (x) = −2, x < 1/2 21.
f (x) = 0 for all x = 12 . Rolle’s theorem is not violated because f is not differentiable at x = 12 . 23. Set P (x) = 6x4 − 7x + 1. If there existed three numbers a < b < c at which P (x) = 0, then by Rolle’s theorem P (x) would have to be zero for some x in (a, b) and also for some x in (b, c). This is not the case: P (x) = 24x3 − 7 is zero only at x = (7/24)1/3 . 24. Set P (x) = 6x5 + 13x + 1. Note that P (−1) < 0 and P (0) > 0. By the intermediate-value theorem, the equation P (x) = 0 has at least one real root c. If this equation had another real root d, then by Rolle’s theorem P (x) would have to be zero for some x between c and d. This is not the case: P (x) = 30x4 + 13 is never zero. 25. Set P (x) = x3 + 9x2 + 33x − 8. Note that P (0) < 0 and P (1) > 0. Thus, by the intermediate-value theorem, there exists some number c between 0 and 1 at which P (x) = 0. If the equation P (x) = 0 had an additional real root, then by Rolle’s theorem there would have to be some real number at which P (x) = 0. This is not the case: P (x) = 3x2 + 18x + 33 is never zero since the discriminant b2 − 4ac = (18)2 − 12(33) < 0. 26. (a)
Suppose that f has two zeros, x1 , x2 ∈ (a, b). Then, f is differentiable on (x1 , x2 ) and continuous
on [x1 , x2 ]. By Rolle’s theorem, f has a zero in (x1 , x2 ) which contradicts the hypothesis. (b) If f had three zeros in (a, b), then, by Rolle’s’ theorem, f would have at least two zeros in (a, b) and f would have at least one zero in (a, b) which contradicts the hypothesis.
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121
27. Let c and d be two consecutive roots of the equation P (x) = 0. The equation P (x) = 0 cannot have two or more roots between c and d for then, by Rolle’s theorem, P (x) would have to be zero somewhere between these two roots and thus between c and d. In this case c and d would no longer be consecutive roots of P (x) = 0. 28. If f (x) = 0 at a1 , a2 , . . . , an then by Rolle’s theorem, f (x) is zero at some number b1 ∈ (a1 , a2 ), at some number b2 ∈ (a2 , a3 ), . . . , at some number bn−1 ∈ (an−1 , an ); f (x), in turn, must be zero at some number c1 ∈ (b1 , b2 ), at some number b2 ∈ (b2 , b3 ), . . . , at some number cn−2 ∈ (bn−2 , bn−1 ). 29. Suppose that f has two fixed points a, b ∈ I, with a < b. Let g(x) = f (x) − x. Then g(a) = f (a) − a = 0 and g(b) = f (b) − b = 0. Since f is differentiable on I, we can conclude that g is differentiable on (a, b) and continuous on [a, b]. By Rolle’s theorem, there exists a number c ∈ (a, b) such that g (c) = f (c) − 1 = 0 or f (c) = 1. This contradicts the assumption that f (x) < 1 on I. 30. Set P (x) = x3 + ax + b. It is obvious that for x sufficiently large, P (x) > 0 and for x sufficiently large negative, P (x) < 0. Thus, by the intermediate-value theorem, the equation P (x) = 0 has at least one real root. If a ≥ 0, then P (x) = 3x2 + a is positive, except possibly at 0, where it remains nonnegative. It follows that P is everywhere increasing and therefore it cannot take on the value 0 more than once. √ √ Suppose now that a < 0. Then − 13 3 |a| and 13 3 |a| are consecutive roots of the equation P (x) = 0 and thus, by Exercise 27, P cannot take on the value zero more than once between these two numbers. 31. (a) f (x) = 3x2 − 3 < 0 for all x in (−1, 1). Also, f is differentiable on (−1, 1) and continuous on [−1, 1]. Thus there cannot be a and b in (−1, 1) such that f (a) = f (b) = 0, or they would contradict Rolle’s theorem. (b) When f (x) = 0, b = 3x − x3 = x(3 − x2 ). When x is in (−1, 1), then |x(3 − x2 )| < 2. Thus |b| < 2. 32. f (x) = 3x2 − 3a2 > 0 for all x in(−a, a). Also, f is differentiable on (−a, a) and continuous on [−a, a]. Thus there cannot be b and c in (−a, a) such that f (b) = f (c) = 0, or they would contradict Rolle’s theorem. 33. For p(x) = xn + ax + b,
1 p (x) = nxn−1 + a, which has at most one real zero for n even x = − na n−1 .
If there were more than two distinct real roots of p(x), then by Rolle’s theorem there would be more than one zero of p (x). Thus there are at most two distinct real roots of p(x). 34. For p(x) = xn + ax + b,
p (x) = nxn−1 + a, which has at most two real zeros for n odd. If there were
more than three distinct real roots of p(x), then by Rolle’s theorem there would be more than two zeros of p (x). Thus there are at most three distinct real roots of p(x).
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SECTION 4.1
35. If x1 = x2 , then |f (x1 ) − f (x2 )| and |x1 − x2 | are both 0 and the inequality holds. If x1 = x2 , then by the mean-value theorem f (x1 ) − f (x2 ) = f (c) x1 − x2 for some number c between x1 and x2 . Since |f (c)| ≤ 1: f (x1 ) − f (x2 ) ≤ 1 and thus |f (x1 ) − f (x2 )| ≤ |x1 − x2 |. x1 − x2 36. See the proof of Theorem 4.2.2. 37. Set, for instance, f (x) =
1, a < x < b 0, x = a, b
38. (a) Let f (x) = cos x. Choose any numbers x and y, (assume x < y). By the mean-value theorem, there is a number c between x and y such that f (y) − f (x) = f (c) y−x
⇒
| cos y − cos x| = | − sin c| ≤ 1 |y − x|
⇒
| cos x − cos y| ≤ |x − y|
(b) Repeat the in part (a) with f (x) = sin x. 39. (a) By the mean-value theorem, there exists a number c ∈ (a, b) such that f (b) − f (a) = f (c)(b − a). If f (x) ≤ M for all x ∈ (a, b), then it follows that f (b) ≤ f (a) + M (b − a) (b) If f (x) ≥ m for all x ∈ (a, b), then it follows that f (b) ≥ f (a) + m(b − a) (c) If |f (x)| ≤ K on (a, b), then −K ≤ f (x) ≤ K on (a, b) and the result follows from parts (a) and (b). f (x) . Then h is defined on [a, b] and h(a) = g(x) h(b) = 0. Therefore, by Rolle’s theorem, there exists a number c ∈ (a, b) such that
40. Assume that g(x) = 0 for all x ∈ [a, b] and let h(x) =
h (c) =
g(c)f (c) − f (c)g (c) =0 g 2 (c)
Thus g(c)f (c) − f (c)g (c) = 0 which contradicts the given condition f (x)g (x) − g(x)f (x) = 0 for all x ∈ I. Thus, g has at least one zero in (a, b). By reversing the roles of f and g, the same argument can be used to show that g cannot have two (or more) zeros on (a, b). 41. We show first that the conditions on f and g imply that f and g cannot be simultaneously 0. Set h(x) = f 2 (x) + g(x). Then h (x) = 2f (x)f (x) + 2g(x)g (x) = 2f (x)[g(x)] + 2g(x)[−f (x)] = 0 =⇒ f 2 (x) + g 2 (x) = C constant.
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123
If C = 0, then f 2 (x) = −g 2 (x) =⇒ f (x) ≡ 0, contradicting the assumptions on f . Therefore, f 2 (x) + g 2 (x) = C, C > 0. If f (α) = 0, then g(α = 0. Assume that f (a) = f (b) = 0 and f (x) = 0 on (a, b). Suppose that g(x) = 0 on (a, b). We know also that g(a) = 0 and g(b) = 0. Set h(x) = f (x)/g(x). Then h is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Therefore, by Rolle’s theorem, there exists at least one point c (a, b) such that h (c) = 0. But, h (x) =
g 2 (x) + f 2 (x) C g(x)f (x) − f (x)g (x) = = 2 >0 2 2 g (x) g (x) g (x)
for all x (a, b), and we have a contradiction. Thus g must have at least one zero in (a, b). Since g (x) = −f (x) = 0 in (a, b), g has exactly one zero in (a, b). Simply reverse the roles of f and g to show that f has exactly one zero between two consecutive zeros of g. 42. We prove the result for h > 0. The proof for h < 0 is similar. If f is differentiable on (x, x + h), it is continuous there and thus, by the hypothesis at x and x + h continuous on [x, x + h]. By the mean-value theorem, there exists c in (x, x + h) for which f (x + h) − f (x) = f (c). x+h−x Multiplying through by (x + h) − x = h, we have f (x + h) − f (x) = f (c)h. Since c is between x and x + h, c can be written c = x + θh 43.
with
0 < θ < 1.
f (x0 + h) − f (x0 ) f (x0 + θh)h = lim = lim f (x0 + θh) h→0 h→0 h→0 h h ∧ (by the hint) = lim f (x) = L x→x0 ∧ (by 2.2.6)
f (x0 ) = lim
44. Suppose that f (a) = f (b) = k, and let g(x) = f (x) − k. Then g is differentiable on (a, b), continuous on [a, b], and g(a) = g(b) = 0. Therefore, by Rolle’s theorem, there exists at least one number c ∈ (a, b) such that g (c) = 0. Since g (x) = f (x), it follows that f (c) = 0. 45. Using the hint, F is continuous on [a, b], differentiable on (a, b), and F (a) = F (b). By Exercise 44, there is a number c in (a, b) such that F (c) = 0. Therefore [f (b) − f (a)]g (c) − [g(b) − g(a)]f (c) = 0 and
f (c) f (b) − f (a) = . g(b) − g(a) g (c)
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SECTION 4.1
46. f (x) = 2x3 + 3x2 − 3x − 2 is differentiable on (−2, 1), continuous on [−2, 1], and f (−2) = f (1) = 0. f (x) = 6x2 + 6x − 3 f (c) = 0 at c1 ∼ = −1.366, c2 ∼ = 0.366
47. f (x) = 1 − x3 − cos(π x/2) is differentiable on (0, 1), continuous on [0, 1], and f (0) = f (1) = 0. π f (x) = −3x2 + sin(π x/2) 2 ∼ f (c) = 0 at c = 0.676
48. f (1) = 0; f satisfies Rolle’s theorem on [0, 1]; f (c) = 0 at c ∼ = 0.6058. y
49. b ∼ = 0.5437, c ∼ = 0.3045, ∼ f (0.3045) = −0.1749
0.2
0.4
0.6
f (x) = 0 at x ∼ = 0.3874
50. x-intercepts: x = 0, x = 1; 51. 0 c = 0 52. x-intercepts: x = −2, x = 53. f (x) = x4 − 7x2 + 2; g(x) = 4x3 − 14x −
4 5
x = 2;
f (x) = 0 at x ∼ = −1.1005 and x ∼ = 1.5577
f (x) = 4x3 − 14x
f (3) − f (1) = 4x3 − 14x − 12 3−1
g(c) = 0 at c ∼ = 2.205
0.8
x
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SECTION 4.2 f (x) = cos x − x sin x + 4 cos x = 5 cos x − x sin x and
54. f (x) = x cos x + 4 sin x;
g(x) = 5 cos x − x sin x −
8 f (π/2) − f (−π/2) = 5 cos x − x sin x − π π
g(c) = 0 at c1 ∼ = −0.872, c2 ∼ = 0.872
55. f (x) = x3 − x2 + x − 1; f (x) = 3x2 − 2x + 1;
c = 8/3.
y 60 40 20 x 1
2
3
4
56. f (x) = x4 − 2x3 − x2 − x + 1; f (x) = 4x3 − 6x2 − 2x − 1;
c = 1/2, −0.6180, 1.6180.
y 10 5
−2
x
−1
1
2
3
5
SECTION 4.2
1. f (x) = 3x2 − 3 = 3 x2 − 1 = 3(x + 1)(x − 1) f increases on (−∞, −1] and [1, ∞), decreases on [−1, 1] 2. f (x) = 3x2 − 6x = 3x(x − 2) f increases on (−∞, 0] and [2, ∞), decreases on [0, 2] 1 x2 − 1 (x + 1)(x − 1) = = 2 2 x x x2 f increases on (−∞, −1] and [1, ∞), decreases on [−1, 0) and (0, 1 ]
3. f (x) = 1 −
4. f (x) = 3(x − 3)2 ;
f increases on (−∞, ∞)
(f is not defined at 0)
125
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SECTION 4.2
5. f (x) = 3x2 + 4x3 = x2 (3 + 4x)
f increases on − 34 , ∞ , decreases on −∞, − 34 6. f (x) = 3x2 + 6x + 2 √ √
f increases on −∞, −1 − 13 3 and −1 + 13 3, ∞ ,
√ √ decreases on −1 − 13 3, −1 + 13 3
7. f (x) = 4(x + 1)3 f increases on [−1, ∞),
decreases on (−∞, −1]
3
2(x + 1) x3 f increases on [−∞, −1] and (0, ∞), decreases on [−1, 0) ⎧ ⎧ 1 1 ⎪ ⎪ , x2 (x − 2)2 x−2
8. f (x) =
f increases on (−∞, 2), 10. f (x) =
(1 + x2 ) − x(2x) ; (1 + x2 )2
11. f (x) = −
(x2
decreases on (2, ∞)
(f is not defined at 2)
f increases on [−1, 1],
decreases on (−∞, −1] and [1, ∞)
4x − 1)2
f increases on (−∞, −1) and (−1, 0], decreases on [0, 1) and (1, ∞) 12. f (x) =
(f is not defined at ±1)
(x2 + 1)(2x) − x2 (2x) 2x = 2 (x2 + 1)2 (x + 1)2
f increases on [0, ∞), decreases on (−∞, 0] ⎧ ⎧ √ √ 2 ⎪ ⎪ x − 5, x < − 5 2x, x g(x) on (0, c). (b) Again let h(x) = f (x) − g(x). Then h is increasing on (−c, 0) which implies that h(x) < 0 on this interval since h(0) = 0. Therefore, f (x) < g(x) on (−c, 0). 57. Let f (x) = tan x and g(x) = x for x ∈ [0, π/2). Then f (0) = g(0) = 0 and f (x) = sec2 x > g (x) = 1 for x ∈ (0, π/2). Thus, tan x > x for x ∈ (0, π/2) by Exercise 56(a).
58. Let f (x) = cos x − 1 − 12 x2 for x ∈ [0, ∞). Then f (0) = 0 and f (x) = − sin x + x = x − sin x > 0 for x ∈ (0, ∞) by Exercise 51 (b). Thus, f (x) > 0 for x ∈ (0, ∞) which implies cos x > 1 −
1 2
x2 on
(0, ∞). 59. Choose an integer n > 1. Let f (x) = (1 + x)n and g(x) = 1 + nx,
and f (x) = n(1 + x)
n−1
> g (x) = n since (1 + x)
n−1
x > 0. Then, f (0) = g(0) = 1
> 1 for x > 0. The result follows from Exercise
56(a).
60. Let f (x) = sin x − x −
1 6
x3 . Then f (0) = 0 and f (x) = cos x − 1 −
Therefore, f (x) > f (0) = 0 for all x ∈ (0, ∞) which implies sin x > x −
1 6
x2 > 0 by Exercise 58.
x on (0, ∞).
61. 4◦ ∼ = 0.06981 radians. By Exercises 51 and 60, 0.6981 −
1 2 3
(0.6981)3 = 0.06975 < sin 4◦ < 0.6981 6
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SECTION 4.2
131
1 1 x3 62. (a) Let f (x) = cos x − (1 − x2 + x4 ). Then f (0) = 0 and f (x) = − sin x + x − < 0 by 2 24 6 1 2 1 Exercise 60. Therefore, f (x) < f (0) = 0 on all x ∈ (0, ∞), which implies cos x < 1 − x + x4 2 24 on (0, ∞). π 1 1 1 (b) 6◦ = . Using this for x in 1 − x2 < cos x < 1 − x2 + x4 , 30 2 2 24 =⇒ 0.994517 < cos 6◦ < 0.994522. 63. Let f (x) = 3x4 − 10x3 − 4x2 + 10x + 9, x ∈ [−2, 5]. Then f (x) = 12x3 − 30x2 − 8x + 10. f (x) = 0 at x ∼ = −0.633, 0.5, 2.633 f is decreasing on [−2, −0.633] and [0.5, 2.633] f is increasing on [−0.633, 0.5] and [2.633, 5]
64. Let f (x) = 2x3 − x2 − 13x − 6, x ∈ [−3, 4]. Then f (x) = 6x2 − 2x − 13. f (x) = 0 at x ∼ = −1.315, 1.648 f is decreasing on [−1.315, 1.648] f is increasing on [−3, −1.315] and [1.648, 4]
65. Let f (x) = x cos x − 3 sin 2x, x ∈ [0, 6]. Then f (x) = cos x − x sin x − 6 cos 2x. f (x) = 0 at x ∼ = 0.770, 2.155, 3.798, 5.812 f is decreasing on [0, 0.770], [2.155, 3.798] and [5.812, 6] f is increasing on [0.770, 2.155] and [3.798, 5.812]
66. Let f (x) = x4 + 3x3 − 2x2 + 4x + 4, x ∈ [−5, 3]. Then f (x) = 4x3 + 9x2 − 4x + 4. f (x) = 0 at x ∼ = −2.747 f is decreasing on [−5, −2.747] f is increasing on [−2.747, 3]
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SECTION 4.3
67. (a) f (x) = 0 at x = 0, π2 , π, 3π 2 , 2π
(c) f (x) < 0 on 0, π2 ∪ π2 , π
3π (b) f (x) > 0 on π, 3π 2 ∪ 2 , 2π
68. (b) f (x) > 0 on (−∞, ∞) 69. (a) f (x) = 0 at x = 0
(b) f (x) > 0 on (0, ∞)
(c) f (x) < 0 on (−∞, 0) 70. (a) f (x) = 0 at x = − 12 ,
(c) f (x) < 0 on 85 , 3
8 5,
(b) f (x) > 0 on −∞, − 12 ∪ − 12 , 85 ∪ (3, ∞)
3
71. f = C, constant; f (x) ≡ 0
SECTION 4.3 1. f (x) = 3x2 + 3 > 0;
no critical pts, no local extreme values
2. f (x) = 8x3 − 8x = 8x(x2 − 1); f (x) = 24x2 − 8;
critical pts −1, 0, 1
f (−1) = f (1) = 16 > 0, f (0) = −8 < 0;
f (0) = 6 local max, f (−1) = 4 local min, f (1) = 4 local min 3. f (x) = 1 − f (x) =
1 ; x2
critical pts −1, 1
2 , f (−1) = −2, f (1) = 2 f (−1) = −2 local max, f (1) = 2 local min x3
6 2x4 + 6 = ; x3 x3 no local extreme values
4. f (x) = 2x +
no critical pts (note: 0 is not in the domain of f ),
5. f (x) = 2x − 3x2 = x(2 − 3x); f (x) = 2 − 6x;
critical pts 0, 23
f (0) = 2, f ( 23 ) = −2
f (0) = 0 local min, f ( 23 ) =
4 27
local max
6. f (x) = −2(1 − x)(1 + x) + (1 − x)2 = (x − 1)(3x + 1); critical pts − 13 , 1
f (x) = (1 + 3x) + 3(x − 1) = 2(3x − 1); f − 13 = −4, f (1) = 4
f − 13 = 32 27 local max, f (1) = 0 local min 7. f (x) =
2 ; no critical pts, no local extreme values (1 − x)2
8. f (x) =
(2 + x)(−3) − (2 − 3x)(1) 8 =− ; (2 + x)2 (2 + x)2
in the domain of f ), no local extreme values
no critical pts (note: −2 is not
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SECTION 4.3 9. f (x) = −
2(2x + 1) ; x2 (x + 1)2
⎧ 2 ⎪ ⎪ ⎨ x − 16, 10. f (x) = 16 − x2 , ⎪ ⎪ ⎩ x2 − 16,
critical pt −
x < −4 −4 ≤ x < 4 x≥4
critical pts −4, 0, 4;
1 2
133
f − 12 = −8 local max
⎧ ⎪ x < −4 ⎪ ⎨ 2x, f (x) = −2x, −4 < x < 4 ⎪ ⎪ ⎩ 2x, x>4
f (−4) = f (4) = 0 local minima, f (0) = 16 local max
11. f (x) = x2 (5x − 3)(x − 1);
critical pts 0, 35 , 1
3 2 2 33 f = 5 local max 5 5 f (1) = 0 local min no local extreme at 0
12. f (x) = 3
x−2 x+2
2
4 ≥ 0; (x + 2)2
13. f (x) = (5 − 8x)(x − 1)2 ;
critical pt 2, no local extreme values
critical pts 58 , 1 f
5 8
=
27 2048
local max
no local extreme at 1 14. f (x) = −(1 + x)3 + (1 − x)(3)(1 + x)2 = 2(1 + x)2 (1 − 2x);
critical pts −1, 12
f 12 = 27 16 local max no local extreme at −1
15. f (x) =
x(2 + x) ; (1 + x)2
critical pts −2, 0 f (−2) = −4 local max f (0) = 0 local min
16. f (x) = (1 − x)1/3 −
1 3
x(1 − x)−2/3 =
3 − 4x ; 3(1 − x)2/3
critical pts 34 , 1 f (3/4) =
3 44/3
local max
no local extreme at 1 17. f (x) = 13 x(7x + 12)(x + 2)−2/3 ;
critical pts −2, − 12 7 , 0
f − 12 = 7
144 49
2 1/3 7
f (0) = 0 local min
local max
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SECTION 4.3
18. f (x) =
−1 1 3(2x − 1) + = ; (x + 1)2 (x − 2)2 (x + 1)2 (x − 2)2
critical pt
1 2
f
⎧ ⎪ x ≤ − 12 ⎪ ⎨ 2 − 3x, 19. f (x) = x + 4, − 12 < x < 3 ⎪ ⎪ ⎩ 3x − 2, 3≤x
1 2
=
4 3
local min
⎧ ⎪ x < − 12 ⎪ ⎨ −3, f (x) = 1, − 12 < x < 3 ⎪ ⎪ ⎩ 3, 3 0. f has a local maximum at − 2a 2a 34. Setting f (x) = 3ax2 + 2bx + c = 0 and checking the discriminant, we get (1)
2 local extrema if b2 > 3ac
(2)
1 local extrema if b2 = 3ac
(3)
0 local extrema if b2 < 3ac
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SECTION 4.3 P (x) = x4 − 8x3 + 22x2 − 24x + 4
35.
P (x) = 4x3 − 24x2 + 44x − 24 P (x) = 12x2 − 48x + 44 Since P (1) = 0, P (x) is divisible by x − 1. Division by x − 1 gives
P (x) = (x − 1) 4x2 − 20x + 24 = 4(x − 1)(x − 2)(x − 3). The critical pts are 1, 2, 3. Since P (1) > 0,
P (2) < 0,
P (3) > 0,
P (1) = −5 is a local min, P (2) = −4 is a local max, and P (3) = −5 is a local min. Since P (x) < 0 for x < 0, P decreases on (−∞, 0]. Since P (0) > 0, P does not take on the value 0 on (−∞, 0]. Since P (0) > 0 and P (1) < 0, P takes on the value 0 at least once on (0, 1). Since P (x) < 0 on (0, 1), P decreases on [0, 1]. It follows that P takes on the value zero only once on [0, 1]. Since P (x) > 0 on (1, 2) and P (x) < 0 on (2, 3), P increases on [1, 2] and decreases on [2, 3]. Since P (1), P (2), P (3) are all negative, P cannot take on the value 0 between 1 and 3. Since P (3) < 0 and P (100) > 0, P takes on the value 0 at least once on (3, 100). Since P (x) > 0 on (3, 100), P increases on [3, 100]. It follows that P takes on the value zero only once on [3, 100]. Since P (x) > 0 on (100, ∞), P increases on [100, ∞). Since P (100) > 0, P does not take on the value 0 on [100, ∞). 36. f has a local maximum at x = 0; f has a local minimum at x = −1 and x = 2. 37.
38. Let f (x) = Ax2 + Bx + C. Then f (x) = 2Ax + B. f (−1) = 3
=⇒
A − B + C = 3;
Since f has a minimum at x = 2,
f (3) = −1
=⇒
f (2) = 4A + B = 0
Solving for A, B, C, we get A = 12 , B = −2, C = 12 .
9A + 3B + C = −1
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SECTION 4.3
137
a b2 − x2 ax 39. Let f (x) = 2 . Then f (x) = . Now x + b2 (b2 + x2 )2
b2 b2 − x2 a 2 f (0) = 2 = 1 ⇒ a = b and f (x) = b (b2 + x2 )2
f (−2) =
b2 b 2 − 4 (b2 + 4)2
= 0 ⇒ b = ±2
Thus, a = 4 and b = ±2. 40. (a) f (x) = xp (1 − x)q , p, q ≥ 2;
f (x) = xp−1 (1 − x)q−1 [p − (p + q)x]
f (x) = 0
=⇒
x = 0, x = 1, x =
p p+q
(b) p even, p − 1 odd: f has a local min at x = 0 (c) q even, q − 1 odd: f has a local min at x = 1 (d) f
p p+q
= − (p + q)
p p+q
p−1
q p+q
q−1 0, f has at least one zero in (2, 3). Since f (x) > 0 for x ∈ (2, 3), f is increasing on this interval and so it has exactly one zero. Thus, f has exactly one critical point c in (2, 3).
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SECTION 4.3 x2 − 2x, then f (x) = cos x + x − 2 and f (x) = − sin x + 1. Since f (2) = 2 −0.4161 < 0 and f (3) = 0.01 > 0, f has at least one zero in (2, 3). Since f (x) > 0 for x ∈ (2, 3),
44. If
f (x) = sin x +
f is increasing on this interval and so it has exactly one zero. Thus, f has exactly one critical point c in (2, 3). 45. f (x) =
ax2 + b cx2 + d
f (x) =
and f (x) =
2(ad − bc)x ; (cx2 + d)2
2(ad − bc)(cx2 − 4cx + d) ; (cx2 + d)3
f (0) =
x=0
is a critical number.
2(ad − bc) d2
Therefore, ad − bc > 0 implies that f (0) is a local minimum; ad − bc < 0 implies that f (0) is a local maximum. 46. Let δ be any positive number and consider f on the interval (−δ, δ). Let n be a positive integer such that 0< Then
f
π 2
π 2
1 < δ and 0 < + 2nπ
1 + 2nπ
−π 2
> 0 and f
−π 2
1 < δ. + 2nπ 1 + 2nπ
< 0.
Thus f takes on both positive and negative values in every interval centered at 0 and it follows that f cannot have a local maximum or minimum at 0. 47. (a)
critical points: x1 ∼ = −0.692, x2 ∼ = 2.248 local extreme values:
f (−0.692) ∼ = 29.342, f (2.248) ∼ = −8.766
(b) f is increasing on [−3, −0.692], and [2.248, 4]; 48. (a)
f is decreasing on [−0.692, 2.248]
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SECTION 4.3
139
critical points: x1 ∼ = −2.085, x2 ∼ = −1, x3 ∼ = 0.207, x4 ∼ = 1.096, x5 = 1.544 f (−2.085) ∼ = −6.255, f (−1) = 7, f (0.207) ∼ = 0.621, f (1.096) ∼ = 7.097,
local extreme values: f (1.544) ∼ = 4.635
(b) f is increasing on [−2.085, −1], [0.207, 1.096], and [1.544, 4] f is decreasing on [−4, −2.085], [−1, 0.207], and [1.096, 1.544] 49. (a)
critical points: x1 ∼ = −2.201, x2 ∼ = −0.654, x3 ∼ = 0.654, x4 ∼ = 2.201 f (−2.204) ∼ = 2.226, f (−0.654) ∼ = −6.634, f (0.654) ∼ = 6.634,
local extreme values: f (2.204) ∼ = −2.226
(b) f is increasing on [−3. − 2.204], [−0.654, 0.654], and [2.204, 3] f is decreasing on [−2.204, −0.654], and [0.654, 2.204] 50. f (x) = 0 at x = 2, 3; 51. f (x) > 0 on
2
3, ∞
;
f (2) = 0 is a local minimum. f has no local extrema.
52. f (x) = 0 at the multiples of
1 4 π;
f (0) = 1 is a local maximum, f (π/4) = 0 is a local minimum,
f (π/2) = 1 is a local maximum, and so on. 53.
critical points of f : f (−1.326) ∼ = −4 < 0
f (0) = 4 > 0 f (1.816) ∼ = −4
x1 ∼ = −1.326, x2 = 0, x3 ∼ = 1.816 ⇒ f has a local maximum at x = −1.326
⇒ f has a local minimum at x = 0 ⇒ f has a local maximum at x = 1.816
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SECTION 4.4
54.
critical number of f : x1 ∼ = −1.935 ∼ 14.60 > 0 ⇒ f has a local minimum at x = −1.935 f (−1.935) =
SECTION 4.4 1. f (x) = 12 (x + 2)−1/2 , x > −2; f (−2) = 0 endpt and abs min; 2. f (x) = 2x − 3;
critical pt.
3 2;
as x → ∞, f (x) → ∞; f
3 2
so no abs max
= − 14 local and abs min
3. f (x) = 2x − 4, x ∈ (0, 3); critical pt. 2; f (0) = 1 endpt and abs max, f (2) = −3 local and abs min, f (3) = −2 endpt max 4. f (x) = 4x + 5, x ∈ (−2, 0); critical pt. − 54 ;
f (−2) = −3 endpt max, f − 54 = − 33 8 local and abs min, f (0) = −1 endpt and abs max 1 2x3 − 1 = , x = 0; f (x) = 0 at x = 2−1/3 2 x x2 2 critical pt. 2−1/3 ; f (x) = 2 + 3 , f 2−1/3 = 6 x
−1/3 −2/3 1/3 −2/3 f 2 +2 =2 + 2 · 2−2/3 = 3 · 2−2/3 local min =2
5. f (x) = 2x −
6. f (x) = 1 −
2 x3
critical pt. 21/3 ; 2x3 − 1 , 7. f (x) = x2
f 21/3 = 3(2)−2/3 local min x∈
1 ,2 ; 10
critical pt. 2−1/3 ;
1
1 f 10 endpt and abs max, f 2−1/3 = 3 · 2−2/3 local and abs min, = 10 100 f (2) = 4 12 endpt max
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SECTION 4.4 8. f (x) = 1 −
2 x3 ,
√ x ∈ (1, 2)
critical pt. 21/3 ; f (1) = 2 endpt and abs max
√ √ f 21/3 = 3(2)−2/3 local and abs min, f 2 = 2+
1 2
endpt max
9. f (x) = 2x − 3, x ∈ (0, 2); critical pt. 32 ; f (0) = 2 endpt and abs max, f
3 2
= − 14 local and abs min,
f (2) = 0 endpt max 10. f (x) = 2(x − 1)(x − 2)(2x − 3);
x ∈ (0, 4)
critical pts. 1, 32 , 2, ; f (0) = 4 endpt max, f (1) = 0 local and abs min,
1 f 32 = 16 local max, f (2) = 0 local and abs min, f (4) = 36 endpt and abs max (2 − x)(2 + x) , x ∈ (−3, 1); (4 + x2 )2 critical pt. −2;
11. f (x) =
3 f (−3) = − 13 endpt max, f (−2) = − 14 local and abs min,
f (1) = 12. f (x) =
1 5
endpt and abs max
2x , (1 + x2 )2
critical pt. 0; f (2) =
x ∈ (−1, 2)
f (−1) =
1 2
endpt max, f (0) = 0 local and abs min,
4 5
endpt and abs max √ 1 13. f (x) = 2 (x − x) 1 − √ , x > 0; 2 x critical pts.
1 4,
1;
f (0) = 0 endpt and abs min, f as x → ∞, 14. f (x) =
f (x) → ∞;
2(2 − x2 ) , (4 − x2 )1/2
1 4
=
1 16
local max, f (1) = 0 local and abs min;
so no abs max
x ∈ (−2, 2)
√ √
√ critical pts. − 2, 2; f (−2) = 0 endpt max, f − 2 = −2 local and abs min,
√ f 2 = 2 local and abs max, f (2) = 0 endpt min 3(2 − x) , x0
no extreme values.
17. f (x) = − 13 (x − 1)−2/3 ,
x = 1;
critical pt. 1; no local extremes;
18. f (x) =
as
x → ∞,
f (x) → −∞
as
x → −∞, f (x) → ∞
no abs extremes
3x − 1 8 ; 3 (4x − 1)2/3 (2x − 1)1/3
critical pts. 14 , 13 , 12 ; no extreme value at
f 13 = 13 local max, f 12 = 0 local min
1 4
√
19. f (x) = sin x 2 cos x + 3 , x ∈ (0, π); 5 critical pt. π ; 6 √ √
7 f (0) = − 3 endpt and abs min, f 56 π = local and abs max, f (π) = 3 endpt min 4 20. f (x) = − csc2 x + 1, critical pt.
x ∈ (0, 2π/3) ;
1 2π
(note: 0 is not in the domain) √
No extreme value at 12 π, f 23 π = 13 2π − 3 endpt and abs max
21. f (x) = −3 sin x 2 cos2 x + 1 < 0, x ∈ (0, π); no critical pts. f (0) = 5 endpt and abs max, f (π) = −5 endpt and abs min 22. f (x) = 2 cos 2x − 1,
x ∈ (0, π) :
√
critical pts. 16 π, 56 π ; f (0) = 0 endpt min, f 16 π = 12 3 − 16 π local and abs max, √
f 56 π = − 12 3 − 56 π local and abs min, f (π) = −π endpt max
23. f (x) = sec2 x − 1 ≥ 0, x ∈ − 13 π, 12 π ; critical pt. 0 ; √
f − 13 π = 13 π − 3 endpt and abs min, no abs max
24. f (x) = 2 sin x cos x(2 sin2 x − 1), x ∈ 0,
2 3
π ;
critical pts. 14 π, 12 π; f (0) = 0 endpt and abs max, f 14 π = − 14 local and abs min,
3 f 12 π = 0 local and abs max, f 23 π = − 16 endpt min
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SECTION 4.4 25. ⎧ ⎪ ⎪ ⎨ −2, f (x) = 1, ⎪ ⎪ ⎩ −1,
0 3. 36 + y 2
√ Thus, the minimum must occur at one of the endpoints: S(0) = 15, S(3) = 2 45 < S(0).
To minimize the sum of the distances, take P = (0, 3). 29.
Minimize L L2 = y 2 + (x + 1)2 . 8 8 y = , y = (x + 1). By similar triangles x+1 x x 2 64 8 2 2 2 L = (x + 1) + (x + 1) = (x + 1) +1 x x2
Since L is minimal when L2 is minimal, we consider the function 64 f (x) = (x + 1)2 + 1 , x > 0. x2 −128 64 2 f (x) = 2(x + 1) + 1 + (x + 1) x2 x3 =
2(x + 1) 3 x − 64 , f (x) = 0 3 x
=⇒
x = 4.
Since f decreases on (0, 4] and increases on [4, ∞), the abs min of f occurs when x = 4. √ The shortest ladder is 5 5 ft long.
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SECTION 4.5
30. Maximize L = x + y. By similar triangles,
y x =√ 2 6 x − 64
6x , x>8 − 64 384 L (x) = 1 − 2 (x − 64)3/2 L (x) = 0 =⇒ x = 64 + (384)2/3 ∼ = 10.81 L(x) = x + √
x2
L(10.81) ∼ = 19.73; the longest ladder is approximately 19.7 ft. Maximize A
31.
(We use feet rather than inches to reduce arithmetic.)
A = (L − 1) W − 43 27 =⇒ W = L 27 4 85 27 4 − = − − L A = (L − 1) L 3 3 L 3
LW = 27
85 27 4 81 − − L, 1 ≤ L ≤ . 3 L 3 4 27 4 9 A (L) = 2 − , A (L) = 0 =⇒ L = . L 3 2 Since A (L) = −54/L3 < 0 for 1 < L < 81 4 , the max at L = A(L) =
9 2
is the abs max.
The banner has length 9/2 ft = 54 in. and height 6 ft = 72 in. 1 2 3 πr h = 1, or h = 2 . 3 πr √ 2 2 Minimize surface area S = πr r + h = πr r2 +
32. Assume V =
2π 2 r6 − 9 dS = √ =0 dr r2 π2 r6 + 9 =⇒
h= π
3 9 2π 2
=⇒
1/3
=
r=
9 2π 2
1/6 =
9 = 2 π r4
√
π2 r6 + 9 . r
31/3 21/6 π 1/3
√ 31/3 21/3 = r 2. 1/3 π
33.
Find the extreme values of A A = πr2 + x2 2πr + 4x = 28
2 1 14 . A(r) = πr + 7 − πr , 0 ≤ r ≤ 2 π 2
=⇒
x = 7 − 12 πr.
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157
Note: the endpoints of the domain correspond to the instances when the string is not cut: r = 0 when no circle is formed, r = 14/π when no square is formed. 1 14 A (r) = 2πr − π 7 − πr , A (r) = 0 =⇒ r = . 2 4+π Since A (r) = 2π + π 2 /2 > 0 on (0, 14/π) , the abs min of A occurs when r = 14/(4 + π) and the abs max of A occurs at one of the endpts: A(0) = 49,
A(14/π) = 196/π > 49.
(a)
To maximize the sum of the two areas, use all of the string to form the circle.
(b)
To minimize the sum of the two areas, use 2πr = 28π/(4 + π) ∼ = 12.32 inches of string for the
circle. 34. Maximize V = x2 h given that x2 + 4xh = 12 =⇒ √ 12 − x2 2 = 3x − 14 x3 , 0 < x ≤ 12. V (x) = x 4x V (x) = 3 −
3 4
x2 ,
V (x) = 0
=⇒
h=
12 − x2 . 4x
x = 2. Since V increases on (0, 2] and decreases on [2,
√
12),
V has an abs max at x = 2; the maximum volume is V (2) = 4 cu ft. 35.
Maximize V V = πr2 h By similar triangles 8 h 8 = or h = (5 − r). 5 5−r 5 8π 2 r (5 − r), 0 ≤ r ≤ 5. 5 8π
10r − 3r2 , V (r) = 0 =⇒ r = 10/3. V (r) = 5 Since V increases on (0, 10/3] and decreases on [10/3, 5), the abs max of V occurs when r = 10/3.
V (r) =
The cylinder with maximal volume has radius 10/3 and height 8/3. 36. Maximize A = 2πrh =
16π r(5 − r), 0 ≤ r ≤ 5, 5
h = 85 (5 − r) from Exercise 33 .
16π (5 − 2r), A (r) = 0 =⇒ r = 52 5 The curved surface is a maximum when r = 52 , h = 4. A (r) =
37.
Minimize C In dollars, C = cost base + cost top + cost sides
= .35 x2 + .15 x2 + .20(4xy) = 12 x2 + 45 xy Volume = x2 y = 1250
y=
1250 x2
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SECTION 4.5 1 2 1000 x + , x > 0. 2 x 1000 C (x) = x − 2 , C (x) = 0 =⇒ x = 10. x Since C (x) = 1 + 2000/x3 > 0 for x > 0, the local min of C at x = 10 is the abs min.
C(x) =
The least expensive box is 12.5 ft tall with a square base 10 ft on a side.
38. Maximize A = xy. By similar triangles y h = , so b−x b A(x) =
h x(b − x), 0 ≤ x ≤ b. b
A (x) =
h (b − 2x), b
A (x) = 0
⇒
x=
b . 2
Since A is increasing on [0, b/2] and decreasing on [b/2, b], A has an abs max at x = b/2 A(b/2) =
1 4
hb =
1 2
area of triangle ABC.
39. Minimize A A = 12 (h)(2x) = hx Triangles ADC and ABE are similar: AD AB = DC BE
or
h AB = . x r
Pythagorean Theorem: r2 + (AB)2 = (h − r)2 . Thus 2 hr 2 r + = (h − r)2 . x Solving this equation for h we find that h= A(x) =
A (x) =
2x3 r , x > r. x2 − r2
2 x − r2 6x2 r − 2x3 r(2x)
A (x) = 0
(x2 − r2 )2 √ =⇒ x = r 3.
=
2x2 r . − r2
x2
2x2 r x2 − 3r2 (x2 − r2 )2
,
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159
√ √ √ r, r 3 and increases on r 3, ∞ , the local min at x = r 3 is the abs √ √ √ min of A. When x = r 3, we get h = 3r so that F C = 2r 3 and AF = F C = h2 + x2 = √ √ 2r 3. The triangle of least area is equilateral with side of length 2r 3. Since A decreases on
40. Maximize A(x) =
1 2
√ (r + x)2 r2 − x2
√ = (r + x) r2 − x2 , 0 ≤ x ≤ r. A (x) =
r2 − rx − 2x2 √ ; r2 − x2
A (x) = 0
=⇒
r . 2
x=
Since A increases on [0, r/2] and decreases on [r/2, r], A has an abs max at x = r/2; √ 3 3 2 A(r/2) = r . 4 41.
Maximize V V = πr2 h By the Pythagorean Theorem, (2r)2 + h2 = (2R)2 so √ h = 2 R2 − r 2 . √ V (r) = 2πr2 R2 − r2 , 0 ≤ r ≤ R.
√ 2πr 2R2 − 3r2 r3 2 2 √ = V (r) = 2π 2r R − r − √ R2 − r 2 R2 − r 2 √ V (r) = 0 =⇒ r = 13 R 6. Since V increases on
√ √ √ 0, 13 R 6 and decreases on 13 R 6, R , the local max at r = 13 R 6 is
the abs max. The cylinder of maximal volume has base radius √ 42. Maximize A = 2πrh = 4πr R2 − r2 , 0 ≤ r ≤ R, 4π(R2 − 2r2 ) √ , R2 − r 2
1 3R
√
6 and height
A (r) = 0
=⇒
√
3.
√ h = 2 R2 − r2 from Exercise 39 .
R r=√ . 2 √ R The curved surface is a maximum when r = √ , h = R 2. 2
A (r) =
2 3R
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SECTION 4.5
43.
Maximize V V = 13 πr2 h Pythagorean Theorem
Case 1 : h ≥ R
Case 2 : h ≤ R
Case 1 :
(h − R)2 + r2 = R2
Case 2 :
(R − h)2 + r2 = R2
In both cases r2 = R2 − (R − h)2 = 2hR − h2 .
V (h) = 13 π 2h2 R − h3 ,
0 ≤ h ≤ 2R.
V (h) = 13 π 4hR − 3h2 ,
V (h) = 0
Since V increases on
0, 43 R
at h =
and decreases on
4R . 3 4
3 R, 2R
, the local max at h = 43 R is the abs
max. √ The cone of maximal volume has height 43 R and radius 23 R 2. 44. Maximize V = V (h) = V (h) =
1 3 1 3
1 3
πr2 h, where r2 + h2 = a2 .
π(a2 − h2 )h,
0 ≤ h ≤ a,
π(a2 − 3h2 ),
V (h) = 0
√ Maximum volume V a/ 3 =
2 27
⇒
√ πa3 3.
45.
a h= √ . 3
Minimize C In units of $10, 000, cost of cable cost of cable + underground under water √ = 3(4 − x) + 5 x2 + 1.
C=
Clearly, the cost is unnecessarily high if x > 4 or x < 0. √ C(x) = 12 − 3x + 5 x2 + 1, 0 ≤ x ≤ 4. C (x) = −3 + √
5x , C (x) = 0 x2 + 1
=⇒
x = 3/4.
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161
Since the domain of C is closed, the abs min can be identified by evaluating C at each critical point: C(0) = 17,
C
3 4
= 16,
√ C(4) = 5 17 ∼ = 20.6.
The minimum cost is $160, 000. 46. Maximize α − θ. Since the tangent function is an increasing function on [0, π/2), it suffices to maximize tan(α − θ). tan α − tan θ 16 9 ; tan α = , tan θ = . 1 + tan α tan θ x x 16 9 − 7x x = Thus, we maximize T (x) = x , x > 0. 16 9 x2 + 144 1+ · x x 7(144 − x2 ) T (x) = 2 , T (x) = 0 =⇒ x = 12. (x + 144)2 tan(α − θ) =
Stand 12 ft from the wall for the most favorable view. 47. P (θ) =
−mW (m cos θ − sin θ) ; (m sin θ + cos θ)2
48. R(θ) =
2v 2 cos θ sin(θ − α), g cos2 α R (θ) = =
R (θ) = 0
=⇒
49. Minimize I = I (x) = −
2θ − α =
1 2
π
P is minimized when tan θ = m.
0 0 on (0, 1/2) ∪ (3/2, ∞). F has local minima at x = 0
and x = 3/2; F has a local maximum at x = 1/2. F (0) = f (0) = 0, F (1) = f (1) = −1, F (2) = f (2) = 4;
n = 1 minimizes f (n).
54. Let x bethe number of passengers and R the revenue in dollars. 37x, 16 ≤ x ≤ 35 R(x) =
1 37 − 2 (x − 35) x, 35 < x ≤ 48;
37,
R (x) =
109 2
16 < x < 35
− x, 35 < x < 48.
The critical number is 35. From R(16) = 592, R(35) = 1295, and R(48) = 1464 we conclude that the revenue is maximized by taking a full load of 48 passengers.
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163
55. Let x be the number of customers and P the net profit in dollars. Then 0 ≤ x ≤ 250 and 12x, 0 ≤ x ≤ 50 P (x) = [12 − 0.06(x − 50)] x, 50 < x ≤ 250; 12, 0 ≤ x ≤ 50 P (x) = 15 − 0.12x, 50 < x ≤ 250. The critical points are: x = 50, x = 125. From
P (0) = 0, P (50) = 600, P (125) = 937.50, and
P (250) = 0, we conclude that the net profit is maximized by servicing 125 customers. 56. Maximize P (x) = 2cy + cx, where c is the price of low-grade steel and y =
40 − 5x 10 − x
.
40 − 5x ) + cx 10 − x (10 − x)(−5) − (40 − 5x)(−1) P (x) = 2c + c, (10 − x)2 √ 1 P (x) = 0 =⇒ x = 10 − 20 or about 5 tons. 2 P (x) = 2c(
57. y = mx −
1 (m2 + 1)x2 . When y = 0, 400
Differentiating x with respect to m, =⇒
800m . m2 + 1 800 − 800m2 =0 x = (m2 + 1)2 x=
m = 1.
225 2 (m + 1). 2 Differentiating y with respect to m, y = 300 − 225m = 0 4 =⇒ m = . 3 300 1 2 300 59. Driving at ν mph, the trip takes hours and uses 1 + ν gallons of fuel. ν 400 ν Thus, the expenses are: 300 1 2 300 6780 E(ν) = 2.60 1 + ν + 20 = 1.95ν + , 35 ≤ x ≤ 70. 400 ν ν ν 58. When x = 300,
y = 300m −
Differentiating, we get E (ν) = 1.95 −
6780 , ν2
and E (ν) = 0 at ν ∼ = 59
E is decreasing on [35, 59] and increasing on [59, 70]; the minimal expenses occur when the truck is driven at 59 mph. 60. Let ν be the speed of the boat measured in kilometers per hour. A 100 kilometer trip will take 100/ν hours.
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SECTION 4.5 Fixed costs: F (ν) = 2500
100 ν
2
dollars.
2
Fuel costs: G(ν) = kν ;
2
100 ν
400 = k(10) implies k = 4. Therefore, G(ν) = 4ν dollars. 100 100 2500 Total cost: C(ν) = 2500 + 4ν 2 = 100 + 4ν , 0 < ν < ∞. Differentiating, we ν ν ν get 2 2500 4ν − 2500 C (ν) = 100 − 2 + 4 = 100 and C (ν) = 0 at ν = 25. ν ν2
C is decreasing on (0, 25] and increasing on [25, ∞); the speed that minimizes the expenses is 25 kilometers per hour. If we replace the 100 kilometers by M kilometers, the total cost will be 2500 + 4ν C(ν) = M ν and we will get exactly the same result. The minimizing speed is independent of the length of the trip. 16 61. Minimize SA = 2πrh + 2πr2 , where 2r ≤ h < 6 and πr2 h = 16π (hence h = 2 ). r 32π 32π + 2πr2 . Differentiating, SA = − 2 + 4πr = 0 Thus SA = r r =⇒ r3 = 8, so r = 2 feet and h = 4 feet. Thus no minimum exists. 200, 000n income = cost 1, 000, 000n + 100, 000(1 + 2 + · · · + n − 1) + 5, 000, 000 4n 2n = 2 . = n + 19n + 100 10n + 12 (n − 1)n + 50 4x , x>0 Let f (x) = 2 x + 19x + 100
62. We want to maximize the ratio
(x2 + 19x + 100)4 − 4x(2x + 19) 4(100 − x2 ) = 2 =0 2 2 (x + 19x + 100) (x + 19x + 100)2 x = 10.
Then f (x) = =⇒
Since f (x) > 0 for x < 0,
f (x) < 0 for x > 10, f has an absolute maximum at x = 10.
A ten story building provides the greatest return on investment. 63. Swimming at 2 miles per hour, Maggie will reach point B in 1 hour; walking along the shore (a distance of π miles), she will reach point B in π/5 ∼ = 0.63 hours. Suppose that she swims to a point C and then walks to B. Let θ be the central angle (measured in radians) determined by the points B and C. See the figure. By the law of cosines, the square of the distance from A to C is given by d2 = 12 + 12 − 2(1)(1) cos (π − θ) = 2 + 2 cos θ. Therefore the distance d from A to C is d =
√
2 + 2 cos θ. The distance from C to B is θ.
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165
Now, the total length of time to swim to C and then walk to B is √ T (θ) = 2 2 + 2 cos θ + 5θ Maggie wants to minimize T . −2 sin θ T (θ) = √ + 5. 2 2 + 2 cos θ
C
Setting T (θ) = 0, we get −2 sin θ √ +5=0 2 2 + 2 cos θ
θ
A
B
which reduces to 2 cos2 θ + 25 cos θ + 23 = 0 and factors into (cos θ + 1)(2 cos θ + 23) = 0. This implies
θ = π. Maggie should walk the entire distance!
64. We modify the solution of Exercise 63, replacing the walking rate of 2 miles per hour by the rowing rate of 3 miles per hour. The total length of time to row to C and then walk to B is √ T (θ) = 3 2 + 2 cos θ + 5θ The derivative of T is −3 sin θ T (θ) = √ + 5. 2 2 + 2 cos θ Setting T (θ) = 0, we get −3 sin θ √ + 5 = 0, 2 2 + 2 cos θ which reduces to 9 cos2 θ + 50 cos θ + 41 = (cos θ + 1)(9 cos θ + 41) = 0. This implies
θ = π. Again, Maggie should walk the entire distance!
65. (b)
The point on the graph of f√that is closest to P is: √ 1− 2
√
√ x− 1+ 2 (c) lP Q : y − 2 + 2 = 3− 2 (d) and (e) lP Q = lN .
1+
√
2, 2 +
√ 2
66. The point on the graph of f (x) = x − x3 that is closest to (1, 8) is (approximately) (−2.1474, 7.7548). 67. D(x) =
x2 + (7 − 3x)2 ;
the point on the line that is closest to the origin is:
21
7 10 , 10
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SECTION 4.6
68. The point on the graph of f that minimizes the distance to the point (4, 3) is (approximately) (1.3918, 2.0629).
PROJECT 4.5 1. Distance over water:
√
36 + x2 .
Distance over land: 12 − x. √ Total energy: E(x) = W 36 + x2 + L(12 − x), 2.
0 ≤ x ≤ 12.
√ W = 1.5L, so E(x) = 1.5L 36 + x2 + L(12 − x), for 0 ≤ x ≤ 12. 1.5Lx 12 − L = 0 =⇒ x = √ 5.36. E (x) = √ 5 36 + x2 12 12 12 E (x) < 0 on (0, √ ) and E (x) > 0 on ( √ , 12), so E has an absolute minimum at √ . 5 5 5
√ 3. (a) W = kL, k > 1, so E(x) = kL 36 + x2 + L(12 − x), for 0 ≤ x ≤ 12. kLx 6 E (x) = √ − L = 0 =⇒ x = √ . 36 + x2 k2 − 1 6 6 ) and E (x) > 0 on ( √ , 12), so E has an absolute minimum E (x) < 0 on (0, √ k2 − 1 k2 − 1 6 at √ . 2 k −1 (b) As k increases, x decreases. As k → 1+ , x increases. √ 5 1.12. (c) x = 12 =⇒ k = 2 (d) No
SECTION 4.6 1. (a) f is increasing on [a, b], [d, n];
f is decreasing on [b, d], [n, p].
(b) The graph of f is concave up on (c, k), (l, m); The graph of f is concave down on (a, c), (k, l), (m, p). The x-coordinates of the points of inflection are: x = c, x = k, x = l, x = m. 2. (a) g is increasing on [a, b], [c, e], [m, n]; g is decreasing on [b, c], [e, m]. (b) The graph of g is concave up on (a, b), (b, d); The graph of g is concave down on (d, m), (m, n). The x-coordinate of the point of inflection is: x = d. 3. (i) f ,
(ii) f,
(iii) f .
4. (a) x = −2 local max,
no local minima.
(b) points of inflection at x = 0, 2.
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SECTION 4.6 (c)
y
x
−2
5. f (x) = −x−2 ,
2
f (x) = 2x−3 ;
concave down on (−∞, 0), concave up on (0, ∞); 6. f (x) = 1 − x−2 ,
f (x) = 2x−3 ;
concave down on (−∞, 0), concave up on (0, ∞); 7. f (x) = 3x2 − 3,
no pts of inflection
f (x) = 6x;
concave down on (−∞, 0), concave up on (0, ∞); 8. f (x) = 4x − 5,
no pts of inflection
f (x) = 4;
pt of inflection (0, 2)
concave up on (−∞, ∞)
9. f (x) = x3 − x, f (x) = 3x2 − 1; √ √
√
√ concave up on −∞, − 13 3 and 13 3, ∞ , concave down on − 13 3, 13 3 ;
√
√ 5 5 pts of inflection − 13 3, − 36 and 13 3, − 36 10. f (x) = 3x2 − 4x3 ,
f (x) = 6x − 12x2 = 6x(1 − 2x);
concave down on (−∞, 0), and 12 , ∞ , concave up on 0, 12 ;
1 pts of inflection (0, 0), 12 , 16
x2 + 1 11. f (x) = − , (x2 − 1)2
f (x) =
2x x2 + 3 (x2 − 1)3
;
concave down on (−∞, −1) and (0, 1), concave up on (−1, 0) and on (1, ∞); pts of inflection (0, 0) 12. f (x) =
−4 , (x − 2)2
f (x) =
8 ; (x − 2)3
concave down on (−∞, 2), concave up on (2, ∞);
no pts of inflection
13. f (x) = 4x3 − 4x, f (x) = 12x2 − 4; √ √
√
√ concave up on −∞, − 13 3 and 13 3, ∞ , concave down on − 13 3, 13 3 ;
√
√ pts of inflection − 13 3, 49 and 13 3, 49
167
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SECTION 4.6 6(1 − x2 ) 12x(x2 − 3) , f (x) = ; (x2 + 1)2 (x2 + 1)3 √
√
√
√ concave down on −∞, − 3 and 0, 3 , concave up on − 3, 0 and 3, ∞ ; √
√
√ 3 √ pts of inflection − 3, − 32 3 , (0, 0), 3, 2 3
14. f (x) =
√ −1 1+3 x , f (x) = √ 2 √ √ 3; x (1 + x ) 2x x (1 + x ) concave up on (0, ∞); no pts of inflection
15. f (x) = √
16. f (x) =
1 5
(x − 3)−4/5 ,
4 f (x) = − 25 (x − 3)−9/5 ;
concave up on (−∞, 3), concave down on (3, ∞); 17. f (x) = 53 (x + 2)2/3 ,
f (x) =
10 9 (x
pt of inflection (3, 0)
+ 2)−1/3 ;
concave down on (−∞, −2), concave up on (−2, ∞); 18. f (x) =
4 − 2x2 , (4 − x2 )1/2
f (x) =
2x(x2 − 6) (4 − x2 )3/2
concave up on (−2, 0), concave down on (0, 2);
pt of inflection (−2, 0)
Note: dom (f ) = [−2, 2] pt of inflection (0, 0)
19. f (x) = 2 sin x cos x = sin 2x, f (x) = 2 cos 2x;
concave up on 0, 14 π and 34 π, π , concave down on 14 π, 34 π ;
pts of inflection 14 π, 12 and 34 π, 12 20. f (x) = −4 cos x sin x − 2x, f (x) = −4(cos2 x − sin2 x) − 2 = −4 cos 2x − 2;
concave down on 0, 13 π and 23 π, π , concave up on 13 π, 23 π ; 2 9 − 8π 2 1 9 − 2π 2 π, and π, pts of inflection 3 18 3 18 21. f (x) = 2x + 2 cos 2x, f (x) = 2 − 4 sin 2x;
1
5
1 concave up on 0, 12 π and on 12 π, π , concave down on 12 π, 2 2 5 1 72 + π 72 + 25π π, and π, pts of inflection 12 144 12 144
5 12 π
;
22. f (x) = 4 sin3 x cos x, f (x) = 4 sin2 x[3 cos2 x − sin2 x];
concave up on 0, 13 π and 23 π, π , concave down on 13 π, 23 π ;
9 9 pts of inflection 13 π, 16 and 23 π, 16 23. points of inflection: (±3.94822, 10.39228) 24. f (x) = 0 at x ∼ = ±0.94, ±2.57, ±3.71, ±5.35 25. points of inflection: (−3, 0), (−2.11652, 2, 39953), (−0.28349, −18.43523) 26. f (x) > 0 for all x ∈ dom f
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SECTION 4.6 27. f (x) = x3 − 9x (a) f (x) = 3x2 − 9 = 3(x2 − 3) √ √ f (x) ≥ 0 ⇒ x ≤ − 3 or x ≥ 3; √ √ f (x) ≤ 0 ⇒ − 3 ≤ x ≤ 3. √ √ Thus, f is increasing on (−∞, − 3] ∪ [ 3, ∞) √ √ and decreasing on [− 3, 3]. √ (b) f (− 3) ∼ = 10.39 is a local maximum; √ ∼ f ( 3) = −10.39 is a local minimum. (c) f (x) = 6x; The graph of f is concave up on (0, ∞) and concave down on (−∞, 0). (d) point of inflection: (0, 0) 28. f (x) = 3x4 + 4x3 + 1 (a) f (x) = 12x3 + 12x2 = 12x2 (x + 1) f (x) ≥ 0 ⇒ x ≥ −1; f (x) ≤ 0 ⇒ x ≤ −1. Thus, f is increasing on [−1, ∞) and decreasing on (−∞, −1]. (b) f (−1) = 0 is a local minimum; no local maxima.
(c) f (x) = 36x2 + 24x = 36x x + 23 ;
The graph of f is concave up on −∞, − 23 and (0, ∞); and concave down on − 23 , 0 .
(d) points of inflection: − 23 , 11 27 , (0, 1) 29. f (x) =
2x +1
x2
2(x + 1)(x − 1) (x2 + 1)2 f (x) ≥ 0 ⇒ −1 ≤ x ≤ 1;
(a) f (x) = −
f (x) ≤ 0 ⇒ x ≤ −1 or x ≥ 1. Thus, f is increasing on [−1, 1]; and decreasing on (−∞, −1] ∪ [1, ∞). (b) f (−1) = −1 is a local minimum; f (1) = 1 is a local maximum. √ √ 4x(x + 3)(x − 3) (c) f (x) = 3 (x2 + 1) √ √ f (x) > 0 ⇒ x ≤ − 3 or x ≥ 3; √ √ f (x) < 0 ⇒ − 3 < x < 3. √ √ The graph of f is concave up on (− 3 , 0) ∪ ( 3, ∞) and concave down √ √ on (−∞, − 3) ∪ (0, 3). √ √ √ √ (d) points of inflection: (− 3, − 3/2), (0, 0), ( 3, 3/2)
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SECTION 4.6
30. f (x) = x1/3 (x − 6)2/3 x−2 (a) f (x) = 2/3 x (x − 6)1/3 f (x) ≥ 0 ⇒ x ≤ 2, x = 0, or x > 6; f (x) ≤ 0 ⇒ 2 ≤ x < 6. Thus, f is increasing on (−∞, 2] ∪ [6, ∞) and decreasing on [2, 6]. (b) f (2) = 2(4)1/3 is a local maximum; f (6) = 0 is a local minimum. −8 (c) f (x) = 5/3 ; x (x − 6)4/3 The graph of f is concave down on (0, ∞) and concave down up (−∞, 0). (d) point of inflection: (0, 0) x ∈ [−π, π]
31. f (x) = x + sin x,
(a) f (x) = 1 + cos x f (x) > 0 on (−π, π) Thus, f is increasing on [−π, π]. (b) No local extrema (c) f (x) = − sin x f (x) > 0 for x ∈ (−π, 0); f (x) < 0 for x ∈ (0, π). The graph of f is concave up on (−π, 0) and concave down on (0, π). (d) point of inflection: (0, 0) 32. f (x) = sin x + cos x,
x ∈ [0, 2π]
(a) f (x) = cos x − sin x f (x) ≥ 0 f (x) ≤ 0
⇒ ⇒
0 ≤ x ≤ 14 π or
5 4π
≤ x ≤ 2π
≤ x ≤ 54 π
Thus, f is increasing on 0, 14 π ∪ 54 π, 2π ;
f is decreasing on 14 π, 54 π . √ (b) f (π/4) = 2 is a local maximum; √ f (5π/4) = − 2 is a local minimum. 1 4π
(c) f (x) = − sin x − cos x f (x) > 0
⇒
3 4π
f (x) < 0
⇒
0 < x < 34 π or
< x < 74 π;
7 π < x < 2π.
43 7
The graph of f is concave up on 4 π, 4 π and concave down on 0, 34 π ∪ 74 π, 2π .
(d) points of inflection: 34 π, 0 , 74 π, 0
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SECTION 4.6 33.
f (x) = (a) f (x) =
x3 ,
x −1.
−2x,
x > −1;
2
x ≤ −1
f (x) ≥ 0 on (−∞, 0]; f (x) ≤ 0 on [0, ∞). Thus, f is increasing on (−∞, 0] and decreasing on [0, ∞). (b) f (0) = 3 is a local maximum. 0, x < −1 (c) f (x) = −2, x > −1; f (x) < 0 for x ∈ (−1, ∞). Thus, the graph of f is concave down on (−1, ∞). The graph of f is a straight line for x ≤ −1. (d) There are no points of inflection. 35.
36.
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SECTION 4.6
37.
38.
39. Since f (x) = 6x − 2(a + b + c), set d = 13 (a + b + c). Note that f (d) = 0 and that f is concave down on (−∞, d) and concave up on (d, ∞); (d, f (d)) is a point of inflection. 40. f (x) = 2cx − 2x−3 , f (1) = 0
=⇒
f (x) = 2c + 6x−4 . To have a point of inflection at 1 we need
2c + 6 = 0
=⇒
c = −3
41. Since (−1, 1) lies on the graph, 1 = −a + b. Since f (x) exists for all x and there is a pt of inflection at x = 13 , we must have f
1 3
Therefore 0 = 2a + 2b. Solving these two equations, we find a = − 12 and b = 12 . Verification: the function 1 1 f (x) = − x3 + x2 2 2 has second derivative f (x) = −3x + 1. This does change sign at x = 13 . 42. f (x) = Ax1/2 + Bx−1/2 ;
f (x) =
1 2
−1/2
Ax
−
1 2
f (1) = 4 −3/2
Bx
,
=⇒
A + B = 4.
f (x) = − Ax−3/2 + 1 4
3 4
To have a point of inflection at (1, 4), we need f (1) = 0
Bx−5/2 . =⇒
− 14 A +
3 4
B = 0.
Solving the two equations gives A = 3, B = 1. 43. First, we require that
1 2
Next we require that
6 π, 5
lie on the curve: 5 = 12 A + B.
d y = −4A cos 2x − 9B sin 3x be zero (and change sign) at x = 16 π : dx2 0 = −2A − 9B.
Solving these two equations, we find A = 18, B = −4. Verification: the function f (x) = 18 cos 2x − 4 sin 3x has second derivative f (x) = −72 cos 2x + 36 sin 3x. This does change sign at x = 16 π.
= 0.
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SECTION 4.6 f (x) = 2Ax + B;
44. f (x) = Ax2 + Bx + C; (a) Concave up
f (x) > 0
=⇒
f (x) = 2A.
=⇒
A > 0;
to decrease between A and B we need
f (x) < 0, for x between A and B (b) Concave down
=⇒
f (x) < 0
173
=⇒
A < 0;
=⇒
B ≤ −2A2 .
to have f (x) > 0 for x between A and B we
need 2A2 + B ≥ 0 and 2AB + B ≥ 0, that is, B ≥ −2A2 and B(2A + 1) ≥ 0. If A > − 12 , then we need B ≥ 0 (and automatically B ≥ −2A2 ). If A ≤ − 12 , then we need B ≤ 0 and B ≥ −2A2 . The conditions are: − 12 < A < 0, B ≥ 0,
A ≤ − 12 , −2A2 ≤ B ≤ 0.
or
45. Let f (x) = 3x2 − 6x + 3. Then we must have f (x) = x3 − 3x2 + 3x + c for some constant c. Note that f (x) = 6x − 6
and f (1) = 0. Since (1, −2) is a point of inflection of the graph of f, (1, −2)
must lie on the graph. Therefore, 13 − 3(1)2 + 3(1) + c = −2
which implies c = −3
and so f (x) = x3 − 3x2 + 3x − 3. 46. f (x) = cos x and f (x) = − sin x = −f (x). Thus f is concave down when f (x) < 0
Similarly, f is concave up when f (x) > 0
=⇒
f (x) > 0.
=⇒
f (x) < 0.
g (x) = − sin x and g (x) = − cos x = −g(x), so g(x) has the same property. 47. (a) p (x) = 6x + 2a is negative for x < −a/3, and positive for x > −a/3. Therefore, the graph of p has a point of inflection at x = −a/3. (b) p (x) = 3x2 + 2ax + b. The discriminant of this quadratic is 4a2 − 12b = 4(a2 − 3b). Thus, p has two real zeros iff a2 > 3b. (c) If a2 ≤ 3b, then b ≥ a2 /3 and p (x) = 3x2 + 2ax + b ≥ 3x2 + 2ax + a2 /3 = 3(x + 13 a)2 ≥ 0 for all x; p is increasing in this case. 48. It is sufficient to show that the x-coordinate of the point of inflection is the x-coordinate of the midpoint of the line segment connecting the local extrema. It is easy to show that the x-coordinate of the point of inflection is x0 = − 13 a. Now suppose that p has local extrema at x1 and x2 , x1 = x2 . Then p (x1 ) = p (x2 ) = 0 Thus, 49. (a)
⇒
3x21 + 2ax1 + b − (3x22 + 2ax2 + b) = 0
⇒
x1 + x2 = − 23 a.
x1 + x2 = − 13 a = x0 . 2 (b) No.
If f (x) < 0 and f (x) < 0 for all x, then
f (x) < f (0)x + f (0) on (0, ∞), which implies that f (x) → −∞ as x → ∞.
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SECTION 4.6
50. Let f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 . Then f (x) = n(n − 1)an xn−2 + · · · + 2a2 is a degree n − 2 polynomial which can have at most n − 2 roots.
Hence f has at most n − 2 points of inflection. 52.
51.
(a) up: (−2π, −3.64), (−1.077, 1.077), (3.64, 2π)
(a) concave up on (−4, −0.913) ∪ (0.913, 4)
down: (−3.64, −1.077) ∪ (1.077, 3.64)
concave down on (−0.913, 0.913)
(b) pts of inflection at x = ±3.64, ±1.077
(b) pts of inflection at x = −0.913, 0.913 53.
54.
(a) concave up on:
(a) concave up on (−5, 5)
(−π, −1.996) ∪ (−.0345, 2.550) concave down on: (−1.996, −0.345) ∪ (2.550, π) (b) pts of inflection at:
(b) no pts of inflection
x = −1.996, −0.345, 2, 550 55. (a) f (x) = 0 at x = 0.68824, 2.27492, 4.00827, 5.59494 (b) f (x) > 0 on (0.68824, 2.27492) ∪ (4.00827, 5.59494) (c) f (x) < 0 on (0, 0.68824) ∪ (2.27492, 4.00827) ∪ (5.59494, 2π) 56. (a)
f (x) = 0 for all x
(b) f (x) > 0 on (−∞, −1) ∪ (1, ∞)
(c) f (x) < 0 on (−1, 1) 57. (a)
f (x) = 0 at x = ±1, 0, ±0.32654, ±0.71523
(b) f (x) > 0 on (−0.71523, −0.32654) ∪ (0, 0.32654) ∪ (0.71523, 1) ∪ (1, ∞) (c) f (x) < 0 on (−∞, −1) ∪ (−1, −0.71523) ∪ (−0.32654, 0) ∪ (0, 32654, 0.71523)
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SECTION 4.7 58. (a) f (x) = 0 at x = 0
(b) f (x) > 0 on (−4, 0)
(c) f (x) < 0 on (0, 4)
SECTION 4.7 1. (a) ∞ (e) 0
(b) −∞
(c) ∞
(f) x = −1, x = 1
(g) y = 0, y = 1
2. (a) d (d) y = d 1 3
3. vertical: x =
; horizontal: y =
(d) 1
(b) c
(c) x = a, x = b
(e) p
(f) q 4. vertical: x = −2; horizontal: none
1 3
5. vertical: x = 2 ; horizontal: none
6. vertical: none;
7. vertical: x = ±3 ;
horizontal: y = 0
8. vertical: x = 16;
9. vertical: x = − 43 ;
horizontal: y =
4 9
horizontal: y = 0 horizontal: y = 0
10. vertical: x = 13 ;
horizontal: y =
4 9
12. vertical: x = 12 ;
horizontal: y = − 18
13. vertical: none ; horizontal: y = ± 32
14. vertical: x = 8;
horizontal: y = 0
15. vertical: x = 1 ; horizontal: y = 0
16. vertical: x = ± 1;
17. vertical: none ; horizontal: y = 0
18. vertical: none;
19. vertical: x = 2n + 12 π ; horizontal: none
20. vertical: x = 2nπ;
21. f (x) = 43 (x + 3)1/3 ; neither
22. f (x) =
23. f (x) = − 45 (2 − x)−1/5 ; cusp
24. neither; f (−1) is not defined
25. f (x) = 65 x−2/5 1 − x3/5 ; tangent
26. f (x) = 75 (x − 5)2/5 ; neither
27. f (−2) undefined; neither
28. f (x) = 37 (2 − x)−4/7 ; tangent
5 2
11. vertical: x =
29. f (x) =
⎧ ⎨
; horizontal: y = 0
− 1)−1/2 ,
x>1
⎩ − 1 (1 − x)−1/2 ,
x < 1;
1 2 (x 2
30. f (x) = (4x − 3)(x − 1)−2/3 ;
tangent
cusp
2 5
horizontal: y = ± 2
horizontal: y = 0 horizontal: none
x−3/5 ; cusp
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SECTION 4.7
31. f (x) =
⎧ ⎨
+ 8)−23 ,
x > −8
+ 8)−2/3 ,
x < −8;
1 3 (x
⎩ − 1 (x 3
cusp
32. f is not defined for x > 2; neither
33. f not continuous at 0; neither
34. f is not continuous at 0; neither 35.
36.
37.
38.
39. f (x) = x − 3x1/3
1 x2/3 f is increasing on (−∞, −1] ∪ [1, ∞)
(a) f (x) = 1 −
f is decreasing on [−1, 1] (b) f (x) =
2 3
x−5/3
concave up on (0, ∞); concave down on (−∞, 0) vertical tangent at (0, 0) 40. f (x) = x2/3 − x1/3 (a) f (x) =
x−2/3 = 1 f is increasing on 8 , ∞
f is decreasing on −∞, 18 2 3
x−1/3 −
1 3
(b) f (x) = − 29 x−4/3 +
2 9
2x1/3 − 1 3x2/3
x−5/3 =
concave up on (0, 1)
2(1 − x1/3 ) 9x5/3
concave down on (−∞, 0) ∪ (1, ∞) vertical tangent at (0, 0)
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SECTION 4.7 41. f (x) =
3 5
x5/3 − 3x2/3
x−2 x1/3 f is increasing on (−∞, 0] ∪ [2, ∞)
(a) f (x) = x2/3 − 2x−1/3 =
f is decreasing on [0, 2] 2x + 2 3x4/3 concave up on (−1, 0) ∪ (0, ∞); concave down on
(b) f (x) =
2 3
x−1/3 +
2 3
x−4/3 =
(−∞, −1) vertical cusp at (0, 0)
42. f (x) = =
|x| x1/2 ,
x≥0
1/2
x < 0.
(−x)
(a) f (x) =
,
1 2
x−1/2 , x > 0
− 12 (−x)−1/2 , x < 0;
f is increasing on [0, ∞) f is decreasing on (−∞, 0] − 14 x−3/2 , x > 0 (b) f (x) = − 14 (−x)−3/2 , x < 0; concave down on (−∞, 0) ∪ (0, ∞) vertical cusp at (0, 0)
43.
44.
no asymptotes
vertical asymptotes: x = 1, x = −1
vertical cusp at (0, 1)
horizontal asymptote: y = 1 no vertical tangents or cusps
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SECTION 4.7
45.
46.
horizontal asymptote: y = 4 horizontal asymptotes: y = −1, y = 1
vertical tangent at (0, 1)
vertical tangent at (0, 0) 47. (a) p odd; (b) p even. 48. [r(x) − (ax + b)] =
Q(x) → 0 as x → ± ∞, since deg [Q(x)] < deg [q(x)]. q(x) 50.
49.
vertical asymptote: x = −1
vertical asymptote: x = 0
oblique asymptote: y = 2x + 1
oblique asymptote: y = x 51.
52.
vertical asymptote: x = 1
vertical asymptote: x = 0
oblique asymptote: y = x
oblique asymptote: y = −3x + 1
53. oblique asymptote: y = 3x − 4
54. oblique asymptote: y = 5x − 3
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SECTION 4.8 55. y = 1 is a horizontal asymptote. √ √
√ x2 + 2x + x 2x f (x) = x2 + 2x − x = x2 + 2x − x · √ =√ → 1. x2 + 2x + x x2 + 2x + x 56. y = − 12 is a horizontal asymptote. √ √
√ x4 − x2 + x2 −x2 2 2 4 2 4 2 x −x −x · √ =√ → − 12 . f (x) = x − x − x = 4 2 2 4 x −x +x x − x2 + x2 SECTION 4.8 [Rough sketches; not scale drawings] 1.
f (x) = (x − 2)2 f (x) = 2(x − 2) f (x) = 2
2.
f (x) = 1 − (x − 2)2 f (x) = −2(x − 2) f (x) = −2
3.
f (x) = x3 − 2x2 + x + 1 f (x) = (3x − 1)(x − 1) f (x) = 6x − 4
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November 25, 2006
SECTION 4.8 f (x) = x3 − 9x2 + 24x − 7 f (x) = 3x2 − 18x + 24 f (x) = 6x − 18
5.
f (x) = x3 + 6x2 ,
x ∈ [−4, 4]
f (x) = 3x(x + 4) f (x) = 6x + 12
6.
f (x) = x4 − 8x2 ,
x ∈ [0, ∞)
f (x) = 4x3 − 16x f (x) = 12x2 − 16
7.
f (x) = 23 x3 − 12 x2 − 10x − 1 f (x) = (2x − 5)(x + 2) f (x) = 4x − 1
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SECTION 4.8 8.
f (x) = x(x2 + 4)2 = x5 + 8x3 + 16x f (x) = 5x4 + 24x2 + 16 f (x) = 20x3 + 48x = 4x(5x2 + 12)
9.
f (x) = x2 + 2x−1
f (x) = 2x − 2x−2 = 2 x3 − 1 /x2 f (x) = 2 + 4x−3
asymptote:
10.
x=0
f (x) = x − x−1 , f (x) = 1 + x−2 f (x) = − x−3
asymptotes:
x = 0, y = x
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SECTION 4.8 f (x) = (x − 4)/x2 f (x) = (8 − x)/x3 f (x) = (2x − 24)/x4
asymptotes:
12.
f (x) =
f (x) =
x = 0, y = 0
x+2 1 2 = 2+ 3 x3 x x
f (x) = −
2 6 −2x − 6 − 4 = x3 x x4
6 24 6(x + 4) + 5 = 4 x x x5
asymptotes: 13.
x = 0, y = 0
f (x) = 2x1/2 − x, x ∈ [0, 4]
f (x) = x−1/2 1 − x1/2 f (x) = − 12 x−3/2
14.
√
f (x) =
1 4
x−
f (x) =
1 4
−
f (x) =
1 4
x−3/2
1 2
T1: PBU
x,
x−1/2
x ∈ [0, 9]
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SECTION 4.8 15.
f (x) = 2 + (x + 1)6/5 f (x) = 65 (x + 1)1/5 f (x) =
16.
6 25 (x
+ 1)−4/5
f (x) = 2 + (x + 1)7/5 f (x) = 75 (x + 1)2/5 f (x) =
17.
18.
14 25 (x
+ 1)−3/5
f (x) = 3x5 + 5x3
f (x) = 15x2 x2 + 1
f (x) = 30x 2x2 + 1
f (x) = 3x4 + 4x3 f (x) = 12x3 + 12x2 = 12x2 (x + 1) f (x) = 36x2 + 24x = 12x(3x + 2)
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SECTION 4.8 f (x) = 1 + (x − 2)5/3 f (x) = 53 (x − 2)2/3 f (x) =
20.
10 9 (x
− 2)−1/3
f (x) = 1 + (x − 2)4/3 f (x) = 43 (x − 2)1/3 f (x) = 49 (x − 2)−2/3
21.
f (x) =
8x (x2 + 4)2
f (x) = 48
8(4 − 3x2 ) (x2 + 4)3
f (x) < 0 on (−∞, 0) f (x) > 0 on (0, ∞) √ √
f (x) < 0 on −∞, −2/ 3 ∪ 2/ 3, ∞ ; √ √
f (x) > 0 on −2/ 3, 2/ 3 ; asymptote: y = 1
22.
2x2 x+1 2x(x + 2) f (x) = (x + 1)2 4 f (x) = (x + 1)3 f (x) =
asymptote: y = 2x − 2
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SECTION 4.8 23.
f (x) =
x (x + 3)2
f (x) =
3−x (x + 3)3
f (x) =
2x − 12 (x + 3)4
asymptotes:
24.
f (x) =
x x2 + 1
f (x) =
1 − x2 (x2 + 1)2
f (x) =
2x(x2 − 3) (x2 + 1)3
asymptote:
25.
x = −3, y = 0
y=0
f (x) =
x2 x2 − 4
f (x) =
−8x (x2 − 4)2
f (x) =
8(3x2 + 4) (x2 − 4)3
asymptotes:
x = −2, x = 2, y = 1
185
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SECTION 4.8 f (x) =
1 x3 − x
1 − 3x2 (x3 − x)2 √
√
increasing on −1/ 3, 0 ∪ 0, 1/ 3 ; f (x) =
decreasing otherwise
2 6x4 − 3x2 + 1 f (x) = (x3 − x)3 concave up on (−1, 0) ∪ (1, ∞) concave down on (−∞, 0) ∪ (0, 1) asymptotes: 27.
x = −1, x = 0, x = 1, y = 0
f (x) = x(1 − x)1/2 f (x) = 12 (1 − x)−1/2 (2 − 3x) f (x) = 14 (1 − x)−3/2 (3x − 4)
28.
f (x) = (x − 1)4 − 2(x − 1)2 f (x) = 4(x − 1)3 − 4(x − 1) f (x) = 12(x − 1)2 − 4
29.
f (x) = x + sin 2x, f (x) = 1 + 2 cos 2x f (x) = −4 sin 2x
x ∈ [0, π]
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SECTION 4.8 30.
x ∈ [0, π]
f (x) = cos3 x + 6 cos x,
f (x) = −3 sin x(cos2 x + 2) f (x) = −9 cos3 x
31.
f (x) = cos4 x,
x ∈ [0, π]
f (x) = −4 cos3 x sin x
f (x) = 4 cos2 x 3 sin2 x − cos2 x
32.
f (x) = f (x) =
√ √
3 x − cos 2x,
x ∈ [0, π]
3 + 2 sin 2x
f (x) = 4 cos 2x
33.
f (x) = 2 sin3 x + 3 sin x, x ∈ [0, π]
f (x) = 3 cos x 2 sin2 x + 1
f (x) = 9 sin x 1 − 2 sin2 x
187
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SECTION 4.8 f (x) = sin4 x,
x ∈ [0, π]
f (x) = 4 sin3 x cos x
f (x) = 4 sin2 x 3 cos2 x − sin2 x
35.
f (x) = [(x + 1) − 1]3 + 1 f (x) = 3x2 f (x) = 6x
36.
f (x) = x3 (x + 5)2 f (x) = 5x2 (x + 3)(x + 5) f (x) = 10x(2x2 + 12x + 15)
37.
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f (x) = x2 (5 − x)3 f (x) = 5x(2 − x)(5 − x)2
f (x) = 10(5 − x) 2x2 − 8x + 5
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SECTION 4.8 38. f (x) =
4 − 2x + x2 ,
4 + 2x − x , x ≤ 0, x ≥ 2 −2 + 2x, 0 < x < 2
f (x) =
2 − 2x,
f (x) =
39. f (x) = f (x) =
2,
x < 0, x > 2
0 1
x + 2, −1 ≤ x ≤ 1 −2x, |x| > 1 2
f (x) =
40.
0 2 x−2 −1/2 x (x − 2)−2 f (x) = − x−2 −3/2 x f (x) = (2x − 1) (x − 2)−4 x−2 f (x) =
asymptotes:
x = 2, y = 1
46.
1/2 x f (x) = ; x < −4, x ≥ 0 x+4 −1/2 x (x + 4)−2 f (x) = 2 x+4 f (x) =
−4(x + 1) (x + 4)5/2 x3/2
asymptotes:
x = −4, y = 1
191
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SECTION 4.8 √ −1/2
f (x) = x2 x2 − 2 , |x| > 2
−3/2 f (x) = x x2 − 4 x2 − 2
−5/2 f (x) = 2 x2 + 4 x2 − 2
asymptotes: 48.
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√ √ x = − 2, x = 2
f (x) = 3 cos 4x,
x ∈ [0, π]
f (x) = −12 sin 4x f (x) = −48 cos 4x
49.
f (x) = 2 sin 3x,
x ∈ [0, π]
f (x) = 6 cos 3x f (x) = −18 sin 3x
50.
f (x) = 3 + 2 cot x + csc2 x,
x ∈ (0, π/2)
f (x) = −2 csc2 x (1 + cot x)
f (x) = 2 csc2 x 3 cot2 x + 2 cot x + 1
asymptote: x = 0
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SECTION 4.8 51.
f (x) = 2 tan x − sec2 x,
x ∈ (0, π/2)
= −(1 − tan x)2 f (x) = 2 sec2 x (1 − tan x)
f (x) = −2 sec2 x 3 tan2 x − 2 tan x + 1
asymptote:
52.
x = 12 π
f (x) = 2 cos x + sin2 x f (x) = 2 sin x (cos x − 1) f (x) = 2(2 cos2 x − cos x − 1)
sin x , x ∈ (−π, π) 1 − sin x cos x f (x) = (1 − sin x)2
53. f (x) =
f (x) =
1 − sin x + cos2 x (1 − sin x)3
asymptote:
x = 12 π
193
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SECTION 4.8 1 , x ∈ (−π, π) 1 − cos x − sin x f (x) = (1 − cos x)2
54. f (x) =
f (x) =
2 − cos x − cos2 x (1 − cos x)3
asymptote: x = 0
55. (a) f increases on (−∞, −1] ∪ (0, 1] ∪ [3, ∞); f decreases on [−1, 0) ∪ [1, 3]; critical points: x = −1, 0, 1, 3. (b)
concave up on (−∞, −3) ∪ (2, ∞) concave down on (−3, 0) ∪ (0, 2).
(c)
The graph does not necessarily have a horizontal asymptote.
56. (a)
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SECTION 4.8 (b)
195
(c)
(d) F is discontinuous at 0;
lim sin(1/x) does not exist
x→0
G is continuous at 0: |x sin(1/x)| = |x| | sin(1/x)| ≤ |x| → 0, so that lim G(x) = 0 = G(0). x→0 H is continuous at 0: x2 sin(1/x) = |x|2 | sin(1/x)| ≤ |x|2 → 0, so that lim H(x) = 0 = H(0). x→0
(e) F is not differentiable at 0 since it is not continuous at 0. G(h) − G(0) = lim sin(1/h) does not exist. h→0 h H(h) − H(0) H is differentiable at 0: H (0) = lim = lim h sin(1/h) = 0. h→0 h→0 h
G is not differentiable at 0:
lim
h→0
1 x3 − x1/3 x3 − x1/3 − x3 − x2 = = − 2/3 → 0 as x → ±∞. x x x The figure shows the graphs of f and g(x) = x2 for x > 0. Since f and g are even functions,
57. f (x) − x2 =
the graphs are symmetric about the y-axis. y
x
58. (a)
y
b −a
b 2 b (b) x − a2 = x a a
x
a
b 2 b x − a2 − x → 0 as x → ∞. a a 2 b a b (c) If x < 0, then the second quadrant arc is given by y = − x 1 − 2 → − x as x → −∞. a x a 1−
a2 b → x as 2 x a
x → ∞. Therefore,
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SECTION 4.9
SECTION 4.9 1. x(5) = −6; v(t) = 3 − 2t so v(5) = −7 and speed = 7; a(t) = −2 so a(5) = −2. 2. x(3) = −12; v(t) = 5 − 3t2 so v(3) = −22 and speed = 22; a(t) = −6t so a(3) = −18. 3. x(1) = 6; v(t) = −18/(t + 2)2 so v(1) = −2 and speed = 2, a(t) = 36/(t + 2)3 so a(1) = 4/3. 4. x(3) = 1; v(t) =
6 so v(3) = 1/6 = speed; a(t) = −12/(t + 3)3 so a(3) = −1/18. (t + 3)2
5. x(1) = 0, v(t) = 4t3 + 18t2 + 6t − 10 so v(1) = 18 and speed = 18, a(t) = 12t2 + 36t + 6 so a(1) = 54. 6. x(2) = −20; v(t) = 4t3 − 18t so v(2) = −4 and speed = 4, a(t) = 12t2 − 18 so a(2) = 30. 7. x(t) = 5t + 1,
x (t) = v(t) = 5,
x (t) = a(t) = 0.
v(t) = 0, a(t) = 0 for all t. 8. x(t) = 4t2 − t + 3,
x (t) = v(t) = 8t − 1,
v(t) = 0 at t = 1/8;
x (t) = a(t) = 8.
a(t) = 0 for all t.
9. x(t) = t3 − 6t2 + 9t − 1,
x (t) = v(t) = 3t2 − 12t + 9,
v(t) = 3(t − 1)(t − 3), v(t) = 0 at t = 1, 3; 10. x(t) = t4 − 4t3 + 4t2 + 2,
x (t) = a(t) = 6t − 12.
a(t) = 0 at t = 2.
x (t) = v(t) = 4t3 − 12t2 + 8t,
v(t) = 4t(t − 1)(t − 2), v(t) = 0 at t = 0, 1, 2;
x (t) = a(t) = 12t2 − 24t + 8. √ a(t) = 0 at t = 1 + 3/3.
11. A
12. C
13. A
14. C
15. A and B
16. B
17. A
18. A
19. A and C
20. B
21. The object is moving right when v(t) > 0. Here, v(t) = 4t3 − 36t2 + 56t = 4t(t − 2)(t − 7);
v(t) > 0 when 0 < t < 2 and 7 < t.
22. The object is moving left when v(t) < 0. Here, v(t) = 3t2 − 24t + 21 = 3(t − 7)(t − 1);
v(t) < 0 when 1 < t < 7.
23. The object is speeding up when v(t) and a(t) have the same sign. v(t) = 5t3 (4 − t)
sign of v(t) :
a(t) = 20t (3 − t)
sign of a(t) :
2
Thus,
0 0 and a(t) < 0. v(t) = 3(t − 5)(t + 1)
sign of v(t) :
a(t) = 6(t − 2)
sign of a(t) :
This never happens. 27. The object is moving right and speeding up when v(t) > 0 and a(t) > 0. v(t) = 4t(t − 2)(t − 4)
sign of v(t) :
a(t) = 4(3t − 12t + 8) √ Thus, 0 < t < 2 − 23 3
sign of a(t) :
2
and
4 < t.
28. The object is moving left and speeding up when v(t) < 0 and a(t) < 0. v(t) = 4t(t − 2)(t − 4)
sign of v(t) :
a(t) = 4(3t2 − 12t + 8) √ Thus, 2 < t < 2 + 23 3.
sign of a(t) :
29. x(t) = (t + 1)2 (t − 9)3 ,
x (t) = v(t) = 3(t + 1)2 (t − 9)2 + 2(t + 1)(t − 9)3 = 5(t + 1)(t − 9)2 (t − 3).
The object changes direction at time t = 3. 30. x(t) = t(t − 8)3 ,
x (t) = v(t) = (t − 8)3 + 3t(t − 8)2 = 4(t − 8)2 (t − 2).
The object changes direction at time t = 2. √ √ x/(t) = v(t) = 4(t3 − 12t)3 (3t2 − 12) = 12t3 (t + 2 3)3 (t − 2 3)3 (t + 2)(t − 2). √ The object changes direction at times t = 2, 2 3.
31. x(t) = (t3 − 12t)4 ,
32. x(t) = (t2 − 8t + 15)3 ,
x (t) = v(t) = 3(t2 − 8t + 15)2 (2t − 8) = 6(t2 − 8t + 15)2 (t − 4).
The object changes direction at time t = 4. Exercises 33–38. The object is moving right with increasing speed when both v(t) and a(t) are positive. 33. x(t) = sin 3t,
x (t) = v(t) = 3 cos 3t,
x (t) = a(t) = −9 sin 3t.
v(t) > 0 on (0, π/6) ∪ (π/2, 5π/6) ∪ (7π/6, 3π/2) ∪ (11π/6, 2π). a(t) > 0 on (π/3, 2π/3) ∪ (π, 4π/3) ∪ (5π/3, 2π). v(t) and a(t) are positive on (π/2, 2π/3) ∪ (7π/6, 4π/3) ∪ (11π/6, 2π).
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SECTION 4.9
34. x(t) = cos 2t,
x (t) = v(t) = −2 sin 2t,
v(t) > 0 on (π/2, π) ∪ (3π/2, 2π);
x (t) = a(t) = −4 cos 2t.
a(t) > 0 on (π/4, 3π/4) ∪ (5π/4, 7π/4).
v(t) and a(t) are positive on (π/2, 3π/4) ∪ (3π/2, 7π/4). x (t) = v(t) = cos t + sin t,
35. x(t) = sin t − cos t,
v(t) > 0 on (0, 3π/4) ∪ (7π/4, 2π);
x (t) = a(t) = − sin t + cos t.
a(t) > 0 on (0, π/4) ∪ (5π/4, 2π).
v(t) and a(t) are positive on (0, π/4) ∪ (7π/4, 2π). x (t) = v(t) = cos t − sin t,
36. x(t) = sin t + cos t,
v(t) > 0 on (0, π/4) ∪ (5π/4, 2π);
x (t) = a(t) = − sin t − cos t.
a(t) > 0 on (3π/4, 7π/4).
v(t) and a(t) are positive on (5π/4, 7π/4). 37. x(t) = t + 2 cos t,
x (t) = v(t) = 1 − 2 sin t,
v(t) > 0 on (0, π/6) ∪ (5π/6, 2π);
x (t) = a(t) = −2 cos t.
a(t) > 0 on (π/2, 3π/2).
v(t) and a(t) are positive on (5π/6, 3π/2). 38. x(t) = t −
√ 2 sin t,
x (t) = v(t) = 1 −
v(t) > 0 on (π/4, 7π/4);
√ 2 cos t,
x (t) = a(t) =
√
2 sin t.
a(t) > 0 on (0, π).
v(t) and a(t) are positive on (π/4, π). 39. Since v0 = 0 the equation of motion is y(t) = −16t2 + y0 . We want to find y0 so that y(6) = 0. From 0 = −16(6)2 + y0 we get y0 = 576 feet. 40. The equation of motion is: y(t) = −4.9t2 + y0 . Therefore, the velocity is given by v(t) = −9.8t. Since the object hits the ground at 98 m/sec., we have −9.8t = −98, and t = 10. Therefore, y(10) = 0 = −4.9(10)2 + y0 and y0 = 490 meters. 41. The object’s height and velocity at time t are given by y(t) = −
1 2 gt + v0 t 2
and
v(t) = −gt + v0
Since the object’s velocity at its maximum height is 0, it takes v0 /g seconds to reach maximum height, and y(v0 /g) = − 12 g(v0 /g)2 + v0 (v0 /g) = v02 /2g
or
v02 /19.6 (meters)
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199
42. Since y0 = 0, we have y(t) = −16t2 + v0 t = t(−16t + v0 ). Now, y(8) = 0
=⇒
v0 = (16)8 = 128
=⇒
the initial velocity was 128 ft/sec.
43. At time t, the object’s height is y(t) = − 12 gt2 + v0 t + y0 , and its velocity is v(t) = −gt + v0 . Suppose that y(t1 ) = y(t2 ), t1 = t2 . Then − 12 gt21 + v0 t1 + y0 = − 12 gt22 + v0 t2 + y0 1 2
g(t22 − t21 ) = v0 (t2 − t1 ) gt2 + gt1 = 2v0
From this equation, we get −(−gt1 + v0 ) = −gt2 + v0 and so |v(t1 )| = |v(t2 )|. 44. Since y0 = 0, we have y(t) = −4.9t2 + v0 t = t(v0 − 4.9t) The object hits the ground at t = v0 /4.9 sec., that is, the object is in the air for v0 /4.9 sec. At its maximum height, the velocity of the object is 0. Since v(t) = −9.8t + v0 , we have −9.8t + v0 = 0 and t = v0 /9.8 =
1 2
(v0 /4.9). The result follows from
this. 45. In the equation y(t) = −16t2 + v0 t + y0 we take v0 = −80 and y0 = 224. The ball first strikes the ground when −16t2 − 80t + 224 = 0; that is, at t = 2. Since v(t) = y (t) = −32t − 80, we have v(2) = −144 so that the speed of the ball the first time it strikes the ground is 144 ft/sec.Thus, the speed of the ball the third time it strikes the 1 1 ground is (144) = 9 ft/sec. 4 4 46. Since y0 = 0, we have y(t) = −16t2 + v0 t. y(2) = 64
=⇒
−16(2)2 + 2v0 = 64
=⇒
Now, at the maximum height, v(t) = −32t + 64 = 0
v0 = 64 =⇒
and y(t) = −16t2 + 64t
t = 2. We already know
the height at t = 2, namely 64 ft. 47. The equation is y(t) = −16t2 + 32t.
(Here y0 = 0 and v0 = 32.)
(a) We solve y(t) = 0 to find that the stone strikes the ground at t = 2 seconds. (b) The stone attains its maximum height when v(t) = 0. Solving v(t) = −32t + 32 = 0,
we get t = 1
and, thus, the maximum height is y(1) = 16 feet.
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SECTION 4.9 (c) We want to choose v0 in y(t) = −16t2 + v0 t so that From
y(t0 ) = 36
when
v(t0 ) = 0 for some time t0 .
v(t) = −32t + v0 = 0
we get t0 = v0 /32 so that v v 2 v0 2 0 0 + v0 = 36, or = 36. −16 32 32 64
Thus, v0 = 48 ft/sec. 48. (a) Measuring height from the water surface, we have y(t) = −16t2 + y0 , since v0 (0) = 0. If the stone hits the water 3 seconds later, then y(3) = −16(3)2 + y0 = 0. so y0 = 144. (b) It takes y0 /1080 seconds for the sound of the splash to reach the man so the stone hits the at time t = 3 − y0 /1080. Thus, y0 2 + y0 = 0 y(t) = −16 3 − 1080
=⇒
y0 ∼ = 132.47 ft.
49. For all three parts of the problem the basic equation is y(t) = −16t2 + v0 t + y0 with (∗)
y(t0 ) = 100
and y(t0 + 2) = 16
for some time t0 > 0. We are asked to find y0 for a given value of v0 . From (∗) we get 16 − 100 = y(t0 + 2) − y(t0 ) = [−16(t0 + 2)2 + v0 (t0 + 2) + y0 ] − [−16t0 2 + v0 t0 + y0 ] = −64t0 − 64 + 2v0 so that t0 =
1 32 (v0
+ 10).
Substituting this result in the basic equation and noting that y(t0 ) = 100, we have 2 v0 + 10 v0 + 10 + y0 = 100 + v0 −16 32 32 and therefore v0 2 25 + . 64 16 We use (∗∗) to find the answer to each part of the problem. y0 = 100 −
(∗∗)
(a) v0 = 0 so y0 =
1625 16
ft
(b) v0 = −5 so y0 =
6475 64
ft
(c) v0 = 10 so y0 = 100 ft
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201
50. Let v0 > 0 be the initial velocity. The equation of motion prior to the impact is: y(t) = −16t2 − v0 t + 4. v02 + 256 − v0 The ball hits the ground at time t = with velocity v = v02 + 256. The equation of 32 v02 + 256 2 t. It reaches its maximum height at time motion following the impact is: y(t) = −16t + 2 √ v02 + 256 T = . Now, y(T ) = 4 =⇒ v0 = 16 3. 64 51. Let y0 > 0 be the initial height. The equation of motion becomes: 0 = −16(8)2 + 5(8) + y0 ,
so y0 = 984 ft.
52. Using 0 = −16t2 − 5t + 984,
yields t =
123 16
or about 7.7 sec.
53. Let f1 (t) and f2 (t) be the positions of the horses at time t. Consider f (t) = f1 (t) − f2 (t). Let T be the time the horses finish the race. Then f (0) = f (T ) = 0. By the mean-value theorem there is a c in (0, T ) such that f (c) = 0. Hence f1 (t) = f2 (t), so the horses had the same speed at time c. 54. Apply the argument given in Exercise 53 to the velocity functions v1 (t), v2 (t). 55. The driver must have exceeded the speed limit at some time during the trip. Let s(t) denote the car’s position at time t, with s(0) = 0 and s(5/3) = 120. Then, by the mean-value theorem, there exists at least one number (time) c such that v(c) = s (c) =
120 s(5/3) − s(0) 360 = 5 = = 72 mi./hr. 5 5 − 0 3 3
1 56. Let f (t) be the function that gives the car’s velocity after t hours. Then f (0) = 30 and f ( ) = 60. f 4 1 1 1 is differentiable on (1, ) and continuous on [1, ], so by the mean-value theorem there is a c in (1, ) 4 4 4 60 − 30 such that f (c) = = 120. i.e. The acceleration at time c was 120 mph. 0 − t1 57. If the speed s(t) of the car is less than 60 mi/hr= 1 mi/min, then the distance traveled in 20 minutes is less than 20 miles. Therefore, the car must have gone at least 1 mi/min at some time t < 20. Let t1 be the first instant the car’s speed is 1 mi/min. Then the speed s(t) is less than 1 mi/min on the interval [0, t1 ) and the distance r traveled in t1 minutes is less than t1 miles. Now, by the mean-value theorem, there is a time c ∈ [t1 , 20] such that s (c) =
20 − r 20 − t1 > = 1 (= 60 mi/hr). 20 − t1 20 − t1
58. Let s(t) denote the distance that the car has traveled in t seconds since applying the brakes, 0 ≤ t ≤ 6. Then s(0) = 0 and s(6) = 280. Assume that s is differentiable on (0, 6) and continuous on [0, 6]. Then, by the mean-value theorem, there exists a time c ∈ (0, 6) such that s (c) = v(c) =
280 ∼ s(6) − s(0) = = 46.67 ft/sec 6−0 6
Now v(0) ≥ v(c) = 46.7 ft/sec. Thus, the driver must have been exceeding the speed limit (44 ft/sec) at the instant he applied his brakes.
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SECTION 4.9
59. y(t) = A sin(ωt + φ0 ),
y (t) = v(t) = ωA cos(ωt + φ0 ),
y (t) = a(t) = −ω 2 A sin(ωt + φ0 )
(a) y (t) + ω 2 y(t) = −ω 2 A sin(ωt + φ0 ) + ω 2 A sin(ωt + φ0 ) = 0. (b) v (t) = a(t) = −ω 2 A sin(ωt + φ0 ) = 0 when ωt + φ0 = nπ; t =
nπ − φ0 . ω
The extreme values of v occur at these times. Now nπ − φ0 nπ − φ0 y = A sin ω + φ0 = A sin(nπ) = 0. ω ω The bob attains maximum speed at the equilibrium position. (c) a (t) − ω 3 A cos(ωt + φ0 ) = 0 when ωt + φ0 = (2n − 1)π/2; t =
(2n − 1)π/2 − φ0 . ω
The extreme values of a occur at these times. Now (2n − 1)π/2 − φ0 (2n − 1)π/2 − φ0 2 a = −ω A sin + φ0 = ±ω 2 A. ω ω The bob attains these values at (2n − 1)π/2 − φ0 (2n − 1)π/2 − φ0 y = A sin ω + φ0 = A sin (2n − 1)π/2 = ±A. ω ω 60. (a) x(t) = t3 − 7t2 + 10t + 5,
v(t) = 3t2 − 14t + 10, 0 ≤ t ≤ 5
(b) The object is moving to the right when 0 < t < 0.88 and when
3.79 < t < 5.
The object is moving to the left when 0.88 < t < 3.79 (c) The object stops at times t ∼ = 0.88 and t ∼ = 3.79. ∼ ∼ The maximum speed is v = 6.33 at t = 2.33. (d) a(t) = 6t − 14 The object is speeding up when v(t) and a(t) have the same sign: 0.88 < t < 2.33 and 3.79 < t < 5. The object is slowing down when v(t) and a(t) have opposite sign: 0 < t < 0.88 and 2.33 < t < 3.79. PROJECT 4.9a 1. length of arc = rθ, speed =
d dθ [rθ] = r = rω dt dt
2. v = rω so KE = 12 mr2 ω 2 . 3. We know that dθ/dt = ω and, at time t = 0, θ = θ0 . x(t) = r cos (ωt + θ0 ) x(t) = r cos(ωt + θ0 ),
y(t) = r sin(ωt + θ0 )
v(t) = x (t) = − rω sin(ωt + θ0 ) = − ω y(t) a(t) = − rω 2 cos(ωt + θ0 ) = − ω 2 x(t) v(t) = y (t) = rω cos(ωt + θ0 ) = ω x(t) a(t) = − rω 2 sin(ωt + θ0 ) = − ω 2 y(t)
and
Therefore θ = ωt + θ0 . y(t) = r sin (ωt + θ0 ).
It follows that
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SECTION 4.10 4. For the sector
A=
For triangle T
1 dθ 1 dA = r2 = r2 ω dt 2 dt 2
1 2 r θ, 2
is constant.
A = 12 (2r sin 12 θ)(r cos 12 θ) = 12 r2 (2 sin 12 θ cos 12 θ) = 12 r2 sin θ, 1 1 dθ dA = r2 cos θ = r2 ω cos θ dt 2 dt 2
varies with θ.
For segment S A = 12 r2 θ − 12 r2 sin θ = 12 r2 (θ − sin θ), dA 1 2 dθ dθ 1 = r − cos = r2 ω(1 − cos θ) dt 2 dt dt 2
varies with θ.
dAT dAS = 12 r2 ω cos θ and = 12 r2 ω − 12 r2 ω cos θ dt dt
5. From Exercise 4, Now,
1 2 1 1 r ω cos θ = r2 ω − r2 ω cos θ 2 2 2
=⇒
cos θ =
1 2
=⇒
θ=
π . 3
PROJECT 4.9B d mgy + dt
1.
1 2
dy 1 d 2 + m (v ) mv 2 = mg dt 2 dt dv 1 = mgv + m 2v 2 dt = mgv + mv
dv dt
= mgv + mv(−g)
(since
dv/dt = a = −g)
= mgv − mgv = 0 2. By Problem 1, mgy +
1 2
mv 2 = C (constant). Since v = 0 at height y = y0 , we have C = mgy0 .
Thus, mgy0 = mgy + 3. y(t) = − 12 gt2 + y0
=⇒
gt =
v(t) = y (t) = −gt Therefore,
1 2
mv 2
and |v| =
2g(y0 − y)
2g(y0 − y) |v(t)| = 2g(y0 − y).
SECTION 4.10 dy dx +2 =0 dt dt dy dx = 4, then = −2 units/sec. If dt dt
1. x + 2y = 2, (a)
(b)
If
dy = −2, dt
then
dx = 4 units/sec. dt
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SECTION 4.10 dy dx y dy dx + 2y = 0 and =− . dt dt dt x dt dx 8 dy = −2. Therefore, = ; the x-coordinate is At the point (3, 4), dt dt 3 increasing at the rate of 8/3 units per second.
2. x2 + y 2 = 25,
2x
dx dx dy dy =4 and = 12 y dt dt dt dt dx dy = 3. Therefore = 12 · 6 · 3 = 9 units/sec. At the point (7, 6), dt dt
3. y 2 = 4(x + 2),
2y
dy dx dx dy dx = 2. Also 4y = (x + 2)2 so 4 = 2(x + 2) or = 12 (x + 2) . 4. We are given dt dt dt dt dt dy = 4. The distance from a point on the parabola to the point (−2, 0) is given by At x = 2, dt S = (x + 2)2 + y 2 = 4y + y 2 since (x + 2)2 = 4y. Now 1 2 + y dy dy dS = (4y + y 2 )−1/2 (4 + 2y) = . dt 2 dt 4y + y 2 dt √ 6 dS = √ 4 = 3 2. Therefore, at the point (2, 4), dt 32 5. Let s =
x2 + y 2 denote the distance to the origin at time t. Since x = 4 cos t and y = 2 sin t,
we have s(t) =
At t = π/4,
16 cos2 t + 4 sin2 t =
12 cos2 t + 4
ds 1 = (12 cos2 t + 4)−1/2 (−24 cos t sin t) dt 2 −12 cos t sin t = √ 12 cos2 t + 4 √ ds −12 cos(π/4) sin(π/4) = = − 35 10. 2 dt 12 cos (π/4) + 4
√ 6. y = x x = x3/2 ;
dx 3 dy = x1/2 . dt 2 dt
dy dx 8 = = z = 0, =⇒ 32 x1/2 = 1 =⇒ x = 49 and y = 27 . dt dt Both coordinates are changing at the same rate at the point (4/9, 8/27). Now
dS dx and when V = 27m3 dt dt dV = −2m3 /min. given that dt
7. Find
(∗)
V = x3 ,
S = 6x2
Differentiation of equations (∗) gives dV dx = 3x2 dt dt
and
dS dx = 12x . dt dt
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SECTION 4.10 When V = 27, x = 3. Substituting x = 3 and dV /dt = −2, we get dx −2 = 27 dt
dx = −2/27 dt
so that
and
−2/27 m/min;
The rate of change of an edge is
dS = 12(3) dt
−2 27
= −8/3.
the rate of change of the surface area is
−8/3 m /min. 2
dS dr and when r = 10 ft dt dt dV = 8 ft3 /min. given that dt (∗) V = 43 πr3 , S = 4πr2
8.
Find
Differentiation of equations (∗) with respect to t gives dV dr = 4πr2 dt dt Substituting
r = 10 and dV /dt = 8, dr dt
and
dS dr = 8πr . dt dt
we get
dS 1 8 = 8π(10) = . dt 50π 5 8 2 1 ft/min; the surface area is increasing ft /min. The radius is increasing 50π 5 8 = 4π(10)2
so that
dr 1 = dt 50π
and
9. The area of an equilateral triangle of side x is given by √ √ 3 2 1 x 3 A= x = x . Thus 2 2 4 When x = α,
dx dA = k, and = dt dt
√
3 2
√ 3 dx dA = x . dt 2 dt
αkcm2 /min.
10. We have 2x + 2y = 24, or x + y = 12. Thus, A = xy = x(12 − x) = 12x − x2 . dx dx dA = (12 − 2x) = ±4 . When A = 32, x = 4 or x = 8, and it follows that dt dt dt dA when l = 6 in. 11. Find dt dl = −2 in./sec. given that dt By the Pythagorean theorem l2 + w2 = 100. Also, A = lw. Thus, A = l
√ 100 − l2 .
Differentiation with respect to t gives −l dl dl dA =l √ + 100 − l2 . 2 dt dt dt 100 − l
Substituting l = 6 and dl/dt = −2, we get −6 dA =6 (−2) + (8)(−2) = −7. dt 8 The area is decreasing at the rate of 7 in.2 /sec.
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SECTION 4.10 dy dx = 2y dt dt dx = 8 ft/min, then If y = 30 ft and dt (30)2 − 36 dy x dy 16 √ 6 ft/min = = 8= dt y dt 30 5
12.
x2 + 62 = y 2 ; 2x
13.
Compare
dy dx to = −13 mph dt dt dz = −17 given that z = 16 and dt when x = y.
By the Pythagorean theorem x2 + y 2 = z 2 . Thus, dx dy dz + 2y = 2z . dt dt dt √ Since x = y when z = 16, we have x = y = 8 2 and √ √ dy 2(8 2 )(−13) + 2(8 2 ) = 2(16)(−17). dt 2x
Solving for dy/dt, we get √ √ dy = −34 −13 2 + 2 dt
or
√ dy 1 = √ (13 2 − 34) ∼ = −11. dt 2
Thus, boat A wins the race. 14. V =
4 3 πr , 3
so
dr dV 1 = 4πr2 = 4πr2 (− ). dt dt 5
Thus at r = 12
we have
dV 576 =− π cm3 /min. dt 5
dx = 2 and x = 12. dt 1 dx dA x2 1 2 2 = 169 − x − √ A = x 169 − x , so 2 2 dt 2 2 169 − x dt dx dA 119 2 When = 2 and x = 12, =− f t /sec. dt dt 5 dx dy dy x 16. x2 + y 2 = (13)2 ; 2x + 2y = 0 and =− dt dt dt 2y dx since = 0.5. When x = 5, y = 12 and dt dy 5 = − ; the top of the ladder is dt 24 dropping 5/24 ft/sec. 15. We want to find
dA dt
when
17. We want to find dV /dt when V = 1000 ft3 and P = 5 lb/in.2 given that dP/dt = −0.05 lb/in.2 /hr. Differentiating P V = C with respect to t, we get P
dP dV +V =0 dt dt
so that
5
dV dV + 1000(−0.05) = 0. Thus, = 10. dt dt
The volume increases at the rate of 10 ft3 /hr.
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SECTION 4.10 dP dV dP 1.4P dV + (1.4)P V 0.4 = 0 and =− . dt dt dt V dt dV dP (1.4)50 With V = 10, P = 50 and = −1, we have =− (−1) = 7 dt dt 10 The pressure is increasing 7 lb/in2 /sec. ds when x = 3 ft (and s = 4 ft) 19. Find dt dx = 400 ft/min. given that dt By similar triangles L 6 = . x+s s
18. P V 1.4 = C; V 1.4
Substitution of x = 3 and s = 4 gives us L = 10.5 ft tall. Rewriting
6 L = so that the lamp post is 7 4
10.5 6 = x+s s
as
s=
4 x 3
and differentiating with respect to t, we find that ds 4 dx 1600 = = . dt 3 dt 3 The shadow lengthens at the rate of 1600/3 ft/min. If the tip of his shadow is at the point z, then z =x+s
and
dz dx ds 1600 2800 = + = 400 + = . dt dt dt 3 3
The tip of his shadow is moving at the rate of 2800/3 ft./min. dy = −32t dt 64 y . By similar triangles, = x 20 + x 64x Thus, y = . 20 + x y(t) = 64 − 16t2 ;
20.
Now,
(20 + x)(64) − 64x dx 1280 dy = = dt (20 + x)2 dt (20 + x)2
At t = 1, y = 48, x = 60, and
21. Let W (t) = 150 1 +
1 4000
r
−2
=⇒
(20 + x)2 dy dx = . dt 1280 dt
dy dx (80)2 = −32. Therefore, = (−32) = −160 ft/sec. dt dt 1280
. We want to find dW/dt when r = 400 given that
dr/dt = 10 mi/sec. Differentiating with respect to t, we get −3 dW 1 1 dr = −300 1 + r dt 4000 4000 dt Now set r = 400 and dr/dt = 10. Then −3 dW 10 ∼ 400 = − 300 1 + = −0.5634 lbs/sec dt 4000 4000
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SECTION 4.10
−3/2
dv dv mv dM = m − 12 1 − v 2 /c2 = . −2v/c2 3/2 dt 2 2 2 dt dt 1 − v 2 /c2 c (1 − v /c ) √ c c 3 dv c dM m(c/2) If v = and = , then = = m. 3/2 100 2 2 2 2 dt 100 dt 225 c (1 − c /4c )
22. M =
m
;
dh dt given that
when h = 3 in. 1 dV = − cu in./min. dt 2 By similar triangles
23.
Find
r = 13 h.
Thus V = 13 πr2 h =
1 3 27 πh .
Differentiating with respect to t , we get dV dh 1 = πh2 . dt 9 dt
When h = 3, −
1 1 dh = π(9) 2 9 dt
and
dh 1 =− . dt 2π
The water level is dropping at the rate of 1/2π inches per minute. 1 2 4 r πr h and = (similar triangles) 3 h 6 dh 4 4 dV 3 πh . Thus = πh2 , so V = 27 dt 9 dt dh = 0.5 and h = 2, so at dt dV 8π = cubic ft per sec. dt 9
24.
25.
V =
dr dSA dr dSA dV = 4πr2 and = 8πr . Thus when =4 dt dt dt dt dt dV 10 5 and = cm3 /min. we get r = π dt π
26. V = πrh2 − 13 πh3 ; (a) When h = 7/2, 27.
and
dr = 0.1 dt
dh dh dh 2 dV dV = 2πrh − πh2 , and = since r = 7 and = 2. dt dt dt dt π(14h − h2 ) dt 8 dh 2 dh = in./sec. (b) When h = 7, = in./sec. dt 147π dt 49π dθ when x = 4 ft dt dx = 2 in./min. given that dt 3 θ (∗) tan = 2 x
Find
Differentiation of (∗) with respect to t gives 1 θ dθ 3 dx sec2 =− 2 2 2 dt x dt
or
θ dx dθ 6 = − 2 cos2 . dt x 2 dt
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SECTION 4.10 Note that dx/dt = 2 in./min=1/6 ft/min. When x = 4, we have cos θ/2 = 4/5 and thus 2 1 6 4 1 dθ =− =− . dt 16 5 6 25 The vertex angle decreases at the rate of 0.04 rad/min. 28.
tan α =
60 ; x
Now, sec2 α =
With y = 100, and
sec2 α y 2 x
dα 60 dx =− 2 dt x dt so
60 dx dα =− 2 . dt y dt
dα 60 dx = −10, we have =− (−10) = 0.06 rad/min. dt dt (100)2 dx when x = 1 mi dt dθ = 2π rad/min. given that dt x (∗) tan θ = = 2x 1/2
29.
Find
Differentiation of (∗) with respect to t gives sec2 θ When x = 1, we get sec θ =
√
5 and thus
The light is traveling at 5π mi/min. 30. (a) We have tan θ = x, so sec2 θ
dx dθ =2 . dt dt
dx = 5π. dt
dx dθ = . dt dt
Switching to radians,
dθ = 4π. dt
dx = 8π mi/min. dt dx = 4π mi/min. (b) At θ = 0, dt 1 dx dx x dθ , so sec2 θ = , and = 4. 31. We have tan θ = 40 dt 40 dt dt √ 5200 dθ 2 ∼ , so = At t = 15, x = 60 and sec θ = = 0.031 rad/sec. 40 dt 65 500 dh −1000 dr 32. We have V = hπr2 , so V = 500π cm3 . Thus h = 2 , and = . r dt r3 dt 1 dh dr = , we have = −4 cm/sec. At r = 5 and dt 2 dt 3 + 2x + 3 4 x 4 4 so tan θ = = . Thus x = h. V = 12 h = 36h + 16h2 . 33. We have sin θ = 5 3 h 3 2 dh dV dV = (36 + 32h) , so at = 10 and h = 2, Thus dt dt dt dh 1 = ft/min. dt 10 Thus at θ =
π , 4
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SECTION 4.10
dx dy −y x y 2 dα dt dt 34. tan α = ; sec α = x dt x2 dy dx x −y 2 dα dt dt · x = 2 2 dt x x + y2 dy dx x −y dt dt = x2 + y 2 dy dx = −30 mph = −44 ft/sec and = −22.5 mph = −33 ft/sec. Now dt dt At x = 300, y = 400, we have 300(−33) − 400(−44) dα = = 0.0308 the angle is increasing 0.0308 rad/sec. dt (300)2 + (400)2 dθ when y = 4 ft dt dx = 3 ft/sec. given that dt 16 tan θ = , x2 + (16)2 = (16 + y)2 x
35. Find
Differentiating tan θ = 16/x with respect to t, we obtain sec2 θ
−16 dx dθ = 2 dt x dt
and thus
dθ −16 dx = 2 cos2 θ . dt x dt
From x2 + (16)2 = (16 + y)2 we conclude that x = 12, when y = 4. Thus 2 12 3 dθ −16 3 −3 x = = and = . (3) = cos θ = 2 16 + y 20 5 dt (12) 5 25 The angle decreases at the rate of 0.12 rad/sec. 36. tan(α/2) = and
3 ; x
sec2 (α/2) 12
dα 3 dx =− 2 , dt x dt
6 dx dα =− 2 dt y dt
Initially, y = 5 so x = 4. 8 seconds later x = 12 so y =
√ (12)2 + (3)2 = 153. Therefore,
dα 6 dx 6 6 =− 2 =− (1) = − ; the angle is decreasing − 6/153 rad/sec. dt y dt 153 153 37. Find
dθ dt
when
t = 6 min.
4t 100t = 500 + 75t 20 + 3t Differentiation with respect to t gives tan θ =
sec2 θ
(20 + 3t)4 − 4t(3) dθ 80 = = . dt (20 + 3t)2 (20 + 3t)2
When t = 6 tan θ =
24 38
=
12 19
and
sec2 θ = 1 +
12 2 19
=
505 361
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SECTION 4.10
211
so that dθ 1 361 80 80 4 · · = = = . dt (20 + 3t)2 sec2 θ (38)2 505 101 The angle increases at the rate of 4/101 rad/min. 1 dx dα x ; sec2 α = H dt H dt dα dx 1 dx H = cos2 α = 2 dt H dt H + x2 dt 2 dx 2 We have H = 2 mi. and = 400 mph. After 2 seconds, x = 400 = miles. dt 3600 9
38. tan α =
dα 200(81) 2 9 (400) = = 2 rad/hr = rad/sec. 2 dt 2 + (2/9) 82 164 39. Let x be the distance from third base to the player. Then the distance from home plate to the player is given by: y = (90)2 + x2 . Differentiation with respect to t gives dx x dy = 2 2 dt (90) + x dt We are given that dx/dt = −15. Therefore, dy 15x = − dt (90)2 + x2
and
dy −15 =√ ∼ = 1.66 dt x=10 82
The distance between home plate to the player is decreasing 1.66 ft./sec. 40. The plane is flying at an altitude 6 miles. If y denotes the horizontal distance between the radar station and the plane, then tan θ =
6 y
=⇒
sec2 θ
−6 dy dθ = dt y dt
Therefore, dy y 2 sec2 θ dθ = dt −6 dt √ At the instant the plane is 12 miles from the station, y = 108, θ = 16 π and the speed is dy 108 4 π 4π ∼ = dt −6 3 360 = 60 mi/sec. = 754 mi/hr. 41. Let x be the position of the runner on the track, let y be the distance between the runner and the spectator, and let θ be the central angle indicated in the figure. The runner’s speed is 5 meters per second so dθ 1 dθ = 5 and = . 50 dt dt 10 By the law of cosines: y 2 = (50)2 + (200)2 − 2(50)(200) cos θ. Differentiating with respect to t, we get dy = 20, 000 sin θ dθ 2y dt dt
R θ
S
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SECTION 4.11 dy 10, 000 sin θ = dt y
1 10
dy 1, 000 = sin θ dt y √ (200)2 − (25)2 39375 dy Now, =5 =5 = 4.96. dt 200 200 y=200
The runner is approaching the spectator at the rate of 4.96 meters per second. 42.
23 dx = dt 4
43.
dx =4 dt
44.
dy 1 =− √ dt 2 3
SECTION 4.11 1.
ΔV = (x + h)3 − x3 = (x3 + 3x2 h + 3xh2 + h3 ) − x3 = 3x2 h + 3xh2 + h3 , dV = 3x2 h, ΔV − dV = 3xh2 + h3
(see figure)
2. The area of the ring can be thought of as the increase of the area of a disk as the radius increases from r to r + h. A = πr2 ;
so
dA = 2πr dr = 2πrh
The exact area is: π(r + h)2 − πr2 = π(r2 + 2rh + h2 ) − πr2 = 2πrh + πh2 . 3. f (x) = x1/3 , x = 1000, h = 2, f (x) = 13 x−2/3 √ √ −2/3 3 3 1002 = f (x + h) ∼ ] = 10 + = f (x) + hf (x) = 1000 + 2 13 (1000
1 150
∼ = 10.0067
−1 1 4. f (x) = √ , x = 25, h = −0.5, f (x) = 3/2 x 2x 1 1 −1 1 101 √ = 0.202 = = f (x + h) ∼ = f (x) + hf (x) = + (− ) 5 2 2(5)3 500 24.5 5. f (x) = x1/4 ,
x = 16,
h = −0.5,
f (x) = 14 x−3/4
(15.5)1/4 = f (x + h) ∼ = f (x) + hf (x) = (16)1/4 + (−0.5) 6. f (x) = x2/3 ,
x = 27,
h = −1,
f (x) = 23 x−1/3
(26)2/3 = f (x + h) ∼ = f (x) + hf (x) = (27)2/3 + (−1)
2
1
−3/4 4 (16)
−1/3 3 (27)
=
∼ = 1 63 64 = 1.9844
79 ∼ = 8.778 9
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SECTION 4.11 7. f (x) = x3/5 ,
x = 32,
f (x) = 35 x−2/5
h = 1,
(33)3/5 = f (x + h) ∼ = f (x) + hf (x) = (32)3/5 + (1) 8. f (x) = x−1/5 ,
x = 32,
h = 1,
3
−2/5 5 (32)
f (x) = − 15 x−6/5
(33)−1/5 = f (x + h) ∼ = f (x) + hf (x) = (32)−1/5 − (1) 9. f (x) = sin x,
x=
π , 4
h=
π , 3
h=
π , 180
f (x) = cos x
π , 90
f (x) = − sin x
1
= 8.15
−6/5 5 (32)
=
159 ∼ = 0.497 320
√ 2 π π π ∼ π ∼ cos = 1+ sin 46 = f (x + h) = f (x) + hf (x) = sin + = 0.719 4 180 4 2 180 ◦
10. f (x) = cos x,
x=
π π 1 π π − sin = − cos 62 = f (x + h) ∼ = f (x) + hf (x) = cos + 3 90 3 2 90 ◦
√ 3 ∼ = 0.470 2
−π , 90
11. f (x) = tan x,
x=
π , 6
h=
12. f (x) = sin x,
x=
π , 4
h=−
f (x) = sec2 x √ −π 4 3 2π ∼ π ◦ ∼ = − tan 28 = f (x + h) = f (x) + hf (x) = tan + = 0.531 6 90 3 3 135 π , 90
f (x) = cos x
√ π √2 2 π π π ∼ cos = − sin 43 = f (x + h) ∼ = f (x) + hf (x) = sin + − = 0.682 4 180 4 2 180 2 ◦
13. f (2.8) ∼ = f (3) + (−0.2)f (3) = 2 + (−0.2)(2) = 1.6 14. f (5.4) ∼ = f (5) + (0.4)f (5) = 1 + (0.4)(3) = 2.2 volume = V (r + t) − V (r) ∼ = tV (r) = 2πrht
15. V (r) = πr2 h;
16. Error in diameter = 0.3
=⇒ error in radius = 0.15. ∼ 9.6π cm2 (a) dS = 8πrh = 8π(8)(0.15) =
(b) dV = 4πr2 h = 4π(8)2 (0.15) ∼ = 38.4 cm3 17. V (x) = x3 , |dV | ≤ 3
V (x) = 3x2 , =⇒
|300h| ≤ 3
ΔV ∼ = dV = V (10)h = 300h =⇒
|h| ≤ 0.01,
error ≤ 0.01 feet
√ √ √ 1 1 x. Then f (x) = √ and x + 1 − x ∼ = (1)f (x) = √ . 2 x 2 x √ 1 1 √ < 0.01 = =⇒ x > 50 =⇒ x > 2500. Now, 100 2 x √ √ (b) Let f (x) = x1/4 . Then f (x) = 14 x−3/4 and 4 x + 1 − 4 x ∼ = (1)f (x) = 14 x−3/4 . 2 =⇒ x3/4 > 125 =⇒ x > 625. Now, 14 x−3/4 < 0.002 = 1000
18. (a) Let f (x) =
213
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SECTION 4.11
19. V (r) =
2 3
πr3
and dr = 0.01. V (r + 0.01) − V (r) ∼ = V (r)(0.01) = 2πr2 (0.01) = 2π(600)2 (0.01) 3
= 22619.5 in
(50 ft = 600 in)
or
98 gallons (approx.)
dV = 4πr2 . dr dV h = 4πr2 h. Therefore, Now, V (r + h) − V (r) = 8 × (10)6 ∼ = dr
20. V (r) =
4 3
πr3 ;
h=
8 × (10)6 ∼ = 0.0398 (miles) ∼ = 210 (feet) 4π(4000)2
L L implies P 2 = 4π 2 g g Differentiating with respect to t, we have
21. P = 2π
2P Thus 22.
4π 2 dL P 2 dL dP = · = · dt g dt L dt
since
P2 4π 2 = . L g
dP 1 dL = · P 2 L
15 dP = −15 sec/hour = − . By Exercise 21, P 3600 1 dL 15 30 1 =− and dL = − L=− L 2 L 3600 3600 120 With L = 90, we have dL = −90/120 = − 0.75; the pendulum should be shortened 0.75 cm to 89.25 cm.
23. L = 3.26 ft, P = 2 sec, and
dL = 0.01 ft
dP 1 dL = · P 2 L 1 0.01 1 dL ·P = · · 2 dP ∼ dP = · = 0.00307 sec 2 L 2 3.26 24. Each edge increases by 0.1%; take h = 0.001x. 12x(0.001x) dS = = 0.002 = 0.2%. S 6x2 3x2 (0.001x) dA = = 0.003 = 0.3%. A = x3 , dA = 3x2 h, and A x3
S = 6x2 , dS = 12xh, and
25. A(x) =
1 2 πx , 4
dA ≤ 0.01 A
dA =
⇐⇒
2
1 πxh, 2
dA h =2 A x
h ≤ 0.01 x
⇐⇒
h ≤ 0.005 x
within
1 % 2
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SECTION 4.11 26. (a) Let y = xn . Then dy = nxn−1 h. To get
dy nxn−1 h 0.01 h = , that is, within < 0.01, we need < n y x x n
(b) Let y = x1/n . Then dy = To get
dy = y
1 (1−n)/n h nx 1/n x
29.
lim
h→0
1 (1−n)/n h. nx
< 0.01, we need
h < (0.01)n, that is, within n%. x
g(h) 28. lim g(h) = lim ·h= h→0 h→0 h
27. (a) and (b)
1 n %.
g(h) lim h→0 h
lim h
h→0
=0
g1 (h) + g2 (h) g1 (h) g2 (h) = lim + lim =0+0=0 h→0 h→0 h h h
g1 (h)g2 (h) g1 (h)g2 (h) lim = lim h = h→0 h→0 h h2
lim h
h→0
g1 (h) lim h→0 h
g2 (h) lim h→0 h
= (0)(0)(0) = 0
30. (a) g(h) = f (x + h) − f (x) − mh. f (x + h) − f (x) g(h) (b) lim = lim − m = f (x) − m = 0 iff m = f (x). h→0 h h→0 h
31. Suppose that f is differentiable at x. Then there is a number m such that
lim
h→0
f (x + h) − f (x) =m h
or
lim
h→0
f (x + h) − f (x) − m = 0. h
Let g(h) = f (x + h) − f (x) − mh. Then g(h) f (x + h) − f (x) = −m h h
and
g(h) f (x + h) − f (x) = lim − m = 0. h→0 h h
lim
h→0
Now suppose that f (x + h) − f (x) − mh = g(h) where g(h) = o(h). Then lim
h→0
f (x + h) − f (x) −m h
= lim
h→0
Therefore lim
h→0
and f is differentiable at x. m = f (x).
f (x + h) − f (x) =m h
g(h) = 0. h
215
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SECTION 4.11
PROJECT 4.11 x2 . 20 x Marginal cost: C (x) = 50 − ; C (20) = 48 10 Exact cost of the 21st component: C(21) − C(20) = 3027.95 − 2980 = 47.95.
1. C(x) = 2000 + 50x −
2. Profit function: P (x) = R(x) − C(x) = 650x − 5x2 − (12, 000 + 30x) = 620x − 5x2 − 12000. Breakeven points:
P (x) = 0, implies x2 − 124x + 2400 = (x − 24)(x − 100) = 0; x = 24, x = 100
units. Maximum profit: P (x) = 620 − 10x;
P (x) = 0 at x = 62 units. x2 x2 − (4x + 1400) = 16x − − 1400. 3. (a) Profit function: P (x) = R(x) − C(x) = 20x − 50 50 2 x − 1400 = 0 or x2 − 800x + 70, 000 = 0 Break-even points: 16x − 50 Thus x = 100, or x = 700 units. x (b) P (x) = 16 − ; P (x) = 0 at x = 400 units. 25 (c)
4. (a)
(b)
Break-even points at x = 81.11 and x = 631.19 5. (a)
Maximum profit at x = 336.11 units 10x (b) P (x) = − 4 − 0.75x 1 + 0.25x2
Breakeven points at x ∼ = 0.46, and x ∼ = 4.53 6. A(x) =
C(x) , x
A (x) = 0
A (x) =
implies
Maximum profit at 175 units
xC (x) − C(x) . x2
xC (x) − C(x) = 0
or
C (x) =
the points where C (x) = C(x)/x. A (x) =
x2 [xC (x)] − [xC (x) − C(x)](2x) x4
At the points where xC (x) − C(x) = 0, is a minimum.
A (x) =
C(x) = A(x). The critical points of A are x
C (x) . If C (x) > 0 at such a point, then A(x) x
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SECTION 4.12 7. B(x) =
R(x) , x
B (x) = 0
B (x) =
implies
xP (x) − P (x) . x2
xP (x) − P (x) = 0
or
P (x) =
the points where P (x) = P (x)/x. B (x) =
x2 [xP (x)] − [xP (x) − P (x)](2x) x4
At the points where xP (x) − P (x) = 0, is a maximum.
SECTION 4.12 1. (a) xn+1
1 = xn + 12 2
2. (a) xn+1 =
1 xn
B (x) =
P (x) = B(x). The critical points of B are x
P (x) . If P (x) < 0 at such a point, then B(x) x
(b) x4 ∼ = 4.89898
2x3n − 1 3x2n − 4
(b) x4 ∼ = 1.86081
2 2 25 1 xn + 3 3 xn 4 1 4 = xn + 6 5 xn
3. (a) xn+1 =
(b) x4 ∼ = 2.92402
4. (a) xn+1
(b) x4 ∼ = 1.97435
5. (a) xn+1 =
xn sin xn + cos xn sin xn + 1
(b) x4 ∼ = 0.73909
6. (a) xn+1 =
xn cos xn − sin xn − x2n cos xn − 2xn
(b) x4 ∼ = 0.87673
6 + xn 7. (a) xn+1 = √ 2 xn + 3 − 1 8. (a) xn+1 =
(b) x4 ∼ = 2.30278
xn sec2 xn − tan xn 1 + sec2 xn
9. Let f (x) = x1/3 . Then f (x) =
1 3
(b) x4 ∼ = 2.30278 x−2/3 . The Newton-Raphson method applied to
this function gives: xn+1 = xn −
f (xn ) = xn − f (xn )
1/3
xn 1 3
−2/3
xn 0. Then x2 = −2x1 , x3 = −2x2 = 4x1 , · · · , Choose any x1 =
= −2xn .
xn = −2xn−1 = (−1)n−1 2n x1 , · · · . 10. xn+1 = xn for all n. 11. (a) f is a continuous function, and f (1) = −2 < 0, f (2) = 3 > 0. Thus, f has a root in (1, 2). (b) f (x) = 6x2 − 6x and f (1) = 0. Therefore, x1 = 1 will fail to generate values that will approach the root in (1, 2). 2x3n − 3x2n − 1 ; 6x2n − 6xn x1 = 2, x2 = 1.75, x3 = 1.68254, x4 = 1.67768, f (x4 ) ∼ = 0.00020.
(c) xn+1 = xn −
217
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SECTION 4.12
12. (a) Let f (x) = x4 − 2x2 −
17 16 .
Then f (x) = 4x3 − 4x. The Newton-Raphson method applied to this
function gives: xn+1 = xn −
x4n − 2x2n − 17 16 4x3n − 4xn
If x1 = 12 , then x2 = − 12 , x3 = 12 , · · · xn = (−1)n−1 12 , · · · . (b) x1 = 2,
x2 = 1.71094,
x3 = 1.58569,
x4 = 1.56165;
f (x4 ) = 0.00748
f (x) = 2x. Substituting into (4.12.1), we have x2n − a x2n + a 1 a , n≥1 = = xn + xn+1 = xn − 2xn 2xn 2 xn 1 a (b) Let a = 5, x1 = 2, and xn+1 = xn + , n ≥ 1. 2 xn Then x2 = 2.25, x3 = 2.23611, x4 = 2.23607, and f (x4 ) ∼ = 0.000009045.
13. (a) f (x) = x2 − a;
14. (a) Let f (x) = xk − a. Then f (x) = kxk−1 . The Newton-Raphson method applied to this function gives: xn+1 = xn −
xkn − a 1 1 a 1 a x (k − 1)x = x − + = + n n n kxk−1 k k xk−1 k xk−1 n n n
(b) Let a = 23, k = 3 and x1 = 3. Then x1 = 3,
x2 = 2.85185,
15. (a) Let f (x) =
1 x
x3 = 2.84389,
x4 = 2.84382;
f (x4 ) = −0.00114
− a. Then f (x) = − x12 . The Newton-Raphson method applied to
this function gives: xn+1 = xn −
1 xn
−a
− x12
= xn + xn − ax2n = 2xn − ax2n
n
(b) Let a = 2.7153, and x1 = 0.3. x2 = 0.35562, Thus
1 2.7153
x3 = 0.36785,
Then x4 = x5 = 0.36828,
0.36828.
16. (a) f (x) = x4 − 7x2 − 8x − 3,
f (x) = 4x3 − 14x − 8,
f (x) = 12x2 − 14.
f (2) = −4 < 0 and f (3) = 58 > 0; f has a zero in (2, 3). f (x) = 12x2 − 14 > 0 on (2, 3). Therefore, f has exactly one zero in this interval. (b) xn+1 = xn − mum at c.
4x3n − 14xn − 8 4x3n + 4 = ; 12x2n − 14 6x2n − 7
17. (a) f (x) = sin x +
1 2
x2 − 2x,
x3 ∼ = 2.1091. Since f (x3 ) > 0, f has a local mini-
f (x) = cos x + x − 2,
f (x) = − sin x + 1.
f (2) = cos 2 ∼ = −0.4161 < 0 and f (3) = cos 3 + 1 ∼ = 0.0100 > 0; f has a zero in (2, 3). f (x) = − sin x + 1 > 0 on (2, 3). Therefore, f has exactly one zero in this interval. cos xn + xn − 2 2 − xn sin xn − cos xn = ; 1 − sin xn 1 − sin xn f has a local minimum at c.
(b) xn+1 = xn −
x3 ∼ = 2.9883. Since f (x3 ) > 0,
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REVIEW EXERCISES 18. (a) xn+1 = xn − (b) x4 ∼ = 6.28319 19. (a) xn+1 = xn − (b) x1 = 5π/3;
sin xn = xn − tan xn , x1 = 3; cos xn xn + tan xn , x1 = 2π/3; 1 + sec2 xn r2 ∼ = 4.913
219
x4 ∼ = 3.14159
r1 ∼ = 2.029
REVIEW EXERCISES 1. f is differentiable on (−1, 1) and continuous on [−1, 1]; f (1) = f (−1) = 0. √ 3 2 2 . f (x) = 3x − 1; 3c − 1 = 0 =⇒ c = ± 3 2. f is differentiable on (0, 2π) and continuous on [0, 2π];f (0) = f (2π) = 0. f (x) = cos x − sin x;
cos c − sin c = 0
=⇒
c = 14 π,
5 4 π.
3. f is differentiable on (−2, 3) and continuous on [−2, 3]. f (3) − f (−2) f (c) = 3c − 2 = =5 5
2
=⇒
2
3c = 7
c=±
=⇒
7 3
4. f is differentiable on (2, 5) and continuous on [2, 5]. f (5) − f (2) 1 1 = = f (c) = √ 3 3 2 c−1
13 . 4
=⇒
c=
=⇒
c=1+
=⇒
c=
5. f is differentiable on (2, 4) and continuous on [2, 4]. f (c) = −
2 2 f (4) − f (2) =− = 2 (c − 1) 2 3
√ 3.
6. f is differentiable on (0, 16) and continuous on [0, 16]. f (c) =
7. f (x) = (−1, 1).
1 3 f (16) − f (0) = = 16 2 4c1/4
81 . 16
1 + x2/3 = 0 for all x ∈ (−1, 1). f (0) does not exist. Therefore f is not differentiable on 3x2/3
−3 < 0 for all x ∈ (1, 4); (x − 2)2 not differentiable on (1, 4).
8. f (x) =
9 f (4) − f (1) = > 0. f (2) does not exist. Therefore f is 3 6
9. No. Reason: If such a function did exist, then, by the mean-value theorem, there is a number c ∈ (1, 4) such that f (c) = − 43 < −1. 10. (a) f (x) = 3x2 − 3 = 3(x2 − 1) < 0 on (−1, 1). Therefore, by Rolle’s theorem, f can have at most one zero in [−1, 1]. (b) Since f decreases on [−1, 1], we must have f (−1) = −1 + 3 + k ≥ 0 and f (1) = 1 − 3 + k ≤ 0. These conditions imply that k ∈ [−2, 2].
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REVIEW EXERCISES
11. f (x) = 6x(x + 1).
f increases on (−∞, −1] ∪ [0, ∞) and decreases on [−1, 0]. The critical points are x = −1, 0; f (−1) = 2 is a local max and f (0) = 1 is a local min. 12. f (x) = 4(x − 1)(x2 + x + 1).
f increases on [1, ∞) and decreases on (−∞, 1]. The critical point is x = 1; f (1) = 0 is a local min. 13. f = (x + 2)(x − 1)2 (5x + 4).
f increases on (−∞, −2] ∪ [− 45 , ∞) and decreases on [−2, − 45 ].
The critical points are x = −2, 1, − 45 . f (−2) = 0 is a local max and f (− 45 ) ∼ = −8.981 is a local min. 14. f (x) = 1 −
8 x3 − 8 = . x3 x3
f increases on (−∞, 0) ∪ [2, ∞) and decreases on (0, 2]. x = 2 is the only critical point; 0 is not a critical point. f (2) = 3 is a local min. 15. f (x) =
1 − x2 . (1 + x2 )2
f increases on [−1, 1] and decreases on (−∞, −1] ∪ [1, ∞). The critical points are x = −1, 1. f (−1) = − 12 is a local min and f (1) =
1 2
is a local max.
16. f (x) = sin x + cos x.
7π 3π 7π f increases on [0, 3π 4 ] ∪ [ 4 , 2π] and decreases on [ 4 , 4 ]. The critical points are x = √ √ 2 is a local max and f ( 7π 4 ) = − 2 is a local min.
3π 7π 4 , 4 .
f ( 3π 4 )=
17. f (x) = 3x2 + 4x + 1 = (3x + 1)(x + 1); critical points: x = − 13 , −1.
f (−2) = −1 endpoint and abs. min; f (−1) = 1 local max; f (− 13 ) = and abs. max.
23 27
local min; f (1) = 5 endpoint
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REVIEW EXERCISES
221
18. f (x) = 4x3 − 16x = 4x(x − 2)(x + 2); critical points: x = 0, 2.
f (−1) = −5 endpoint min; f (0) = 2 local max; f (2) = −14 local and abs. min; f (3) = 11 endpoint and abs. max. 19. f (x) = 2x −
√ 8 2(x4 − 4) 1/4 = ; critical point: x = 4 = 2. x3 x3
√ f (1) = 5 endpoint max; f ( 2) = 4 local and abs. min; f (4) = 20. f (x) = cos x(1 − 2 sin x); critical points: x =
65 4
endpoint and abs. max.
π π 5π 3π 6, 2, 6 , 2 .
f (0) = 1 endpoint min; f ( π6 ) = 5/4 local and abs. max; f ( π2 ) = 1 local min; f ( 5π 6 )= max;
f ( 3π 2 )
5 4
local and abs.
= −1 local and abs. min; f (2π) = 1 endpoint max.
√ −x 2 − 3x 21. f (x) = √ + 1−x= √ ; critical point: x = 23 . 2 1−x 2 1−x f (1) = 0 endpoint min; f ( 23 ) = 22. f (x) =
√ 2 3 9
local and abs. max.
(x − 2)2x − x2 x(x − 4) = ; critical point: x = 4; (x − 2)2 (x − 2)2
f (4) = 8 local and abs. min. 23. f (x) =
3x(x − 3) . Vertical asymptotes: x = −3 and x = 4; horizontal asymptote: y = 3. (x − 4)(x + 3)
24. f (x) =
(x − 2)(x + 2) x+2 = , x = 2. Vertical asymptote: x = 3; horizontal asymptote: y = 1. (x − 2)(x − 3) x−3
25. f (x) = x −
26. f (x) =
x . Vertical asymptote: x = 1; oblique asymptote: y = x. (x − 1)(x2 + x + 1)
3 , vertical tangent. 5(x − 1)2/5
27. f (x) = x2/5 − 2x−3/5 =
x−2 , vertical cusp. x3/5
28. f (x) = 2x−2/3 + 4x1/3 =
2 + 4x , vertical tangent. x2/3
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REVIEW EXERCISES
29. f (x) = 6 + 4x3 − 3x4 , domain: (−∞, ∞) f (x) = 12x2 (1 − x) critical pts.
x = 0, 1
f (x) = 12x(2 − 3x)
y
f (x) > 0 on (−∞, 1),
6
f (x) < 0 on (1, ∞)
4
f (x) > 0 on (0, 2/3)
2
f (x) < 0 on (−∞, 0) ∪ (2/3, ∞)
−1
1
x
30. f (x) = 3x5 − 5x3 + 1, domain: (−∞, ∞) f (x) = 15x2 (x − 1)(x + 1) x = −1, 0, 1 f (x) = 30x 2x2 − 1
critical pts.
y 5
f (x) > 0 on (−∞, −1) ∪ (1, ∞) f (x) < 0 on (−1, 1) √ √ f (x) > 0 on − 22 , 0 ∪ 22 , ∞ √ √ f (x) < 0 on −∞, − 22 ∪ 0, 22
3 1 −1
1
−1
x
2x , domain: (−∞, ∞) +4 symmetric with respect to the origin.
2 4 − x2 f (x) = (x2 + 4)2
31. f (x) =
x2
x = 2, x = −2 4x x2 − 12
critical pts. f (x) =
y
3
(x2 + 4)
0.5
f (x) > 0 on (−2, 2), f (x) < 0 on (−∞, 2) ∪ (2, ∞)
√ √ f (x) > 0 on − 12, 0 ∪ 12, ∞ √ √
f (x) < 0 on −∞, − 12 ∪ 0, 12
−2 −0.5
2
x
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REVIEW EXERCISES
223
32. f (x) = x2/3 (x − 10), domain: (−∞, ∞) 5(x − 4) 3x1/3 critical pts. x = 4, x = 0 10(x + 2) f (x) = 9x4/3 f (x) > 0 on (−∞, 0) ∪ (4, ∞), f (x) =
y 10 −2
4
x
10
−10
f (x) < 0 on (0, 4) f (x) > 0 on (−2, 0) ∪ (0, ∞) f (x) < 0 on (−∞, −2) √ 33. f (x) = x 4 − x, domain: (−∞, 4] 8 − 3x f (x) = √ 2 4−x 3x − 16 f (x) = 4(4 − x)3/2
f (x) > 0 on −∞, 83 , f (x) < 0 on 83 , 4
y 2 −2
2
4
x
−2
f (x) < 0 on (−∞, 4) 34. f (x) = x4 − 2x2 + 3, domain (−∞, ∞); symmetric with respect to the y-axis f (x) = 4x3 − 4x = 4x(x − 1)(x + 1) critical pts.
y
x = −1, 0, 1
√ √ f (x) = 12x2 − 4 = 12(x − 1/ 3)(x + 1/ 3)
3
f (x) > 0 on [−1, 0] ∪ [1, ∞)
2
f (x) < 0 on (−∞, −1] ∪ [0, 1] √ √ f (x) > 0 on (∞, −1/ 3) ∪ (1/ 3, ∞) √ √ f (x) < 0 on [−1/ 3, 1/ 3) 35. f (x) = sin x +
1
-2
-1
1
√
3 cos x, domain [0, 2π] √ f (x) = cos x − 3 sin x x = 16 π, x = √ f (x) = − sin x − 3 cos x
critical pts.
7 6
y
π
2
f (x) > 0 on [0, π/6) ∪ (7π/6, 2π]
π
f (x) < 0 on (π/6, 7π/6) f (x) > 0 on (2π/3, 5π/3) f (x) < 0 on [0, π/3) ∪ (5π/3, 2π]
−2
2π
x
2
x
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REVIEW EXERCISES
36. f (x) = sin2 x − cos x, x ∈ [0, 2π] f (x) = 2 sin x cos x + sin x critical pts.
x = 23 π, π,
y 1
4 3π
f (x) = 4 cos2 x + cos x − 2
f (x) > 0 on 0, 23 π ∪ π, 43 π ,
f (x) < 0 on 23 π, π ∪ 43 π, 2π 37.
1
2
3
4
5
6
x
-1
y
−1
x
2
38. Let r be the radius of the sphere and x the side length of the cube. Then the sum of the surface areas equals constant implies
2
2
4πr + 6x = C The sum of the volumes is: 4 V = πr3 + 3 Now
C − 4πr2 6
Set V (r) = 0: 1/2 C − 4πr2 2r = 6
=⇒
x=
C − 4πr2 . 6
3/2 ,
0≤r≤
C . 4π
1/2 C − 4πr2 4 − πr 6 3 1/2 C − 4πr2 = 4πr2 − 2πr 6
3 V = 4πr + 2
=⇒
2
C − 4πr2 4r = 6 2
=⇒
r=
C , and x = 2 24 + 4π
C . 24 + 4π
Without loss of generality, set C = 1. Then √ √ V (0) = (1/6)3/2 ∼ = 0.055, V (1/ 4π) ∼ = 0.09. = 0.07, V (1/ 24 + 4π) ∼ Thus, the sum of the volumes is a minimum when the side length of the cube equals the diameter of the sphere. The sum of the volumes is a maximum when the side length of the cube is 0. 39. Let x be the side length of the square base. Then, since the volume is 27, the height y = 27/x2 . The √ √ conditions: x2 ≤ 18, 0 ≤ y ≤ 4 imply 3 2 3 ≤ x ≤ 3 2. Thus, the surface area S of the box is
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REVIEW EXERCISES given by
√ √ 3 3 ≤x≤3 2 2
108 S(x) = 2x + , x 2
S (x) = 4x − Now, S
√ 3 3 2
=
27 2
108 ; x2
S (x) = 0
225
=⇒
4x3 = 27
=⇒
x=3
√ + 24 3 ∼ = 55.07,
S(3) = 54 √ √ S(3 2) = 36 + 18 2 ∼ = 61.46 (a) The minimal surface area is 54 sq. ft. at x = 3.
√ (b) The maximal surface area is 61.46 sq. ft. (approx.) at x = 3 2. x y + =1 a b
40. An equation for the line with x-intercept a and y-intercept b is: Since the line passes through (1, 2),
1 2 + =1 a b
S = 12 ab =
The area of the right triangle OAB is: Now, S (a) =
=⇒
a2 − 2a (a − 1)2
b=
2a a−1
a2 a−1
and S (a) = 0 at a = 0, a = 2
The triangle OAB will have minimum area when A = (2, 0) and B = (0, 4). 41. Introduce a rectangular coordinate system so that the rectangle is in the first quadrant and the base and left side lie on the coordinate axes. Let x denote the length of the base, and y the height. Let P = 2x + 2y be the given perimeter. If the rectangle is rotated about the y-axis, then the volume V is given by: V (x) =
πP 2 x − πx3 , 0 ≤ x ≤ P/2. 2
Now, V (x) = πP x − 3πx2
and V (x) = 0
=⇒
x = 0 or x = 13 P.
At x = 0, V = 0; at x = P/3, y = P/6, V = πP 3 /54; at x = P/2, V = 0. The dimensions that generate the cylinder of maximum volume are x = P/3, y = P/6. 42. Let x be the width of the page and y the height. Then xy = 80. The area available for print is: 80 A = (x − 2)(y − 2.5) = (x − 2) − 2.5 x = 85 − 2.5x − A (x) = −2.5 +
160 ; x2
A (x) = 0 :
160 , 0 < x < ∞. x
2.5x2 = 160,
x2 = 64,
x=8
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REVIEW EXERCISES Since A (x) < 0, A has an absolute maximum at x = 8. The dimensions of the page that will maximize the printed area are: width 8, height 10.
43. x (t) = 1 − 2 sin t,
x (t) = −2 cos t.
The object is slowing down when x (t)x (t) < 0 44. x(t) = (4t − 1)(t − 1)2 , t ≥ 0;
t ∈ [0, 16 π) ∪ ( 12 π, 56 π) ∪ ( 32 π, 2π].
=⇒
x (t) = v(t) = 2(4t − 1)(t − 1) + 4(t − 1)2 = 6(2t2 − 3t + 1)
= 6(2t − 1)(t − 1) (a) The object is moving to the right: v(t) > 0 when 0 ≤ t < left: v(t) < 0 when
1 2
1 2
and when t > 1;
moving to the
1 2
< t < 1;
the object changes direction at t = and t = 1.
(b) Speed when moving left: ν = |v| = 6(3t − 2t2 − 1) on 12 , 1 . dν/dt = 6(3 − 4t); dν/dt = 0 at t = 3/4. The maximum speed when moving left is ν(3/4) = 3/4 units per unit time. 45. (a) x(t) =
√
t+1
1 x (t) = v(t) = √ 2 t+1 1 x (t) = a(t) = − = −2 [v(t)]3 , 4(t + 1)3/2 3 x (t) = a (t) = 8(t + 1)5/2 Position: x(17) = x(15 + 2) ∼ = x(15) + 2x (15) = 4.25;
(b)
15 ∼ = 0.1172; 128 Acceleration: a(17) = a(15 + 2) ∼ = a(15) + 2a (15) ∼ = 0.0032.
Velocity:
v(17) = v(15 + 2) ∼ = v(15) + 2v (15) =
46. The height of the rocket at time t is given by: y(t) = 128 t − 16 t2 . (a) At maximum height,
v(t) = y (t) = 128 − 32t = 0
=⇒
t = 4. The rocket reaches its
maximum height at 4 seconds; the maximum height is: y(4) = 256 feet. (b) y(t) = 0 seconds;
=⇒
16t(8 − t) = 0
=⇒
t = 0, t = 8; the rocket hits the ground at t = 8
v(8) = −128 ft/sec.
47. The height at time t is: y(t) = −16t2 + 8t + y0 . y(10) = 0 = −1600 + 80 + y0
=⇒
48. The height at time t is: y(t) = −16t2 + v0 t. y(1) = 24 = −16 + v0
v0 = 40. Therefore,
=⇒
y0 = 1520 ft.
y(t) = −16t + 40t. The ball achieves maximum height at the instant v(t) = −32t + 40 = 0 2
t=
5 4.
The maximum height is y(5/4) = −16(5/4) + 40(5/4) = 25. 2
49. Let x be the position of the boy and let y be the length of his shadow. Then dx x+y y = 168 and = =⇒ dt 12 5 Differentiating implicitly with respect to t, we get 7
dy dx =5 dt dt
=⇒
dy = dt
The shadow is lengthening at the rate of 120 ft/min.
7y = 5x and y = 5 (168) = 120 7
5 x 7
=⇒
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REVIEW EXERCISES 50. V = 13 πr2 h,
V constant 0=
d d V = dt dt
1 2 πr h 3
227
2πrh dr πr2 dh + 3 dt 3 dt
=
dh dh 2h dr we get: =− · . dt dt r dt 30 dh = − (0.3) = −2.25; At the instant r = 4 and h = 15, dt 4 2.25 in/min. Solving for
the height is decreasing at the rate of
51. Let x be the position of the locomotive at time t and let y be the position of the car. By the law of cosines the distance between the locomotive and the car is given by s2 = x2 + y 2 − 2xy cos 60◦ = x2 + y 2 − xy. Differentiating with respect to t, we get 2s
dx dy dy dx ds = 2x + 2y −x −y . dt dt dt dt dt
Setting dx/dt = 60, dy/dt = −30 and converting to feet, gives 2s
ds = 2x(60)(5280) − 2y(30)(5280) + x(30)(5280) − y(60)(5280) = 5280(150x − 120y). dt
When x = y = 500, s = 500 and 2(500)
ds = 5280(30)(500) dt
ds = 15(5280). dt
=⇒
The distance between the locomotive and the car is increasing at the rate of 15 miles per hour. √ 52. Let r be the radius of the circle. Then dr/dt = 5. The square has side length r 2 and area A = 2r2 . Differentiating with respect to t, we get dA dr = 4r = 20r; dt dt
and
dA = 200. dt r=10
53. The cross-section of the water at time t is an isosceles triangle with base x and height y where x and 1 x y are related by 2 = 12 (similar triangles). Thus x = y. y The volume of water in the trough at time t is V = 12( 12 xy) = 6y 2 . Differentiating with respect to t gives dy dV = 12y dt dt Since dV /dt = −3, 1/6 feet per minute.
dy 1 =− dt 4y
and
=⇒
dy 1 dV = . dt 12y dt
dy 1 = − . The water level is falling at the rate of dt t=1.5 6
54. f (3.8) ∼ = f (4) + f (4)(−0.2) = 2 + 2(−0.2) = 1.6.
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REVIEW EXERCISES
55. f (x) =
√
√ x + 1/ x,
f (x) = 12 x−1/2 − 12 x−3/2
f (4.2) ∼ = f (4) + f (4)(0.2) =
5 2
+
= 2.5375.
f (x) = 14 x−3/4
56. f (x) = x1/4 ,
f (83) ∼ = f (81) + f (81)(2) = 3 + 57. f (x) = tan x,
1 54
∼ = 3.01852.
f (x) = sec2 x
f (43◦ ) = f ( 14 π − 58. V = 43 πr3 ,
3 16 (0.2)
1 90 π)
∼ = f (π/4) + f (π/4)(−π/90) = 1 − 2(π/90) = 1 − (π/45) ∼ = 0.9302.
dV = 4πr2 dr. Calculate dV at r = 10 ft. and dr = 0.05 in. ∼ = 0.00417 ft.
dV |r = 10, dr = 0.00417 = 4π (10)2 (0.00417) ∼ = 5.240 cu. ft. = 9055.025 cu. in.; = 9055.025/231 ∼ = 39.20 gal. 59. f (x) = x3 − 10
x3n − 10 2x3n + 10 = 3x2n 3x2n (b) x4 ∼ = 2.15443; f (2.15443) ∼ = −0.00007 (a) xn+1 = xn −
60. f (x) = x sin x − cos x (a) xn+1 =
x2n cos xn + xn sin xn + cos xn 2 sin xn + xn cos xn
(b) x4 ∼ = 0.86057;
f (0.86057) ∼ = 0.00049
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SECTION 5.2 CHAPTER 5 SECTION 5.2 1. Lf (P ) = 0( 14 ) + 12 ( 14 ) + 1( 12 ) = 58 ,
Uf (P ) = 12 ( 14 ) + 1( 14 ) + 2( 12 ) =
11 8
5 97 2. Lf (P ) = 23 ( 13 ) + 14 ( 12 ) + 0( 14 ) + (−1)(1) = − 144 , 5 Uf (P ) = 1( 13 ) + 23 ( 12 ) + 14 ( 14 ) + 0(1) =
3. Lf (P ) = 14 ( 12 ) + 4. Lf (P ) =
15 1 16 ( 4 )
1 1 16 ( 4 )
97 144
+ 0( 14 ) =
9 64 ,
Uf (P ) = 1( 12 ) + 14 ( 14 ) +
+ 34 ( 14 ) + 0( 12 ) =
27 64 ,
Uf (P ) = 1( 14 ) +
5. Lf (P ) = 1( 12 ) + 98 ( 12 ) =
17 16 ,
1 3 5 7 9 Uf (P ) = 15 ( 25 ) + 25 ( 25 ) + 35 ( 25 ) + 45 ( 25 ) + 1( 25 )= 1 3 16 ( 4 )
+ 0( 12 ) +
8. Lf (P ) =
9 1 16 ( 4 )
+
Uf (P ) = 1( 14 ) +
1 1 16 ( 2 )
9 1 16 ( 2 )
1 1 16 ( 4 )
+ 14 ( 12 ) =
+ 0( 12 ) +
+
1 1 16 ( 2 )
1 1 16 ( 4 )
3 16 ,
37 64
+ 34 ( 12 ) =
55 64
+ 14 ( 12 ) =
π 6,
19 25
11. (a) Lf (P ) ≤ Uf (P ) but 3 ≤ 2. 1 (b) Lf (P ) ≤ f (x) dx ≤ Uf (P ) but
1 1 16 ( 2 )
+ 14 ( 14 ) + 1( 12 ) =
43 32
5 16 ,
9 8
Uf (P ) =
10. Lf (P ) = 12 ( π3 ) + 0( π6 ) + (−1)( π2 ) = − π3 ,
25 16
14 25 ,
Uf (P ) = 1( 34 ) +
+ 14 ( 14 ) + 1( 12 ) =
9. Lf (P ) = 0 π6 + 12 π3 + 0 π2 =
−1 1
=
Uf (P ) = 98 ( 12 ) + 2( 12 ) =
1 3 5 7 9 6. Lf (P ) = 0( 25 ) + 15 ( 25 ) + 25 ( 25 ) + 35 ( 25 ) + 45 ( 25 )=
7. Lf (P ) =
15 1 16 ( 4 )
1 1 16 ( 4 )
1 π 2 6
+ 1 π3 + 1 π2 =
Uf (P ) = 1( π3 ) + 12 ( π6 ) + 0( π2 ) =
11π 12 5π 12
3 ≤ 2 ≤ 6.
(c) Lf (P ) ≤
−1
f (x) dx ≤ Uf (P )
but
3 ≤ 10 ≤ 6.
12. (a) Lf (P ) = (x0 + 3)(x1 − x0 ) + (x1 + 3)(x2 − x1 ) + · · · + (xn−1 + 3)(xn − xn−1 ), Uf (P ) = (x1 + 3)(x1 − x0 ) + (x2 + 3)(x2 − x1 ) + · · · + (xn + 3)(xn − xn−1 ) (b) For each index i xi−1 + 3 ≤
1 (xi−1 + xi ) + 3 ≤ xi + 3 2
Multiplying by Δxi = xi − xi−1 gives 1 (xi−1 + 3)Δxi ≤ x2i − x2i−1 + 3(xi − xi−1 ) ≤ (xi + 3)Δxi . 2 Summing from i = 1 to i = n, we find that 1 1 Lf (P ) ≤ x21 − x20 + 3(x1 − x0 ) · · · + x2n − xn−1 2 + 3(xn − xn−1 ≤ Uf (P ) 2 2
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SECTION 5.2 The middle sum collapses to 1 2 1 xn − x0 2 + 3(xn − x0 ) = (b2 − a2 ) + 3(b − a) 2 2 Thus
b
(x + 3)dx = a
1 2 (b − a2 ) + 3(b − a) 2
13. (a) Lf (P ) = −3x1 (x1 − x0 ) − 3x2 (x2 − x1 ) − · · · − 3xn (xn − xn−1 ), Uf (P ) = −3x0 (x1 − x0 ) − 3x1 (x2 − x1 ) − · · · − 3xn−1 (xn − xn−1 ) (b) For each index i −3xi ≤ − 32 xi + xi−1 ≤ −3xi−1 . Multiplying by Δxi = xi − xi−1 gives −3xi Δxi ≤ − 32 xi 2 − x2i−1 ≤ −3xi−1 Δxi . Summing from i = 1 to i = n, we find that Lf (P ) ≤ − 32 x1 2 − x0 2 − · · · − 32 xn 2 − x2n−1 ≤ Uf (P ). The middle sum collapses to − 32 xn 2 − x0 2 = − 32 (b2 − a2 ). Thus 3 Lf (P ) ≤ − (b2 − a2 ) ≤ Uf (P ) 2
so that a
b
3 −3x dx = − (b2 − a2 ). 2
14. (a) Lf (P ) = (1 + 2x0 )(x1 − x0 ) + (1 + 2x1 )(x2 − x1 ) + · · · + (1 + 2xn−1 )(xn − xn−1 ), Uf (P ) = (1 + 2x1 )(x1 − x0 ) + (1 + 2x2 )(x2 − x1 ) + · · · + (1 + 2xn )(xn − xn−1 ) (b) For each index i 1 + 2xi−1 ≤ 1 + (xi−1 + xi ) ≤ 1 + 2xi Multiplying by Δxi = xi − xi−1 gives (1 + 2xi−1 ) Δxi ≤ (xi − xi−1 ) + xi 2 − x2i−1 ≤ (1 + 2xi ) Δxi . Proceeding as before, we get
b
(1 + 2x) dx = (b − a) + (b2 − a2 )
a
15.
2
−1
(x2 + 2x − 3) dx
2π
0
(x3 − 3x) dx
0
2
t sin(2t + 1) dt
17.
3
16.
18. 1
4
√
t dt t2 + 1
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SECTION 5.2 19.
20.
21. Δx1 = Δx2 = 18 , m1 = 0,
Δx3 = Δx4 = Δx5 =
m2 = 14 ,
m3 = 12 ,
1 4
m4 = 1,
m5 =
3 2
f (x∗1 ) = 18 , f (x∗2 ) = 38 , f (x∗3 ) = 34 , f (x∗4 ) = 54 , f (x∗5 ) = M1 = 14 ,
M2 = 12 ,
(a) Lf (P ) = 22.
25 32
M3 = 1,
M4 = 32 ,
(b) S ∗ (P ) =
15 16
3 2
M5 = 2
(c) Uf (P ) =
39 32
1
2x dx = 1. 0
23.
Lf (P ) = x0 3 (x1 − x0 ) + x1 3 (x2 − x1 ) + · · · + x3n−1 (xn − xn−1 ) Uf (P ) = x1 3 (x1 − x0 ) + x2 3 (x2 − x1 ) + · · · + xn 3 (xn − xn−1 ) For each index i x3i−1 ≤ 14 xi 3 + xi 2 xi−1 + xi x2i−1 + x3i−1 ≤ xi 3 and thus by the hint x3i−1 (xi − xi−1 ) ≤
1 4
xi 4 − x4i−1 ≤ xi 3 (xi − xi−1 ).
Adding up these inequalities, we find that Lf (P ) ≤ 14 xn 4 − x0 4 ≤ Uf (P ). 1 1 1 Since xn = 1 and x0 = 0, the middle term is : x3 dx = . 4 4 0 24. (a) Lf (P ) = x0 4 (x1 − x0 ) + x1 4 (x2 − x1 ) + · · · + xn−1 4 (xn − xn−1 ), Uf (P ) = x1 4 (x1 − x0 ) + x2 4 (x2 − x1 ) + · · · + xn 4 (xn − xn−1 ) (b) For each index i xi−1 4 ≤
xi 4 + xi 3 xi−1 + xi 2 xi−1 2 + xi xi−1 3 + xi−1 4 ≤ xi 4 5
Multiplying by Δxi = xi − xi−1 gives xi−1 4 Δxi ≤
1 5 xi − xi−1 5 ≤ xi 4 Δxi . 5
231
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SECTION 5.2 Summing and collapsing the middle sum gives 1 Lf (P ) ≤ xn 5 − x0 5 ≤ Uf (P ), 5 Thus 1 1 1 x4 dx = (15 − 05 ) = . 5 5 0
25. Necessarily holds: Lg (P ) ≤
b a
g(x) dx
0 for x ∈ [0, 2]. Thus, f is increasing on [0, 2]. 1 + x2 (b) Let P = {x0 , x1 , . . . , xn } be a regular partition of [0, 2] and let Δx = 2/n
33. (a) f (x) = √
By Exercise 30,
√ 2( 5 − 1) ∼ 2.47 2 f (x) dx − Lf (P ) ≤ |f (2) − f (0)| = = n n n 0 2 It now follows that 0 f (x) dx − Lf (P ) < 0.1 if n > 25. 2 f (x) dx ∼ (c) = 2.96 2
0
34. (a) f (x) =
−2x < 0 on (0, 1) (1 + x2 )2
⇒
f is decreasing.
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SECTION 5.2 (b) Uf (P ) − so need
1 0
f (x) dx ≤ |f (1) − f (0)|Δx = | 12 − 1| n1 =
1 = 0.05, or n = 10. 2n
1
(c) Using Uf (P ) with n = 10, we have 0
233
1 . 2n
1 dx ∼ = 0.78 1 + x2
35. Let S be the set of positive integers for which the statement is true. Since that k ∈ S. Then
1(2) = 1, 1 ∈ S. Assume 2
k(k + 1) +k+1 2 (k + 1)(k + 2) = 2
1 + 2 + · · · + k + k + 1 = (1 + 2 + · · · + k) + k + 1 =
Thus, k + 1 ∈ S and so S is the set of positive integers. 36. See Exercise 5 in section 1.8. 37. Let f (x) = x and let P = {x0 , x1 , x2 , . . . , xn } be a regular partition of [0, b]. Then Δx = b/n and ib xi = , i = 0, 1, 2, . . . , n. n (a) Since f is increasing on [0, b], 2b (n − 1)b b b Lf (P ) = f (0) + f +f + ··· + f n n n n b 2b (n − 1)b b = 0+ + + ··· + n n n n b2 [1 + 2 + · · · + (n − 1)] n2 b 2b (n − 1)b b Uf (P ) = f +f + ··· + f + f (b) n n n n b b 2b (n − 1)b = + + ··· + +b n n n n =
(b)
=
b2 [1 + 2 + · · · + (n − 1) + n] n2
(c) By Exercise 35,
1 2 b 1− 2 2 2 1 2 n +n b n(n + 1) 1 2 = b = b 1+ Lf (P ) = 2 · n 2 2 n2 2
Lf (P ) =
b2 (n − 1)n 1 · = b2 2 n 2 2
n2 − n n2
=
1 n 1 n
(d) For any partition P, Lf (P ) ≤ §∗ (P ) ≤ Uf (P ). Since lim Lf (P ) = lim Uf (P ) =
||P ||→0
lim S ∗ (P ) =
||P ||→0
1 2 b by the pinching theorem. 2
||P ||→0
1 2 b , 2
=
1 2 b (1 − ||P ||) 2
=
1 2 b (1 + ||P ||) 2
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SECTION 5.2
38. Let f (x) = x2 and let P = {x0 , x1 , x2 , . . . , xn } be a regular partition of [0, b]. Then Δx = b/n and ib xi = , i = 0, 1, 2, . . . , n. n (a) Since f is increasing on [0, b], 2b (n − 1)b b b Lf (P ) = f (0) + f +f + ··· + f n n n n 4b2 (n − 1)2 b2 b b2 = 0 + 2 + 2 + ··· + n n n n =
b3 [1 + 22 + · · · + (n − 1)2 ] n3
2b (n − 1)b b b +f + ··· + f + f (b) Uf (P ) = f n n n n 2 b 4b2 n2 b 2 b + 2 + ··· + 2 = n2 n n n
(b)
=
b3 [1 + 22 + · · · + n2 ] n3
(c) By Exercise 36, Lf (P ) =
b3 (n − 1)n(2n − 1) = b3 · n3 6
Uf (P ) =
b3 n(n + 1)(2n − 1) · = b3 n3 6
2n3 − 3n2 + n 6n3 2n3 + 3n2 + n 6n3
=
1 3 b = (2 − 3||P || + ||P ||2 ) 6
=
1 3 b = (2 + 3||P || + ||P ||2 ) 6
(d) For any partition P, Lf (P ) ≤ §∗ (P ) ≤ Uf (P ). Since lim Lf (P ) = lim Uf (P ) =
||P ||→0
lim S ∗ (P ) =
||P ||→0
||P ||→0
1 3 b , 3
1 3 b by the pinching theorem. 3
39. Choose each x∗i so that f (x∗i ) = mi . Then Si∗ (P ) = Lf (P ). Similarly, choosing each x∗i so that f (x∗i ) = Mi gives Si∗ (P ) = Uf (P ). 1 Also, choosing each x∗i so that f (x∗i ) = (mi + Mi ) (they exist by the intermediate value theorem) 2 gives 1 1 Si∗ (P ) = (m1 + M1 )Δx1 + · · · + (mn + Mn )Δxn 2 2 1 = [m1 Δx1 + · · · + mn Δxn + M1 Δx1 + · · · + Mn Δxn ] 2 1 = [Lf (P ) + Uf (P )]. 2 40. (a) Lf (P ) = (b)
181 ∼ = 7.24, 25
Uf (P ) =
1 402 ∼ [Lf (P ) + Uf (P )] = = 8.04 2 50
221 ∼ = 8.84 25 (c) S ∗ (P ) ∼ = 7.98
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SECTION 5.3 41. (a) Lf (P ) ∼ = 0.6105, (b)
Uf (P ) ∼ = 0.7105
1 [Lf (P ) + Uf (P )] ∼ = 0.6605 2
42. (a) Lf (P ) ∼ = 1.0224, (b)
Uf (P ) ∼ = 1.1824
1 [Lf (P ) + Uf (P )] ∼ = 1.1024 2
43. (a) Lf (P ) ∼ = 0.53138, (b)
(c) S ∗ (P ) ∼ = 0.6684
(c) S ∗ (P ) ∼ = 1.1074
Uf (P ) ∼ = 0.73138
1 [Lf (P ) + Uf (P )] ∼ = 0.63138 2
(c) S ∗ (P ) ∼ = 0.63926
SECTION 5.3
5
1. (a)
2
2 2
f (x) dx = 1
5
5
0
(e) 1
f (x) dx = 1
8
4
8
(b)
4
f (x) dx = −
3
4
f (x) dx = 1
f (x) dx = −7
8
8
4
f (x) dx = 11 − (−2) = 13
1
8
f (x) dx = −
8
3
f (x) dx −
1
(e)
f (x) dx = 5 − 7 = −2
3
f (x) dx = 3
4
f (x) dx −
1
(d)
f (x) dx = 11 − 5 = 6
3
(c)
(f)
4
f (x) dx − 1
4
1 3
5
f (x) dx = −
f (x) dx =
f (x) dx = −4
0
1
2. (a)
2
f (x) dx = −
5
f (x) dx = 5 − 6 = −1
0
2
(f)
1
f (x) dx −
0
(d) 0
f (x) dx = 4 − 6 = −2
0
f (x) dx = 1
1
f (x) dx −
0
(c)
f (x) dx = 4 + 1 = 5
0
(b)
5
f (x) dx +
0
2
f (x) dx =
f (x) dx = −6
4 4
f (x) dx = 0 4
3. With P =
3 1, , 2 2
and f (x) =
1 , we have x
7 = Lf (P ) ≤ 0.5 ≤ 12
1
2
dx 5 ≤ Uf (P ) = < 1. x 6
235
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SECTION 5.3
4. Using P = {0,
1 2 , 1},
1
we have 0.6 < 0.65 = Lf (P ) ≤
1 dx ≤ Uf (P ) = 0.9 < 1. 1 + x2
0
√ (c) F (2) = 2 3
√ (b) F (x) = x x + 1
5. (a) F (0) = 0
2
(d) F (2) =
√ t t + 1 dt
(e) −F (x) =
0
π
(b) By Theorem 5.3.5, F (x) = x sin x. 2π (d) F (2π) = t sin t dt
t sin t dt = 0 π
sin π2 = π2 π (e) −F (x) = t sin t dt. (c)
=
√ t t + 1 dt
x
6. (a) F (π) = F ( π2 )
0
π 2
π
x
1 7. F (x) = 2 ; x +9
(a)
√ 8. F (x) = − x2 + 1
√ (a) − 2
√ 9. F (x) = −x x2 + 1;
(a)
1 10
√
(b)
2
1 9
(c)
4 37
(b) −1
√ (c) − 12 5
(b) 0
√ (c) − 14 5
(d)
−2x (x2 + 9)2
−x x2 + 1 √ x2 2 (d) − x +1+ √ x2 + 1 (d) √
10. F (x) = sin πx
(a) 0
(b) 0
(c) 1
(d) π cos πx
11. F (x) = cos πx;
(a) −1
(b) 1
(c) 0
(d) −π sin πx
12. F (x) = (x + 1)3
(a) 0
(b) 1
(c)
13. (a) Since P1 ⊆ P2 , Uf (P2 ) ≤ Uf (P1 )
but
5 ≤ 4.
(b) Since P1 ⊆ P2 , Lf (P1 ) ≤ Lf (P2 )
but
5 ≤ 4.
14. (a) constant functions.
27 8
(d) 3(x + 1)2
(b) constant functions.
15. constant functions 16. We know this is true for a < c < b. Assume a < b. If c = a or c = b, the equality becomes b b f (x) dt = f (x) dt, trivially true. If c < a, we get a
a
c
f (t) dt + a
c
b
f (t) dt = −
a
c
b
f (t) dt +
f (t) dt = c
b
f (t) dt, as desired a
The other possible cases are proved in a similar manner. 17.
F (x) =
x−1 =0 1 + x2
F (x) =
(1 + x2 ) − 2x(x − 1) 1 , so F (1) = > 0 means x = 1 is a local minimum. 2 2 (1 + x ) 2
=⇒
x = 1 is a critical number.
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SECTION 5.3 18.
F (x) =
x−4 =0 1 + x2
F (x) =
(1 + x2 ) − 2x(x − 4) 1 , so F (4) = > 0 means x = 4 (1 + x2 )2 17
=⇒
237
x = 4 is a critical number.
is a local minimum. 1 > 0 for x > 0. x Thus, F is increasing on (0, ∞);
1 < 0 for x > 0. x2 The graph of F is concave down on (0, ∞);
19. (a) F (x) =
(b) F (x) = −
there are no critical numbers.
there are no points of inflection.
(c)
20. (a) F (x) = x(x − 3)2 ,
(b) F (x) = (x − 3)2 + 2x(x − 3) = 3(x − 3)(x − 1).
F is increasing on [0, ∞);
The graph of F is concave up on (−∞, 1) ∪ (3, ∞);
F is decreasing on (−∞, 0];
The graph of F is concave down on (1, 3);
critical numbers 0, −3.
Inflection points at x = 1, x = 3.
(c)
y 6 4 2 −1
1
2
3
x
21. (a) F is differentiable, therefore continuous (c) F (1) = f (1) = 0
(b) F (x) = f (x) f is differentiable; F (x) = f (x) (d) F (1) = f (1) > 0
(e) f (1) = 0 and f increasing (f > 0) implies f < 0 on (0, 1) and f > 0 on (1, ∞). Since F = f, F is decreasing on (0, 1) and increasing on (1, ∞); F (0) = 0 implies F (1) < 0. 22. (a) G is differentiable, therefore continuous (c) G (1) = g(1) = 0 (e) G (x) = g(x) > 0 for all x = 0.
(b) G (x) = g(x) and g is differentiable; G (x) = g (x) (d) G (x) = g (x) < 0 for x < 1 G (x) = g (x) > 0 for x > 1
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SECTION 5.3
23. (a)
(b)
y
y
4 6 2
4
−1
1
2
3
2
x −1
F (x) =
⎧ ⎨ 2x − 1 x2 + 2
5 2
−1 ≤ x ≤ 0
⎩ 2x + 1 x2 + 2
5 2
0 0. Then (f −1 ) (y) = The length of the graph of f −1 is given by:
f (b)
# 1+
[(f −1 ) (y)]2
f (b)
1 + 1/[f (x)]2 dy
dy =
f (a)
1 , where y = f (x), dy = f (x) dx. f (x)
f (a) b
1 (f (x))2 + 1 f (x) dx f (x)
b
= a
=
a
1 + (f (x))2 dx
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REVIEW EXERCISES 60. Solve the differential equations x = x y = 2y with initial conditions x(0) = 4, y(0) = 2. We get x(t) = 4et , y(t) = 2e2t . √ √ √ Initial speed: ν0 = 16 2; terminal speed: ν1 = 16e2 + 16e4 = 4e 1 + e2 . 1 1 2t 4t s= 16e + 16e dt = 4 et 1 + e2t dt 0
0
√ = 2 e 1 + e2 + ln e + 1 + e2 − 1 − ln 1 + 2 61. If
x y = cos3 θ and = sin3 θ, then a a x 2/3 y 2/3 + = cos2 θ + sin2 θ = 1 a a
and x2/3 + y 2/3 = a2 /3.
62. Assume a ≥ 0. By symmetry, π/2 # L=4 (3a cos2 θ sin θ)2 + (3a sin2 θ cos θ)2 dθ
0
=4 0
π/2
1 sin2 θ 3a cos θ sin θ dθ = 12a 2
π/2 = 6a 0
63. By symmetry, x = y. π/2 π2 L 3a2
3a2 cos5 θ x = a cos3 θ(3a cos θ sin θ) dθ = − = 0 4 5 5 0 2a 2a Therefore, (x, y) = , 5 5 64. (a) The speed μ satisfies |μ|2 = (x (t))2 + (y (t))2 = (4 − π cos πt)2 + (4 − π sin πt)2 = 32 + π − 8π(cos πt + sin πt) √ π = 32 + π − 8π 2 cos(πt − ) 4 √ The particle has minimum speed 32 + π − 8 2π at t = 1/4; √ the particle has maximum speed 32 + 9π at t = 1.
y (1/4) = 1. (b) x (1/4)
587
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SECTION 11.1
CHAPTER 11 SECTION 11.1 1.
lub = 2;
3.
no lub;
5.
lub = 2;
7.
no lub;
9.
lub = 2 12 ;
glb = 0
2.
lub = 2;
glb = 0
4.
lub = 1,
no glb
6.
lub = 3;
glb = −1
8.
lub = 2;
glb = −2
glb = 2
10.
lub = 0;
glb = −1
glb = 0 glb = −2 glb = 2
glb = −2 19
11.
lub = 1;
glb = 0.9
12.
lub = 2 19 ,
13.
lub = e;
glb = 0
14.
no lub,
glb = 1
15.
lub = 12 (−1 +
16.
no lub,
no glb
17.
no lub;
no glb
18.
no lub;
no glb
19.
no lub;
no glb
20.
lub = 0;
no glb
21.
glb S = 0,
22.
glb = 1;
1 ≤ 1 < 1 + 0.0001
√
5);
0≤
23. glb S = 0,
24. glb = 0;
0≤
glb = 12 (−1 −
1 3 11 1 10
√
5)
< 0 + 0.001
2n−1
2k
25. Let > 0. The condition m ≤ s is satisfied by all numbers s in S. All we have to show therefore is that there is some number s in S such that s < m + . Suppose on the contrary that there is no such number in S. We then have m + ≤ x for all x ∈ S. This makes m + a lower bound for S. But this cannot be, for then m + is a lower bound for S that is greater than m, and by assumption, m is the greatest lower bound. 26. (a) Let M = |a1 | + · · · + |an |. Then for any i, |ai | < M , so S is bounded (b) lub S = max {a1 , a2 , . . . , an } ∈ S glb S = min {a1 , a2 , . . . , an } ∈ S
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SECTION 11.1 Since b ∈ S,
27. Let c = lub S.
b ≤ c. Since b is an upper bound for S,
c ≤ b.
589
Thus, b = c.
28. S consists of a single element, equal to lub S. 29. (a) Suppose that K is an upper bound for S and k is a lower bound. Let t be any element of T . Then t ∈ S which implies that k ≤ t ≤ K. Thus K is an upper bound for T and k is a lower bound, and T is bounded. (b) Let a = glb S. Then a ≤ t for all t ∈ T. Therefore, a ≤ glb T. Similarly, if b = lub S, then t ≤ b for all t ∈ T, so lub T ≤ b. It now follows that glb S ≤ glb T ≤ lub T ≤ lub S. 30. (a) Let S = {r : r
2. Choose a positive integer n such that 2x0 1 − 2 < x20 − 2. n n Then,
2 2x0 2x0 1 1 − 2 < x20 − 2 =⇒ 2 < x20 − + 2 = x0 − n1 n n n n
(c) If x20 < 2, then choose a positive integer n such that 2x0 1 + 2 < 2 − x20 . n n Then x20 +
2 2x0 1 + 2 < 2 =⇒ x0 + n1 < 2 n n
34. Assume that there are only finitely many primes, p1 , p2 , · · · , pn and let Q = p1 · p2 · · · pn + 1. Q has a prime divisor p. But Q is not divisible by any of the pi , so p = p1 for all i a contradiction. 35. (a) n = 5 : 2.48832; n = 10 : 2.59374; n = 100 : 2.70481; n = 1000 : 2.71692; n = 10, 000 : 2.71815 (b) lub = e;
glb = 2
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SECTION 11.2
36.
a2
a3
a4
a5
a6
9 4
5 3
6 5
1 2
−3
(b) a20 = 94 , (c) lub S = 4,
a30 = −3,
a40 = 65 ,
a7
a8
a9
a10
4
9 4
5 3
6 5
a50 =
9 4
glb S = −3
a1
a2
a3
a4
a5
1.4142
1.6818
1.8340
1.9152
1.9571
a6
a7
a8
a9
a10
1.9785
1.9892
1.9946
1.9973
1.9986
37. (a)
(b) Let S be the set of positive integers for which an < 2. Then 1 ∈ S since √ a1 = 2 ∼ = 1.4142 < 2. Assume that k ∈ S. Since a2k+1 = 2ak < 4, it follows that ak+1 < 2. Thus k + 1 ∈ S and S is the set of positive integers. (c) 2 is the least upper bound. (d) Let c be a positive number. Then c is the least upper bound of the set
√ √ √ c, c c, c c c, . . . . S= 38. (a) a1 ∼ = 1.4142136
a2 ∼ = 1.8477591
a3 ∼ = 1.9615706
a4 ∼ = 1.9903695
a5 ∼ = 1.9975909
a6 ∼ = 1.9993976 a7 ∼ = 1.9998494 a8 ∼ = 1.9999624 a9 ∼ = 1.9999906 a10 ∼ = 1.9999976 √ √ √ (b) a1 = 2 < 2. Assume true for an . Then an+1 = 2 + an < 2 + 2 = 2. (c) lub S = 2. (d) For any positive number c, lub S is the positive number satisfying √ √ x = c + x, that is, x = (1 + 1 + 4c)/2
SECTION 11.2 1.
an = 2 + 3(n − 1) = 3n − 1, n = 1, 2, 3, . . .
2.
an = 1 − (−1)n , n = 1, 2, 3, . . .
3.
an =
(−1)n−1 , n = 1, 2, 3, . . . 2n − 1
4.
an =
5.
an =
n2 + 1 , n = 1, 2, 3, . . . n
6.
an = (−1)n
7. an =
n 1/n,
if n = 2k − 1 if n = 2k,
where k = 1, 2, 3, . . .
2n − 1 , n = 1, 2, 3, . . . 2n n , n = 1, 2, 3, . . . (n + 1)2
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SECTION 11.2 8. an =
n
if n = 2k 2
1/n ,
9. decreasing;
591
where k = 1, 2, 3, . . .
if n = 2k − 1,
bounded below by 0 and above by 2
10. not monotonic; bounded below by −1 and above by 12 . 11.
1 n + (−1)n = 1 + (−1)n : not monotonic; n n
bounded below by 0 and above by
3 2
12. increasing; bounded below by 1.001 but not bounded above. 13. decreasing; bounded below by 0 and above by 0.9 14. increasing; bounded below by 0 and above by 1 n2 1 1 =n−1+ : increasing; bounded below by but not bounded above n+1 n+1 2 √ √ 16. increasing; bounded below by 2, but not bounded above: n2 + 1 > n
15.
17.
18.
4n 1 2 and decreases to 0: increasing; = 4n2 4n2 + 1 1 + 1/4n2 √ bounded below by 45 5 and above by 2 √
2n 1 2n = n 2 = n +1 (2 ) + 1 2 +
4n
19. increasing; 20.
√
1 2n
decreasing; bounded below by 0 and above by 25 .
.
bounded below by
1 n2 = 1 n3 + 1 n +
1 n4
and
2 51
but not bounded above
1 1 + decreases to 0 =⇒ n n4
√ increasing; bounded below by
2 , 2
but not bounded above. 2 2n =2− increases toward 2: increasing; bounded below by 0 and above by ln 2. n+1 n+1 √ 2 2 22. increasing (n ≥ 2); bounded below by 10 , but not bounded above. 3 21.
23. decreasing; bounded below by 1 and above by 4 24. not monotonic; not bounded below and not bounded above 25. increasing; bounded below by 26. decreasing (since
√
3 and above by 2
n+1 decreases to 1); bounded below by 0 and above by ln 2. n
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SECTION 11.2
√ √ 27. (−1)2n+1 n = − n: decreasing; √ 28.
n+1 √ = n
1+
bounded above by −1 but not bounded below
√ 1 decreasing; bounded below by 1 and above by 2 n
29.
1 2n − 1 = 1 − n : increasing; n 2 2
30.
1 3 1 − = decreasing; bounded below by 0 and above by 2n 2n + 3 2n(2n + 3)
bounded below by
31. consider sin x as x → 0+ : decreasing;
1 and above by 1 2
bounded below by 0 and above by 1
32. not monotonic; bounded below by − 12 and above by 33. decreasing;
3 10 .
bounded below by 0 and above by
1 4
5 6
4 x is an increasing function on [4, ∞).); bounded below by ln x ln 4 but not bounded above.
34. increasing (because
35.
1 1 1 − = : decreasing; n n+1 n (n + 1)
bounded below by 0 and above by
1 2
36. not monotonic; bounded below by −1 and above by 1 37. Set f (x) =
ln x . x
Then, f (x) =
bounded below by 0 and above by
1 − ln x e: decreasing;
ln 3.
38. not monotonic; not bounded below nor above (because exponentials grow faster than polynomials). 39. Set an =
3n . (n + 1)2
bounded below by 40.
1 − ( 12 )n = 2n − 1 ( 12 )n
Then, 3 4
an+1 =3 an
n+1 n+2
2 > 1: increasing;
but not bounded above. increasing; bounded below by 1 but not bounded above.
41. For n ≥ 5 an+1 n! 5n+1 5 · an . Since a2 = 1 + √ Assume that ak = 1 + ak−1 > ak−1 . Then ak+1 = 1 +
√
ak > 1 +
√
√
a1 = 2 > 1, 1 ∈ S.
ak−1 = ak .
Thus, k ∈ S implies k + 1 ∈ S. It now follows that {an } is an increasing sequence.
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SECTION 11.3
595
(b) Since {an } is an increasing sequence, an = 1 +
√
an−1 < 1 +
√
an ,
or
an −
√
an − 1 < 0.
Rewriting the second inequality as
and solving for
√ an
√ 2 √ ( an ) − an − 1 < 0 √ √ it follows that an < 12 (1 + 5). Hence,
an < 12 (3 +
√
5) for all n.
(c) a2 = 2, a3 ∼ = 2.4142, a4 ∼ = 2.5538, a5 ∼ = 2.5981, . . . , a9 ∼ = 2.6179, . . . , a15 ∼ = 2.6180; √ (e) lub {an } = 12 (3 + 5) ∼ = 2.6180 68. (a) We show that an < an+1 for all n.
True for n = 1 since a1 = 1
2 =n 2 n n
32. converges to ln 9 :
33. converges to 0:
2n n 2 1 1 1+ = 1+ → e2 n n
29. converges to e2 :
31. diverges;
diverges:
n5 =n 17n4 + 12
4=2
30. converges to
→ ln 2
√ 1 √ → 0 so e1/ n → e0 = 1 n
√
√
→1
6 n4
n
23. converges to 1:
27.
2n n+1
n2 1 1 →√ = 4 4 2 2n + 1 2 + 1/n
1 − n14 n4 − 1 = n4 + n − 6 1 + n13 −
20. converges to 1:
21.
nπ π → tan = 1 4n + 1 4
tan
(2n + 1)2 4 + 4/n + 1/n2 4 = → (3n − 1)2 9 − 6/n + 1/n2 9
4 : 9
18. converges to ln 2 :
19. converges to
so
9n2 2 n +1
→ ln 9
1 1 1 − = →0 n n+1 n(n + 1)
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SECTION 11.3 34.
n → 1, arctan 1 = π/4 n+1
converges to π/4:
35. converges to 1/2: √ 36. converges to 2:
597
√ n2 + n + n n 1 2 +n−n= n +n−n √ =√ → 2 2 2 n +n+n n +n+n
n2
√ 4n2 + n = 4 + (1/n) → 4 = 2. n
37. converges to π/2 38. converges to 31/9: let s = 3.444 · · ·. Then 10s − s = 31 and s = 31/9. 1−n → −1; arcsin (−1) = π/2. n
39. converges to −π/2:
n2 + 5n + 5 (n + 1)(n + 4) = 2 → 1. (n + 2)(n + 3) n + 5n + 6
40. converges to 1:
41. (a)
√ n
n→1
(b)
n 42. (a) √ →e n n! 43. b
0.
n ≥ N1 ,
if
then
|an − L| < /2.
Since bn → M , there exists N2 such that n ≥ N2 , then
if
|bn − M | < /2.
Now set N = max {N1 , N2 }. Then, for n ≥ N , |(an + bn ) − (L + M )| ≤ |an − L| + |bn − M | < 46. Let > 0, choose k such that n ≥ k =⇒ |an − L| < Then for n ≥ k,
. |α| + 1
|αan − αL| = |α||an − L| ≤ (|α| + 1)|an − L| <
α an → α L.
Therefore 47. Since
+ = . 2 2
1 1+ n
→1
n
and
1 1+ n
n+1
1 1+ n
→ e, =
1 1+ n
n
1 1+ n
→ (e)(1) = e.
√ n
an + bn → b
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SECTION 11.3
48. (a)
(b)
(c)
If k = j,
If k < j,
If k > j,
1 1 + · · · + α0 · k αk n n an = → 1 1 βk βk + βk−1 · + · · · + β0 · k n n αk + αk−1 ·
an =
an =
αk + αk−1 ·
1 1 + · · · + α0 · k n n
→
βj · nj−k + βj−1 · nj−1−k + · · · + β0 ·
1 nk
αk · nk−j + αk−1 · nk−1−j + · · · + α0 ·
1 nj
βj + βj−1 ·
1 1 + · · · + β0 · j n n
0
diverges
49. Suppose that {an } is bounded and non-increasing. If L is the greatest lower bound of the range of this sequence, then an ≥ L for all n. Set > 0. By Theorem 11.1.4 there exists ak such that ak < L + . Since the sequence is non-increasing, an ≤ ak for all n ≥ k. Thus, L ≤ an < L + or and
|an − L| <
for all
n≥k
an → L. If an → L,
50. Let > 0.
then there exists a positive integer k such that |an − L| < for all n ≥ k
If n ≥ k,
then 2n ≥ k and 2n − 1 ≥ k,
and thus
|en − L| = |a2n − L| < and |on − L| = |a2n−1 − L| < It follows that en → L and on → L.
If en → L and on → L, then there exist k1 and k2 such
that if m ≥ k1 ,
then |em − L| = |a2m − L| <
and if m ≥ k2 , Let k = max{2k1 , 2k2 − 1}.
then
|om − L| = |a2m−1 − L| <
If n ≥ k then
either an = a2m with m > k1 or an = a2m−1 with m ≥ k2 In either case, 51. Let > 0.
|an − L| < .
This shows that an → L.
Choose k so that, for n ≥ k, L − < an < L + ,
L − < cn < L +
and an ≤ bn ≤ cn .
For such n, L − < bn < L + . 52. Let M be a bound for {bn }.
Then
|an bn | ≤ |an |M.
Given > 0, choose k such that |an | < /M for n ≥ k. Then |an bn | < for n ≥ k.
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SECTION 11.3
599
53. Let > 0. Since an → L, there exists a positive integer N such that L − < an < L + for all n ≥ N. Now an ≤ M for all n, so L − < M, or L < M + . Since is arbitrary, L ≤ M. 54. The converse is false. For example, let an = (−1)n .
Then |an | → 1,
but {an } diverges.
55. Assume an → 0 as n → ∞. Let > 0. There exists a positive integer N such that |an − 0| < for all n ≥ N . Since | |an | − 0| ≤ |an − 0|, it follows that |an | → 0. Now assume that |an | → 0. Since −|an | ≤ an ≤ |an |, an → 0 by the pinching theorem. 56. Let > 0. There exists a positive integer N1 such that |an − L| < for all n > 2N1 − 1, and there exists a positive integer N2 such that |bn − L| < for all n > 2N2 . The sequence a1 , b1 , a2 , b2 , . . . can be represented by the sequence c1 , c2 , c3 , . . ., where cn =
a(n+1)/2 ,
if n is odd
bn/2 ,
if n is even.
Let N = max{2N1 − 1, 2N2 }. Then n > N =⇒ |cn − L| < =⇒ cn → L. 57. By the continuity of f, f (L) = f 58.
lim an = lim f (an ) = lim an+1 = L.
n→∞
n→∞
n→∞
2 2 2 2 2 4 2n = · · · · · = 2 · · (terms that are ≤ 1) ≤ . n! 1 2 3 n n n 4 →0 n
Since
and
0
1, implies L = 12 (3 + 5). √ (b) an+1 = 3an Suppose that an → L as n → ∞. Then an+1 → L as n → ∞. Therefore √ L = 3L which, since L > 1, implies L = 3. 76. (a)
a2 ∼ = 2.6458,
a3 ∼ = 2.9404,
a4 ∼ = 2.9900,
(b)
True for n = 1.
(c)
an+1 2 − an 2 = 6 + an − an 2 = (3 − an )(2 + an ) ≥ 0 Since ak ≥ 0
Assume true for n.
a5 ∼ = 2.9983, a6 ∼ = 2.9997 √ √ Then an+1 = 6 + an−1 ≤ 6 + 3 = 3 since 0 ≤ an ≤ 3.
this implies an+1 ≥ an
for all k,
(d) an → 3 77. (a)
a2
a3
a4
a5
a6
0.5403
0.8576
0.6543
0.7935
0.7014
a7
a8
0.7640 0.7221 ∼ (b) L is a fixed point of f (x) = cos x, that is, cos L = L; L = 0.739085. a2 ∼ = 1.5403, a3 ∼ = 1.5708, a4 ∼ = a5 ∼ = ··· ∼ = a10 ∼ = 1.5708. (b) L ∼ = 1.570796 Let f (x) = x + cos x. L must satisfy L = f (L), π and cos L = 0. Indeed, the L we found is just ∼ = 1.570796327 2
a9
a10
0.7504
0.7314
78. (a)
so L = L + cos L,
PROJECT 11.3 1. (a)
a2
a3
a4
a5
a6
a7
a8
2.000000 1.750000 1.732143 1.732051 1.732051 1.732051 √ 1 3 (b) L = L+ which implies L2 = 3 or L = 3. 2 L
2. The Newton-Raphson method applied to the function f (x) = x2 − R an = an−1 −
= 3 & 4. (a) f (x) = x3 − 8, (c) f (x) = ln x − 1,
so
gives
a2 − R f (an−1 ) = an−1 − n−1 f (an−1 ) 2an−1
1 1 R 1 an−1 + = 2 2 an−1 2 so
1.732051
xn → 2 xn → e
an−1 +
R an−1
,
n = 2, 3, . . . .
(b) f (x) = sin x − 12 ,
so
xn →
π 6
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SECTION 11.4
601
SECTION 11.4 1.
22/n = (21/n )2 → 12 = 1
converges to 1:
3. converges to 0:
4. converges to 0:
5. converges to 0:
6.
converges to 0:
8.
converges to 1:
3n = 4n
n 3 →0 4
n/(n+2) →1 n1/(n+2) = n1/n n+1 n
11. converges to 0:
0
−n
3n 4n
1 1− n
:
n
n
e−x dx = 1 −
n
integral = 2 0
n
0
n
1 →1 en
→ e−1
16.
(by (11.4.7)
diverges.
π dx −1 =π = 2 tan n → 2 1 + x2 2
1 e−n + →0 dx = − n n 2
−nx
e
18. converges to 0:
(−1) 1+ n
=
0
21. converges to 0:
n 3 = 12 → 12(0) = 0 4
(n + 2)1/n → e0 = 1.
converges to 1:
converges to 1:
nα/n = (n1/n )α → 1α = 1
1 e−2n 1 − → 2 2 2
e2x dx =
19.
converges to 1:
1
it follows that
17. converges to π:
9.
(x100 )n x100n = →0 n! n!
(n + 2)1/n = e n ln(n+2) and, since
ln (n + 2) n+2 1 ln (n + 2) = → (0)(1) = 0, n n+2 n
13. converges to 1:
15.
converges to 0:
→ ln(1) = 0
3n+1 = 12 4n−1
1 : 2
7.
ln
14. converges to e
e−α/n → e0 = 1
1 ln n log10 n = · →0 n ln 10 n
ln (n + 1) ln (n + 1) n+1 = → (0)(1) = 0 n n+1 n
10. converges to 0:
−1
converges to 1:
n n 2 2 0< < →0 n 3
for n > 3,
12. converges to
2.
recall (11.4.6) ln n ln (n2 ) =2 → 2(0) = 0 n n
20.
converges to 0:
n2 sin nπ = 0
for all n
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SECTION 11.4
22. converges to π :
23. diverges:
24.
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1−1/n
−1+1/n
Since
lim
x→0
1 1 dx −1 −1 √ − sin → sin−1 (1) − sin−1 (−1) = π 1− −1 + = sin n n 1 − x2
sin x = 1, x
n π sin (π/n) sin = → 1. Therefore, π n π/n n π π n2 sin = nπ sin → nπ. n π n
diverges
25.
converges to 0:
5n+1 = 20 42n−1
26. converges to e3x :
x 3n x n 3 1+ = 1+ → (ex )3 = e3x n n
27. converges to e−1 :
n+2 (−1) n n 1+ n+1 1 e−1 n+2 = e−1 = 1− = 2 → n+2 n+2 1 (−1) 1+ n+2
5 16
n →0
1
28. converges to 2: 1/n
dx 2 √ = 2 − √ → 2. x n
29. converges to 0:
n+1
0
2 , 1+ ≈e>2 n n n 35. converges to 0:
1/n 1/n 1/n 2 2 2 sin x dx ≤ | sin x | dx ≤ 1 dx = → 0 −1/n n −1/n −1/n
n x n x/t = tn 1 + ; n n
36.
t+
37.
converges:
sin(6/n) →2 sin(3/n)
converges to 0 if t < 1,
38.
converges to ex if t = 1,
converges:
arctan n →0 n
diverges if t > 1.
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SECTION 11.4 39.
40.
√
n+1−
n2
√
603
√ √ √ n + 1 − n √ 1 n= √ n+1+ n = √ √ √ →0 n+1+ n n+1+ n
√ n2 + n − n 2 n 1 1 +n−n= √ → ( n + n + n) = √ = 2 2 2 n +n+n n +n+n 1 + 1 + 1/n
41. (a) The length of each side of the polygon is 2r sin(π/n). Therefore the perimeter, pn , of the polygon is given by: pn = 2rn sin(π/n). (b) 2rn sin(π/n) → 2πr as n → ∞ : The number 2rn sin (π/n) is the perimeter of a regular polygon of n sides inscribed in a circle of radius r. As n tends to ∞, the perimeter of the polygon tends to the circumference of the circle. d < (cn + dn )1/n < (2dn )1/n = 21/n d → d,
42. Since 0 < c < d,
so by the pinching theorem
(cn + dn )1/n → d. 43. By the hint,
44. diverges:
lim
n→∞
1 + 2 + ... + n n(n + 1) 1 + 1/n 1 = . = lim = lim n→∞ n→∞ n2 2n2 2 2
n(n + 1)(2n + 1) 2n3 + 3n2 + n 12 + 22 + · · · + n2 = = 2 →∞ (1 + n)(2 + n) 6(1 + n)(2 + n) 6n + 18n + 12
45. By the hint,
13 + 23 + . . . + n3 n2 (n + 1)2 1 + 2/n + 1/n2 1 = . = lim = lim n→∞ n→∞ 4(2n4 + n − 1) n→∞ 8 + 4/n3 − 4/n4 2n4 + n − 1 8 lim
46. Here we show that every convergent sequence is a Cauchy sequence. Let > 0. If an → L, then there exists a positive integer k such that |ap − L| < With m, n ≥ k
2
for all p ≥ k
we have |am − an | ≤ |am − L| + |L − an | = |am − L| + |an − L| < mn+1 − mn =
47. (a)
+ = . 2 2
1 1 (a1 + · · · + an + an+1 ) − (a1 + · · · + an ) n+1 n n
1 nan+1 − (a1 + · · · + an ) = n(n + 1) >0
since {an } is increasing.
(b) We begin with the hint |a1 + · · · + aj | mn < + n 2
Since j is fixed, |a1 + · · · + aj | →0 n
n−j n
.
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SECTION 11.4 and therefore for n sufficiently large |a1 + · · · + aj | < . n 2 Since 2
n−j n
0 we can find k such that |an − am | < for n, m ≥ k. In particular,
|an+1 − an | < ,
so
lim (an − an−1 ) = 0.
n→∞
(b) {an } does not necessarily converge. For example let an = ln n. This diverges, but n → ln 1 = 0 an − an−1 = ln n − ln(n − 1) = ln n−1 49. (a) Let S be the set of positive integers n (n ≥ 2) for which the inequalities hold. Since √ 2 √ √ 2 √ √ 2 b − 2 ab + a = b − a > 0, it follows that
a+b √ > ab and so a1 > b1 . Now, 2 a1 + b1 < a1 and b2 = a1 b1 > b1 . a2 = 2
Also, by the argument above, a2 =
a1 + b1 > a1 b1 = b2 , 2
and so a1 > a2 > b2 > b1 . Thus 2 ∈ S. Assume that k ∈ S. Then ak + ak ak + b k < = ak , bk+1 = ak bk > b2k = bk , ak+1 = 2 2 and ak+1 =
ak + b k > ak bk = bk+1 . 2
Thus k + 1 ∈ S. Therefore, the inequalities hold for all n ≥ 2. (b) {an } is a decreasing sequence which is bounded below. {bn } is an increasing sequence which is bounded above. Let La = lim an , n→∞
Lb = lim bn . Then n→∞
an−1 + bn−1 an = 2 50.
implies
La =
La + Lb 2
and La = Lb .
1 100 5 100 e5 − 1 + 100 e − 1 + 100 ∼ ∼ = 0.004995 : within 0.01%; = 0.11395 : within 12% e e5 1000 1000 1 5 e5 − 1 + 1000 e − 1 + 1000 ∼ ∼ = 0.0004995 : within 0.05%; = 0.01238 : within 1.3% e e5
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SECTION 11.5
605
51. The numerical work suggests L ∼ = 1. Justification: Set f (x) = sin x − x2 . Note that f (0) = 0 and f (x) = cos x − 2x > 0 for x close to 0. Therefore sin x − x2 > 0 for x close to 0 and sin 1/n − 1/n2 > 0 for n large. Thus, for n large, 1 1 1 (| sin x| ≤ |x| < sin < n2 n n 1/n 1/n 1/n 1 1 1 < sin < n2 n n 1/n 2 1 1 1 < sin < 1/n . 1/n n n n
for all x)
As n → ∞ both bounds tend to 1 and therefore the middle term also tends to 1. 52. Numerical work suggests L ∼ = 1/3.
Conjecture: L = 1/k. (1 + 1/n)1/k − 1 ; 1/n
Proof: (nk + nk−1 )1/k − n = n(1 + 1/n)1/k − n =
(1 + 1/n)1/k − 1 (1 + h)1/k − 1 = lim = f (1), n→∞ h→0 1/n h lim
where f (x) = x1/k . 53. (a) (b)
Since f (x) = (1/k)x(1/k)−1 ,
a3
a4
a5
a6
a7
a8
a9
a10
2
3
5
8
13
21
34
55
r1
r2
r3
r4
r5
r6
1
2
1.5
1.667
1.600
1.625
f (1) = 1/k.
(c) Following the hint, 1+
1 an−1 an + an−1 an+1 = 1 + an = 1 + = = = rn . an an an an−1
1 rn−1
Now, if rn → L, then rn−1 → L and
√ 1+ 5 ∼ which, since L > 1, implies L = = 1.618034. 2
1 1+ =L L
54. With the partition {0, n1 , n2 , . . . , nn } and f (x) = x, we have
n n 1 2 1 1 2 n 1 an = f (xi )Δxi , + + ··· + = f +f + ··· + f = n n n n n n n n i=1
1
x dx,
so it is a Riemann sum for
and therefore lim an = n→∞
0
1
x dx = 0
1 2
SECTION 11.5 (We’ll use to indicate differentiation of numerator and denominator.) 1.
lim
x→0+
√ sin x √ = lim 2 x cos x = 0 x x→0+
2.
lim
x→1
ln x 1/x = −1 = lim 1 − x x→1 −1
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5.
7.
SECTION 11.5
11.
13.
√
ex − 1 = lim (1 + x)ex = 1 x→0 ln (1 + x) x→0
4.
cos x − sin x 1 = = lim sin 2x x→π/2 2 cos 2x 2
6.
lim
lim
x→π/2
2x − 1 = lim 2x ln 2 = ln 2 x→0 x→0 x lim
x1/2 − x1/4 9. lim = lim x→1 x→1 x−1 10.
December 2, 2006
1 −1/2 1 −3/4 − x x 2 4
8. =
x→4
lim
x→a
1
√ x−2 1 2 x = lim = x→4 1 x−4 4
lim
x−a 1 1 = lim = xn − an x→a nxn−1 nan−1
tan−1 x = lim x→0 x→0 x lim
1 1+x2
1
=1
1 4
ex − 1 ex = lim =1 x→0 x(1 + x) x→0 1 + 2x lim
ex − e−x ex + e−x = lim =2 x→0 x→0 cos x sin x
12.
lim
lim
x→0
1 − cos x sin x = lim =0 x→0 x→0 3 3x lim
x + sin πx 1 + π cos πx 1+π = = lim x→0 x − sin πx 1 − π cos πx 1−π
14.
ax − (a + 1)x ax ln a − (a + 1)x ln(a + 1) = ln = lim x→0 x→0 x 1
15.
ex + e−x − 2 ex − e−x ex + e−x 1 = lim = lim = x→0 1 − cos 2x x→0 2 sin 2x x→0 4 cos 2x 2
16.
1 − x+1 x − ln(x + 1) 1 (x+1)2 = = lim = lim x→0 1 − cos 2x x→0 2 sin 2x x→0 4 cos 2x 4
17.
tan πx π sec2 πx = lim =π x x→0 e − 1 x→0 ex
18.
1 cos x − 1 + x2 /2 − sin x + x − cos x + 1 sin x = = lim = lim = lim x→0 x→0 x→0 x→0 24x x4 4x3 12x2 24
19.
1 + x − ex 1 − ex −ex 1 = lim = lim =− x x x x x→0 x(e − 1) x→0 xe + e − 1 x→0 xe + 2ex 2
20.
ln(sec x) tan x sec2 x 1 = = lim = lim 2 x→0 x→0 2x x→0 x 2 2
21.
x − tan x 1 − sec2 x −2 sec2 x tan x −2 sec2 x = lim = −2 = lim = lim x→0 x − sin x x→0 1 − cos x x→0 x→0 sin x cos x
22.
2 xenx − x enx + nxenx − 1 enx (2n + n2 x) = = lim = lim x→0 1 − cos nx x→0 x→0 n sin nx n2 cos nx n
23.
√ 1 − x2 1 − x2 2 1√ = = lim− = 6 lim− √ 3 x→1 x→1 1−x 3 3 1 − x3
lim
a a+1
lim
1
1
lim
lim
lim
lim
lim
lim
lim
since
lim−
x→1
1 − x2 2x 2 = lim = 1 − x3 x→1− 3x2 3
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SECTION 11.5 24.
25.
lim
x→0
2x − sin πx =0 4x2 − 1 ln (sin x) − cot x csc2 x 1 =− = lim = lim (π − 2x)2 x→π/2 4(π − 2x) x→π/2 −8 8
lim
x→π/2
√
26.
27.
lim+ √
x→0
lim
x→0
lim
x→0
29.
1 √ 2 x √ cos√ x 1 √ + 2 x 2 x
= lim
x→0+
1 1 √ = 2 1 + cos x
cos x − cos 3x − sin x + 3 sin 3x − cos x + 9 cos 3x = lim = lim =4 x→0 x→0 2 cos(x2 ) − 4x2 sin(x2 ) sin(x2 ) 2x cos(x2 ) √
28.
x √ = lim x + sin x x→0+
a+x− x
√ a−x
= lim
√1 2 a+x
+
√1 2 a−x
1
x→0
√ a 1 =√ = a a
sec2 x − 2 tan x 2 sec2 x tan x − 2 sec2 x = lim 1 + cos 4x −4 sin 4x x→π/4
lim
x→π/4
= lim
x→π/4
2 sec4 x + 4 sec2 x tan2 x − 4 sec2 x tan x 1 = −16 cos 4x 2
−x 1 1 − √1−x x − arcsin x 2 (1−x2 )3/2 = lim = lim 30. lim x→0 x→0 3 sin2 x cos x x→0 6 sin x cos2 x − 3 sin3 x sin3 x
= lim
x→0
31.
32.
lim
x→0
6 cos3
x − 21 sin x cos x
sin−1 x = lim x→0 x→0 x lim
x→∞
1 6
√ 1 1−x2
1
=1
π/2 − tan−1 x x2 = lim =1 x→∞ 1 + x2 1/x
ln(1 − n1 ) ln(1 − x) 1 = lim = −1 = lim n→∞ sin( 1 ) sin x x→0+ x→0+ (1 − x) cos x n
34. −1 :
35. 1 :
=−
1 tan−1 x 1 1 +˙4x2 = = lim −1 2 x→0 tan 2x 1+x 2 2
lim
33. 1 :
36.
−1−2x2 (1−x2 )5/2 2
lim
lim
1 1/x t = lim = lim = lim (1 + t) = 1 x[ ln (x + 1) − ln x ] x→∞ ln (1 + 1/x) t→0+ ln (1 + t) t→0+
lim
sinh(π/n) − sin(π/n) sinh(π/x) − sin(π/x) sinh u − sin u = lim = lim x→∞ u→0+ sin3 (π/n) sin3 (π/x) sin3 u
x→∞
1 : 3
n→∞
= lim+ u→0
cosh u − cos u sinh u + sin u cosh u + cos u 2 1 = = = lim+ = lim+ 3 2 6 3 u→0 6 sin u cos2 u − 3 sin u u→0 6 cos3 u − 21 sin u cos u 3 sin2 u cos u
607
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x3 =0 x→0 4x − 1
39.
lim
43.
December 2, 2006
lim
x→0 π 2
lim
x→0
38.
x =1 − arccos x
tanh x =1 x
lim (2 + x + sin x) = 0,
x→0
44.
T1: PBU
SECTION 11.5
37.
41.
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1/n
lim n a
n→∞
lim
x→0
4x does not exist sin2 x
40.
√ 2x − 2 =0 lim √ x→2+ x−2
42.
1 + cos 2x =4 x→π/2 1 − sin x lim
lim (x3 + x − cos x) = 0
x→0
a1/x ln a − x12 a1/x − 1 1 = ln a − 1 = lim = lim x→∞ x→∞ 1/x − x2
45. The limit does not exist if b = 1. Therefore, b = 1. lim
x→0
cos ax − 1 −a sin ax −a2 cos ax a2 = lim = lim =− 2 x→0 x→0 2x 4x 4 4
Now, −
46.
a2 = − 4 implies 4
a = ±4.
sin 2x + ax + bx3 2 cos 2x + a + 3bx2 = lim 3 x→0 x→0 x 3x2 lim
−4 sin 2x + 6bx −8 cos 2x + 6b =0 = lim x→0 x→0 6x 6 4 =⇒ a = −2, b = 3
= lim
need a = −2 to keep numerator 0 if 6b = 8
47. Recall lim+ (1 + x)1/x = e: x→0
x − (1 + x) ln (1 + x) (1 + x)1/x − e 1/x = lim (1 + x) lim x x2 + x3 x→0+ x→0+ = e lim
x→0+
= e lim
x→0+
= e lim
x→0+
x − (1 + x) ln (1 + x) x2 + x3 − ln (1 + x) 2x + 3x2 −1/(1 + x) e =− 2 + 6x 2
f (x + h) − f (x − h) f (x + h) − f (x − h)(−1) = lim = f (x) h→0 h→0 2h 2
48. (a) lim
(note that here we differentiated with respect to h, not x. ) f (x + h) − 2f (x) + f (x − h) f (x + h) − f (x − h) = lim 2 h→0 h→0 h 2h
(b) lim
f (x + h) + f (x − h) = f (x) h→0 2
= lim
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SECTION 11.5 49.
1 x→0 x
x
lim
f (t) dt = lim
x→0
0
50. (a) lim
x→0
f (x) = f (0) 1
Si(x) sin x = lim =1 x→0 x x
(b)
lim
x→0
Si(x) − x sin x/x − 1 sin x − x = lim = lim x→0 x→0 x3 3x2 3x3
= lim
x→0
= lim
x→0
cos x − 1 9x2 − sin x 1 =− 18x 18
C(x) cos2 x =1 = lim x→0 x→0 x 1 C(x) − x cos2 x − 1 −2 cos x sin x −2 cos2 x + 2 sin2 x 1 (b) lim = lim = lim = lim =− 3 2 x→0 x→0 x→0 x→0 x 3x 6x 6 3 x f (t) dt f (x) 52. (a) lim a =0 = lim x→a x→a f (x) f (x) x f (t) dt f (x) f (k−1) (x) a (b) Similarly, lim =0 = lim = · · · = lim (k) x→a x→a f (x) x→a f f (x) (x)
51. (a) lim
√
b
x3 53. A(b) = 2 (b − x2 ) dx = 2 bx − 3 0 √ T (b) b b 3 = 4 √ = . Thus, lim b→0 A(b) 4 b b 3 54. T (θ) =
1 (1 − cos θ) sin θ; 2 lim
θ→0+
S(θ) =
√b = 0
4 √ b b 3
and
T (b) =
√ 1 √ 2 b b = b b. 2
θ 1 − sin θ: 2 2
T (θ) (1 − cos θ) sin θ (1 − cos θ) cos θ + sin2 θ cos θ − cos 2θ = lim = lim = lim + + + S(θ) θ − sin θ 1 − cos θ 1 − cos θ θ→0 θ→0 θ→0
=
− sin θ + 4 sin θ cos θ − sin θ + 2 sin 2θ = lim+ =3 sin θ sin θ θ→0 f (x) → ∞ as x → ±∞
y
55. (a)
15 f 10 5
-5
5
x
(b) f (x) → 10 as x → 4 Confirmation: lim √ x→4
x2 − 16 2x = lim = lim 2 x2 + 9 = 10 −1/2 x→4 x2 + 9 − 5 x→4 x (x2 + 9)
609
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SECTION 11.6 y
56. (a)
lim f (x) = 0
x→±∞
0.17
−10
(b) f (x) →
10
1 as x → 0; 6
lim
x→0
x
x − sin x 1 − cos x sin x 1 = lim = lim = x→0 x→0 6x x3 3x2 6 f (x) → 0.7 as x → 0
y
57. (a)
0.8 f
0.6
-2
-1
1
(b) Confirmation: lim
x→0
2
x
2sin x − 1 ln(2) 2sin x cos x = lim = ln 2 ∼ = 0.6931 x→0 x 1 g(x) → −1.6 as x → 0
y
58. (a)
x
−1
−2
(b) Confirmation: lim
x→0
3cos x − 3 3cos x (− sin x) ln 3 sin x = lim = − lim 3cos x ln 3 x→0 x→0 x2 2x 2x =−
3 ln 3 ∼ = −1.6479 2
SECTION 11.6 (We’ll use to indicate differentiation of numerator and denominator.) 1.
3.
lim
x→−∞
lim
x→∞
x2 + 1 2x = lim =∞ x→−∞ 1−x −1
x3 1 = −1 = lim x→∞ 1/x3 − 1 1 − x3
1 5. lim x sin = lim x→∞ x h→0+ 2
sin h 1 =∞ h h
2.
4.
lim
x→∞
20x =0 +1
x2
x3 − 1 = −∞ x→∞ 2 − x lim
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SECTION 11.6 ln(xk ) k/x = lim =0 x→∞ x→∞ 1 x
sin 5x cos x tan 5x 1 7. lim = lim = − π− tan x sin x cos 5x 5 x→ π x→ 2 2 6.
lim
lim π−
x→ 2
8. 9. 10.
11.
12.
lim π−
x→ 2
1 cos x sin x = = lim cos 5x x→ π2 − 5 sin 5x 5
cos x ln | sin x| x (x cos x) = 0 = lim sin1x = lim − x→0 x→0 − 2 x→0 1/x sin x x
x→0
lim x2x = lim+ (xx )2 = 12 = 1
x→0+
x→0
[see (11.6.4)]
sin πt π cos πt π = lim = lim =π + + x t→0 t 1 t→0
lim x sin
x→∞
(ln |x|)2 2 ln |x| 2 = lim 2x = 0 = lim = lim x→0 x→0 x→0 1/x −1/x 1/x x→0
lim x( ln |x| )2 = lim
x→0
ln x 1/x sin2 x sin x = lim = lim+ − = lim+ − · sin x = 0 2 cot x x→0+ − csc x x→0 x x x→0
lim
x→0+
√ lim
x→∞
x
0
2
ex =∞ e dt = lim x→∞ 1
t2
1 + x2 = lim x→∞ x
1 +1=1 x2
1 1 − 2 lim x→0 sin2 x x
15.
= lim
x2 − sin2 x 2x − 2 sin x cos x = lim x→0 2x2 sin x cos x + 2x sin2 x x2 sin2 x
= lim
2x − sin 2x 2 − 2 cos 2x = lim x2 sin 2x + 2x sin2 x x→0 2x2 cos 2x + 4x sin 2x + 2 sin2 x
x→0
x→0
= lim
x→0 −4x2
= lim
x→0
16. Since 17.
and
lim (x ln | sin x|) = lim
1 13. lim x→∞ x 14.
sin 5x =1 sin x
since
−8x2
4 sin 2x sin 2x + 12x cos 2x + 6 sin 2x 8 cos 2x 1 = cos 2x − 32x sin 2x + 24 cos 2x 3
lim ln(| sin x|x ) = lim (x ln | sin x|) = 0
x→0
x→0
lim x1/(x−1) = e
x→1
since
sin x
lim ln x
x→0+
x→0
ln x 1 lim ln x1/(x−1) = lim = lim = 1 x→1 x→1 x − 1 x→1 x
18. Take log:
by Exercise 8, lim | sin x|x = e0 = 1
= lim (sin x ln x) = lim + + x→0
= lim
x→0+
x→0
ln x csc x
1/x − sin2 x = lim = 0, − csc x cot x x→0+ x cos x
so
lim xsin x = e0 = 1
x→0+
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SECTION 11.6
cos
lim
19.
x→∞
1 x
x =1
since
lim ln
cos
x→∞
1 x
x
1 ln cos x = lim x→∞ (1/x) sin (1/x) = lim − =0 x→∞ cos (1/x)
20. Take log: lim ln (| sec x|cos x ) = lim cos x ln | sec x| = lim
x→π/2
x→π/2
x→π/2
ln | sec x| sec x
tan x = lim cos x = 0, x→π/2 sec x tan x x→π/2
= lim
lim
21.
x→0
x→0
so
ln(x2 + a2 ) = lim x→∞ x→∞ x2
2
lim ln(x2 + a2 )(1/x) = lim
x→∞
2x x2 +a2
2x
2
lim
x→0
1 sin x − x cos x x sin x − cot x = lim = lim x→0 x→0 sin x + x cos x x x sin x
= lim
x→0
lim ln
x→∞
lim
x→∞
x2 − 1 x2 + 1
3
= 3 lim ln x→∞
x2 + 2x − x = lim
x→∞
x→∞
27.
x2 − 1 x2 + 1
= lim √
26.
= 0,
lim (x2 + a2 )(1/x) = e0 = 1
25.
1 1 = 1 + 1 + ln (1 + x) 2
x→∞
23.
24.
lim | sec x|cos x = e0 = 1
x→π/2
x − ln (1 + x) x 1 1 − = lim = lim x→0 x ln (1 + x) x→0 x + (1 + x) ln (1 + x) ln (1 + x) x = lim
22. Take log:
so
=0
x2 + 2x − x
x2
sin x + x cos x =0 2 cos x − x sin x
√
x2 + 2x + x √ x2 + 2x + x
2x 2 =1 = lim x→∞ + 2x + x 1 + 2/x + 1
a bx a x b lim 1 + = lim 1 + = (ea )b = eab . x→∞ x→∞ x x 1/ ln x = e3 lim x3 + 1
x→∞
since
1/ ln x ln x3 + 1 3 = lim = lim lim ln x + 1 x→∞ x→∞ x→∞ ln x 28. Take log: so
ln(ex + 1) lim = lim x→∞ x→∞ x x
1/x
lim (e + 1)
x→∞
=e
ex ex +1
1
ex = 1, x→∞ ex
= lim
3x2 x3 + 1 1/x
= lim
x→∞
3 = 3. 1 + 1/x3
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SECTION 11.6 29.
lim (cosh x)1/x = e
x→∞
since lim ln [(cosh x)1/x ] = lim
x→∞
x→∞
1 30. Take log: lim 3x ln 1 + x→∞ x 3x 1 = e3 so lim 1 + x→∞ x 31.
lim
x→0
1 1 − sin x x
613
= lim
x→0
= lim
x→0
= 3 lim
ln (cosh x) sinh x = lim = 1. x→∞ x cosh x
ln 1 + x1 1 x
x→∞
−1/x2 1+1/x = 3 lim x→∞ − 12 x
= 3,
x − sin x 1 − cos x = lim x→0 x sin x sin x + x cos x sin x =0 2 cos x − x sin x ex +3
ln(ex + 3x) x = lim e +3x = 4, so lim (ex + 3x)1/x = e4 x→0 x→0 x→0 x 1 1 x − 1 − x ln x − ln x x − = lim lim = lim x→1 x→1 (x − 1) ln x x→1 (x − 1)(1/x) + ln x ln x x − 1
32. Take log: 33.
lim
= lim
x→1
34.
35.
√
2:
1 ln n 1 ln = − →0 n n n
39. 1 : 40. 1:
41. 0 : 42. 1: 43.
36.
0:
lim
x→0
1 + 2x 2
1/x
1
= e2
ln 2
=
√ 2
nk →0 n→∞ 2n lim
1 ln (ln n)1/n = ln (ln n) → 0 n
37. 1 : 38. 0:
2x ln 2 x ln(1 + 2x ) − ln 2 ln 2 ; lim = lim 1 + 2 = x→0 x→0 x 1 2
take log:
0:
1 − x ln x − ln x − 1 =− = lim x − 1 + x ln x x→1 2 + ln x 2
lim
n→∞
ln n 1/n 1 = lim = lim =0 n→∞ p np−1 n→∞ p np np 1/n
ln n ln (n + 1) 1 ln [n(n + 1)] = + →0 n n n sin(π/n) ln n sin(π/n) lim ln n = 0, = lim [sin(π/n) ln n] = lim n→∞ n→∞ n→∞ 1/n n ln
n2 + n
=
x3 =0 x→∞ ex √ √ √ ln( n − 1) 1/ n √ = take log: ln( n − 1) → 0, n 0≤
n3 n2 ln n < n, n e e
lim (sin x)x = 1
x→0
lim
√ √ so ( n − 1)1/ n → 1
44.
lim (tan x)tan 2x =
x→π/4
1 e
so nsin(π/n) → 1
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SECTION 11.6
45.
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lim
x→0
1 1 − sin x tan x
=0
46.
47.
lim (sinh x)−x = 1
x→0+
48.
vertical asymptote x = 1 horizontal asymptote y = 1
vertical asymptote y-axis 50.
49.
horizontal asymptote x-axis
horizontal asymptote x-axis 52.
51.
horizontal asymptote x-axis
vertical asymptote y-axis horizontal asymptote x-axis
53.
√ x2 − a2 + x b 2 b 2 b −ab 2 x −a − x= √ x − a2 − x = √ →0 2 2 2 a a a x −a +x x − a2 + x
54. cosh x − sinh x = 55. for instance, 56. for instance, 57.
lim+ −
x→0
1 1 x (e + e−x ) − (ex − e−x ) = e−x → 0, 2 2
f (x) = x2 + F (x) = x +
as
x→∞
as x → ∞
(x − 1)(x − 2) x3
sin x 1 + x2
2x 2 = lim . + cos x x→0 − sin x
L’Hospital’s rule does not apply here since
lim cos x = 1.
x→0+
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SECTION 11.6 58. (a) Let S be the set of positive integers for which the statement is true. Since lim
x→∞
Assume that k ∈ S. By L’Hospital’s rule, lim
x→∞
(ln x)k+1 (k + 1)(ln x)k = lim =0 x→∞ x x
ln x = 0, x
k ∈ S).
(since
Thus k + 1 ∈ S, and S is the set of positive integers. (b) Choose any positive number α. Choose a positive integer k > α. Then, for x > e, (ln x)k (ln x)α < x x
0
−1, we can find k such that xr < ex/2 for x ≥ k (ex/2 grows faster than any power ∞ k ∞ ∞ r −x r −x −x/2 of x). Then x e dx < x e dx + e dx, which converges. Thus xr e−x dx 0
0
k
0
converges for all r > −1. ∞ b (b) For n = 1 : xe−x dx = lim −xe−x − e−x 0 = lim −be−b − e−b + 1 = 1 b→∞
0
∞
Assume true for n.
n+1 −x
x 0
n+1 −b
= lim (−b b→∞
b→∞
e
e ) + (n + 1) 0
∞
dx = lim
b→∞
b −xn+1 e−x 0
+ (n + 1)
xn e−x dx = 0 + (n + 1)n! = (n + 1)!
b
n −x
x e 0
dx
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SECTION 11.7 39.
∞
√
0
∞ 1 1 √ dx + dx x (1 + x) x (1 + x) 0 1 1 b 1 1 √ √ = lim+ dx + lim dx b→∞ x (1 + x) x (1 + x) a→0 a 1
1 dx = x (1 + x)
621
1
√
√ 1 2 √ Now, du = 2 arctan u + C = 2 arctan x + C. dx = 1 + u2 x (1 + x) √ u= x 1 √ √ 1 1 π √ Therefore, lim+ dx = lim+ 2 arctan x a = lim+ 2 π/4 − arctan a ] = 2 x (1 + x) a→0 a→0 a→0 a b √ √ b 1 π √ and lim dx = lim 2 arctan x 1 = lim 2 arctan b − π/4 = . b→∞ 1 b→∞ b→∞ 2 x (1 + x) ∞ 1 √ dx = π. Thus, x (1 + x) 0
∞
40. 1
b 1 1 √ √ dx + lim dx 2 2 b→∞ a x x −1 2 x x −1 2 ∞ = lim+ sec−1 x a + lim sec−1 x 2
1 √ dx = lim+ a→1 x x2 − 1
2
b→∞
a→1
= lim (sec−1 2 − sec−1 a) + lim (sec−1 b − sec−1 2) a→1+
b→∞
π π = (sec−1 2 − 0) + ( − sec−1 2) = 2 2
√ ∞ ∞ ∞ x4 + 1 1 1 1 41. surface area S = dx 2π 1 + 4 dx = 2π dx = ∞ by comparison with 3 x x x x 1 1 1
π/2
42. A =
(sec x − tan x) dx = lim − b→π/2
0
0
b
b (sec x − tan x) dx = lim − ln(sec x − tan x) − ln sec x
0
b→π/2
b = lim − ln(1 + sin x) = ln 2 0
b→π/2
1
1 √ dx = lim x a→0+ 0 √ 1 = lim+ 2 x a = 2
(b) A =
43. (a)
a
1
1 √ dx x
a→0
(c) V =
1
π 0
1 √ x
2
dx = π 0
1
1 dx = π lim x a→0+
a
1
1 dx = π lim [ln x]1a x a→0+
diverges
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SECTION 11.7
44. (a)
∞
1 π dx = lim tan−1 b = b→∞ 1 + x2 2
(b) A = 0
b
π 1 x −1 dx = lim x + tan b→∞ 2 (1 + x2 )2 1 + x2 0 0 π b π2 tan−1 b + = lim − 0 = b→∞ 2 1 + b2 4 ∞ b 2πx (d) Vy = dx = lim π ln(1 + x2 ) 0 = ∞ 2 b→∞ 1 + x 0
∞
(c) Vx =
π·
45. (a)
∞
(b) A =
e−x dx = 1
0
(c) Vx =
∞
πe−2x dx = π/2
0
(d)
∞
Vy =
−x
2πxe
b→∞
0
∞
(e) A =
2πe−x
1 + e−2x
0
0
b
2πe−x
0
b 2πxe−x dx = lim 2π(−x − 1)e−x 0 b→∞
(by parts)
b+1 = 2π 1 − lim = 2π(1 − 0) = 2π b→∞ eb b dx = lim 2πe−x 1 + e−2x dx b→∞
1 + e−2x dx = −2π u = e−x
b
dx = lim
0 e−b
1 + u2 du
1
e−b = −π u 1 + u2 + ln u + 1 + u2 1 √ √ −b −2b =π 2 + ln 1 + 2 − e 1+e − ln e−b + 1 + e−2b
Taking the limit of this last expression as b → ∞, we have √ √ A=π 2 + ln 1 + 2 .
∞
xA =1 A 0 ∞ 1 −2x yA 1 1 yA = dx = , y = e = ; centroid: (1, 14 ) 2 4 A 4 0 1 Yes: 2πxA = 2π = Vy , 2πyA = π = Vx 2
46. xA =
xe−x dx = 1,
x=
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SECTION 11.7
∞
(b) Vy =
−x2
2πxe
2
b
dx = lim
∞
is finite.
b→∞
0
b→∞
0
∞
(c) Vy =
e−x dx
2 2 b 2 2πxe−x dx = lim π −e−x = lim π 1 − e−b = π
1 x 1 48. (a) A = − 2 dx = ln 2 x x + 1 2 1 2 ∞ ∞ 2 1 x dx (b) Vx = dx < − 2+1 x x x2 1 1
∞
and 1
b→∞
0
e−x ≤ e−x
47. (a) The interval [0, 1] causes no problem. For x ≥ 1,
623
2πx 1
1 x − x x2 + 1
∞
dx = 2π 1
finite
1 1 dx = π 2 x2 + 1 2
49. (a)
1
(b) A = lim
x
a→0+
dx = lim
a→0+
a
(c) Vx = lim + a→0
1
1
4 3
= a
1 πx−1/2 dx = lim+ 2πx1/2 = 2π a
1
(d) Vy = lim
3/4
2πx
a→0+
4 3/4 x 3
a→0
a
−1/4
dx = lim
a→0+
a
∞
8π 7/4 x 7
1 = a
8 π 7
x
g(x) dx = L. Since f (x) ≥ 0 for x ∈ [a, ∞), f (t) dt is increasing. a a x Therefore it is sufficient to show that f (t) dt is bounded above. For any number
50. (i) Suppose that
M ≥ a, we have
Therefore, (ii) If
a
M
M
f (x) dx ≤
a
g(x) dx ≤
a
x
f (t) dt is bounded and a
∞
f (x) dx diverges,
then
g(x) dx = L a
∞
f (x) dx converges. a
∞
g(x) dx can not converge, by (i)
0
0
∞
dx x3/2
51. converges by comparison with 1
52. converges by comparison with
∞
Converges by comparison with π
56. Diverges by comparison with e
∞
dx x2
∞
e−x dx on [2, ∞).
2
53. diverges since for x large the integrand is greater than
54.
∞
1 and x
1
∞
dx diverges x
55.
dx (x + 1) ln(x + 1)
converges by comparison with 1
∞
dx x3/2
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SECTION 11.7
b 2x 2 ln 1 + x dx = lim =∞ b→∞ 0 1 + x2 b→∞ 0 ∞ 2x Thus, the improper integral dx diverges. 1 + x2 0 b b (b) 2x 2 lim ln 1 + x dx = lim b→∞ −b 1 + x2 b→∞ −b = lim ln 1 + b2 − ln 1 + (−b)2 = lim (0) = 0 b
57. (a) lim
b→∞
∞
sin x dx = lim
b→∞
0
b→∞
b→∞
0
lim
b
sin x dx = lim
58.
b
−b
sin x dx = lim
L= ∧ (10.7.3)
θ1
b − cos x = 1 − lim cos b; the limit does not exist b→∞
0
b − cos x = lim [− cos b + cos(−b)] = lim [ 0 ] = 0 −b
b→∞
r (θ) = acecθ
59. r(θ) = aecθ ,
b→∞
−∞
b→∞
b→∞
a2 e2cθ + a2 c2 e2cθ dθ
= a 1 + c2 lim b→−∞
= a 1 + c2
θ1
b
lim
b→−∞
ecθ c
ecθ dθ θ1 b
√ √ 2 cθ 1 + c a 1 + c2 a = lim e 1 − ecb = ecθ1 b→−∞ c c 60. For all real t, −
∞
61. F (s) =
−sx
e
b→∞
1 ; s
∞
−sx
xe 0
=
∞
−sx
e 0
x
−∞
e−t
2
/2
dt
converges by comparison with
1 dx = lim − e−sx b→∞ s
b = 0
1 s
x
−∞
et+1 dt.
provided s > 0.
dom(F ) = (0, ∞).
F (s) =
63. F (s) =
b
· 1 dx = lim
62.
Therefore
0
Thus, F (s) =
t2 < t + 1. 2
1 s2
−xe−sx e−sx − 2 dx = lim b→∞ s s
if s > 0,
e−sx cos 2x dx = lim
0
Using integration by parts
b→∞
diverges for s ≤ 0,
b
= lim
0
b→∞
−be−sb e−sb 1 − 2 + 2 s s s
so dom(F ) = (0, ∞).
e−sx cos 2x dx
0 −sx
e
b
1 −sx 4 s −sx e cos 2x dx = 2 sin 2x − e cos 2x + C. s +4 2 4
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SECTION 11.7
625
Therefore,
b 1 −sx 4 s −sx F (s) = lim 2 e sin 2x − e cos 2x b→∞ s + 4 2 4 0
1 −sb 4 s −sb s 4 s s = 2 lim e sin 2b − e cos 2b + = 2 · = 2 s + 4 b→∞ 2 4 4 s +4 4 s +4 Thus,
F (s) =
s ; s2 + 4
dom(F ) = (0, ∞).
∞
F (s) =
64.
eax e−sx dx =
0
∞
e(a−s)x dx = lim
b→∞
0
1 s−a
=
diverges if s ≤ a;
if s > a,
65. The function f is nonnegative on (−∞, ∞) and ∞ 0 ∞ f (x) dx = 0 dx + −∞
provided s > 0.
−∞
0
e(a−s)b 1 − a−s a−s
so dom F = (a, ∞)
6x dx = (1 + 3x2 )2
∞
0
6x dx (1 + 3x2 )2
1 6x 2 dx = − 1 + 3x2 + C. 2 (1 + 3x )
Now, Therefore,
∞
1 f (x) dx = lim − b→∞ 1 + 3x2
−∞
b
= lim
b→∞
0
1 1− 1 + 3b2
= 1.
66. f is nonnegative, and ∞ ∞ b f (x) dx = ke−kx dx = lim −e−kx 0 = lim −e−kb + 1 = 1 −∞
b→∞
0
b→∞
So, f is a probability density function. 67.
μ=
∞ −∞
xf (x) dx =
0
−∞
∞
kxe−kx dx = lim
0 dx +
Using integration by parts,
b→∞
0
b
kxe−kx dx
0
kxe−kx dx = −xe−kx −
1 −kx e + C. k
Therefore,
b
∞ 1 1 1 1 μ= = xf (x) dx = lim −xe−kx − e−kx = lim −be−kb − e−kb + b→∞ b→∞ k k k k −∞ 0 68. σ =
∞
−∞
=
∞
(x − μ)2 f (x) dx = kx2 e−kx dx − 2
1 k
x−
∞
1 k
ke−kx dx
∞
e−kx dx =
0
1 k2
t
F (t) =
f (x) dx 1
is increasing, and that
2
xe−kx dx +
0
69. Observe that
0
0
∞
(∗) an ≤
is continuous and increasing, that
t 1
f (x) dx ≤ an+1
an =
for t ∈ [ n, n + 1 ].
n
f (x) dx 1
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REVIEW EXERCISES If
∞
converges, then F , being continuous, is bounded and, by (∗), {an } is bounded
f (x) dx 1
and therefore convergent. If {an } converges, then {an } is bounded and, by (∗), F is bounded. Being ∞ increasing, F is also convergent; i.e., f (x) dx converges. 1
REVIEW EXERCISES 1. |x − 2| ≤ 3 =⇒ −1 ≤ x ≤ 5 : 2. x2 > 3 =⇒ x >
√
lub = 5, glb = −1.
√ 3 or x < − 3;
no lub, no glb.
3. x2 − x − 2 ≤ 0 =⇒ (x − 2)(x + 1) ≤ 0 =⇒ −1 ≤ x ≤ 2 : 4. cos x ≤ 1 for all x;
lub = 2, glb = −1.
no lub, no glb.
5. Since e−x ≤ 1 for all x, 2
e−x ≤ 2 for all x; 2
6. ln x < e =⇒ 0 < x < ee :
no lub, no glb.
lub = ee , glb = 0.
7. increasing; bounded below by
1 2
and above by 23 .
8. increasing; bounded below by 0 but not bounded above:
n2 − 1 1 = n − → ∞ as n → ∞. n n
9. bounded below by 0 and above by 32 ; not monotonic 10. increasing; bounded below by 11.
12.
4 5
and above by 1.
! # " 2n , . . . ; the sequence is not monotonic. = 2, 1, 89 , 1, 32 25 2 n However, it is increasing from a3 on. The sequence is bounded below by 89 ; it is not bounded above. sin (nπ/2) n2
!
" = 1, 0, − 19 , 0,
#
1 25 , · · ·
;
bounded below by − 19 and above by 1; not monotonic
13. the sequence does not converge; n21/n → ∞ as n → ∞
14. converges to 1:
3 1+ + n2 + 3n + 2 n = 7 n2 + 7n + 12 1+ + n
15. converges to 1:
1 lim ln n→∞ n
n n+1
2 n2 → 1. 12 n2
= 0 =⇒ lim
n→∞
n 1+n
1/n =1
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REVIEW EXERCISES 16. converges to 0:
5 4 1 + 2+ 3 4n2 + 5n + 1 n n n = → 0. 1 n3 + 1 1+ 3 n
17. converges to 0:
cos nπ sin nπ → cos 0 sin 0 = 0 as n → ∞
627
18. diverges: (2 + n1 )n > 2n and 2n diverges. 0 = [ln 1]n ≤ [ln(1 +
19. converges to 0:
3 ln 2n − ln(n3 + 1) = ln
20. converges to ln 8:
3 2
21. converges to
1 n ) ] ≤ [ln 2]n ; n
√
:
[ln 2]n → 0 as n → ∞.
8n3 → ln 8. n3 + 1
3 − n12 3n2 − 1 = 4n4 + 2n2 + 3 4 + n22 +
3 n4
→
3 as n → ∞. 2
1
(n2 + 4) 3 (1/n + 4/n3 )1/3 = → 0 as n → ∞. 2n + 1 2 + 1/n
22. converges to 0:
π π cos → 0 cos 0 = 0 as n → ∞. n n
23. converges to 0: 24. converges 0:
(n/π) sin(nπ) = 0 for all positive integers n.
n+1
25. converges to 0: n
26. diverges: 1
n
n+1 1 e−x dx = − e−x = e−n (1 − ) → 0 as n → ∞ n e
√ n √ √ 1 √ dx = 2 x = 2 n − 2 and 2 n − 2 diverges. 1 x
27. Given > 0. Since an → L, there exists a positive integer K such that if n ≥ K, then |an − L| < . Now, if n ≥ K − 1, then n + 1 ≥ K and |an+1 − L| < . Therefore, an+1 → L. 28. Let > 0. Since an → L, there is positive integer K such that if n ≥ K, |an − L|
N , |ai − L| < , i = 1, 2, · · · , K. n 2K Let M = max {K, N }. Then, if n ≥ M , K n a1 + · · · + an |ai − L| |aj − L| ≤ − L + < K( ) + n( ) = . n n n 2K 2n i=1 j=K+1
Therefore mn → L.
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REVIEW EXERCISES
29. As an example, let a =
π 3.
Then cos π/3 = 0.5,
cos cos 0.5 ∼ = 0.87758, · · · .
Using technology (graphing calculator, CAS), we get cos cos · · · cos π/3 → 0.73910. and cos (0.73910) ∼ = 0.73910. Hence, numerically, this sequence converges to 0.73910. 30. Let f (x) = sin (cos x) and let a = π/3. Then f (π/3) ∼ = 0.4794, f (f (π/3)) ∼ = 0.7753, · · · After 14 steps, we get f (f (· · · f (π/3)) ∼ = 0.6948 and sin (cos 0.6948) ∼ = 0.6948. ln x 5+2 5x + 2 ln x x = 5; = lim 31. lim ln x x→∞ x + 3 ln x x→∞ 1+3 x 32.
ln x lim =0 x→∞ x
ex − 1 ex 1 = = lim 2 x→0 tan 2x x→0 2 sec 2x 2 lim
− sin x ln(cos x) −1 sin x 1 33. lim · =− = lim cos x = lim 2 x→0 x→0 x→0 2 cos x x 2x x 2 34. Set y = x1/(x−1) . Then ln y =
ln x , x−1
and
lim
x→1
ln x 1 = lim = 1 x − 1 x→1 x
Therefore, lim x1/(x−1) = e. x→1
35.
36.
lim (1 +
x→∞
2 4 2x 4 ) = lim (1 + )x = e8 x→∞ x x
e2x − e−2x 2e2x + 2e−2x = lim =4 x→0 x→0 sin x cos x lim
1 ln x −x2 =0 = lim+ x = lim+ 37. lim+ x2 ln x = lim+ 1 2 x→0 x→0 x→0 −2 x→0 x2 x3 38.
39.
10x 10x ln 10 10x (ln 10)2 10x (ln 10)10 = lim = lim = · · · = lim →∞ 10 9 8 x→∞ x x→∞ x→∞ (10)(9)x x→∞ 10x 10! lim
ex + e−x − x2 − 2 ex − e−x − 2x ex − e−x − 2 ex + e−x − 2 = lim = lim = lim x→0 x→0 2 sin x cos x − 2x x→0 sin 2x − 2x x→0 2 cos 2x − 2 sin2 x − x2 lim
1 ex − e−x ex + e−x =− = lim x→0 −4 sin x2x x→0 −8 cos 2x 4
= lim
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REVIEW EXERCISES ln x 1/x 1 = lim =− sin πx x→1 π cos πx π x 2 x 2 et dt ex x2 1 2 2 41. lim xe−x = et dt = lim 0 x2 = lim 2 2 = lim 2 2 x x x→∞ x→∞ x→∞ 2x e x→∞ 2x − 1 2 e −e 0 2 x x 40.
lim csc(πx) ln x = lim
x→1
x→1
1 − 2 − n ln |x| . x 1 1 + nx2 ln |x| = −∞ lim − 2 − n ln |x| = lim − x→0 x→0 x x2
e−1/x 42. Consider ln = xn 2
e−1/x → 0. x→0 xn 2
Therefore
∞
43.
√
e− √
1
√
44.
1
45. 0
lim
x
x
b
dx = lim
b→∞
√
e− √
1
x
x
dx = lim
b→∞
√
− 2e−
x
b 1
√
= lim (−2e−
b
b→∞
+ 2e−1 ) = 2e−1
x dx = − 1 − x2 + C 1 − x2 b 1 b x x √ √ dx = lim− dx = lim− − 1 − x2 = lim− (− 1 − b2 + 1) = 1 b→1 b→1 b→1 0 1 − x2 1 − x2 0 0 a 1 1 1 1 − dx (partial fractions) dx = − lim− 2 2 a→1 0 x−1 x+1 0 1−x
a a+1 1 1 = lim− ln(x + 1) − ln(1 − x) = lim− ln = ∞; 0 2 a→1 2 a→1 1−a
1 dx = lim− a→1 1 − x2
a
the integral diverges.
π/2
sec x dx = lim −
46.
c→π/2
0
c
0
c sec x dx = lim − ln(sec x + tan x) = lim − ln(sec c + tan c) = ∞; c→π/2
0
the integral diverges.
∞
47. 1
sin(π/x) dx = lim b→∞ x2
1
b
1 b sin(π/x) 2 dx = lim = cos π/x b→∞ π 1 x2 π
c 9 1 1 1 48. dx = lim− dx + lim+ dx 2/3 2/3 2/3 c→1 c→1 0 (x − 1) 0 (x − 1) c (x − 1) c c 1 1/3 lim− dx = lim 3 (x − 1) =3 2/3 c→1 c→1− 0 0 (x − 1) 9 9 1 1/3 dx = lim 3 (x − 1) =6 lim+ 2/3 c c→1 c→1+ c (x − 1) 9 1 = 3 + 6 = 9. (x − 1)2/3 0
9
c→π/2
629
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REVIEW EXERCISES
49.
∞
0
1 ex + e−x
ex dx = arctan ex + C +1 c c 1 π dx = lim dx = lim arctan ex = x −x c→∞ 0 e + e c→∞ 0 2
1 dx = ex + e−x
e2x
50. Set u = ln x, du = 2
∞
1 dx = x(ln x)k
1 dx; u(2) = ln 2. Then x ∞
ln 2
1 du uk
The integral converges if k > 1 and diverges otherwise. For k > 1,
∞ 2
51.
1 dx = lim c→∞ x(ln x)k
a
a
ln(1/x) dx = 0
0
c
ln 2
c 1 1 1 du = lim = u1−k k c→∞ 1 − k ln 2 u (k − 1) (ln 2)k−1
− ln x dx = lim+ c→0
c
a
a − ln x dx = lim+ − x ln x + x c→0
c
= lim+ [−a ln a + a + c ln c − c] = a ln(1/a) + a c→0
52. y = (a2/3 − x2/3 )3/2 ; y = −x−1/3 (a2/3 − x2/3 )1/2 a a 3a L= 1 + (y )2 dx = a1/3 x−1/3 dx = 2 0 0 53. For any a ∈ S + T , a = s + t for some s ∈ S and t ∈ T . Hence a ≤ lub(S) + lub(T ). Therefore, S + T is bounded above and lub(S) + lub(T ) is an upper bound for S + T . Let M = lub(S + T ) and suppose M < lub(S) + lub(T ). Set = (lub(S) + lub(T )) − M . There exist s ∈ S and t ∈ T such that lub(S) − s < /2,
and
lub(T ) − t < /2
Now, lub(S) + lub(T ) − (s + t) = lub(S) − s + lub(T ) − t < = lub(S) + lub(T ) − M which implies s + t > M , a contradiction. Therefore lub(S) + lub(T ) = lub(S + T ). 54. (a) Since S is bounded below, there is a number b such that b ≤ s for every s ∈ S. Thus b is a lower bound for S and B = ∅. (b) Choose any s ∈ S. Then, for any b ∈ B, b ≤ s. Therefore B is bounded above (each element s ∈ S is an upper bound for B). (c) We show first that glb(S) is an upper bound for B. For if not, there is b ∈ B such that b > glb(S). Then there is an s ∈ S such that glb(S) < s < b, which contradicts the fact that b is a lower bound of S. It follows that lub(B) ≤ glb(S). If lub(B) < glb(S), then there exists a number a such that lub(B) < a < glb(S) which implies that a is a lower bound for S and a ∈ B. Therefore a ≤ lub(B), a contradiction. Thus, lub(B) = glb(S).
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REVIEW EXERCISES 55. (a) If
∞ −∞
f (x)dx = L, then
c
f (x) dx,
lim
c→∞
both exist and
c→∞
and
b→−∞
lim
c→∞
c→∞
−c
x dx = 0, but
−c
f (x) dx = L
∞
−∞
56. (a) Assume that f is nonnegative on (−∞, ∞). ∞ f (x) dx = L =⇒ lim By Exercise 55, −∞
Now assume that lim
c→∞
lim
c→∞
x dx diverges.
c
−c
f (x) dx = L
c
−c
f (x) dx = L. Since f is nonnegative,
x
f (t) dt ≤ L
on
[0, ∞).
0 x
Therefore
f (x) dx = L b
c
c
f (x) dx b
0
f (x) dx + lim
Let c = −b, then
(b) Set f (x) = x. then lim
b→−∞
c
0
0
lim
0
lim
c→∞
631
f (t) dt is a bounded and nondecreasing function, which implies that 0
c
f (x) dx exists. Similarly, lim 0
Therefore,
c→∞
0
−c
f (x) dx exists.
∞ −∞
f (x) dx exists, and, by the uniqueness of the limit,
∞
−∞
f (x) dx = L.
57. Let S be a set of integers which is bounded above. Then there is an integer k ∈ S such that k ≥ n for all n ∈ S, for if not, S is not bounded above. Therefore, k is an upper bound for S. Let M = lub(S). Then M ≥ k since k ∈ S. Also M ≤ k since k is an upper bound for S. Therefore M = k; the least upper bound of S is an element of S. 58. lub [Lf (P )] =
b a
f (x) dx;
glb [Uf (P )] =
b a
f (x) dx.
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SECTION 12.1
CHAPTER 12 SECTION 12.1 1. 1 + 4 + 7 = 12
2. 2 + 5 + 8 + 11 = 26
3. 1 + 2 + 4 + 8 = 15
4.
5. 1 − 2 + 4 − 8 = −5
6. 2 − 4 + 8 − 16 = −10
+
1 9
+
1 27
=
9. 1 +
1 4
+
1 16
+
7.
11.
1 3
11
=
10. 1 −
85 64
(2n − 1)
12.
n=1
14.
n
mk Δxk
10 1 , 2k 10
+
10
1 4
1 8
1 16
+
1 24
−
1 120
+
1 16
−
(−1)k+1 (2k − 1)
15.
n
2 = − 15
1 64
=
13.
k=3
k , k+1
Mk Δxk
16. 10 kk k=3
(−1)i
i=0
35
k(k + 1)
n
f (x∗k ) Δxk
k=1
18. 7
51 64
k=1
7 1 i+3 2 i=0
(−1)k+1
15 16
=
k=1
k=3
19.
1 4
k=1
k=1
17.
+
8. − 16 +
13 27 1 64
1 2
i+3 i+4
20.
10 k=3
k!
,
7 (i + 3)i+3 i=0
1 , 2k − 3
(i + 3)! 7 i=0
1 2i + 3
21. Set k = n + 3. Then n = −1 when k = 2 and n = 7 when k = 10. 10 k=2
22.
7 7 k n+3 n+3 = = k 2 + 1 n=−1 (n + 3)2 + 1 n=−1 n2 + 6n + 10
12 11 11 (−1)n (−1)k+1 (−1)k+1 = = n−1 k+1−1 k n=2 k=1
k=1
23. Set k = n − 3. Then n = 7 when k = 4 and n = 28 when k = 25. 25 k=4
24.
15 32k k=0
k!
=
28 28 1 1 1 = = k 2 − 9 n=7 (n − 3)2 − 9 n=7 n2 − 6n
13 13 32(n+2) 32n = 81 (n + 2)! (n + 2)! n=−2 n=−2
ak a1 an a2 + 2 + ··· + n = 10 10 10 10k n
25. 0.a1 a2 · · · an =
k=1
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SECTION 12.2 26.
n √ 1 1 1 1 1 1 1 1 1 n √ = √ + √ + √ + · · · + √ ≥ √ + √ + √ + · · · + √ = √ = n. n n n n n n 1 2 3 k k=1
27.
50 1 = 1.3333 · · · 4k
50 1 ∼ = 1.62513 k2
28.
k=0
k=1
50 1 29. = 2.71828 · · · k!
50 k 2
30.
k=0
SECTION 12.2 1.
1 ; 2
2.
1 ; 2
3.
11 ; 18
4.
3 ; 4
5.
6.
10 ; 3 5 ; 6
3 7. − ; 2
8.
1 ; 4
9. 24;
∞ k=3
k=0
3
∼ =3
1 1 1 1 sn = + + ··· + 2 1·2 2·3 (n)(n + 1) 1 1 1 1 1 1 1 1 1 = 1− + − + ··· + − = 1− → 2 2 2 3 n n+1 2 n+1 2 ∞
1 = 2 k −k
k=3
1 1 − k−1 k
= lim
n→∞
1 1 − 2 n
=
1 2
1 1 1 + + ··· + 1·4 2·5 n(n + 3) 1 1 1 1 1 1 = 1− + − + ··· + − 3 4 2 5 n n+3 1 1 1 1 1 1 1 1 1 11 = 1+ + − − − → 1+ + = 3 2 3 n+1 n+2 n+3 3 2 3 18
sn =
∞ k=0
∞
1 1 = (k + 1)(k + 3) 2
k=0
1 1 − k+1 k+3
=
1 1 1 3 1 lim 1 + − − = 2 n→∞ 2 n+2 n+3 4
k ∞ ∞ 1 1 3 30 10 = = =3 =3 k 10 10 1 − 1/10 9 3
k=0
k=0
∞ (−1)k
5k
k=0
=
∞ 1 − 2k
3k
k=0 ∞ k=0
1 2k+3
k ∞ 1 1 = − 5 1 + k=0 =
∞ k 1 k=0
3
−
1 5
=
∞ k 2 k=0
3
∞
=
1 1 1 1 = · 8 2k 8 1− k=0
geometric series with
5 6
a=8
1 2
=
=
1 3 3 1 − = −3=− 1 − 1/3 1 − 2/3 2 2
1 4
and
r=
2 , 3
sum =
a = 24 1−r
633
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SECTION 12.2 k ∞ ∞ ∞ 3 3 3 1 3 3k−1 3k+1 = = = 7· = 3 3k+1 3k+7 7 3 4 4 4 4 4 1 − 43 15, 616 k=2 k=0 k=0
3 ; 15, 616
11. Let x = 0. a1 a2 · · · an a1 a2 · · · an · · · . x=
Then
∞ a1 a2 · · · an
(10n )k
k=1
k ∞ 1 = a1 a2 · · · an 10n k=1 1 a1 a2 · · · an = a1 a2 · · · an − 1 = . 1 − 1/10n 10n − 1
12. (a) Denote the partial sums of the first series by sn and those of the second series by tn and observe that sn = (a0 + a1 + · · · + aj ) + tn .
Obviously
sn → L
iff
tn = sn − (a0 + a1 + · · · + aj ) → L − (a0 + a1 + · · · + aj ). Parts (b) and (c) follow from this equation. 13.
14.
15.
16.
17.
∞
∞
k=0
k=0
1 1 (−x)k = (−1)k xk = = 1+x 1 − (−x) ∞
∞
k=0
k=0
1 1 = = (−x2 )k = (−1)k x2k 2 2 1+x 1 − (−x ) x =x 1−x
1 1−x
=x
∞
(xk ) =
k=0
∞
xk+1
k=0
∞
∞
k=0
k=0
x 1 (−x)k = (−1)k xk+1 =x· =x 1+x 1 − (−x) ∞ ∞ 1 x 2 k = x = x (−x ) = (−1)k x2k+1 1 + x2 1 − (−x2 ) k=0
18.
k=0
∞
∞
k=0
k=0
x x 2 k = x = (−4x ) = (−1)k (2x)2k+1 1 + 4x2 1 − (−4x2 ) ∞
19. 1 +
3 9 27 81 + + + + ··· = 2 4 8 16
k=0
k 3 2
This is a geometric series with x = 1 20. ak = 4 21.
lim
k→∞
−5 4
k+1 k
k does not go to zero
k = e = 0
3 2
> 1. Therefore the series diverges.
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SECTION 12.2 22. ak =
k k−2 = 3k
k k 1 2k · 2 > 2 3 k k
for k > 6,
635
ak → ∞
so
23. Rebounds to half its previous height: ∞ 1 3 3 3 3 6 s = 6 + 3 + 3 + + + + + ··· = 6 + 6 =6+ 2 2 4 4 2k 1 − k=0
1 2
= 18 ft.
2 ∞ k h h h 11h + 2h = 21 + · · · = 6 + 2h =6+ 24. s = 6 + 2h + 2h 6 6 6 6−h k=0
=⇒ 11h = 15(6 − h) =⇒ h =
45 13
25. A principal x deposited now at r% interest compounded annually will grow in k years to r k x 1+ . 100 This means that in order to be able to withdraw nk dollars after k years one must place r −k nk 1 + 100 dollars on deposit today. To extend this process in perpetuity as described in the text, the total deposit must be ∞ k=1
26. (a)
∞
5000
k=1
(b) 800 (c)
1
r −k nk 1 + . 100
k−1 k−1 ∞ 1 1 5000 5000 1 ∼ (1.05)−k = = · 1 = $9090.91 2 1.05 2(1.05) 1.05 1 − 2·1 k=1
0.8 1.06 0.8 − 1.06
∼ = $2461.54
1 N · 1 = 20N 1.05 1 − 1.05
n ∞ 9 = 27. 10 n=1
9 10 1−
9 10
= 9 or $9 k 2 1 1 = · 3 3 1− k=0 ∞
28. Total length removed = Some points: 29.
4 1 1 2 + + + ··· = 3 9 27 3
2 3
=1
0, 1, 13 , 23 , 19 , 29 , 79 , 89 . n 2 √ √ 1 A = 42 + (2 2)2 + 22 + ( 2)2 + 12 + · · · + 4 √ + ··· 2 n 2 ∞ ∞ n 1 1 1 4 √ = = 16 = 16 · 1 = 32 2 1 − 2 2 n=0 n=0
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SECTION 12.2
30. (a) If
(ak + bk )
converges, then
(b) If ak = bk = 2k ,
(c) If ak = 2k , bk = −2k ,
ak ,
ak ,
bk and
bk =
(ak + bk ) −
ak
(ak + bk ) diverge, but
bk and
would also converge.
(ak − bk ) = 0 converges.
(ak − bk ) diverge, but
(ak + bk ) =
0
converges. ∞
31. Let L =
ak . Then
k=0
L=
∞
ak =
k=0
n k=0
∞
ak +
ak = sn + Rn .
k=n+1
Therefore, Rn = L − sn and since sn → L as n → ∞, it follows that Rn → 0 as n → ∞. 1 1 diverges, hence diverges. ak ak 1 √ ak diverges and diverges (Example 5) (b) If ak = k, then ak 1 1 ak diverges and = converges. If ak = 2k , then ak 2k
32. (a) By convergence,
33. sn =
n
ln
k=1
34. ak =
k k+1
35. (a) sn =
n
k+1 k
k
ak → 0,
so
= [ ln(n + 1) − ln (n)] + [ ln n − ln(n − 1)] + · · · + [ ln 2 − ln 1] = ln(n + 1) → ∞ ⎛
⎜ =⎝
⎞k 1 ⎟ 1 → = 0 k+1⎠ e k
(dk − dk+1 ) = d1 − dn+1 → d1
k=1
(b) We use part (a). √ ∞ √ ∞ k+1− k 1 1 √ −√ (i) =1 = k+1 k k(k + 1) k=1 k=1 (ii)
∞ k=1
∞ 2k + 1 1 1 1 1 = = − 2 2 2 2 2k (k + 1) 2 k (k + 1) 2 k=1
36. Use induction to verify the hint. Then sn =
1 − (n + 1)xn + nxn+1 1 → 2 (1 − x) (1 − x)2
Since −(n + 1)xn + nxn+1 → 0 for |x| < 1. This last statement follows from observing that nxn → 0.
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SECTION 12.2 To see this, choose > 0 so that (1 + )|x| < 1.
Since n1/n → 1, there exists k so that
n1/n < 1 + for n ≥ k. Then for n ≥ k |nxn | = |n1/n x|n ≤ ((1 + )|x|)n → 0 n+1 1 ∞ 1 1 4 = = ; 37. Rn = k 1 4 3 · 4n k=n+1 1− 4 1 < 0.0001 =⇒ 4n > 3333.33 3 · 4n
=⇒
n>
ln 3333.33 ∼ = 5.85 ln 4
Take N = 6. ∞
38. Rn =
k
(0.9) = (0.9)
k=n+1 ∞
39. Rn = 1 2
k=n+1
n+1
1 = 9(0.9)n < 0.0001 1 − 0.9
=⇒
n≥
ln
< 0.0001 =⇒ n ≥ 9999. Take N = 9999.
2 3
n ln 0.0001 2 2 ∼ =2 < 0.0001 =⇒ n ≥ = 25 3 ln 23
∞ k xn+1 |x|n+1 = ; x = 41. |Rn | = 1 − x 1−x k=n+1
|x|n+1
n>
ln (1 − x) ln |x|
[recall ln |x| < 0]
ln (1 − x) −1 ln |x|
Take N to be smallest integer which is greater than
ln (1 − x) . ln |x|
42. sn = an − a1 . Thus {sn } converges iff {an } converges. (ak+1 − ak ) converges iff {an } converges.
k=1
∼ = 109
k=n+1
k n+1 ∞ 2 2 1 = 40. Rn = 3 3 1− k=n+1
∞
9
ln 0.9
∞ 1 1 1 1 1 1 1 = − = + ; k(k + 2) 2 k k+2 2 n+1 n+2
1 1 + n+1 n+2
Hence
0.0001
637
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SECTION 12.3
SECTION 12.3 1.
converges;
basic comparison with
1 k2
2.
diverges;
3.
converges;
basic comparison with
1 k2
4.
diverges; basic comparison with
5.
diverges; basic comparison with
6.
converges;
basic comparison with
7.
diverges; limit comparison with
8.
converges;
geometric with x =
9. converges;
∞
integral test, 1
1 k+1
1 k
1 k
limit comparison with
1 k 1 k2
2 5
b 1 tan−1 x 3π 2 −1 2 dx = lim x) = (tan b→∞ 2 1 + x2 32 1
10.
converges;
basic comparison with
1 k2
11.
diverges; p-series with p =
2 3
12.
converges;
basic comparison with
1 k3
13.
diverges; divergence test,
( 34 )−k → 0
14.
diverges;
15.
diverges; basic comparison with
basic comparison with
16. converges;
∞
integral test, 2
17. diverges;
1 1 + 2k
1 1 → = 0 −k 2+3 2
divergence test,
limit comparison with
1 k4
19.
limit comparison with
1 k2
21. diverges;
integral test, 2
∞
20.
diverges;
ak → 0.
dx = lim [ ln (ln x)]b2 = ∞ x ln x b→∞
22.
converges;
24.
diverges;
limit comparison with
26.
diverges;
limit comparison with
28. diverges;
1 k
∞ −1 dx 1 = lim = 2 b→∞ ln x 2 x(ln x) ln 2
18. converges; converges;
≤1
limit comparison with
1 2k
23.
converges;
1 k
25.
diverges; limit comparison with
1 √ k
27.
converges;
limit comparison with
1 k
limit comparison with
2k 5k
1 k
limit comparison with
1 k 3/2
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SECTION 12.3 29. converges;
∞
integral test, 1
30. converges;
integral test:
b 1 1 2 2 xe−x dx = lim − e−x = b→∞ 2 2e 1
∞
2 −x3
x 2
dx = lim
b→∞
1
31. converges; 32. diverges;
basic comparison with basic comparison with
639
3 , k2
2−x −3 ln 2 3
∞ = 1
1 6 ln 2
2 + sin k ≤ 3 for all k.
1 √ , k
2 + cos k 1 √ >√ k+1 k
k(k + 1) . Therefore 2 2 1 = . This series converges; 1 + 2 + 3 + ··· + k k(k + 1)
33. Recall that 1 + 2 + 3 + · · · + k =
34.
n 1+
22
+
32
+ ··· +
n2
=
1 6 n(n
n + 1)(n + 2)
direct comparison with
converges: limit comparison with
2 k2 1 n2
35. converges;
basic comparison with
1 1 2k 1 = < : 2 k (2k)! (2k − 1)(2k − 2) · · · 3 · 2 · 1 k2
36. converges;
basic comparison with
1 1 2k! 1 = < : 2 k (2k)! k(2k − 1)(2k − 2) · · · (k + 1) k2
37. Use the integral test: 1 Let u = ln x, du = dx : x
∞
1
1 dx = x(ln x)p
1 dx = lim b→∞ x(ln x)p
1
u−p du = b
u1−p + C. 1−p
1 1 (ln a)1−p dx = lim p b→∞ 1 − p x(ln x)
The series converges for p > 1. ln k
1 diverges. kp p−1 p−1 > 0, so for large k, ln k < k 2 If p > 1, then 2 p−1 1 ln k k 2 1 p+1 Then < = p+1 . Since > 1, p+1 p p k k 2 k 2 k 2 ln k . so converges iff p > 1. Hence so does kp
38. If p ≤ 1,
kp
>
b 1 −αx 1 converges. e dx = lim − e = b→∞ α α 0 0 b ∞ x 1 1 xe−αx dx = lim − −αx − 2 e−αx = 2 b→∞ αe α α 0 0
39. (a) Use the integral test: (b) Use the integral test:
converges
∞
−αx
converges.
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SECTION 12.3 (c) The proof follows by induction using parts (a) and (b) and the reduction formula xn eax dx =
∞
40. n+1
=⇒
41. (a)
n xn eax − a a
xn−1 eax dx
[see Exercise 67, Section 8.2]
∞ ∞ dx 1 dx < < p xp kp n x k=n+1
∞
n
k=1
k=1
1 1 1 1 < − < p−1 p p (p − 1)(n + 1) k k (p − 1)np−1 4 1 ∼ = 1.1777 k3
(b)
k=1
1 1 < R4 < 2 2·5 2 · 42
0.02 < R4 < 0.0313 ∞ 1 (c) 1.1777 + 0.02 = 1.1977 < < 1.1777 + 0.0313 = 1.2090 k3 k=1
4 1 1 1 1 =1+ 4 + 4 + 4 ∼ 42. (a) = 1.0788 k4 2 3 4 k=1
(b)
1 1 < R4 < =⇒ 0.0027 < R4 < 0.0052 3 3(5) 3(4)3
(c) 1.0815
3333
=⇒
n > 14.94 : Take n = 15.
n3 > 333.33
=⇒
n > 6.93 : Take n = 7.
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SECTION 12.3 1 < 0.0001 =⇒ n ≥ 8 4n4 1 < 0.001 =⇒ n3 > 333.33 (b) Rn < (4)n4
641
46. (a) Rn
3.97 : Take n = 4.
∞ 4 1 ∼ 1 ∼ = = 1.0363 k5 k5
k=1
k=1
47. (a) If ak /bk → 0, then ak /bk < 1 for all k ≥ K for some K. But then ak < bk for all k ≥ K and, since bk converges, ak converges. [ The Basic Comparison Theorem 12.3.6.] (b) Similar to (a) except that this time we appeal to part (ii) of Theorem 12.3.6. 1 1 1/k 2 1 converges, bk = converges, = √ →0 (c) ak = 2 3/2 k k 1/k 3/2 k 2 1 1 1/k 1 √ diverges, √ = 3/2 → 0 ak = converges, bk = 2 k k k 1/ k (d)
1 √ diverges, k 1 √ diverges, bk = k
bk =
48. (a) Since ak /bk → ∞,
1 1/k 2 1 √ = 3/2 → 0 converges, k2 k 1/ k 1 1/k 1 √ = √ →0 diverges, ak = k 1/ k k
ak =
bk /ak → 0,
so this follows from Exercise 45(b)
(b) Follows from Exercise 45(a) (c)
(d)
1 √ diverges, ak = k
ak =
1 √ diverges, k
bk =
1 converges, k2
bk =
1 converges, k2
1 converges, bk = k2
√ 1/ k = k 3/2 → ∞ 1/(k 2 ) √ 1/ k √ = k→∞ 1/k
bk =
1 diverges, k
ak =
1 converges, k 3/2
ak =
1 √ diverges, k
√ 1/k 3/2 = k→∞ 1/(k 2 ) √ 1/ k = k 3/2 → ∞ 1/(k 2 )
ak converges, ak → 0. Therefore there exists a positive integer N such that 0 < ak < 1
2 for k ≥ N. Thus, for k ≥ N, a2k < ak and so ak converges by the comparison test.
(b) ak may either converge or diverge: 1/k 4 and 1/k 2 both converge; 1/k 2 converges
and 1/k diverges.
49. (a) Since
2 2 1 1 1 < ak 2 + 2 , ak − converges by comparison with ak − k k k 2 1 ak 1 1 2 2 + But a − = a − 2 , ak + k k k2 k k k2 ak so must converge. k
50. Since 0
tan − 0.001 = 1000 x2 + 1 2 2
xe−x dx = lim 2
b→∞
n
b
xe−x dx 2
n
1 2 = lim − e−x b→∞ 2 1 −n2 e < 0.001 2
∞
f (k)
100 if ln(n + 1) > 100 =⇒ n + 1 > e100 ∼ = 2.7 × 1043 k 1 55. Set f (x) = x1/4 − ln x. Then f (x) =
1 1/4 1 −3/4 1 x (x − 4). − = 4 x 4x
Since f (e12 ) = e3 − 12 > 0 and f (x) > 0 for x > e12 , we have that n1/4 > ln n for sufficiently large n. Since
and therefore
1 n5/4
>
ln n n3/2
1 ln n is a convergent p-series, converges 5/4 n n3/2
by the basic comparison test. 56. The series converges if deg q ≥ deg p + 2 and diverges otherwise.
SECTION 12.4 1. converges;
ratio test:
10 ak+1 = →0 ak k+1
root test: (ak )1/k =
3.
converges;
5.
diverges; ratio test:
7.
diverges; limit comparison with
9.
converges; root test: (ak )1/k =
1 →0 k
ak+1 k+1 = →∞ ak 100 1 k
2 1/k 2 → k 3 3
2. converges; root test:
4.
converges;
6.
diverges;
8.
converges;
10.
diverges;
root test:
1 k2k
1/k
1/k
ak
comparison with root test
=
=
1 k → 2k + 1 2
1 k
(ak )1/k =
comparison with
1 1 → 1/k 2 2k
1 →0 ln k
1 k
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SECTION 12.4 11.
diverges; limit comparison with
13.
diverges; ratio test:
15. converges;
16. converges;
1 √ k
ak+1 k+1 = →∞ ak 104
basic comparison with
ratio test,
12.
converges;
14.
converges;
limit comparison with
root test: (ak )1/k =
1 k 3/2
ak+1 2k+1 (k + 1)! k k =2 = · ak (k + 1)k+1 2k k!
k k+1
k
2 2 = → k+1 k e k
1 17. converges; basic comparison with k2 ∞ dx 2 18. converges; integral test =√ 3/2 x(ln x) ln 2 2 ∞ b 1 (ln x)−1/2 dx = lim 2(ln x)1/2 = ∞ 19. diverges; integral test: b→∞ 2 x 2 20. converges; 21. diverges;
22. converges; 23.
1 k 3/2 k −k k 100 = 1+ → e−100 = 0 k + 100 k
limit comparison with divergence test:
ratio test:
1 (k + 1)2 [(k + 1)!]2 (2k)! · → = 2 (2k + 2)! (k!) (2k + 1)(2k + 2) 4
diverges; limit comparison with
1 k
24.
25. converges; ratio test:
1 ln (k + 1) ak+1 → = ak e ln k e
26. converges; ratio test:
(k + 1)! 1 kk 1 = · k → k+1 k+1 (k + 1) k! e
diverges;
ak → 0
k
27. converges;
basic comparison with
1 k 3/2
1 · 3 . . . · (2k − 1) k+1 1 (k + 1)! · = → 1 · 3 · . . . · (2k + 1) k! 2k + 1 2
28. converges:
ratio test
29. converges;
ratio test:
30. converges;
root test: (ak )1/k =
31. converges;
ratio test:
2(k + 1) ak+1 →0 = ak (2k + 1)(2k + 2) 4 (2k + 1)2 → 5k 2 + 1 5
4 (k + 1)(2k + 1)(2k + 2) ak+1 → = ak (3k + 1)(3k + 2)(3k + 3) 27
643 1 k2
k 2/k 1 → e e
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SECTION 12.4
32. converges by Exercise 38, section 11.2 33. converges;
34. diverges:
ak+1 1 ratio test: = ak (k + 1)1/2 ak =
k+1 k
k/2 →0·
√
e=0
k k → 0 9
35. converges;
root test: (ak )1/k =
√ √ k− k−1= √
36. converges;
root test: (ak )1/k =
k →0 3k
k+
1 √
k+1
→0
∞
37.
2k 1 2 4 8 + 2 + 3 + 4 + ··· = 2 3 4 5 (k + 2)k+1 k=0
converges; 38. converges:
root test: (ak )1/k =
2 →0 (k + 2)1+1/k
ratio test (see Exercise 28) ∞
39.
1 · 3 · · · (1 + 2k) 1 1·3 1·3·5 + + + ··· = 4 4 · 7 4 · 7 · 10 4 · 7 · · · (4 + 3k) k=0
converges;
ratio test:
2 3 + 2k ak+1 → = ak 7 + 3k 3
40. converges; ratio test : 41. By the hint
2 · 4 · 6 · . . . · 2(k + 1) 3 · 7 · 11 · . . . · (4k − 1) 2k + 2 1 ak+1 = · = → ak 3 · 7 · 11 · . . . · [4(k + 1) − 1] 2 · 4 · 6 · . . . · 2k 4k + 3 2
2 k k−1 ∞ ∞ 1 1 1 1 1 10 . k = k = = 10 10 10 10 1 − 1/10 81
k=1
42. (a) If λ > 1,
k=1
then for k sufficiently large ak+1 >1 ak
and thus ak+1 > ak
This shows that the kth term cannot tend to 0 and thus the series cannot converge. (b)
1 ak+1 k diverges, →1 = k ak k+1 1 ak+1 k2 converges, = →1 2 k ak (k + 1)2
43. The series
∞ k! kk
k=0
k! = 0 by Theorem 12.2.5. k→∞ k k
converges (see Exercise 26). Therefore, lim
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SECTION 12.5 44.
rn+1 r n! = · → 0, (n + 1)! rn n+1
rn
so by ratio test
converges, and therefore
n!
rn →0 n!
45. Use the ratio test: [(k + 1)!]2 (pk)! (k + 1)2 [p(k + 1)]! 2 = = = (k + 1) (pk)!(pk + 1) · · · (pk + p) (pk + 1) · · · (pk + p) (k!)2 (pk)!
ak+1 ak
Thus ak+1 ak
⎧ ⎨ 1, → 4 ⎩ 0,
if p = 2 if p > 2.
The series converges for all p ≥ 2. 46. By root test:
(ak )1/k =
If r = 1, we get 47. Set bk = ak rk .
r
=
r
r k 1/k
k r/k
→r
converges if r < 1,
diverges if r > 1.
1 , which diverges. k
If (ak )1/k → ρ
and
ρ
k+1 k < 2 ; (k + 1)2 + 1 k +1
1 1 1 √ = 2 2 k+1 k+1 √
(a p-series with p < 1)
Theorem 12.5.3
14. (a) does not converge absolutely: (b)
k2
k k 1 > 2 = , +1 2k 2k
converges conditionally;
comparison with
Theorem 12.5.3.
15. converges absolutely (terms already positive); basic comparison, π π π 1 ≤ = (| sin x| ≤ |x|) sin 4k 2 4k 2 4 k2 1 1 > 16. (a) does not converges absolutely: k+1 k(k + 1) (b) converges conditionally by Theorem 12.5.3 17. converges absolutely;
ratio test,
1 ak+1 k+1 → = ak 2k 2
1 2k
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SECTION 12.5 1 √ 18. terms all positive, converges absolutely: ak = √ √ , √ k k + 1( k + k + 1) 19. (a) does not converge absolutely; (b) converges conditionally;
limit comparison with
Theorem 12.5.3 k k+2 1 > 2 = , k2 + k 2k 2k Theorem 12.5.3.
(b) converges conditionally; ak =
4k−2 1 = k e 16
1 k 3/2
1 k
comparison with
20. (a) does not converge absolutely:
21. diverges;
comparison with
647
1 2k
k 4 → 0 e
∞
22. converges absolutely by integral test:
x2 2−x dx
converges
1
sin(1/k) → 1 = 0 1/k
23. diverges;
ak = k sin(1/k) =
24. diverges:
k ak+1 (k + 1)k+1 k! k+1 = · k = > 1, ak (k + 1)! k k
25. converges absolutely; 26. (a)
cos πk k
=
ratio test,
(−1)k k
(b) converges conditionally;
so
ak → 0
ak+1 1 (k + 1)e−(k+1) k+1 1 → = = ak k e−k k e e
does not converge absolutely. Theorem 12.5.3.
1 cos πk (−1)k = (−1)k = k k k sin(πk/2) < 1 √ 28. Converges absolutely; |ak | = k k k 3/2 27. diverges;
(−1)k
29. converges absolutely;
30. The series
basic comparison
1 1 − 3k + 2 3k + 3
sin(πk/4) 1 ≤ k2 k2
=
1 (3k + 2)(3k + 3)
converges by comparison with
1 1 1 − − converged, then 3k + 2 3k + 3 3k + 4 1 1 1 1 1 1 = − − − − 3k + 4 3k + 2 3k + 3 3k + 2 3k + 3 3k + 4 If
would converge, which is
not the case. 31.
diverges;
ak → 0
32.
error < a21 =
1 . k2
1 ∼ = 0.04762 21
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SECTION 12.5
33. Use (12.5.4);
1 |s − s80 | < a81 = √ ∼ = 0.1104 82
35. Use (12.5.4);
|s − s9 | < a10 =
36. error < an+1 =
1 10n+1
37.
10 ; 11
38.
(0.9)N +1 < 0.001 N +1
(a)
1 = 0.00001 105
1 = 0.001 103 1
10n+1
< 10−3 =⇒ n ≥ 3
geometric series with a = 1 and r = −
1 , 10
sum =
(b)
1 10n+1
< 10−4 =⇒ n ≥ 4
a 10 = 1−r 11
N ≥ 32
|s − sn | < an+1 = √
39. Use (12.5.4);
34. error < a5 =
1 < 0.005 n+2
=⇒
n ≥ 39, 998
40. The series diverges because among the partial sums are all sums of the form 1 1 1 1 + + + ··· + 2 3 4 n Thus for instance, s1 =
1 , 2
s5 =
1 1 + , 2 3
s11 =
1 1 1 + + , 2 3 4
and so on.
This does not violate the theorem on alternating series because, in the notation of the theorem, it is not true that {ak } decreases. 41. Use (12.5.4). (a) n = 4; (b) n = 6;
1 < 0.01 =⇒ 100 < (n + 1)! (n + 1)! 1 < 0.001 =⇒ 1000 < (n + 1)! (n + 1)!
42. Yes. This can be shown by making slight changes in the proof of Theorem 12.5.3. The even partial sums s2m are now nonnegative. Since s2m+2 ≤ s2m ,
the sequence converges; say,
Since s2m+1 = s2m − a2m+1 and a2m+1 → 0, we have s2m+1 → l. Thus, 43. No.
s2m → l.
sn → l.
For instance, set a2k = 2/k and a2k+1 = 1/k.
44. If ak is absolutely convergent, then |ak | converges. Therefore |bk | by comparison with
|ak |. Thus bk is absolutely convergent. 45. (a) Since (b)
|ak | converges,
|ak |2 =
a2k converges (Exercise 49, Section 12.3).
1 1 converges, (−1)k is not absolutely convergent. 2 k k
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SECTION 12.6
649
46. s2m+1 = a0 − a1 + a2 − a3 + a4 + · · · − a2m+1 = a0 + (−a1 + a2 ) + (−a3 + a4 ) + · · · + (−a2m−1 + a2m ) − a2m+1 = a0 + negative terms. Then s2m+1 < a0 :
and {s2m+1 } is bounded above.
s2m+3 = s2m+1 + (a2m+2 − a2m+3 ) > s2m+1 ,
thus {s2m+1 } is increasing.
47. See the proof of Theorem 12.8.2. 48. (a)
∞ (−1)k−1 (a + b) + (a − b)
2k
k=1
=
∞ (−1)k−1 (a + b) k=1
2k
+
∞ a−b k=1
2k
(b) The series is absolutely convergent if a = b = 0; conditionally convergent if a = b = 0; divergent if a = b.
SECTION 12.6 1.
−1 + x + 12 x2 −
2.
1 + 12 x − 18 x2 +
3.
− 12 x2 −
4.
1 + 12 x2 +
5.
1 − x + x2 − x3 + x4 − x5
6.
x + x2 + 13 x3 −
7.
x + 13 x3 +
8.
x − 12 x5
1 4 24 x
1 4 12 x
2 5 15 x
9. P0 (x) = 1,
P1 (x) = 1 − x,
10. P0 (x) = 1,
P1 (x) = 1 + 3x,
11.
n
(−1)k
k=0
13.
P2 (x) = 1 − x + 3x2 , P2 (x) = 1 + 3x + 3x2 ,
xk k!
12.
m x2k n where m = and n is even (2k)! 2
14.
k=0
(bx)2k
1 5 30 x
P3 (x) = 1 + 3x + 3x2 + x3 m x2k+1 n−1 where m = and n is odd. (2k + 1)! 2
−
n xk k=1
k n rk k=0
(2k)!
where m =
n and n is even. 2
17. |f (1/2) − P5 (1/2)| = |R5 (1/2)| ≤ (1) 18. |f (−2) − P7 (−2)| = |R7 (−2)| ≤
1 (1/2)6 = 6 < 0.00002 6! 2 6!
28 < 0.00635 8!
5 4 128 x
P3 (x) = 1 − x + 3x2 + 5x3
15. f (k) (x) = rk erx and f (k) (0) = rk , k = 0, 1, 2, . . . . Thus, Pn (x) = m (−1)k
−
5 4 24 x
k=0
k=0
16.
1 3 16 x
k!
xk
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SECTION 12.6 2n+1 2n+1 = ; (n + 1)! (n + 1)!
19. |f (2) − Pn (2)| = |Rn (2)| ≤ (1)
the least integer n that satisfies the inequality
n+1
2 < 0.001 (n + 1)!
is n = 9.
20. |f (−4) − Pn (−4)| = |Rn (−4)| ≤ (1) inequality
4n+1 < 0.001 (n + 1)!
4n+1 4n+1 = ; (n + 1)! (n + 1)!
is n = 14.
21. |f (1/2) − Pn (1/2)| = |Rn (1/2)| ≤ (3) inequality
3 < 0.00005 2n+1 (n + 1)!
22. |f (2) − Pn (2)| = |Rn (2)| ≤ (3) 3 · 2n+1 < 0.0005 (n + 1)!
the least integer n that satisfies the
3 (1/2)n+1 = n+1 ; (n + 1)! 2 (n + 1)!
the least integer n that satisfies the
is n = 9.
(2)n+1 3 · 2n+1 = ; (n + 1)! (n + 1)!
the least integer n that satisfies the inequality
is n = 10.
23. |f (x) − P5 (x)| = |R5 (x)| ≤ (3)
|x|6 |x|6 = ; 6! 240
24. |f (x) − P9 (x)| = |R9 (x)| ≤ (3)
|x|10 |x|10 = ; 10! 1, 209, 600
|x|10 < 0.05 1, 209, 600
Pn (0.5) = 1 + (0.5) +
(0.5)2 (0.5)n + ··· + 2! n!
|x| < 3.0072
|x|6 < 0.05 =⇒ |x|6 < 12 =⇒ |x| < 1.513 240 =⇒
|x|10 < 60480
=⇒
25. The Taylor polynomial
estimates e0.5 within |Rn+1 (0.5)| ≤ e0.5
|0.5|n+1 (0.5)n+1 0;
the right branch of the
r (t) = i + 2t j r (t) · r (t) = t i + 1 + t2 j · (i + 2t j) = t 2t2 + 3 = 0 =⇒ t = 0
r (t) and r (t) are perpendicular at (0, 1). (b) and (c) r (t) = α r (t) with α = 0
=⇒
t = α and 1 + t2 = 2tα
=⇒
t = ±1.
If α > 0, then t = 1. r (t) and r (t) have the same direction at (1, 2). If α < 0, then t = −1. r (t) and r (t) have opposite directions at (−1, 2). 12. r(t) = eαt (i + 2 j + 3 k) 13. The tangent line at t = t0 has the form R (u) = r (t0 ) + u r (t0 ). If r (t0 ) = α r (t0 ), then R (u) = r (t0 ) + u αr (t0 ) = (1 + uα) r (t0 ). The tangent line passes through the origin at u = −1/α. 14. r1 (0) = i, θ = cos−1
r2 (0) = 2 i + j + k r1 (0) · r2 (0) 2 = cos−1 √ ∼ = 0.62 radian, or 35.3◦ r1 (0) r2 (0) 6
15. r1 (t) passes through P (0, 0, 0) at t = 0; r2 (u) passes through P (0, 0, 0) at u = −1. 1 k; r1 (0) = i + 2 j + k r1 (t) = et i + 2 cos t j + t+1 r2 (u) = i + 2u j + 3u2 k; r2 (−1) = i − 2 j + 3 k cos θ =
r1 (0) · r2 (1) = 0; r1 (0) r2 (1)
θ=
π ∼ = 1.57, or 90◦ . 2
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SECTION 14.3
16. r1 (0) = −i, −1
θ = cos
r2 (−1) = i − 4 j − 8 k
r1 (0) · r2 (−1) = cos−1 r1 (0) r2 (−1)
17. r1 (t) = r2 (u) implies
−1 9
∼ = 1.68 radians, or 96.4◦
⎧ ⎪ ⎪ ⎨
⎫ et = u⎪ ⎪ ⎬ 1 2 sin t + 2 π = 2 ⎪ ⎪ ⎪ ⎩ t2 − 2 = u2 − 3⎪ ⎭
so that t = 0,
u = 1.
The point of intersection is (1, 2, −2).
π r1 (t) = et i + 2 cos t + j + 2t k, 2
r2 (u) = i + 2u k, cos θ = 18. (a)
r2 (1) = i + 2 k
r1 (0) · r2 (1) 1√ ∼ 5 = 0.447, = r1 (0) r2 (1) 5
θ∼ = 1.11 radians
R(u) = r(t0 ) + ur (t0 ) = (t0 i + f (t0 ) j) + u(i + f (t0 ) j) =⇒ x(u) = t0 + u,
(b)
r1 (0) = i
y(u) = f (t0 ) + uf (t0 )
From (a), we get u = x(u) − t0
y − f (t0 ) = f (t0 )(x − t0 ),
and y(u) − f (t0 ) = f (t0 )u. so
as expected. (b) r (t) = a cos t i − b sin t j
19. (a) r (t) = a cos t i + b sin t j
(d) r (t) = a cos 3t i − b sin 3t j
(c) r (t) = a cos 2t i + b sin 2t j 20. (a) r(t) = −a sin t i + b cos t j
(b) r(t) = a sin t i + b cos t j
(c) r(t) = −a sin 2t i + b cos 2t j
(d) r(t) = a sin 3t i + b cos 3t j
21.
r (t) = t3 i + 2t j
22.
r (t) = 2 i + 2t j
23.
r (t) = 2e2t i − 4e−4t j
24.
r (t) = cos t i + 2 sin t j
25.
r (t) = −2 sin t i + 3 cos t j
26.
r (t) = sec t tan t i + sec2 t j
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SECTION 14.3 27. r (t) = (t2 + 1) i + t j,
t ≥ 1;
or, r (t) = sec2 t i + tan t j,
28. r(t) = cos t(1 − cos t) i + sin t(1 − cos t) j, 29. r (t) = cos t sin 3t i + sin t sin 3t j,
t ∈
1
1 4 π, 2 π
733
t ∈ [0, 2π]
t ∈ [0, π]
t≤0
30. r(t) = t4 i + t3 j, 31. y 3 = x2
There is no tangent vector at the origin. 32. (a) r(0) = i + 2 j = r(1) r (0) 1 (b) T(0) = =√ (−i − j + π k), 2 r (0) π +2
T(1) =
1 r (1) =√ (i + 2 j − π k) 2 r (1) π +5
33. We substitute x = t, y = t2 , z = t3 in the plane equation to obtain 4t + 2t2 + t3 = 24, (t − 2) t2 + 4t + 12 = 0,
t = 2.
The twisted cubic intersects the plane at the tip of r(2), the point (2, 4, 8). The angle between the curve and the normal line at the point of intersection is the angle between the tangent vector r (2) = i + 4 j + 12 k and the normal N = 4 i + 2 j + k : cos θ =
(i + 4 j + 12 k) · (4 i + 2 j + k) 24 ∼ √ =√ = 0.412, i + 4 j + 12 k 4 i + 2 j + k 161 21 1
34. (a) T (t) =
a2
2
sin t + b2 cos2 t 1
θ∼ = 1.15 radians.
(−a sin t i + b cos t j)
(−b cos t i + a sin t j) N (t) = a2 sin2 t + b2 cos2 t √
2 (b) r(u) = (a i + b j) + u(−a i + b j), 2
35. r (t) = 2 j + 2t k, r (t) = 2 1 + t2
√ 2 R(u) = (a i + b j) + u(−b i − a j) 2
1 r (t) =√ (j + t k) , r (t) 1 + t2 1 [−t j + k] T (t) = (1 + t2 )3/2
T (t) =
at t = 1:
tip of r = (1, 2, 1),
1 1 T = T(1) = √ j + √ k; 2 2
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SECTION 14.3 T (1) 1 1 1 1 1 = −√ j + √ k T (1) = − √ j + √ k; T (1) = ; N = N(1) = 2 T (1) 2 2 2 2 2 2 normal for osculating plane:
1 1 1 1 1 T × N = √ j + √ k × −√ j + √ k = i 2 2 2 2 2 equation for osculating plane: 1 (x − 1) + 0(y − 2) + 0(z − 1) = 0, 2
36. T(t) = N(t) =
r (t) i + 2t j + 4t k = √ r (t) 20t2 + 1
x−1=0
which gives
1 T(1) = √ (i + 2 j + 4 k) 21
−20t i + 2 j + 4 k T (t) = √ T (t) 400t2 + 20
N(1) = √
1 1 (−20 i + 2 j + 4 k) = √ (−10 i + j + 2 k) 420 105
1 T(1) × N(1) = √ (−2 j + k). 5 Osculating plane at (1, 1, 2) :
−2(y − 1) + (z − 2) = 0 =⇒ −2y + z = 0
r (t) = −2 sin 2t i + 2 cos 2t j + k,
37.
T (t) =
r (t) =
√ 5
r (t) 1√ = 5 (−2 sin 2t i + 2 cos 2t j + k) r (t) 5
T (t) = −
4√ 5 (cos 2t i + sin 2t j), 5
T (t) =
4√ 5 5
T (t) = −(cos 2t i + sin 2t j) T (t) 1√ 5 (−2 i + k), N = −j at t = π/4 : tip of r = (0, 1, π/4), T = 5 normal for osculating plane: 1√ 1√ 2√ T×N = 5 (−2 i + k) × (−j) = 5i + 5k 5 5 5 equation for osculating plane: 1√ π 2√ π = 0, which gives x + 2z = 5 (x − 0) + 5 z− 5 5 4 2 N (t) =
38. T(t) = N(t) =
r (t) i + 2 j + 2t k = √ r (t) 4t2 + 5
1 T(2) = √ (i + 2 j + 4 k) 21
−2t i − 4t j + 5 k T (t) √ = T (t) 20t2 + 25
N(2) = √
1 (−4 i − 8 j + 5 k) 105
1 T(2) × N(2) = √ (2 i − j) 5 Osculating plane at (2, 4, 2) :
2(x − 2) − (y − 4) = 0 =⇒ 2x − y = 0
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SECTION 14.3 39.
r (t) = cosh t i + sinh t j + k, T (t) =
r (t) =
√ cosh2 t + sinh2 t + 1 = 2 cosh t
r (t) 1 = √ (i + tanh t j + sech t k) , r (t) 2
1 T (t) = √ sech2 t j − sech t tanh t k 2 1 1 at t = 0: tip of r = (0, 1, 0), T = √ (i + k), T (0) = √ j; N = j 2 2 normal for osculating plane:
1 1 √ (−i + k) × j = √ (−i + k) or i − k T×N = 2 2 equation for osculating plane:
40. T(t) =
x−z =0
−3 sin 3t i + j − 3 cos 3t k r (t) √ = r (t) 10
T (t) −9 cos 3t i + 9 sin 3t k = T (t) 9 π π ×N = 3j − k T 3 3
N(t) =
Osculating plane at (−1, π3 , 0) :
41.
T N
π 3
π
3(y − π3 ) − z = 0
3
1 = √ (j + 3 k) 10
=i
or 3y − z − π = 0
r (t) = et [(sin t + cos t) i + (cos t − sin t) j + k] , T (t) =
√ r (t) = et 3
r (t) 1 = √ [(sin t + cos t) i + (cos t − sin t) j + k] , r (t) 3
1 T (t) = √ [(cos t − sin t) i − (sin t + cos t) j] 3 1 at t = 0: tip of r = (0, 1, 1), T = T(0) = √ (i + j + k); 3 √ 2 1 1 T (0) = √ (i − j) T (0) = √ (i − j); T (0) = √ ; N = N(0) = T (0) 3 3 2 normal for osculating plane: 1 1 1 T × N = √ (i + j + k) × √ (i − j) = √ (i + j − 2 k) 3 2 6 equation for osculating plane: 1 2 1 √ (x − 0) + √ (y − 1) − √ (z − 1) = 0, 6 6 6
or
x + y − 2z + 1 = 0
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SECTION 14.4
√ π √2 2 r (t) t cos t i + t sin t j 42. T(t) = = = cos t i + sin t j T = i+ j r (t) t 4 2 2 √ √ π 2 2 T (t) = − sin t i + cos t j N =− i+ j N(t) = T (t) 4 2 2 π π ×N =k T 4 4 √ √ 2 2 π π Osculating plane at [1 + ], [1 − ], 2 : z − 2 = 0 or z = 2 2 4 2 4 43. T1 =
r (a + b − u) R (u) =− = −T. R (u) r (a + b − u)
Therefore T1 (u) = T (a + b − u) and N1 = N. √ √ 44. (a) r (t) = − 2 sin t i + 2 cos t j + k tangent line (t = π/4) :
x = 1 − t,
y = 1 + t,
z = 14 π + t
(c) Since 14 π + t > 0, t ∈ [0, 2π], the tangent line is never parallel to the x, y-plane. √ √ 45. (a) r (t) = − 2 sin t i + 2 cos t j + 5 cos t k
√ √ 2 5 2 − t tangent line (t = π/4) : x = 1 − t, y = 1 + t, z = − 2 2 (2n + 1)π (c) The tangent line is parallel to the x, y-plane at the points where t = , n = 0, 1, 2, . . . , 9. 10
√ √ 1 46. (a) r (t) = − 2 sin t i + 2 cos t j + k t tangent line (t = π/4) : x = 1 − t,
y = 1 + t,
z = ln(π/4) +
4 +t π
SECTION 14.4 1. r (t) = i + t1/2 j, L= 0
8
√
r (t) =
√ 1+t
2 (1 + t)3/2 1 + t dt = 3
2. r (t) = (t2 − 1)i + 2t j; 8 0
52 = 3
3. r (t) = −a sin t i + a cos t j + b k, r (t) = 2π
L= a2 + b2 dt = 2π a2 + b2
L=
2
(t2 + 1) dt =
0
√ a2 + b2
0
√ 4. r (t) = i + 2t1/2 j + t k; 2 L= (t + 1) dt = 4
r (t) = t + 1
0
√ 5. r (t) = i + tan t j, r (t) = 1 + tan2 t = | sec t| π/4 π/4 √ π/4 | sec t | dt = sec t dt = [ ln | sec t + tan t| ]0 = ln (1 + 2 ) L= 0
0
r (t) = t2 + 1 14 3
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SECTION 14.4 1 t 1 i+ j; r (t) = √ 1 + t2 1 + t2 1 + t2 1 1
√ dt √ L= = ln |t + t2 + 1| = ln(1 + 2) 2 0 1+t 0
6. r (t) =
√ √ 7. r (t) = 3t2 i + 2t j, r (t) = 9t4 + 4t2 = |t| 4 + 9t2 1 1 1 1 1 √ 2 3/2 2 2 4 + 9t 13 13 − 8 L= t 4 + 9t dt = = t 4 + 9t dt = 27 27 0 0 0
1 2 1 −2 1 1 t − t k; r (t) = t2 + t−2 2 2 2 2
3 1 2 1 −2 14 dt = t + t L= 2 2 3 1
8. r (t) = i +
9. r (t) = (cos t − sin t)et i + (sin t + cos t)et j, π√ √ L= 2 et dt = 2 (eπ − 1)
r (t) =
√ t 2e
0
10. r (t) = 2t i + 2t j − 2t k;
√ r (t) = 2t 3 =⇒ L =
2
√ √ 2t 3 dt = 4 3
0
1 1 i + 2 j + 2t k, r (t) = + 4 + 4t2 t t2
e e e 1 1 2 + 2t dt = ln |t| + t2 1 = e2 L= + 4 + 4t dt = 2 t t 1 1
11. r (t) =
12. r (t) = t cos t i − t sin t j;
r (t) = t =⇒ L =
2
t dt = 2 0
√ 13. r (t) = t cos t i + t sin t j + 3 t k, 2π 2π L= 2t dt = t2 = 4π 2 0
r (t) =
√ t2 cos2 t + t2 sin2 t + 3t2 = 4t2 = 2t
0
14. r (t) = (1 + t)1/2 i + (1 − t)1/2 j + 1/2 2 dt = 2 L=
√ 2 k,
r (t) =
√ (1 + t) + (1 − t) + 2 = 4 = 2
−1/2
√ 15. r (t) = 2 i + 2t j − 2t k, r (t) = 2 1 + 2t2 √ 2 √ tan−1 (2 2) 2 L= 2 1 + 2t dt = 2 sec3 u du 0
∧
0
√
(t 2 = tan u) tan−1 (2√2) √ √ √ = 12 2 sec u tan u + ln | sec u + tan u| = 6 + 12 2 ln (3 + 2 2 ) 0
737
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SECTION 14.4
16. r (t) = (3 cos t − 3t sin t)i + (3 sin t + 3t cos t)j + 4 k; r (t) = 9t2 + 25 4 4 25 25 25 25 3 3 25 3 t2 + dt = t t2 + + · ln t + t2 + = 26 + ln 5. L= 9 2 9 2 9 9 6 0 0 t s = s(t) = r (u) du 17. a
s (t) = r (t) = x (t) i + y (t) j + z (t) k
= [x (t)]2 + [y (t)]2 + [z (t)]2 . In the Leibniz notation this translates to
2 2 2 dx dy dz ds + + . = dt dt dt dt 18. Parameterize the graph of f by setting r(x) = x i + f (x) j, x ∈ [a, b]. x 1 + [f (t)]2 dt s = s(x) = 19. a
s (x) = In the Leibniz notation this translates to
ds = dx 20. L1 =
e
r (t) dt =
e
1
1
Setting t = ex ,
2 dt; t2 √ L1 = 2L2 2+
we get
1 + [f (x)]2 .
1+
1
dy dx
2 .
1 + e2x dx
L2 = 0
21. r (t) = − sin t i + cos t j. Since r ≡ 1, the parametrization is by arc length. 22. r (t) = −a + b; r = b − a. t s= b − a du = b − a t; t = 0
R(s) =
s 1− b − a
a+
s . b − a
s b, b − a
0 ≤ s ≤ b − a.
√ 23. r (t) = t sin t i + t cos t j + t k; r = t 2. √ t √ √ 2 2 t ; t = 21/4 s. u 2 du = s= 2 0 √ √ √ √ √ √ 1 R(s) = sin 21/4 s − 21/4 s cos 21/4 s i + cos 21/4 s + 21/4 s sin 21/4 s j + √ s k. 2 24. r (t) = (et cos t − et sin t) i + (et sin t + et cos t) j; r = t√ √ √ s= 2 eu du = 2(et − 1); t = ln (1 + s/ 2). 0
R(s) = ev [cos v i + sin v j]
√ t 2e .
√ √ where v = ln (1 + s/ 2), 0 ≤ s ≤ 2 (eπ − 1).
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SECTION 14.4 25. r (t) = t3/2 j + k, r (t) = 1/2 t3 + 1 dt ∼ s= = 0.5077
t3/2
2
+1=
√
739
t3 + 1
0
26. r (t) = i + t2 j;
r (t) =
√ 1 + t4
L=
2
1 + t4 ∼ = 3.6535
0
27. r (t) = −3 sin t i + 4 cos t j, r (t) = 9 sin2 t + 16 cos2 t dt 2π 9 sin2 t + 16 cos2 t dt ∼ s= = 22.0939 0
1 28. r (t) = i + 2tj + k; t
r (t) =
1+
4t2
1 + 2 t
4
1 + 4t2 +
L= 1
29. (a)
2π
(b) s =
1 dt ∼ = 15.4480 t2
1 + 16 cos2 4t dt ∼ = 17.6286
0
30. (a)
(b) s =
2π
1+
0
1 1+t
2
dt ∼ = 6.6818
PROJECT 14.4 1. Given the differentiable curve r = r(t), t ∈ I. Let t = φ(u) be a continuously differentiable one-to-one function that maps the interval J onto the interval I, and let R(u) = r(φ(u)), u ∈ J. Suppose that φ (u) > 0 on J. Then, as u increases across J, t = φ(u) increases across I. As a result, R(u) takes
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SECTION 14.5
on exactly the same values in exactly the same order as r. If φ (u) < 0 on J, then R(u) takes on exactly the same values as r but in the reverse order. 2. Let r = r(t), t ∈ I be a differentiable curve. Let t = φ(u) be a continuously differentiable one-to-one function that maps the interval J onto the interval I with φ (u) > 0. Set R(u) = r(φ(u)), u ∈ J. r (φ(u))φ (u) r (φ(u))φ (u) r (t) R (u) = = = . ||R (u)|| ||r (φ(u))φ (u)|| φ (u)||r (φ(u))|| ||r (t)|| ∧ since φ (u) > 0
R (u) = [r(φ(u))] = r (φ(u))φ (u);
Therefore, the unit tangent is left unchanged by a sense-preserving change of parameter. The invariance of the principal normal and the osculating plane follows from the invariance of the unit tangent. 3. Suppose that φ (u) < 0 on J. Then r (φ(u))φ (u) r (φ(u))φ (u) r (t) R (u) = =− =− . R (u) ||r (φ(u))φ (u)|| φ (u)||r (φ(u))|| ||r (t)|| ∧ since φ (u) < 0
R (u) = [r(φ(u))] = r (φ(u))φ (u);
Thus, the unit tangent is reversed by a sense-reversing change of parameter. That is, if TR and Tr are the respective unit tangents, then TR = −Tr . Now consider the principal normals: TR Tr (φ(u)φ (u) Tr (φ(u)φ (u) Tr (t) = − = = . || ||TR ||Tr (φ(u)φ (u)|| φ (u)||Tr (φ(u)φ (u)|| ||Tr (t)||
∧
since φ (u) < 0
Thus the principal normal is unchanged by a sense-reversing change of parameter. The osculating plane is also unchanged since T × N and −T × N are each normal to the osculating plane. 4. s =
t
{r(v)|| dv = ψ(t). ψ is an continuously differentiable increasing function on I; t = ψ −1 (s).
0
5. Let L be the length as computed from r and L∗ the length as computed from R. Then d d b L∗ = R (u) du = r (φ (u)) φ (u) du = r (t) dt = L. c
c
a
∧ t = φ (u)
SECTION 14.5 1. r (t) = a[cos θ(t) i + sin θ(t) j], r (t) = v
=⇒
a|θ (t)| = v
r (t) = a[− sin θ(t) i + cos θ(t) j)]θ (t) =⇒
r (t) = a[− cos θ(t) i − sin θ(t) j] [θ (t)]2 ,
|θ (t)| = v/a r (t) = a[θ (t)]2 = v2 /a
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SECTION 14.5 2. r (t) = (−πa sin πt + 2bt)i + (πa cos πt − 2bt)j r (t) = (−π 2 a cos πt + 2b)i + (−π 2 a sin πt − 2b)j
At t = 1, v = 2b i − (aπ + 2b)j, v = v = 4b2 + (aπ 2 + 2b)2
a = (aπ 2 + 2b)i − 2b j, a = 4b2 + (aπ 2 + 2b)2 3. r (t) = at i + b sin at j, r (t) = −a2 b sin at j,
= a i + ab cos at j r (t) = a2 |b sin at| = a2 |y(t)|
4. r (t) = 2t j + 2(t − 1) k;
speed is minimum when r (t) 2 is minimum 1 4t2 + 4(t − 1)2 = 8t2 − 8t + 4 is minimum at t = 2
5.
y = cos πx,
7.
x=
0≤x≤2
1 + y2 ,
y ≥ −1
6.
x = y3 ,
8.
x = sin πy,
all real x
0≤y≤2
9. (a) initial position is tip of r (0) = x0 i + y0 j + z0 k (b) r (t) = (α cos θ) j + (α sin θ − 32t) k, (c) |r (0)| = |α|
r (0) = (α cos θ) j + (α sin θ) k (d) r (t) = −32 k
(e) a parabolic arc from the parabola z = z0 + (tan θ ) (y − y0 ) − 16 in the plane x = x0
(y − y0 )2 α2 cos2 θ
741
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SECTION 14.5
10. (a) x(t) = 2 cos 2t = 4 cos2 t − 2 and y(t) = 3 cos t =⇒ x = −2 ≤ x(t) ≤ 2,
4 2 y − 2. 9
Since
−3 ≤ y(t) ≤ 3,
the path consists only of the bounded arc x=
4 2 y − 2, 9
−3 ≤ y ≤ 3.
The motion traces out this arc twice on every t-interval of length 2π. (b) included in part (d) r (t) = −4 sin 2t i − 3 sin t j,
(c) r(t) = 2 cos 2t i + 3 cos t j,
The velocity r (t) is 0 when t = nπ. the acceleration =
r (t) = −8 cos 2t i − 3 cos t j
At such points ! −8 i − 3 j, if n is even −8 i + 3 j,
if n is odd
(b) and (d)
r(t) = C iff r(t)2 = r(t) · r(t) = C
11.
iff
d r(t) = 2r(t) · r (t) = 0 dt
iff r(t) ⊥ r (t) 12. Problem 11 with v in place of r. 13. κ =
e−x (1 + e−2x )3/2
14. y = 3x2 ,
y = 6x;
κ=
6|x| (1 + 9x4 )3/2 −1/4x3/2 2 2 3/2 = (1 + 4x)3/2 1 + 1/2x1/2
1 15. y = 1/2 ; 2x
−1 y = 3/2 κ = 4x
16. y = 1 − 2x,
y = −2;
17. κ =
sec2 x (1 + tan2 x)
18. y = sec2 x,
3/2
κ=
√ 2 2 = [1 + (1 − 2x)2 ]3/2 2(1 − 2x + 2x2 )3/2
= | cos x |
y = 2 sec2 x tan x;
κ=
|2 sec2 x tan x| (1 + sec4 x)3/2
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SECTION 14.5 19. κ =
| sin x| (1 + cos2 x)3/2
x 20. 2x − 2yy = 0 =⇒ y = , y
21. κ =
|x| (1 +
22. y = x,
23. κ =
x4 /4)3/2
;
y = 1,
(1 +
1/y 2 )3/2
y =
4 2, 3
at
κ=
− 1/y 3
24. y = 4 cos 2x,
y2
1 ; (1 + x2 )3/2
(1 +
y 2 )3/2
;
y = −8 sin 2x;
a2 = − 3; y
1 , x+1
at (0, 0),
at (2, 2),
at x =
π , 4
κ=1
1 κ= √ 5 5 y = 0,
y (2) =
π , 4
y = sec x tan x = √ 2 3 2 =⇒ κ = = 3 (1 + 2)3/2
26. At x =
κ(x) =
− 1/x2 1/x2 )3/2
(1 + 1 − 2x2
=
a2 /y 3 a2 = κ= [1 + ( xy )2 ]3/2 (x2 + y 2 )3/2
2 κ= √ 5 5
,
1
=
y − x( xy )
1 −1 , ; y (x) = 3 (x + 1)2 1 − 9 3 √ At x = 2, κ = 2 3/2 = 10 10 1 1+ 3
25. y (x) =
27.
√
(x2 + 1)3/2
y (2) = −
1 . 9
y = sec x tan2 x + sec3 x =
2,
x
y = −8 =⇒ κ = 8
,
√
√ 2 + 23/2 = 3 2
x>0
1√ 2 2 (x2 + 1) √ √ √ Since κ increases on 0, 12 2 and decreases on 12 2, ∞ , κ is maximal at 12 2,
κ (x) =
5/2
28. y = 3 − 3x2 , 29. x(t) = t, 30. x = et , κ=
743
,
κ (x) = 0
y = −6x.
x (t) = 1, x = et ,
=⇒
x=
local max at x = 1 :
x (t) = 0; y = −e−t ,
y(t) = 12 t2 , y = e−t
|x y − x y | 2 = 2t [(x )2 + (y )2 ]3/2 (e + e−2t )3/2
y = 0, y (t) = t,
y = −6, y (t) = 1
κ=
1 2
ln 12 .
6 =6 (1 + 0)3/2
κ=
1 (1 + t2 )3/2
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SECTION 14.5
31. x(t) = 2t,
32. x = 2t,
x (t) = 2,
x = 2,
x (t) = 0;
y = 3t2 ,
y = 6t;
κ=
x (t) = et (cos t − sin t),
33. x(t) = et cos t,
y (t) = 3t2 ,
y(t) = t3 ,
y (t) = 6t;
κ=
12| t | (4 + 9t4 )3/2
|12t2 − 6t2 | 6 = (4t2 + 9t4 )3/2 |t|(4 + 9t2 )3/2
x (t) = −2 et sin t
y(t) = et sin t, y (t) = et (sin t + cos t), y (t) = 2 et cos t 2t 2e cos t (cos t − sin t) + 2e2t sin t (cos t + sin t) 2e2t 1 √ −t κ= = = 2e 3/2 3/2 2t 2 2t 2 2t 2 [ e (cos t − sin t) + e (cos t + sin t) ] (2e ) 34. x = −2 sin t,
x = −2 cos t,
y = 3 cos t,
35. x(t) = t cos t,
x (t) = cos t − t sin t,
y = −3 sin t;
κ=
6 (4 sin2 t + 9 cos2 t)3/2
x (t) = −2 sin t − t cos t
y(t) = t sin t, y (t) = sin t + t cos t, y (t) = 2 cos t − t sin t (cos t − t sin t)(2 cos t − t sin t) − (sin t + t cos t)(−2 sin t − t cos t) 2 + t2 κ= = [ (cos t − t sin t)2 + (sin t + t cos t)2 ]3/2 [1 + t2 ]3/2 36. x = t cos t, x = cos t − t sin t, y = t sin t, y = sin t + t cos t t cos t(sin t + t cos t) − t sin t(cos t − t sin t) t2 1 = κ = t3 = t . (t > 0) (t2 cos2 t + t2 sin2 t)3/2 37. κ =
2/x3 [1 + 1/x4 ]3/2
=
38. From Exercise 20,
2x3 (x4 + 1)3/2 κ=
;
at x = ±1,
1 . (x2 + y 2 )3/2
√ 2 κ= 2
At (±1, 0),
κ=1
39. We use (14.5.3) and the hint to obtain
a 2 b 2 y − x ab sinh2 t − ab cosh2 t b a κ= 3/2 = 2 3/2 2 2 2 2 a sinh t + b cosh t bx ay 2 + b a a b a3 b3 y 2 − x2 a4 b4 b a = = . 3/2 [a4 y 2 + b4 x2 ]3/2 a4 y 2 + b4 x2
40. x = r(1 − cos t),
x = r sin t,
y = r sin t,
Highest point when t = π =⇒ x = 2r, κ=
1 2r2 = 4r (4r2 )3/2
y = r cos t
x = 0,
y = 0,
y = −r
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SECTION 14.5 41. r (t) = et (cos t − sin t) i + et (sin t + cos t) j + et k √ ds d2 s √ t = r (t) = 3 et , = 3e dt dt2 r (t) 1 T (t) = = √ [(cos t − sin t) i + (sin t + cos t) j + k] r (t) 3
1 T (t) = √ [(− sin t − cos t) i + (cos t − sin t) j]; T (t) = 2/3 3
2 √ 2/3 ds 1 √ −t d2 s √ T (t) = 2e ; aT = 2 = 3 et , aN = κ = 2 et = √ κ= ds/dt 3 dt dt 3 et 42. r (t) = cosh t i + sinh t j + k;
√ ds = ||r || = 2 cosh t; dt
d2 s √ = 2 sinh t. dt2
r (t) 1 = √ (i + tanh t j + sech t k). ||r (t)|| 2 1 1 T (t) = √ sech2 t j − sech t tanh t k ; ||T (t)|| = √ sech t 2 2 2 2 √ ds d s 1 ||T (t)|| 2 aT = 2 = 2 sinh t, aN = κ =1 = sech t; κ= ds/dt 2 dt dt
T(t) =
43. r (t) = −2 sin 2t i + 2 cos 2t j; T (t) =
ds = ||r (t)|| = 2, dt
r (t) = − sin 2t i + cos 2t j ||r (t)||
T (t) = −2 (cos 2t i + sin 2t j) ; κ=
d2 s =0 dt2
2 ||T (t)|| = = 1; ds/dt 2
||T (t)|| = 2
aT =
d2 s = 0, dt2
aN = κ
ds dt
2 = 1 · 4 = 4.
1 a(t) = 2 j − 2 k t √
ds 4t4 + t2 + 1 4 1 = v = 1 + 4t2 + 1/t2 = ; v×a = − i + 2 j + 2k dt t t t √ t 4t4 + 16t2 + 1 ||v × a|| = κ= (ds/dt)3 (4t4 + t2 + 1)3/2 2 ds d2 s 4t4 − 1 1 4t4 + 16t2 + 1 aT = 2 = √ = ; aN = κ dt dt t 4t4 + t2 + 1 t2 4t4 + t2 + 1
44. r (t) = v(t) = i + 2t j +
1 k, t
ds = ||r || = 5; dt r (t) 3 4 3 T(t) = = − sin 3t i + j − cos 3t k ||r (t)|| 5 5 5 9 9 9 T (t) = − cos 3t i + sin 3t k; ||T (t)|| = 5 5 5
45. r (t) = −3 sin 3t i + 4 j − 3 cos 3t k;
κ=
9/5 9 ||T (t)|| = = ; ds/dt 5 25
aT =
d2 s = 0, dt2
d2 s = 0. dt2
aN = κ
ds dt
2 =
9 · 25 = 9. 25
745
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SECTION 14.5 √ ds = ||r (t)|| = 4t2 = 2t, dt √ r (t) T (t) = = 12 cos t i + 12 sin t j + 12 3 k ||r (t)| 1 T (t) = − 12 sin t i + 12 cos t j, T (t) = 2 T (t) 1/2 1 Then, κ = = = ds/dt 2t 4t 2 ds d2 s 1 = (4t2 ) = t. aT = 2 = 2, aN = κ dt dt 4t
√ 46. r (t) = t cos t i + t sin t j + t 3 k,
d2 s =2 dt2
ds = r (t) = (1 + t) + (1 − t) + 2 = 2, dt √ √ √ 1+t 1−t 2 r (t) = i− j+ k T (t) = r (t)| 2 2 2 1 1 T (t) = √ i+ √ j. 4 1+t 4 1−t 1 2 1 1 + = T (t) = 16(1 + t) 16(1 − t) 4 1 − t2 2 1 T (t) = Then, κ = ds/dt 8 1 − t2 2 ds 2 d2 s 1 aT = 2 = 0, aN = κ = . dt dt 2 1 − t2
47. r (t) =
√
1 + ti −
√ 48. (b) κ(t) =
√
1 − tj +
1 + 4t2 + t4
6 (1 + t2 + t4 )3/2
49. tangential component:
√
d2 s =0 dt2
2 k,
maximum curvature occurs at x ∼ = ±0.2715
6t + 12t3 aT = √ ; 1 + t2 + t4
normal component:
aN = 6
1 + 4t2 + t4 1 + t2 + t4
50. Set r(θ) = cos θf (θ)i + sin θf (θ)j 51. By Exercise 50
52. f (θ) = aθ,
f (θ) = a,
aθ 2 2 e + 2 aeaθ − eaθ a2 eaθ e−aθ √ κ= = . 3/2 2 2 1 + a2 (eaθ ) + (aeaθ )
f (θ) = 0 =⇒ κ =
a2 θ2 + 2a2 θ2 + 2 = (a2 θ2 + a2 )3/2 |a|(θ2 + 1)3/2
53. By Exercise 50, 2 a (1 − cos θ)2 + 2a2 sin2 θ − a2 (1 − cos θ)(cos θ) 3a2 (1 − cos θ) 3ar 3 κ= = = = √ . 3/2 3/2 3/2 2 2 [2ar] 2 2ar [2a (1 − cos θ)] a2 (1 − cos θ)2 + a2 sin θ
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SECTION 14.4 54. By Exercise 50, κ=
=
747
(a sin 2θ)2 + 2(2a cos 2θ)2 − (a sin 2θ)(2a cos 2θ) 3
[(a sin 2θ)2 + (2a cos 2θ)2 ] 2
2 2 a sin 2θ + 8a2 cos2 2θ − 3a2 sin 2θ cos 2θ 3
(a2 sin2 2θ + 4a2 cos2 2θ) 2 2 2 √ √ r + 8(a2 − r2 ) + 3r a2 − r2 8a − 7r2 + 3r a2 − r2 = = . 3 3 [r2 + 4(a2 − r2 )] 2 (4a2 − 3r2 ) 2 PROJECT 14.5A 1. The system of equations generated by the specified conditions is: a+b+c+d=3 27a + 9b + 3c + d = 7 6a + 2b = 0
27α + 9β + 3γ + δ = 7
729α + 81β + 9γ + δ = −2
54α + 2β = 0
27a + 6b + c = 27α + 6β + γ a∼ b∼ = −0.1094 = 0.3281
18a + 2b = 18α + 2β c∼ = 2.1094
α∼ = 0.0365
β∼ = −0.9844
γ∼ = 6.0469
d∼ = 0.6719 ∼ δ = −3.2656
2. Clearly p and q are continuous on their respective intervals. The conditions p(3) = q(3), p (3) = q (3) and p (3) = q (3) imply that F, F , and F are continuous on [1, 9].
3. (a), (b) The system of equations generated by the specified conditions (and the derivative conditions of Problem 1) is: 27α + 9b + 3c + d = 10
64α + 16b + 4c + d = 15
18a + 2b = 0
64α + 64β + 4γ + δ = 15
216α + 36β + 6γ + δ = 35
36α + 2β = 0
48α + 8b + c = 48α + 8β + γ
24α + 2b = 24α + 2β
(c) a = b = 0, c = 5, d = −5; α = 1.25, β = −15, γ = 65, δ = −85 4. κ =
Cn(n − 1)xn−2 |y | = 2 3/2 [1 + (y ) ] [1 + (Cnxn−1 )2 ]3/2
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SECTION 14.6
At x = 0, κ = 0 as desired. At x = 1, want κ = y=
1 3 x 4
Cn(n − 1) 96 = 125 (1 + C 2 n2 )3/2
works.
PROJECT 14.5B 1.
dT dT ds dT dT/dt T (t) T/(t) = =⇒ = = = κ N. dt ds dt ds ds/dt T (t) ds/dt
2.
dB d = (T × N) = ds ds
dT dN dN T× + ×N = T× . ds ds ds
Therefore
dB ⊥ T. ds
dB dB ⊥ B. Being perpendicular to both T and B, ds ds is parallel to N and is therefore a scalar multiple of N. Since B has constant length,
3.
dN d = (B × T) = ds ds
dT B× ds
+
dB ×T ds
= (B × κN) + τ (N × T)
= −κ(N × B) − τ (T × N) = −κT − τ B 4. We know that dB/ds = τ N.
It follows that dB/ds = |τ | N = |τ |.
Thus |τ | is the magnitude
of the change in direction of B per unit arc length, or equivalently, the rate per unit arc length at which the curve tends away from the osculating plane. SECTION 14.6 aω ωt bω ωt e − e−ωt i + e + e−ωt j, r (0) = bωj 2 2 aω 2 ωt bω 2 ωt (b) r (t) = e + e−ωt i + e − e−ωt j = ω 2 r (t) 2 2 (c) The torque τ is 0 : τ (t) = r (t) × ma (t) = r (t) × mω 2 r (t) = 0.
1. (a) r (t) =
The angular momentum L (t) is constant since L (t) = τ (t) = 0. 2. (a) F (t) = mr (t) = mb2 r(t), (b) F(t) = −mr(t),
mb2 > 0
−m < 0
(c) L(t) = r(t) × mv(t) = m(i + j − k) 3. We begin with the force equation F(t) = αk. In general, F (t) = m a (t), so that here a (t) =
α k. m
Integration gives v (t) = C1 i + C2 j +
α m
t + C3 k.
Since v (0) = 2 j, we can conclude that C1 = 0, C2 = 2, C3 = 0. Thus v (t) = 2 j +
α tk. m
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SECTION 14.6 Another integration gives
α t2 + D3 k. 2m Since r (0) = y0 j + z0 k, we have D1 = 0, D2 = y0 , D3 = z0 , and therefore α t2 + z0 k. r (t) = (2t + y0 ) j + 2m The conditions of the problem require that t be restricted to nonnegative values. r (t) = D1 i + (2t + D2 ) j +
To obtain an equation for the path in Cartesian coordinates, we write out the components α 2 t + z0 . (t ≥ 0) x(t) = 0, y(t) = 2t + y0 , z(t) = 2m From the second equation we have t=
1 2
[y(t) − y0 ].
(y(t) ≥ y0 )
Substituting this into the third equation, we get α (y(t) ≥ y0 ) [y(t) − y0 ]2 + z0 . z(t) = 8m Eliminating t altogether, we have α (y − y0 )2 + z0 . (y ≥ y0 ) z= 8m Since x = 0, the path of the object is a parabolic arc in the yz-plane. Answers to (a) through (d): α t k. m p(t) = 2m j + α t k.
(b)
(a) velocity: v(t) = 2 j + (c) momentum:
(d) path in vector form: r (t) = (2t + y0 ) j + path in Cartesian coordinates: z = 4. F(t) · v(t) = 0
speed: v(t) =
α t2 + z0 k, 2m
α (y − y0 )2 + z0 , 8m
1 2 4m + α2 t2 . m
t ≥ 0. y ≥ y0 ,
x = 0.
for all t
1 d [v(t) · v(t)] = 0 for all t 2 dt =⇒ v(t) · v(t) = [v(t)]2 is constant =⇒ v(t) is constant
=⇒ a(t) · v(t) = v (t) · v(t) =
5. F (t) = m a (t) = m r (t) = 2mk 6. If v (t) = 0, then L (t) = r (t) × mv(t) + r(t) × mv (t) = v(t) × mv(t) + r(t) × 0 = 0 7. From F (t) = m a (t) we obtain a (t) = π 2 [a cos πt i + b sin πt j]. By direct calculation using v (0) = −πbj + k and r (0) = bj we obtain v (t) = aπ sin πt i − bπ cos πt j + k r (t) = a(1 − cos πt) i + b(1 − sin πt) j + tk.
749
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SECTION 14.6 (b) v (1) =
(a) v (1) = bπj + k (c)
a (1) = −π 2 ai
(e)
L (1) = r (1) × m v(1) = [2ai + bj + k] × [m (bπj + k)]
√ π 2 b2 + 1
(d) m v(1) = m (πbj + k)
= m [b(1 − π)i − 2aj + 2abπk]
(f) τ (1) = r (1) × F (1) = [2ai + bj + k] × −mπ 2 ai = −mπ 2 a [j − bk] d dt
8.
1 m[v(t)]2 2
=
1 d 1 m [v(t) · v(t)] = m[2v (t) · v(t)] 2 dt 2
= ma(t) · v(t) = F(t) · v(t) 9. We have
mv = mv1 + mv2
and v2 =
1 1 2 2 2 mv = 2 mv1 v1 2 + v2 2 .
+ 12 mv2 2 .
Therefore
v = v1 + v2
Since
v2 = v · v = (v1 + v2 ) · (v1 + v2 ) = v1 2 + v2 2 + 2(v1 · v2 ),
we have
v1 · v2 = 0
and and
v1 ⊥ v2 .
10. That the path is of the form r(t) = cos ωtA + sin ωtB can be seen by applying the hint to each component of the equation F(t) = −mω 2 r(t). A = r(0) = the initial position B = r (0)/ω = the initial velocity divided by ω The path is circular if A = B and A ⊥ B. 11. r (t) = a,
r (t) = v (0) + ta,
r (t) = r (0) + t v(0) +
1 2 2 t
a.
If neither v (0) nor a is zero, the displacement r (t) − r (0) is a linear combination of v (0) and a and thus remains on the plane determined by these vectors. The equation of this plane can be written [a × v (0)] · [r − r (0)] = 0. (If either v (0) or a is zero, the motion is restricted to a straight line; if both of these vectors are zero, the particle remains at its initial position r (0). ) 12. Clearly we can take φ1 ∈ [0, 2π).
With ω = −1, the path takes the form
r(t) = [A1 cos(−t + φ1 ) + D1 ]i − [A1 sin(−t + φ1 ) + D2 ]j + [Ct + D3 ]k. Differentiation gives v(t) = A1 sin(−t + φ1 )i + A1 cos(−t + φ1 )j + Ck. r(0) = a i =⇒ A1 cos φ1 + D1 = a,
A sin φ1 + D2 = 0,
v(0) = a j + b k =⇒ A1 sin φ1 = 0,
A1 cos φ1 = a,
D3 = 0
C=b
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SECTION 14.7
751
From these equations we have D1 = D2 = D3 = 0,
c = b,
A1 sin φ1 = 0,
A1 cos φ1 = a.
The last two equations give φ1 = 0
and A1 = a
or
φ1 = π and A1 = −a.
The first possibility gives r(t) = a cos(−t) i − a sin(−t) j + bt k = a cos t i + a sin t j + bt k. The second possibility gives the same path: r(t) = −a cos(−t + π) i − a sin(−t + π) j + bt k = a cos t i + a sin t j + bt k 13. r(t) = i + t j +
qE0 2m
t2 k
14. Since v has magnitude w r , lies in the plane of the wheel, and makes an angle of 90◦ counterclockwise with r, we have v = ω × r 15. r(t) =
t3 1+ 6m
17.
d dt
Therefore d/dt
t4 i+ j + tk 12m
1 mv2 2
1
2 2 mv
= mv
− r4 = 0
1 16. r(t) = sin ωt + π 2
and
k
dv dv dv dr =m v· =m ·v=F · dt dt dt dt
dr dr dr d 4 = 4r2 r · = 4r2 r = 4r3 = r . dt dt dt dt and
find that E = 2m. Thus 12 mv2 − r4 = 2m
v=
1 2 2 mv
− r4 is a constant E.
Evaluating E from t = 0,
4 + (2/m) r4 .
SECTION 14.7 1. On Earth: year of length T , On Venus: year of length α T,
average distance from sun d. average distance from sun 0.72d.
Therefore
This gives
α2 = (0.72)3 ∼ = 0.372
(α T )2 (0.72d)3 = . T2 d3 ∼ 0.615. Answer: and α =
about 61.5% of an Earth year.
we
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November 30, 2006
REVIEW EXERCISES
1 d dE d = m (v 2 ) − mρ (r−1 ) dt 2 dt dt d 2 d (v ) = (v · v) = 2(a · v) dt dt
d −1 dr 1 dr 1 1 (r ) = − 2 =− 3 r = − 3 (r · v) (using 13.2.3) dt r dt r dt r ρr mρ dE ρr = m(a · v) + 3 (r · v) = (a · mv) + 3 · mv = a + 3 · mv = 0 dt r r r
3.
4. Using
dx dt
2
r˙ =
+
dr ˙ dθ θ
dy dt
2
and θ˙ =
we have
2 d (r sin θ) dt 2 2 dθ dr dθ dr cos θ + r cos θ + sin θ = r (− sin θ ) + dt dt dt dt 2 2 2 2 dθ dr dθ dr = r2 sin2 θ + cos2 θ + r2 cos2 θ + sin2 θ dt dt dt dt 2 2 dθ dr 2 +r = dt dt =
d (r cos θ ) dt
2
+
L mr 2
2 dr 1 mρ 1 = m(r˙ 2 + r2 θ˙2 ) = E+ θ˙2 + r2 θ˙2 r 2 2 dθ
2 2 dr L2 1 L2 1 dr L2 1 = m + 2 2 = + 2 2 dθ m2 r 4 m r 2m r4 dθ r
and therefore
2 L2 1 mρ 1 dr E= − + 4 2m r2 r dθ r
5. Substitute r=
a , 1 + e cos θ
dr dθ
2
=
−a · (−e sin θ) (1 + e cos θ)2
2 =
(ae sin θ)2 (1 + e cos θ)4
into the right side of the equation and you will see that, with a and e2 as given, the expression reduces to E. REVIEW EXERCISES 1. f (t) = 6t i − 15t2 j, 2. f (t) = 2e2t i +
2t j, 1 + t2
f (t) = 6 i − 30t j f (t) = 4e2t i +
3. f (t) = (et cos t − et sin t) i + 2 sin 2t j,
2 − 2t2 j (1 + t2 )2
f (t) = −2et sin t i + 4 cos 2t j
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REVIEW EXERCISES 4. f (t) = cosh t i − (2t − t2 )e−t j + sinh tk,
2
2t i + (t − 1) j dt = t i +
5.
2
2
0
π
6.
[sin 2t i + 2 cos t j + 0
7.
f (t) = sinh t i + (t2 − 4t + 2)e−t j + cosh t k
2 2 j = 4i + j 3 0
1 3 t −t 3
π √ 2 1 2 t k] dt = − cos 2t i + 2 sin t j + t3/2 k = π 3/2 k 2 3 3 0 8.
y 50
100
753
150
200
y
x
12 −5
6
−10
1
9.
2
3
4
x
10.
π π i + 4 sin t + j 11. (a) r(t) = 2 cos t + 2 2 12. direction vector: d = (2, 4, 6);
(b) r(t) = −2 cos 2t i + 4 sin 2t j
r(t) = (1 + 2t) i + (1 + 4t) j + (−2 + 6t) k,
13. f (t) = 13 t3 i + ( 12 e2t + t) j + 13 (2t + 1)3/2 k + C. f (0) = i − 3 j + 3k =⇒ C = i −
7 2
j + 83 k; f (t) =
1
3t
3
+ 1 i + 12 e2t + t − 72 j + 13 (2t + 1)3/2 + 83 k
14. f (t) = −f (t) =⇒ f (t) = f 0 e−t f (0) = i + 2k =⇒ f 0 = i + 2k and so f (t) = e−t i + 2e−t k 15. f (t) = (6 i + 12t3 j) + (8t i − 12 k) = (6 + 8t) i + 12t3 j − 12k 16. Note: f is not a vector function.
0≤t≤1
f (t) = et + 1 =⇒ f (t) = et
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REVIEW EXERCISES
1 17. f (t) = (t + 2t ) i − 2t + 2 t 2
3
j + (t − t) k,
2
4
2 f (t) = (2t + 6t ) i − 4t − 3 t
j + (4t3 − 1) k
2
f (t) = t3 cos t + t2 sin t + 3t cos t = (t3 + 3t) cos t + t2 sin t
18. Note: f is not a vector function.
f (t) = −(t3 + 3t) sin t + (3t2 + 3) cos t + 2t sin t + t2 cos t = (4t2 + 3) cos t − (t3 + t) sin t 19. r (t) = 2r(t) =⇒ r(t) = r0 e2t r(0) = (1, 2, 1) =⇒ r0 = (1, 2, 1) 20. F (t) = e2t i + e−2t j,
and
r(t) = (e2t , 2e2t , e2t )
F (t) = 2e2t i − 2e−2t j,
Since F (t) = 4F (t) for all t, F and F F and F 2t
−2t
e i+e
F (t) = 4e2t i + 4e−2t j are parallel.
will have the same direction for some value of t iff there is a number k > 0 such that j = k(2e2t i − 2e−2t j). No such value of k exists.
21. The tip of r(t) is P (1, 1, 1) when t = 0. r (t) = (2t + 2) i + 3 j + (3t2 + 1) k, r (0) = 2 i + 3 j + k Scalar parametric equations for the tangent line are: x = 1 + 2t, y = 1 + 3t, z = 1 + t. √ 22. r(π/3) = ( 3/2, −1/2, π/3)
√ r (π/3) = (−1, − 3, 1)
r (t) = (2 cos 2t, −2 sin 2t, 1);
√
Scalar parametric equations for the tangent line are: x = 23. r1 (t) = (2, 1, 1) at t = 1;
3 2
− t,
y = − 12 −
√
3 t,
z=
π 3
+t
r2 (u) = (2, 1, 1) at u = −1. Therefore the curves intersect at the point
(2, 1, 1). r1 (t) = (2 i + 2t j + k, r1 (1) = 2 i + 2 j + k;
r2 (u) = −i − 2u j + 2u k, r2 (−1) = −i + 2 j − 2 k.
Since r1 (1) · r2 (−1) = 0, the angle of intersection is π/2 radians 24. r(t) = 2t i + (1 − t2 ) j − t2 k,
r (t) = 2t i − 2t j − 2t k.
(2t i + (1 − t2 ) j − t2 k) · (2t i − 2t wj − 2t k) = 4t3 ;
4t3 = 0 =⇒ t = 0
The curve and the tangent line meet at right angles at the point where t = 0; (0, 1, 0). 25. r(t) = t i + e2t j,
√ r(π/6) = i − 32 3 j √ r (t) = 2 cos t i + 3 sin t j r (π/6) = 3 i + 32 j
26. r(t) = 2 sin t i − 3 cos t j,
r(0) = j;
r (t) = i + 2e2t j,
r (0) = i + 2 j
3
y
y
8 6
−2
2
4 2 −1
−3
1
x
x
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REVIEW EXERCISES 27. r (t) = t cos t i + t sin t j +
√ 3tk;
||r || =
755
ds = 2t dt
√ r (t) 1 = (cos t i + sin t j + 3 k). ||r (t)|| 2 1 1 1 T (t) = − sin t i + cos t j; ||T (t)|| = . 2 2 2 T (t) principal normal vector: N = = − sin t i + cos t j ||T (t)||
unit tangent vector: T =
28. r (t) = (−a sin t i + a cos t j + b k.
|b| r · k =√ = constant. ||r || ||k|| a2 + b2
The cosine of the angle θ between r (t) and k is: Therefore θ is a constant. 1 1 + 2t2 j − 2tk; ||r (t)|| = . t t r (t) 2t 1 2t2 T(t) = = 2 i+ 2 j− 2 k; ||r (t)|| 2t + 1 2t + 1 2t + 1
29. r (t) = 2 i +
T (t) =
2 − 4t2 4t 4t i− j− k; 2 2 2 2 2 (2t + 1) (2t + 1) (2t + 1)2
N(1) =
T (1) ||T (1)||
= − 13 i −
j−
2 3
2 3
T(1) =
2 1 i+ j− 3 3
T (1) = − 29 i −
4 9
j−
2 3
k
4 9
k,
||T (1)|| =
2 3
k
A normal vector for the osculating plane is: (2 i + j − 2 k) × (i + 2 j + 2 k) = 6 i − 6 j + 3 k. Since r(1) = 2 i − k, an equation for the osculating plane is 6(x − 2) − 6y + 3(z + 1) = 0 or 2x − 2y + z = 3. √ ||r (t)|| = 2. √ r (t) 1 1 1 1 T(t) = = √ (− sin t i − sin t j − 2 cos t k); T(π/4) = − i − j − √ k ||r (t)|| 2 2 2 2 √ 1 1 1 1 T (t) = √ (− cos t i − cos t j + 2 sin t k); T (π/4) = − i − j + √ k, ||T (π/4)|| = 1 2 2 2 2 T (π/4) 1 1 1 N(π/4) = =− i− j+ √ k ||T (π/4)|| 2 2 2 √ √ √ √ A normal vector for the osculating plane is: (i + j + 2 k) × (i + j − 2 k) = −2 2 i + 2 2 j. √ 2 (i + j − 2 k, an equation for the osculating plane is Since r(π/4) = 2 " " √ # √ # √ √ 2 2 +2 2 y− = 0 or x − y = 0. −2 2 x − 2 2
30. r (t) = − sin t i − sin t j −
√ 2 cos t k;
1/2
31. r (t) = 2 i + t
j;
L= 0
32. r (t) = et i − e−t j −
√ 2k;
5
||r (t)||dt =
5
√
4 + tdt =
0
||r (t)|| = et + e−t ;
L= 0
38 3 ln 3
ln 3 8 (et + e−t )dt = et − e−t = 0 3
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REVIEW EXERCISES
√ 33. r (t) = cosh t i + sinh t j + k; ||r (t)|| = cosh2 t + sinh2 t + 1 = 2 cosh t; 1√ √ 1 √ L= 2 cosh t dt = 2 sinh t = 2 sinh 1. 0
0
34. r (t) = − sin t i − cos t j + sinh t k; ln 2 ln 2 3 cosh t dt = sinh t = 0 4 0
||r || =
1 + sinh2 t = cosh t;
√ 35. (a) r (t) = − sin t i + cos t j + t1/2 k; ||r (t)|| = 1 + t. t t t 2 √ 2 2 s= ||r (u)|| du = 1 + t dt = (1 + u)3/2 = (1 + t)3/2 − 0 3 3 3 0 0
2/3 3 2 − 1 = φ(s); R(s) = cos φ(s) i + sin φ(s) j + [φ(s)]3/2 k s+1 (b) t = 2 3 (c) R (s) = − sin φ(s) i + cos φ(s) j + φ(s)1/2 k φ (s) −1/3
2/3
3 3 2 3 ||R (s)|| = φ (s) 1 + φ(s) = s+1 s+1 =1 3 2 2 2
speed: s = ||v|| = 1 + 4 sin2 2t √ acceleration: a = r (t) = − cos t i − sin t j + 4 cos 2t k; ||a|| = 1 + 16 cos2 2t
36. velocity: v = r (t) = − sin t i + cos t j + 2 sin 2t k
37. r (t) = − cos t i − sin t j Thus:
and r (0) = k =⇒ r (t) = − sin t i + (cos t − 1) j + k. √ velocity v = − sin t i + (cos t − 1) j + k and speed ||v|| = 3 − 2 cos t.
r (t) = − sin t i + (cos t − 1) j + k and r(0) = i =⇒ r(t) = cos t i + (sin t − t) j + t k. 38. The acceleration vector remains perpendicular to the path means: r · r = 0. r (t) = f (t) i + 2f (t)f (t) j,
r (t) = f (t) i + [2f (t)2 + 2f (t)f (t)] j
r · r = 0 =⇒ f (t)f (t) + 4f (t)[f (t)]3 + 4f 2 (t)f (t)f (t) = 0 3 1/2 3 x , y = x−1/2 ; 2 4 3 −1/2 6 |y | 4x √ = κ= 9 3/2 = [1 + (y )2 ]3/2 x(4 + 9x)3/2 [1 + 4 x]
39. y =
40. y = −2 sin 2x,
y = −4 cos 2x;
κ=
4| cos 2x| [1 + 4 sin2 2x]3/2
41. x(t) = 2e−t , y(t) = e−2t =⇒ x (t) = −2e−t , y (t) = −2e−2t =⇒ x (t) = 2e−t , y (t) = 4e−2t κ=
|(−2e−t )(4e−2t ) − (−2e−2t )(2e−t )| 1 e3t = = −2t −4t 3/2 −2t 3/2 2t [4e + 4e ] 2(1 + e ) 2(e + 1)3/2
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REVIEW EXERCISES 42. x(t) = 13 t3 , κ=
y(t) = 12 t2 =⇒ x (t) = t2 ,
|t2 − 2t2 | (t2
+
3 t4 ) 2
=
y (t) = t =⇒ x (t) = 2t,
757
y (t) = 1
1 3
|t|(1 + t2 ) 2
ds = |r (t)| = 5 dt 9 4 3 9 3 T(t) = − sin 3t i − j + cos 3t k, T (t) = − cos 3t i − sin 3t k; 5 5 5 5 5 ||T (t)|| 9 κ= = ds/dt 25
43. r (t) = −3 sin 3t i − 4 j + 3 cos 3t k,
√ 1/2 ds = ||r (t)|| = 1 + t 2t j + t k, dt √ 2t −1 t 1 i+ j+ k T (t) = i+ T= 1+t 1+t 1+t (1 + t2 )
||T (t)|| = 9/5
44. r (t) = 1 i +
||T (t)|| = √
1 ; 2t(1 + t)
45. y = sinh (x/a),
y =
κ=
√
√ t − t) 1 j+ k (1 + t)2 (1 + t)2
2 2 (1/
√
1 ||T (t)|| =√ ds/dt 2t(1 + t)2
1 cosh (x/a); a
κ=
|y | [1 +
3 (y )2 ] 2
=
1 a = 2 2 y a cosh (x/a)
46. (a) Suppose ||v|| = c constant. Then ||v||2 = c2 =⇒ r · r = c2 =⇒ 2r r = 0 =⇒ a ⊥ v. (b) T =
v , ||v||
T =
a |T | |a| v = since ||v|| is constant. κ = = ||v|| ||v|| |v| |v|2
3 ds 4 = ||r (t)|| = 1 47. r (t) = − sin t i + sin t j + cos tk; 5 5 dt 3 3 4 4 T = r (t) = − sin t i + sin t j + cos tk, T (t) = − cos t i + cos t j − sin t k; 5 5 5 5 2 ds d2 s =1 κ = 1; aT = 2 = 0, aN = κ dt dt ds = ||r (t)|| = 2 + t2 dt 1 1 T(t) = (2 i + 2t j + t2 k), T (t) = (−4t i + (4 − 2t2 ) j + 4t k); 2 2+t (2 + t2 )2 2 ds 2 d2 s κ= ; aT = 2 = 2t, aN = κ = 2. (2 + t2 )2 dt dt
||T (t)|| = 1
48. r (t) = 2 i + 2t j + t2 k;
||T (t)|| =
2 2 + t2
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SECTION 15.1
CHAPTER 15 SECTION 15.1 1. dom (f ) = the first and third quadrants, including the axes;
range (f ) = [0, ∞)
2. dom (f ) = the set of all points (x, y) with xy ≤ 1; the two branches of the hyperbola xy = 1 and all points in between; range (f ) = [0, ∞) 3. dom (f ) = the set of all points (x, y) except those on the line y = −x; range (f ) = (−∞, 0) ∪ (0, ∞) 4. dom (f ) = the set of all points (x, y) other than the origin; 5. dom (f ) = the entire plane;
range (f ) = (0, ∞)
range (f ) = (−1, 1) since ex − ey ex + ey − 2ey 2 = = 1 − x−y x y x y e +e e +e e +1
and the last quotient takes on all values between 0 and 2. 6. dom (f ) = the set of all points (x, y) other than the origin;
range (f ) = [0, 1]
7. dom (f ) = the first and third quadrants, excluding the axes; range (f ) = (−∞, ∞) 8. dom (f ) =the set of all points (x, y) between the branches of the hyperbola xy = 1; range (f ) = (−∞, ∞) 9. dom (f ) = the set of all points (x, y) with x2 < y —in other words, the set of all points of the plane above the parabola y = x2 ; range (f ) = (0, ∞) 10. dom (f ) = the range (f ) = [0, 3]
set
of
all
points
(x, y)
with
−3 ≤ x ≤ 3,
11. dom (f ) = the set of all points (x, y) with −3 ≤ x ≤ 3, −2 ≤ y ≤ 2
−1 ≤ y ≤ 1
(a rectangle);
(a rectangle);
range (f ) = [−2, 3] 12. dom (f ) = all of space;
range (f ) = [−3, 3]
13. dom (f ) = the set of all points (x, y, z) not on the plane x + y + z = 0;
range (f ) = {−1, 1}
14. dom (f ) = the set of all points (x, y, z) with x2 = y 2 —that is, all points of space except for those which lie on the plane x − y = 0 or on the plane x + y = 0; range (f ) = (−∞, ∞) 15. dom (f ) = the set of all points (x, y, z) with |y| < |x|;
range (f ) = (−∞, 0 ]
16. dom (f ) =the set of all points (x, y, z) not on the plane x − y = 0;
range (f ) = (−∞, ∞)
17. dom (f ) = the set of all points (x, y) with x2 + y 2 < 9 —in other words, the set of all points of the plane inside the circle x2 + y 2 = 9; range (f ) = [ 2/3, ∞) 18. dom (f ) = all of space;
range (f ) = [0, ∞)
19. dom (f ) = the set of all points (x, y, z) with x + 2y + 3z > 0 — in other words, the set of all points in space that lie on the same side of the plane x + 2y + 3z = 0 as the point (1, 1, 1); range (f ) = (−∞, ∞) 20. dom (f ) = the set of all points (x, y, z) with x2 + y 2 + z 2 ≤ 4 — in other words, the set of all points inside and on the sphere x2 + y 2 + z 2 = 4; range (f ) = [1, e2 ] 21. dom (f ) = all of space;
range (f ) = (0, 1]
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SECTION 15.1 22. dom (f ) = the set of all points (x, y, z) with −1 ≤ x ≤ 1, −2 ≤ y ≤ 2, −3 ≤ z ≤ 3 solid); range (f ) = [0, 3]
759
(a rectangular
23. dom (f ) = {x : x ≥ 0}; range (f ) = [0, ∞) dom (g) = {(x, y) : x ≥ 0, y real}; range (g) = [0, ∞) dom (h) = {(x, y, z) : x ≥ 0, y, z real}; range (h) = [0, ∞)
24. dom (f ) = the entire plane, dom (g) = all of space, 25.
range (f ) = [−1, 1]
range (g) = [−1, 1]
f (x + h, y) − f (x, y) 2(x + h)2 − y − (2x2 − y) 4xh + 2h2 = lim = lim = 4x h→0 h→0 h→0 h h h lim
f (x, y + h) − f (x, y) 2x2 − (y + h) − (2x2 − y) = lim = −1 h→0 h→0 h h lim
f (x + h, y) − f (x, y) xy + hy + 2y − (xy + 2y) = lim = lim y = y. h→0 h→0 h h f (x, y + h) − f (x, y) xy + xh + 2y + 2h − (xy + 2y) lim = lim = lim (x + 2) = x + 2 h→0 h→0 h→0 h h
26.
27.
lim
h→0
f (x + h, y) − f (x, y) 3(x + h) − (x + h)y + 2y 2 − (3x − xy + 2y 2 ) 3h − hy = lim = lim =3−y h→0 h→0 h→0 h h h lim
f (x, y + h) − f (x, y) 3x − x(y + h) + 2(y + h)2 − (3x − xy + 2y 2 ) = lim h→0 h→0 h h lim
−xh + 4yh + 2h2 = −x + 4y h→0 h
= lim
28.
lim
f (x + h, y) − f (x, y) x sin y + h sin y − x sin y = lim = lim sin y = sin y. h→0 h→0 h h
lim
f (x, y + h) − f (x, y) x sin(y + h) − x sin y sin(y + h) − sin y = lim = x lim = x cos y h→0 h→0 h h h
h→0
h→0
29.
lim
h→0
f (x + h, y) − f (x, y) cos[(x + h)y] − cos[xy] = lim h→0 h h cos[xy] cos[hy] − sin[xy] sin[hy] − cos[xy] = lim h→0 h cos[hy] − 1 sin hy − sin[xy] lim = cos[xy] lim h→0 h→0 h h cos[hy] − 1 sin hy = y cos[xy] lim − y sin[xy] lim h→0 h→0 hy hy = −y sin[xy]
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SECTION 15.2 and lim
h→0
f (x, y + h) − f (x, y) cos[x(y + h)] − cos[xy] = lim h→0 h h cos[xy] cos[hx] − sin[xy] sin[hx] − cos[xy] h cos[hx] − 1 sin hx − sin[xy] lim = cos[xy] lim h→0 h→0 h h cos[hx] − 1 sin hx = x cos[xy] lim − x sin[xy] lim h→0 h→0 hx hx
= lim
h→0
= −x sin[xy] 30.
f (x + h, y) − f (x, y) (x2 + 2xh + h2 )ey − x2 ey = lim = lim (2x + h)ey = 2xey . h→0 h→0 h→0 h h lim
f (x, y + h) − f (x, y) x2 ey+h − x2 ey ey+h − ey = lim = x2 lim = x2 ey . h→0 h→0 h→0 h h h lim
(b) f (x, y) = |2 i × (x i + y j)| = 2|y|
(b) f (x, y) = πx2 y
31. (a) f (x, y) =Ay
32. (a) f (x, y, z) = xy + 2xz + 2yz (b) f (x, y, z) = cos−1
(i + j) · (x i + y j + z k) x+y = cos−1 √ i + jx i + y j + z k 2 x2 + y 2 + z 2
(c) f (x, y, z) = [i × (i + j)] · (x i + y j + z k) = z 33. Surface area: S = 2lw + 2lh + 2hw = 20 =⇒ w = Volume: V = lwh = 34. wlh = 12
=⇒
35. V = πr2 h + 36. A =
lh(10 − lh) l+h
h=
12 ; wl
10 − lh 20 − 2lh = 2l + 2h l+h
C = 4wl + 2(2wh + 2lh) = 4wl +
48 48 + l w
4 3 πr 3
1 [2(12 − 2x) + 2x cos θ] · x sin θ = (12 − 2x + x cos θ) · x sin θ 2
SECTION 15.2 1.
an elliptic cone
2.
an ellipsoid
3.
a parabolic cylinder
4.
a hyperbolic paraboloid
5.
a hyperboloid of one sheet
6.
an elliptic cylinder
7.
a sphere
8.
a hyperboloid of two sheets
9.
an elliptic paraboloid
11. a hyperbolic paraboloid
10.
a hyperbolic cylinder
12.
an elliptic paraboloid
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SECTION 15.2 13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
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SECTION 15.2
25. elliptic paraboloid xy-trace: the origin xz-trace: the parabola x2 = 4z yz-trace: the parabola y 2 = 9z surface has the form of Figure 15.2.5
26. ellipsoid xy-trace: the ellipse 9x2 + 4y 2 = 36 xz-trace: the ellipse x2 + 4z 2 = 4 yz-trace: the ellipse y 2 + 9z 2 = 9 surface has the form of Figure 15.2.1
27. elliptic cone xy-trace: the origin xz-trace: the lines x = ±2z yz-trace: the lines y = ±3z surface has the form of Figure 15.2.4
28. hyperboloid of one sheet xy-trace: the ellipse 9x2 + 4y 2 = 36 xz-trace: the hyperbola x2 − 4z 2 = 4 yz-trace: the hyperbola y 2 − 9z 2 = 9 surface has the form of Figure 15.2.2
29. hyperboloid of two sheets xy-trace: none xz-trace: the hyperbola 4z 2 − x2 = 4 yz-trace: the hyperbola 9z 2 − y 2 = 9 surface has the form of Figure 15.2.3
30. hyperbolic paraboloid xy-trace: the lines y = ± 32 x xz-trace: the parabola x2 = 4z yz-trace: the parabola y 2 = −9z surface has the form Figure 15.2.6
31. hyperboloid of two sheets xy-trace: the hyperbola 9x2 − 4y 2 = 36 xz-trace: the hyperbola x2 − 4z 2 = 4 yz-trace: none see Figure 15.2.3
32.
33. elliptic paraboloid xy-trace: the parabola x2 = 9y xz-trace: the origin yz-trace: the parabola z 2 = 4y surface has the form of Figure 15.2.5
34. elliptic cone xy-trace: the lines x = ±2y xz-trace: the origin yz-trace: the lines z = ±3y surface has the form of Figure 15.2.4, rotated 90◦ about the x-axis.
35. hyperboloid of two sheets xy-trace: the hyperbola 9y 2 − 4x2 = 36 xz-trace: none yz-trace: the hyperbola y 2 − 4z 2 = 4 see Figure 15.2.3
36.
37. paraboloid of revolution xy-trace: the origin xz-trace: the parabola x2 = 4z yz-trace: the parabola y 2 = 4z surface has the form of Figure 15.2.5
38. ellipsoid xy-trace: the ellipse 4x2 + y 2 = 4 xz-trace: the ellipse 9x2 + z 2 = 9 yz-trace: the ellipse 9y 2 + 4z 2 = 36 the surface has the form of Figure 15.2.1, rotated 90◦ about the x-axis.
hyperboloid of one sheet xy-trace: the hyperbola x2 − 9y 2 = 9 xz-trace: the circle x2 + y 2 = 9 yz-trace: the hyperbola z 2 − 9y 2 = 9 surface has the form of Figure 15.2.2, rotated 90◦ about the x-axis.
elliptic paraboloid xy-trace: the parabola y 2 = 4x xz-trace: the parabola z 2 = 9x yz-trace: the origin surface has the form of Figure 15.2.5, but opening along the positive x-axis.
39. (a) an elliptic paraboloid (vertex down if A and B are both positive, vertex up if A and B are both negative) (b) a hyperbolic paraboloid (c) the xy-plane if A and B are both zero; otherwise a parabolic cylinder 40. The xz-plane and all planes parallel to the xy-plane.
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SECTION 15.2 41. x2 + y 2 − 4z = 0
(paraboloid of revolution)
42. c2 x2 + c2 y 2 − b2 z 2 = b2 c2
(hyperboloid of revolution, one sheet)
43. (a) a circle (b) (i) x2 + y 2 = −3z
(ii)
√
x2 + z 2 = 13 y
44. (a) the ellipse b2 x2 + y 2 = b2 (b) ellipse approaches parallel lines x = ±1 in the plane z = 1 (c) paraboloid approaches parabolic cylinder z = x2 45. x + 2y + 3
x+y−6 2
=6
46. 3x + y − 2(4 − x + 2y) = 1, 47.
x2 + y 2 + (z − 1)2 =
or or
5x + 7y = 30, 5x − 3y = 9,
3 2
(z 2 + 1) + (z − 1)2 =
x2 + y 2 − z 2 = 1
3 ; 2
a line a line (2z − 1)2 = 0,
z=
48. z 2 + (z − 2)2 = 2 =⇒ 2(z − 1)2 = 0 =⇒ z = 1 =⇒ x2 + y 2 = 1, 49. x2 + y 2 + x2 + 3y 2 = 4
or
x2 + 2y 2 = 2,
50. y 2 + (x2 + 3y 2 ) = 4 =⇒ x2 + 4y 2 = 4, 51. x2 + y 2 = (2 − y)2 2
2
52. x + y =
2−y 2
or
x2 = −4(y − 1),
1 5 so that x2 + y 2 = 2 4
a circle.
an ellipse
an ellipse. a parabola
2 =⇒ 4x2 + 3y 2 + 4y = 4,
an ellipse.
53. (a) Set x = a cos u cos v, y = b cos u sin v, z = c sin u. Then: (b)
4 2 0 -2 -4 2 1 0 -1 -2 -2 0 2
y2 z2 x2 + + = 1. a2 b2 c2
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SECTION 15.3 (a) Set x = a sec u cos v,
55.
(a) Set x = av cos u,
y = b sec u sin v,
y = bv sin u,
z = c tan u
z = cv
(b) 10
4
-10-5 0 5 10
2
0
0
-10 10
-2 20
-4 6 0
4 2
-20
0 -2
-1
0
1
2
SECTION 15.3 1. lines of slope 1: y = x − c
3.
parabolas: y = x2 − c
2. lines of slope 2 : y = 2x − c
4.
parabolas: x − y 2 =
1 c
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SECTION 15.3 5. the y-axis and the lines y =
1−c c
x with
the origin omitted throughout
7. the cubics y = x3 − c
9. the lines y = ±x and the hyperbolas
6. the x-axis and the parabolas y = cx2 with the origin omitted throughout
8. the coordinate axes and the hyperbolas xy = ln c
√ 10. pairs of vertical lines x = ± c and the y-axis
x2 − y 2 = c
√ 11. pairs of horizontal lines y = ± c and the x-axis
765
12. the lines x = 0, y = 1 and the hyperbolas c y = +1 x
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SECTION 15.3
13. the circles x2 + y 2 = ec , c real
14. the parabolas y = ec x2 with the origin omitted throughout
15.
2
the curves y = ecx with the point (0, 1) omitted
17. the coordinate axes and pairs of lines √ 1−c x, with the origin omitted y=± √ c throughout
19. x + 2y + 3z = 0,
x2 + y 2 ,
22. ellipsoid
18.
the curves y = ecx with the point (0, 1) omitted
plane through the origin
20. circular cylinder x2 + y 2 = 4 21. z =
16. the coordinate axes and pairs of hyperbolas √ xy = ± c
(Figure 15.2.8)
the upper nappe of the circular cone z 2 = x2 + y 2
y2 z2 x2 + + =1 4 6 9
23. the elliptic paraboloid
(Figure 15.2.1)
y2 x2 + =z 18 8
(Figure 15.2.5)
(Figure 15.2.4)
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SECTION 15.3 x2 y2 + − z 2 = −1 (1/6)2 (1/3)2
24. hyperboloid of two sheets 25. (i) hyperboloid of two sheets (ii) circular cone
767
(Figure 15.2.3)
(Figure 15.2.3)
(Figure 15.2.4)
(iii) hyperboloid of one sheet
(Figure 15.2.2)
26. (i) hyperboloid of two sheets (ii) elliptic cone (iii) hyperboloid of one sheet 27. The level curves of f are: 1 − 4x2 − y 2 = c. Substituting P (0, 1) into this equation, we have 1 − 4(0)2 − (1)2 = c
=⇒
c=0
The level curve that contains P is: 1 − 4x2 − y 2 = 0, or 4x2 + y 2 = 1. 28. (x2 + y 2 )exy = 1 29. The level curves of f are: y 2 arctan x = c. Substituting P (1, 2) into this equation, we have 4 arctan 1 = c
=⇒
c=π −1
2
The level curve that contains P is: y tan
x = π.
30. (x2 + y) ln(2 − x + ey ) = 5 31. The level surfaces of f are: x2 + 2y 2 − 2xyz = c. Substituting P (−1, 2, 1) into this equation, we have (−1)2 + 2(2)2 − 2(−1)(2)(1) = c
=⇒
c = 13
The level surface that contains P is: x2 + 2y 2 − 2xyz = 13. 32.
x2 + y 2 − ln z = 4
33. (a)
4
2
0
50 0 -50 -4
4 2 0
-2
-2 -2
0 2 4 -4
-4 -4
-2
0
2
4
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SECTION 15.3 (b) 4
2
0 15 10 5 0
5
-2
0
-5
-4 0 -5
5
-4
34. (a) the level surfaces are planes
35. (a)
-2
0
2
4
(b)
3 3x + 2y + 1 = 4x2 + 9 5
(b) x2 + 2y 2 − z 2 = 21
36. (a) 2
(b)
1
0
-1
-2 -2
37.
-1
GmM = c =⇒ + y2 + z2 concentric spheres. x2
0
1
2
x2 + y 2 + z 2 =
GmM ; the surfaces of constant gravitational force are c
38. Circular cylinders around the positive y-axis:
x2 + z 2 =
k2 c2
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SECTION 15.4 k
39. (a) T (x, y, z) =
x2 + y 2 + z 2
(b)
k x2 + y 2 + z 2
=c
=⇒
,
769
where k is a constant. k2 ; the level surfaces are concentric spheres. c2 √ √ 50 6 k = 50 6 =⇒ T (x, y, z) = x2 + y 2 + z 2
x2 + y 2 + z 2 =
k = 50 =⇒ + 22 + 12 √ √ 50 6 Now, T (4, 0, 3) = √ = 10 6 degrees 2 2 3 +4
(c) T (1, 2, 1) = √
40. x2 + y 2 = r2 −
12
k2 ; c2
circles about the origin, for |c| >
41.
f (x, y) = y 2 − y 3 ; F
42.
D.
44.
B.
45.
f (x, y) = xye−(x
2
k . r
+y 2 )/2
; E
43.
f (x, y) = cos
46.
C.
PROJECT 15.3 1.
2.
3.
4.
SECTION 15.4 1.
3.
∂f = 6x − y, ∂x ∂ρ = cos φ cos θ, ∂φ
∂f =1−x ∂y ∂ρ = − sin φ sin θ ∂θ
2.
∂g = 2xe−y , ∂x
∂g = −x2 e−y ∂y
x2 + y 2 ; A
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SECTION 15.4 ∂ρ = −2 sin(θ − φ) cos(θ − φ) ∂φ
∂ρ = 2 sin(θ − φ) cos(θ − φ), ∂θ
5.
∂f = ex−y + ey−x , ∂x
7.
∂g (AD − BC)y , = ∂x (Cx + Dy)2
9.
∂u = y + z, ∂x
∂f = −ex−y − ey−x ∂y ∂g (BC − AD)x = ∂y (Cx + Dy)2
∂u = x + z, ∂y
∂u =x+y ∂z
6.
∂z x , = ∂x x2 − 3y
8.
∂u −ez = 2 2, ∂x x y
10.
∂z −3 = ∂y 2 x2 − 3y
∂u −2ez , = ∂y xy 3
∂z = 2Ax + By, ∂x
∂z = Bx + 2Cy ∂y
11.
∂f = z cos (x − y), ∂x
12.
∂g 2u = 2 , ∂u u + vw − w2
13.
∂ρ = eθ+φ [cos (θ − φ) − sin (θ − φ)], ∂θ
∂ρ = eθ+φ [cos (θ − φ) + sin (θ − φ)] ∂φ
14.
∂f = (x + y) cos(x − y) + sin(x − y), ∂x
∂f = −(x + y) cos(x − y) + sin(x − y) ∂y
15.
∂f = 2xy sec xy + x2 y(sec xy)(tan xy)y = 2xy sec xy + x2 y 2 sec xy tan xy ∂x
∂f = −z cos (x − y), ∂y w ∂g = 2 , ∂v u + vw − w2
∂f = sin (x − y) ∂z v − 2w ∂g = 2 ∂w u + vw − w2
∂f = x2 sec xy + x2 y(sec xy)(tan xy)x = x2 sec xy + x3 y sec xy tan xy ∂y 16.
∂g 2 = , ∂x 1 + (2x + y)2
17.
x2 + y 2 − x(2x) ∂h y 2 − x2 = = 2 ∂x (x2 + y 2 ) (x2 + y 2 )2
18.
x ∂z = 2 , ∂x x + y2
19.
(y cos x) sin y − (x sin y)(−y sin x) ∂f sin y(cos x + x sin x) = = ∂x (y cos x)2 y cos2 x
1 ∂g = ∂y 1 + (2x + y)2 ∂h −2xy = ∂y (x2 + y 2 )2
y ∂z = 2 ∂y x + y2
∂f (y cos x)(x cos y) − (x sin y) cos x x(y cos y − sin y) = = ∂y (y cos x)2 y 2 cos x 20.
∂f = exy (y sin xz + z cos xz), ∂x
21.
∂h = 2f (x)f (x)g(y), ∂x
∂f = xexy sin xz, ∂y
∂h = [f (x)]2 g (y) ∂y
∂f = xexy cos xz ∂z
∂u ez = 2 ∂z xy
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SECTION 15.4 22.
∂h = f (x)g(y)ef (x)g(x) , ∂x
23.
∂f 2 = (y 2 ln z)z xy , ∂x
24.
25.
∂h = f (x)g (y)ef (x)g(y) ∂y
∂f 2 = (2xy ln z)z xy , ∂y
∂f 2 = xy 2 z xy −1 ∂z
∂h ∂g ∂f = 2[f (x, y)]3 g(x, z) + 3[f (x, y)]2 [g(x, z)]2 ∂x ∂x ∂x ∂h ∂h ∂g ∂f = 3[f (x, y)]2 [g(x, z)]2 , = 2[f (x, y)]3 g(x, y) ∂y ∂y ∂z ∂z ∂h = 2re2t cos(θ − t) ∂r
∂h = −r2 e2t sin(θ − t) ∂θ
∂h = 2r2 e2t cos(θ − t) + r2 e2t sin(θ − t) = r2 e2t [2 cos(θ − t) + sin(θ − t)] ∂t 26.
27.
28.
∂u 1 = − − exz , ∂y y
1 ∂u = − yzexz , ∂x x 1 ∂f =z ∂x 1 + (y/x)2 ∂f = arctan (y/x) ∂x
−y x2
∂w = y sin z − yz cos x, ∂x
29. fx (x, y) = ex ln y,
∂u = −xyexz ∂z
∂w = x sin z − z sin x, ∂y
fx (0, e) = 1;
fy (x, y) =
30. gx = e−x [− sin(x + 2y) + cos(x + 2y)], gy = 2e−x cos(x + 2y), 31. fx (x, y) =
y , (x + y)2
32. gx (x, y) =
y2 , (x + y 2 )2
33. fx (x, y) = lim
h→0
h→0
34. fx (x, y) = 0,
1 xz = 2 x x + y2
∂w = xy cos z − y sin x ∂z
1 x e , y
fy (0, e) = e−1
gx (0, 14 π) = −1
gy (0, 14 π) = 0 fx (1, 2) =
2 ; 9
gx (1, 2) =
fy (x, y) =
4 25
(x + h)2 y − x2 y = lim y h→0 h
−x , (x + y)2
gy (x, y) =
2xh + h2 h
fy (1, 2) = −
−2xy , (x + y 2 )2
1 9
gy (1, 2) = −
= y lim (2x + h) = 2xy h→0
x (y + h) − x y x h = lim = lim x2 = x2 h→0 h h→0 h 2
fx (x, y) = lim
∂f 1 =z ∂y 1 + (y/x)2
yz =− 2 x + y2
2
2
fy (x, y) = 2y
ln y(x + h)2 − ln x2 y ln y + 2 ln(x + h) − 2 ln x − ln y 35. fx (x, y) = lim = lim h→0 h→0 h h ln(x + h) − ln x d 2 = 2 lim = 2 (ln x) = h→0 h dx x
4 25
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SECTION 15.4 ln x2 (y + h) − ln x2 y ln x2 + ln(y + h) − ln x2 − ln y = lim fy (x, y) = lim h→0 h→0 h h = lim
h→0
36. fx (x, y) = −(x + 4y)−2 ,
ln(y + h) − ln y d 1 = (ln y) = h dy y
fy (x, y) = −4(x + 4y)−2
1 −h 1 1 − = lim h→0 h (x + h) − y x − y (x + h − y) (x − y) −1 −1 = = lim h→0 (x + h − y) (x − y) (x − y)2 1 h 1 1 1 − = lim fy (x, y) = lim h→0 h h→0 h x − (y + h) x − y (x − y − h) (x − y) 1 1 = = lim h→0 (x − y − h) (x − y) (x − y)2
1 37. fy (x, y) = lim h→0 h
38. fx (x, y) = 2e2x+3y , 39. fx (x, y, z) = lim
h→0
fy (x, y, z) = lim
h→0
fy (x, y) = 3e2x+3y
(x + h)y 2 z − xy 2 z = lim y 2 z = y 2 z h→0 h x(y + h)2 z − xy 2 z xz(2yh + h2 ) = lim h→0 h h
= lim xz(2y + h) = 2xyz h→0
fz (x, y, z) = lim
h→0
40. fx (x, y, z) =
xy 2 (z + h) − xy 2 z = lim xy 2 = xy 2 h→0 h
2xy , z
fy (x, y, z) =
x2 , z
fz (x, y, z) = −
x2 y z2
41. (b) The slope of the tangent line to C at the point P (x0 , y0 , f (x0 , y0 )) is fy (x0 , y0 ) Thus, equations for the tangent line are: x = x0 , 42. fx (x, y) = 2x,
fx (1, 3) = 2,
z − z0 = fy (x0 , y0 )(y − y0 )
equations for the tangent line are:
43. Let z = f (x, y) = x2 + y 2 . Then f (2, 1) = 5, equations for the tangent line are:
x = 2,
fy (x, y) = 2y
y = 3,
z − 10 = 2(x − 1).
and fy (2, 1) = 2;
z − 5 = 2(y − 1)
−x2 2y x2 , fy (x, y) = 2 −3 (y − 3)2 Tangent line: x = x0 , z − z0 = fy (x0 , y0 )(y − y0 ) =⇒ x = 3, z − 9 = −36(y − 2)
44. f (x, y) =
y2
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SECTION 15.4 x2 2x . Then f (3, 2) = 9, fx (x, y) = 2 y2 − 3 y −3 equations for the tangent line are: y = 2, z − 9 = 6(x − 3)
45. Let z = f (x, y) =
46. f (x, y) = (4 − x2 − y 2 )1/2 , fx (x, y) = −x(4 − x2 − y 2 )−1/2 , √ √ √ 2 2 (a) fy (1, 1) = − =⇒ x = 1, z − 2 = − (y − 1) 2 2 √ √ √ 2 2 =⇒ y = 1, z − 2 = − (x − 1) (b) fx (1, 1) = − 2 2 √
and fx (3, 2) = 6;
fy (x, y) = −y(4 − x2 − y 2 )−1/2
√
(c) l1 and l2 have direction vectors j − 22 k, i − 22 k respectively. The normal to the plane is
√ √ √ √
2 2 2 2 k × i− k =− i− j − k, so the tangent plane is j− 2 2 2 2 √ √ √ √ √ 2 2 (x − 1) − (y − 1) − (z − 2) = 0, or (x − 1) + (y − 1) + 2(z − 2) = 0 − 2 2 47. (a) mx = −6;
tangent line:
y = 2,
z = −6x + 13
(b) my = 18;
tangent line:
x = 1,
z = 18y − 29
tangent line:
y = 2,
z=
tangent line:
x = 1,
z=
48. (a) mx = (b) my =
7 25 ; 1 − 25 ;
uy (x, y) = −2y = −vx (x, y)
49. ux (x, y) = 2x = vy (x, y); 50. ux = ex cos y, 51. ux (x, y) =
1 25 (7x − 12) 1 − 25 (y + 3)
uy = −ex sin y,
vx = ex sin y = −uy ,
1 1 x 2x = 2 ; 2 x2 + y 2 x + y2
vy = ex cos y = ux
vy (x, y) =
1 1 + (y/x)2
vx (x, y) =
1 1 + (y/x)2
Thus, ux (x, y) = vy (x, y). uy (x, y) =
1 y 1 2y = 2 ; 2 x2 + y 2 x + y2
1 x = 2 x x + y2
−y x2
=
−y x2 + y 2
Thus, uy (x, y) = −vx (x, y). 52. ux =
y 2 − x2 , (x2 + y 2 )2
uy =
53. (a) f depends only on y.
−2xy , 2 (x + y 2 )2
vx =
2xy = −uy , 2 (x + y 2 )2
vy =
y 2 − x2 = ux (x2 + y 2 )2
(b) f depends only on x.
√ 1/2 54. (a) a0 = b20 + c20 − 2b0 c0 cos θ0 =5 7 √ 1 7 ∂a 2 2 −1/2 = (2b − 2c cos θ) (b + c − 2bc cos θ) = (b) ∂b 2 14 √ √ √ √ 7 7 7 ∼ (b − b0 ) = 5 7 + · (−1) decreases by about inches. (c) a = a0 + 14 14 14 1 15 √ ∂a = 2bc sin θ( )(b2 + c2 − 2bc cos θ)−1/2 = 21 (d) ∂θ 2 7
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SECTION 15.4 (e) Differentiate implicitly:
0 = 2c
∂c bc sin θ 15 √ = =− 3 ∂θ b cos θ − c 2 √ 75 3 in.2 55. (a) 2 ∂A 1 (b) = c sin θ; ∂b 2
at time t0 ,
∂c ∂c − 2b cos θ + 2bc sin θ ∂θ ∂θ
√ ∂A 15 3 = ∂b 4
1 75 ∂A ∂A = bc cos θ; at time t0 , = ∂θ 2 ∂θ 2 ∂A π 75 5π π , A (b, c, θ + h) − A (b, c, θ) ∼ = = in.2 (d) with h = =h 180 ∂θ 180 2 24 1 ∂c ∂c −c 3 (e) 0 = sin θ b + c ; at time t0 , = =− 2 ∂b ∂b b 2
(c)
56. By theorem 7.6.1, f (x, y) = Cekx where C is independent of x. Since C may depend on y, we write C = g(y). 57. (a) y0 -section: r(x) = xi + y0 j + f (x, y0 )k
∂f (x0 , y0 )k tangent line: R(t) = [x0 i + y0 j + f (x0 , y0 )k ] + t i + ∂x
(b) x0 -section: r(y) = x0 i + yj + f (x0 , y)k
∂f (x0 , y0 )k tangent line: R(t) = [x0 i + y0 j + f (x0 , y0 ) k] + t j + ∂y (c) For (x, y, z) in the plane ∂f ∂f (x0 , y0 )k × j + (x0 , y0 )k [(x − x0 )i + (y − y0 )j + (z − f (x0 , y0 ))k ]. i + = 0. ∂x ∂y From this it follows that ∂f ∂f z − f (x0 , y0 ) = (x − x0 ) (x0 , y0 ) + (y − y0 ) (x0 , y0 ). ∂x ∂y 58. Fix y and set F (x) = f (x, y). Then, for that value of y, h(x, y) = g(F (x)) and thus d hx (x, y) = [g(F (x))] = g (F (x))F (x) = g (f (x, y))fx (x, y). dx The second formula can be derived in the same manner. 59. (a) Set u = ax + by.
Then b
(b) Set u = xm y n . nx
∂w ∂w −a = b(a g (u)) − a(b g (u)) = 0. ∂x ∂y
Then ∂w ∂w − my = nx mxm−1 y n g (u) − my nxm y n−1 g (u) = 0. ∂x ∂y
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SECTION 15.5 ∂x ∂y ∂x ∂y − = (cos θ) (r cos θ) − (−r sin θ) (sin θ) = r ∂r ∂θ ∂θ ∂r k ∂P T kT ∂P ∂P T 61. V = V − 2 = −k = −P ; V +T = −k + T =0 ∂V V V ∂V ∂T V V 60.
R1 R2 R3 ; 62. R = R1 R2 + R1 R3 + R2 R3
∂R = ∂R1
R2 R 3 R1 R2 + R1 R3 + R2 R3
SECTION 15.5 1. interior = {(x, y) : 2 < x < 4,
1 < y < 3} (the
inside of the rectangle), boundary = the union of the four boundary line segments; set is closed.
2. same interior and same boundary as in Exercise 1; set is open
3. interior = the entire set (region between the two concentric circles), boundary = the two circles, one of radius 1, the other of radius 2; set is open.
4. interior = {(x, y) : 1 < x2 < 4} = {(x, y) : −2 < x < − 1} ∪ {(x, y) : 1 < x < 2} (two vertical stripes without the boundary lines), boundary = {(x, y) : x = −2, x = −1, x = 1, or x = 2} (four vertical lines);
set is closed.
5. interior = {(x, y) : 1 < x2 < 4} = {(x, y) : −2 < x < −1} ∪ {(x, y) : 1 < x < 2} (two vertical strips without the boundary lines), boundary = {(x, y) : x = −2, x = −1, x = 1, or x = 2} (four vertical lines); set is neither open nor closed.
2
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SECTION 15.5
6. interior = the entire set (region below the parabola y = x2 ), boundary = the parabola y = x2 ;
the set
is open.
7. interior = region below the parabola y = x2 , boundary = the parabola y = x2 ; the set is closed.
8. interior = the inside of the cube; boundary = the faces of the cube; set is neither open nor closed (upper face of cube is omitted)
9. interior = { (x, y, z) : x2 + y 2 < 1, 0 < z ≤ 4} (the inside of the cylinder), boundary = the total surface of the cylinder (the curved part, the top, and the bottom); the set is closed.
10. interior = the entire set (the inside of the ball of radius 12 , centered at (1,1,1)), boundary = the spherical surface; set is open. 11. (a)
φ
(b)
S
(c)
closed
12. interior = the entire set, boundary = {1, 3};
set is open.
13. interior = {x : 1 < x < 3}, boundary = {1, 3}; set is closed. 14. interior = {x : 1 < x < 3},
boundary = {1, 3};
15. interior = the entire set, boundary = {1}; 16. interior ={x : x < −1},
boundary = {−1};
set is neither open nor closed.
set is open. set is closed.
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SECTION 15.6 17. interior = {x : |x| > 1}, boundary = {1, −1}; 18. interior = φ,
boundary = the entire set;
777
set is neither open nor closed.
set is closed.
19. interior = φ, boundary = {the entire set} ∪ {0};
the set is neither open nor closed.
20. (a) φ is open because it contains no boundary points, φ is closed because it contains its boundary (the boundary is empty). (b) X is open because it contains a neighborhood of each of its points, X is closed because it contains its boundary (the boundary is empty). (c) Suppose that U is open. Let x be a boundary point of X − U . Then every neighborhood of x contains points from X − U . The point x can not be in U because U contains a neighborhood of each of its points. Thus x ∈ X − U . This shows that X − U contains its boundary and is therefore closed. Suppose now that X − U is closed. Let x be a point of U . If no neighborhood of x lies entirely in U , then every neighborhood of x contains points from X − U . This makes x a boundary point of X − U and, since X − U is closed, places x in X − U . This contradiction shows that some neighborhood of x lies entirely in U . Thus U contains a neighborhood of each of its points and is therefore open. (d) Set U = X − F and note that F = X − U . By (c) F = X − U is closed
iff
X − F = U is open.
SECTION 15.6
1.
∂2f = 2A, ∂x2
∂2f = 2C, ∂y 2
2.
∂2f = 6Ax + 2By, ∂x2
3.
∂2f = Cy 2 exy , ∂x2
4.
∂2f = 2 cos y − y 2 sin x, ∂x2
5.
∂2f = 2, ∂x2
∂2f = 2Cx, ∂y 2
∂2f = Cx2 exy , ∂y 2
∂2f 1 =− , ∂x2 4(x + y 2 )3/2
∂2f ∂2f = = 2Bx + 2Cy ∂y∂x ∂x∂y ∂2f ∂2f = = Cexy (xy + 1) ∂y∂x ∂x∂y
∂2f = 2 sin x − x2 cos y, ∂y 2
∂2f = 4(x + 3y 2 + z 3 ), ∂y 2
∂2f ∂2f = = 4y, ∂x∂y ∂y∂x 6.
∂2f ∂2f = = 2B ∂y∂x ∂x∂y
∂2f ∂2f = = 2(y cos x − x sin y) ∂y∂x ∂x∂y
∂2f = 6z(2x + 2y 2 + 5z 3 ) ∂z 2
∂2f ∂2f = = 6z 2 , ∂z∂x ∂x∂z ∂2f x = , ∂y 2 (x + y 2 )3/2
∂2f ∂2f = = 12yz 2 ∂z∂y ∂y∂z ∂2f ∂2f y = =− ∂x∂y ∂y∂x 2(x + y 2 )3/2
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SECTION 15.6
7.
∂2f 1 1 = − 2, ∂x2 (x + y)2 x
∂2f 1 = , ∂y 2 (x + y)2
8.
∂2f 2C(AD − BC)y =− , 2 ∂x (Cx + Dy)3
9.
∂2f = 2(y + z), ∂x2
∂2f ∂2f 1 = = ∂y∂x ∂x∂y (x + y)2
∂2f 2D(AD − BC)x = , 2 ∂y (Cx + Dy)3
∂2f = 2(x + z), ∂y 2
∂2f (AD − BC)(Cx − Dy) ∂2f = = ∂y∂x ∂x∂y (Cx + Dy)3
∂2f = 2(x + y) ∂z 2
all the second mixed partials are 2(x + y + z) 10.
2xy 3 z 3 ∂2f =− , 2 ∂x (1 + x2 y 2 z 2 )2
∂2f 2yx3 z 3 =− 2 ∂y (1 + x2 y 2 z 2 )2
∂2f 2zx3 y 3 = − , ∂z 2 (1 + x2 y 2 z 2 )2
∂2f ∂2f z(1 − x2 y 2 z 2 ) = = ∂y∂x ∂x∂y (1 + x2 y 2 z 2 )2
∂2f ∂2f y(1 − x2 y 2 z 2 ) = = , ∂z∂x ∂x∂z (1 + x2 y 2 z 2 )2
∂2f x(1 − x2 y 2 z 2 ) ∂2f = = ∂y∂z ∂z∂y (1 + x2 y 2 z 2 )2
11.
∂2f = y(y − 1)xy−2 , ∂x2
∂2f = (ln x)2 xy , ∂y 2
∂2f ∂2f = = xy−1 (1 + y ln x) ∂y∂x ∂x∂y
12.
∂2f = − sin(x + z y ), ∂x2
∂2f = z y (ln z)2 [cos(x + z y ) − z y sin(x + z y )] ∂y 2
∂2f = y(y − 1)z y−2 cos(x + z y ) − y 2 z 2y−2 sin(x + z y ) ∂z 2 ∂2f ∂2f = = −z y ln z sin(x + z y ), ∂y∂x ∂x∂y
∂2f ∂2f = = −yz y−1 sin(x + z y ) ∂z∂x ∂x∂z
∂2f ∂2f = = z y−1 (1 + y ln z) cos(x + z y ) − yz 2y−1 (ln z) sin(x + z y ) ∂z∂y ∂y∂z 13.
∂2f = yex , ∂x2
∂2f = xey , ∂y 2
∂2f ∂2f = = ey + ex ∂y∂x ∂x∂y
14.
∂2f 2xy = 2 , 2 ∂x (x + y 2 )2
∂2f −2xy = 2 , 2 ∂y (x + y 2 )2
∂2f ∂2f y 2 − x2 = = 2 ∂y∂x ∂x∂y (x + y 2 )2
15.
∂2f y 2 − x2 = , 2 ∂x (x2 + y 2 )2
∂2f x2 − y 2 = , 2 ∂y (x2 + y 2 )2
∂2f 2xy ∂2f = =− ∂y∂x ∂x∂y (x2 + y 2 )2
16.
∂2f = 6xy 2 cos(x3 y 2 ) − 9x4 y 4 sin(x3 y 2 ), ∂x2
∂2f = 2x3 cos(x3 y 2 ) − 4x6 y 2 sin(x3 y 2 ) ∂y 2
∂2f ∂2f = = 6x2 y cos(x3 y 2 ) − 6x5 y 3 sin(x3 y 2 ) ∂y∂x ∂x∂y 17.
∂2f = −2 y 2 cos 2xy, ∂x2
∂2f = −2 x2 cos 2xy, ∂y 2
∂2f ∂2f = = −[sin 2xy + 2xy cos 2xy] ∂y∂x ∂x∂y
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SECTION 15.6 18.
∂2f 2 = y 4 exy , ∂x2
19.
∂2f = 0, ∂x2
∂2f 2 = exy (2x + 4x2 y 2 ), ∂y 2
∂2f = xz sin y, ∂y 2
∂2f = zex , ∂x2
∂2f = xey , ∂y 2
∂2f ∂2f = = ex , ∂z∂x ∂x∂z 21. x2
∂2f ∂2f 2 = = exy (2y + 2xy 3 ) ∂y∂x ∂x∂y
∂2f = − xy sin z, ∂z 2
∂2f ∂2f = = sin z − z cos y, ∂y∂x ∂x∂y 20.
779
∂2f ∂2f = = y cos z − sin y, ∂x∂z ∂z∂x
∂2f = yez , ∂z 2
∂2f ∂2f = = x cos z − x cos y ∂y∂z ∂z∂y
∂2f ∂2f = = ey , ∂y∂x ∂x∂y
∂2f ∂2f = = ez ∂z∂y ∂y∂z
∂2u ∂2u ∂2u + y 2 2 = x2 + 2xy 2 ∂x ∂x∂y ∂y
−2y 2 (x + y)3
+ 2xy
2xy (x + y)3
+ y2
−2x2 (x + y)3
=0
22. (a) mixed partials are 0 (b) mixed partials are g (x) h (y) (c) by the hint mixed partials for each term xm y n are mnxm−1 y n−1 23. (a) no, since
24.
∂2f ∂2f = ∂y∂x ∂x∂y
∂h = g (x + y) + g (x − y), ∂x ∂2h = g (x + y) + g (x − y), ∂x2
25.
(b)
no, since
∂2f ∂2f = for x = y ∂y∂x ∂x∂y
∂h = g (x + y) − g (x − y) ∂y ∂2h ∂2h = g (x + y) + g (x − y) = 2 ∂y ∂x2
2 2 ∂ f ∂ f ∂f ∂f ∂ ∂ ∂2 ∂2 ∂ ∂2f ∂3f ∂3f = = = = = = ∂x2 ∂y ∂x ∂x∂y ∂x ∂y∂x ∂x∂y ∂x ∂y∂x ∂x ∂y ∂x2 ∂y∂x2 ∧ ∧ ∧ ∧ ∧ ∧ by def. (15.6.5) by def. (15.6.5) by def. by def.
26. (a) as (x, y) tends to (0, 0) along the x-axis, f (x, y) = f (x, 0) = 1 tends to 1; as (x, y) tends to (0, 0) along the line y = x, f (x, y) = f (x, x) = 0 tends to 0; (b) as (x, y) tends to (0, 0) along the x-axis, f (x, y) = f (x, 0) = 0 tends to 0; as (x, y) tends to (0, 0) along the line y = x, f (x, y) = f (x, x) = 27. (a) lim
x→0
(c) lim
x→0
(x)(0) = lim 0 = 0 x2 + 0 x→0
lim
y→0
(0)(y) = lim 0 = 0 y→0 0 + y2
(x)(mx) m m = lim = x→0 1 + m2 x2 + (mx)2 1 + m2
(d) lim+ θ→0
(b)
(θ cos θ)(θ sin θ) = lim+ cos θ sin θ = 0 (θ cos θ)2 + (θ sin θ)2 θ→0
1 2
tends to 12 ;
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SECTION 15.6 (e) By L’Hospital’s rule
lim
x→0
f (x) = lim f (x) = f (0). x→0 x
Thus
xf (x) f (x)/x f (0) = lim = . x→0 x2 + [ f (x) ]2 x→0 1 + [ f (x)/x ]2 1 + [ f (0) ]2 lim
(f)
lim
θ→(π/3)−
=
(cos θ sin 3θ)(sin θ sin 3θ) 1√ = lim − cos θ sin θ = 3 2 2 (cos θ sin 3θ) + (sin θ sin 3θ) 4 θ→(π/3)
(1/t)(sin t)/t sin t 2 ; = lim 2 2 t→∞ 1/t + sin t /t t→∞ 1 + sin2 t
(g) lim
28. (a) lim
x→0 (x2
(c) lim
x→0 (x2
(d) lim
θ→0+
x(0)2 = lim 0 = 0 x→0 + 02 )3/2
does not exist
0 · y2 = lim 0 = 0 y→0 (0 + y 2 )3/2 y→0
(b)
lim
xm2 x2 m2 x3 m2 x = lim 3 = lim ; 2 2 3/2 2 3/2 x→0 |x| (1 + m ) x→0 |x| (1 + m2 )3/2 +m x )
does not exist
(θ cos θ)(θ sin θ)2 = lim+ cos θ sin2 θ = 0 θ→0 [(θ cos θ)2 + (θ sin θ)2 ]3/2
x[f (x)]2 [f (x)/x]2 x3 [f (0)]2 = lim = lim ; x→0 (x2 + [f (x)]2 )3/2 x→0 (1 + [f (x)/x]2 )3/2 x→0 |x|3 (1 + [f (0)]2 )3/2
(e) lim
(f) lim π− θ→ 3
(cos θ sin 3θ)(sin θ sin 3θ)2 3 = lim cos θ sin2 θ = 2 2 3/2 − π 8 [(cos θ sin 3θ) + (sin θ sin 3θ) ] θ→ 3 (1/t) (sin t/t)2
(g) lim t→∞
sin2 t
lim 3/2 = t→∞
(1/t2 ) + (sin2 t/t2 )
29. (a)
does not exist
3/2 ;
1 + sin2 t
does not exist
g(h, 0) − g(0, 0) ∂g (0, 0) = lim = lim 0 = 0, h→0 h→0 ∂x h g(0, h) − g(0, 0) ∂g (0, 0) = lim = lim 0 = 0 h→0 h→0 ∂y h
(b) as (x, y) tends to (0, 0) along the x-axis,
g(x, y) = g(x, 0) = 0
tends to 0;
as (x, y) tends to (0, 0) along the line y = x, g(x, y) = g(x, x) =
1 2
tends to
1 2
30. No; as (x, y) tends to (1, 1) along the line x = 1, f (x, y) = 1 tends to 1; as (x, y) tends to (1, 1) along the line y = 1, f (x, y) =
31. For y = 0,
Since
x−1 1 = 2 x3 − 1 x +x+1
tends to
1 3
f (h, y) − f (0, y) y(y 2 − h2 ) ∂f (0, y) = lim = lim = y. h→0 h→0 h2 + y 2 ∂x h f (h, 0) − f (0, 0) ∂f (0, 0) = lim = lim 0 = 0, h→0 h→0 ∂x h
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SECTION 15.6 we have For x = 0,
781
∂f (0, y) = y for all y. ∂x f (x, h) − f (x, 0) x(h2 − x2 ) ∂f (x, 0) = lim = lim = −x. h→0 h→0 x2 + h2 ∂y h f (0, h) − f (0, 0) ∂f (0, 0) = lim = lim 0 = 0, h→0 h→0 ∂y h
Since we have Therefore In particular
∂f (x, 0) = −x for all x. ∂y ∂2f (0, y) = 1 for all y and ∂y∂x ∂2f (0, 0) = 1 ∂y∂x
∂2f (x, 0) = −1 for all x. ∂x∂y
∂2f (0, 0) = −1. ∂x∂y
while
f (x0 + h, y0 ) − f (x0 , y0 ) lim [f (x0 + h, y0 ) − f (x0 , y0 )] = lim (h) h→0 h→0 h f (x0 + h, y0 ) − f (x0 , y0 ) = lim h lim h→0 h→0 h
32.
=0·
∂f ∂x (x0 , y0 )
=0
33. Since fxy (x, y) = 0, fx (x, y) must be a function of x alone, and fy (x, y) must be a function of y alone. Then f must be of the form f (x, y) = g(x) + h(y). 34.
35.
0.5 0 -0.5
2
2
0.5
-2
0 -2
0 0
0 0
2 -2
2 -2
36.
0.5 2
0 -0.5 -2
0 0 2 -2
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SECTION 15.6
PROJECT 15.6 1. (a)
∂u x2 y 2 + 2xy 3 = , ∂x (x + y)2 x
(b)
x2 y 2 + 2x3 y ∂u = ∂y (x + y)2
∂u 3x2 y 2 (x + y) ∂u +y = = 3u ∂x ∂y (x + y)2
∂u ∂u ∂u = 2xy + z 2 , = 2yz + x2 , = 2xz + y 2 ∂x ∂y ∂z ∂u ∂u ∂u + + = 2xy + z 2 + 2yz + x2 + 2xz + y 2 = (x + y + z)2 ∂x ∂y ∂z
2. (a) (i)
∂2f ∂2f + 2 = 6x − 6x = 0 2 ∂x ∂y
(ii)
∂2f ∂2f + 2 = (− cos x sinh y − sin x cosh y) + (cos x sinh y + sin x cosh y) = 0 2 ∂x ∂y
(iii)
y 2 − x2 ∂2f = , ∂x2 (x2 + y 2 )2
∂2f x2 − y 2 = ∂y 2 (x2 + y 2 )2
∂2f ∂2f y 2 − x2 x2 − y 2 + = + =0 ∂x2 ∂y 2 (x2 + y 2 )2 (x2 + y 2 )2 (b) (i) (ii)
3. (i) (ii)
∂2f ∂2f ∂2f 2x2 − y 2 − z 2 2y 2 − x2 − z 2 2z 2 − x2 − y 2 + 2 + 2 = 2 + 2 + 2 =0 2 2 2 5/2 2 2 5/2 ∂x ∂y ∂z (x + y + z ) (x + y + z ) (x + y 2 + z 2 )5/2 √ ∂2f x+y = e cos 2z , ∂x2
√ √ ∂2f ∂2f x+y x+y = e cos 2 z , = − 2e cos 2z ∂y 2 ∂z 2 √ √ √ ∂2f ∂2f ∂2f x+y x+y x+y + + = e cos 2 z + e cos 2 z + − 2e cos 2z = 0 ∂x2 ∂y 2 ∂z 2
∂2f ∂2f ∂2f ∂2f = = 0 =⇒ − c2 2 = 0 2 2 2 ∂t ∂x ∂t ∂x ∂2f = − 5c2 sin(x + ct) cos(2x + 2ct) − 4c2 cos(x + ct) sin(2x + 2ct) ∂t2 ∂2f = − 5 sin(x + ct) cos(2x + 2ct) − 4 cos(x + ct) sin(2x + 2ct) ∂x2 It now follows that
(iii)
(iv)
4.
∂2f c2 = − , ∂t2 (x + ct)2
∂2f ∂2f − c2 2 = 0 2 ∂t ∂x
2 ∂2f 1 ∂2f 2∂ f = − =⇒ − c =0 ∂x2 (x + ct)2 ∂t2 ∂x2
kx ckt ∂2f ∂2f 2 2 −kx −ckt Ae Ce , = c k + Be + De = k 2 Aekx + Be−kx Ceckt + De−ckt 2 2 ∂t ∂x 2 ∂2f 2∂ f −c =0 It now follows that ∂t2 ∂x2
∂2f ∂2f − c2 2 = c2 g (x + ct) + c2 h (x − ct)] − c2 [g (x + ct) + h (x − ct) = 0 2 ∂t ∂x
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REVIEW EXERCISES REVIEW EXERCISES 1. domain {(x, y) : y > x2 },
range (0, ∞)
2. domain {(x, y) : x, y ∈ R},
range (0, ∞)
3. domain {(x, y, x) : z ≥ x2 + y 2 },
dange [0, +∞)
4. domain {(x, y, z) : x + 2y + z > 0},
range R
1 5. (a) f (x, y) = πx2 y; 3 1 (b) f (x, y) = yx2 ; 2 x + 2y (c) θ = arccos √ 5 x2 + y 2 6. Assume one of the vertices is (x, y, z), x > 0, y > 0, z > 0. x2 y2 V = 8cxy 1 − 2 − 2 a b 7. ellipsoid xy−trace: ellipse 4x2 + 9y 2 = 36 xz−trace: ellipse 4x2 + 36z 2 = 36 yz−trace: ellipse 9y 2 + 36z 2 = 36 9. hyperbolic paraboloid xy−trace: lines x = ±y xz−trace: parabola z = −x2 yz−trace: parabola z = y 2
8. hyperboloid of two sheets xy−trace: none xz−trace: hyperbola 4z 2 − x2 = 4 yz−trace: hyperbola 4z 2 − y 2 = 4 10. elliptic paraboloid xy−trace: parabola 4x2 = y xz−trace: (0, 0) yz−trace: parabola 9z 2 = y
11. cone xy−trace: lines x = ±y xz−trace: lines x = ±z yz−trace: (0, 0)
12. hyperboloid of one sheet xy−trace: ellipse 9x2 + 4y 2 = 36 xz−trace: hyperbola z 2 = 9x2 − 36 yz−trace: hyperbola z 2 = 4y 2 − 36
13.
14. 4 2 2
0
-2 -2 0 2
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REVIEW EXERCISES
15.
16.
2 2
0
-2 2 -2 0 2
0 1 2
17. c = 0, =⇒ 0 = 2x2 + 3y 2 =⇒ (0, 0)
18. c = 0, =⇒ 0 = x2 + y 2 − 4, circle
c = 6, =⇒ 6 = 2x2 + 3y 2 , ellipse
c = 1, =⇒ 5 = x2 + y 2 , circle
c = 12, =⇒ 6 = 2x2 + 3y 2 , ellipse
c = 2, =⇒ 8 = x2 + y 2 , circle √ c = 5, =⇒ 9 = x2 + y 2 , circle
19. c = −4, =⇒ x = −4y 2 , parabola
20. c = 1, =⇒ x2 + y 2 = 1 circle
c = −1, =⇒ x = −y , parabola
c = 4, =⇒ x2 + y 2 = 4 circle
c = 1, =⇒ x = y 2 , parabola
c = 9, =⇒ x2 + y 2 = 9 circle
2
c = 4, =⇒ x = 4y 2 , parabola the origin is omitted 21.
c = 6, 2x + y + 3z = 6, plane
23. (a) f (0, 0) = 1, level curve: f (x, y) = 1
22.
c = 16, x2 + y 2 + 4z 2 = 16, ellipsoid
24. (a) f (2, 0, 1) = 4, level surface: f (x, y, z) = 4 (b) f (1, π, −1) = −1, level surface:
(b) f (ln 2, 1) = 4, level curve: f (x, y) = 4 (c) f (1, −1) = 2e, level curve: f (x, y) = 2e
f (x, y, z) = −1 (c) f (4, π, 1/2) = 0, level surface: f (x, y, z) = 0
25.
f (x + h, y) − f (x, y) (x + h)2 + 2(x + h)y − x2 − 2xy = lim h→0 h→0 h h = lim (2x + h + 2y) = 2x + 2y
fx = lim
h→0
f (x, y + h) − f (x, y) x + 2x(y + h) − x2 − 2xy = lim = 2x h→0 h→0 h h 2
fy = lim 26.
f (x + h, y) − f (x, y) y 2 cos 2(x + h) − y 2 cos 2x = lim h→0 h→0 h h cos 2(x + h) − cos 2x = y 2 lim h→0 h = y 2 cos 2x = −2y 2 sin 2x
fx = lim
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REVIEW EXERCISES f (x, y + h) − f (x, y) (y + h)2 cos 2x − y 2 cos 2x = lim h→0 h→0 h h (y + h)2 − y 2 = cos 2x lim = 2y cos 2x h→0 h
fy = lim
27. fx = 2xy − 2y 3 ;
fy = x2 − 6xy 2
28. gx = (x2 + y 2 )−1/2 − x2 (x2 + y 2 )−3/2 ; 29.
∂z = 2x sin(xy 2 ) + x2 y 2 cos(xy 2 ); ∂x
30. fx = yexy ln(y/x) −
1 xy e x
gy = −xy(x2 + y 2 )−3/2 ∂z = 2x3 y cos(xy 2 ). ∂y
fy = xexy ln(y/x) +
31. hx = −e−x cos(2x − y) − 2e−x sin(2x − y) 32. ux = y 2 sec x tan x + 2x tan y 33. fx =
2y 2 + 2yz ; (x + y + z)2
fy =
34. wx = arctan(y − z)
35.
2x2 + 2xz ; (x + y + z)2
gy =
x2 , y−x
−x 1 + (y − z)2
∂g z . = 2 ∂z x + y2 + z2
hv = ueuv sin uw;
hw = ueuv cos uw
fy = 2x3 y − 12xy 2 − 1;
fyy = 2x3 − 24xy, x2 , y−x
gyy = −
−2xy (x + y + z)2
wz =
∂g y ; = 2 ∂y x + y2 + z2
37. fx = 3x2 y 2 − 4y 3 + 2,
38. gx = 2x ln(y − x) −
fz =
x 1 + (y − z)2
36. hu = veuv sin uw + weuv cos uw;
fxx = 6xy 2 ,
hy = e−x sin(2x − y)
uy = 2y sec x + x2 sec2 y
wy =
∂g x ; = 2 ∂x x + y2 + z2
exy y
gxx = 2 ln(y − x) −
x2 , (y − x)2
39. gx = y sin xy + xy 2 cos xy, gy = x sin xy + x2 y cos xy,
fyx = fxy = 6x2 y − 12y 2
gxy = gyx =
4x x2 ; − y − x (y − x)2
2x x2 + y − x (y − x)2
gxx = 2y 2 cos xy − xy 3 sin xy; gyy = 2x2 cos xy − yx3 sin xy,
gxy = gyx = sin xy + 3xy cos xy − x2 y 2 sin xy 40. fx = 2xex/y + fy = −
x3 x/y e , y2
x2 x/y e , y fyy =
fxx = 2ex/y +
4x x/y x2 x/y e + 2e ; y y
2x3 x/y x4 x/y e + 4e , y3 y
fxy = fyx = −
3x2 x/y x3 x/y e − 3e y2 y
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REVIEW EXERCISES
41. fx = 2xe2y cos(2z + 1), fy = 2x2 e2y cos(2z + 1), fz = −2x2 e2y sin(2z + 1); fxx = 2e2y cos(2z + 1), fyy = 4x2 e2y cos(2z + 1), fzz = −4x2 e2y cos(2z + 1); fxy = fyx = 4xe2y cos(2z + 1), fxz = fzx = −4xe2y sin(2z + 1), fyz = fzy = −4x2 e2y sin(2z + 1) 42. gx = 4xyz 3 + yzexyz ,
gy = 2x2 z 3 + xzexyz ,
gxx = 4yz 3 + y 2 z 2 exyz ,
gyy = x2 z 2 exyz ,
gxy = gyx = 4xz 3 + zexyz + xyz 2 exyz ,
gz = 6x2 yz 2 + xyexyz ; gzz = 12x2 yz + x2 y 2 exyz ;
gxz = gzx = 12xyz 2 + yexyz + xy 2 zexyz
gyz = gzy = 6x2 z 2 + xexyz + x2 yzexyz 43.
∂z = 4x + 6|(1,2,8) = 10 ∂x x = 1 + t,
y = 2,
44.
∂z = 3x|(2,−1,2) = 6 ∂y x = 2, y = −1 + t, z = 2 + 6t
z = 8 + 10t
−6y 45. (a) zy (2, 1) = (2, 1) = −1; the equation for l1 is: 2 20 − 2x2 − 3y 2 x = 2;
y = 1 − t;
z =3+t
−4x 4 (b) zx (2, 1) = (2, 1) = − ; the equation for l2 is: 2 2 3 2 20 − 2x − 3y 3 x = 2 − t; 4
y = 1;
z =3+t 3 3 j − k or 4 i + 3 j + 3 k; 4 4 4(x − 2) + 3(y − 1) + 3(z − 3) = 0.
(c) The normal vector for this plane is: −i − an equation for the plane is: 46. Neither. interior: {(x, y) : 0 < x < 3, 2 < y < 5}
boundary: {(x, y) : x = 0 or x = 3, 2 ≤ y ≤ 5} ∪ {(x, y) : y = 2 or y = 5, 0 ≤ x ≤ 3} 47. Open. interior: {(x, y) : 0 < x2 + y 2 < 4} boundary: {(0, 0)} ∪ {(x, y) : x2 + y 2 = 4} 48. Closed. interior: {(x, y) : x + y > 4} boundary: {(x, y) : x + y = 4} 49. Closed. interior: {(x, y, z) : 0 < x < 2, 0 < y, 0 < z, y 2 + z 2 < 4} boundary: the quarter disks x = 0, y 2 + z 2 ≤ 4; the squares z = 0, 0 ≤ x, y ≤ 2;
x = 2, y 2 + z 2 ≤ 4;
y = 0, 0 ≤ x, z ≤ 2;
the cylindrical surface y + z = 4, 0 ≤ x ≤ 2, y, z ≥ 0 2
2
and
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REVIEW EXERCISES 50. Neither. interior: {(x, y, z) : 0 < x2 + y 2 < z < 4} boundary: the cone z = x2 + y 2 and the disk x2 + y 2 ≤ 4, z = 4 51. (a) fx = yg (xy),
fy = xg (xy);
(b) fxx = y 2 g (xy), 52. fx = −
y , x2 + y 2
xfx − yfy = xyg − xyg = 0
fyy = x2 g (xy);
fy =
x ; x2 + y 2
x2 fxx − y 2 fyy = x2 y 2 g − x2 y 2 g = 0
fxx =
2xy , (x2 + y 2 )2
fyy =
−2xy (x2 + y 2 )2
fxx + fyy = 0 ∂2f ∂2f = x2 exy = y 2 exy = ∂y∂x ∂x∂y
53. No.
54. (a) lim f (x, 0) = lim 0 = 0 x→0
x→0
2x2 mx 2mx (c) lim 4 = lim 2 =0 2 2 x→0 x + m x x→0 m + x2 lim(x,y)→(0,0) f (x, y) does not exist.
(b) lim f (0, y) = lim 0 = 0 y→0
y→0
2x2 ax2 2a (d) lim 4 = 2 4 x→0 x + a x 1 + a2
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SECTION 16.1
CHAPTER 16 SECTION 16.1 1. ∇f = (6x − y) i + (1 − x) j
2. ∇f = (2Ax + By)i + (Bx + 2Cy)j
3. ∇f = exy [ (xy + 1) i + x2 j] 4. ∇f =
(x2
1 [(y 2 − x2 + 2xy)i + (y 2 − x2 − 2xy)j] + y 2 )2
5. ∇f = 2y 2 sin(x2 + 1) + 4x2 y 2 cos(x2 + 1) i + 4xy sin(x2 + 1) j 6. ∇f =
2x 2y i+ 2 j 2 +y x + y2
x2
7. ∇f = (ex−y + ey−x ) i + (−ex−y − ey−x ) j = (ex−y + ey−x )(i − j) 8. ∇f =
AD − BC [ yi − xj ] (Cx + Dy)2
9. ∇f = (z 2 + 2xy) i + (x2 + 2yz) j + (y 2 + 2zx) k x
10. ∇f =
x2 + y 2 + z 2
y
i+
x2 + y 2 + z 2
j+
z x2 + y 2 + z 2
k
11. ∇f = e−z (2xy i + x2 j − x2 y k) xyz xyz + yz ln(x + y + z) i + + xz ln(x + y + z) j 12. ∇f = x+y+z x+y+z xyz + + xy ln(x + y + z) k x+y+z
13. ∇f = ex+2y cos z 2 + 1 i + 2ex+2y cos z 2 + 1 j − 2zex+2y sin z 2 + 1 k 14. ∇f = eyz
2
/x3
−
3yz 2 z2 2yz i + 3j + 3 k 4 x x x
2 1 15. ∇f = 2y cos(2xy) + i + 2x cos(2xy) j + k x z 16. ∇f =
2xy − 3z 4 z
x2 j− i+ z
17. ∇f = (4x − 3y) i + (8y − 3x) j;
x2 y + 12xz 3 z2
k
at (2, 3), ∇f = −i + 18j
18. ∇f =
1 (−2yi + 2xj), (x − y)2
3 1 ∇f (3, 1) = − i + j 2 2
19. ∇f =
2y 2x i+ 2 j; x2 + y 2 x + y2
at (2, 1), ∇f =
2 4 i+ j 5 5
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SECTION 16.1 20. ∇f = tan−1 (y/x) −
xy x2 + y 2
i+
x2 x2 + y 2
21. ∇f = (sin xy + xy cos xy) i + x2 cos xy j; 22. ∇f = e−(x
2
+y 2 )
∇f (1, 1) =
j,
π 1 − 4 2
i+
1 j 2
at (1, π/2), ∇f = i ∇f (1, −1) = e−2 (i − j)
[(y − 2x2 y)i + (x − 2xy 2 )j],
23. ∇f = −e−x sin (z + 2y) i + 2e−x cos (z + 2y) j + e−x cos (z + 2y) k; √ at (0, π/4, π/4), ∇f = − 12 2 (i + 2j + k) 24. ∇f = cos πzi − cos πzj − π(x − y) sin πzk, 25. ∇f = i −
y y2
+
z2
j−
z y2
+
z2
k;
27. (a) ∇f (0, 2) = 4 i
2
2
(b) ∇f
1
1, 0,
at (2, −3, 4),
26. ∇f = − sin(xyz )(yz i + xz j + 2xyzk), 2
∇f
1 4 π, 6 π
1 2
= −πk
∇f = i +
3 4 j− k 5 5
√
2 1 π 1 i+πj− k ∇f π, , −1 = − 4 2 4 2 =
√ √ −1 + 3 1 −1 + 3 √ √ −1 − i+ − − j 2 2 2 2
(c) ∇f (1, e) = (1 − 2e) i − 2 j 1 1 27 28. (a) ∇f (1, 2, −3) = √ i + √ j − √ k 8 2 2 2 8 2 √ 3 π i+ j+k (c) ∇f (1, e2 , π/6) = 2 12e2
(b) ∇f (1, −2, 3) = −
1 1 5 i+ j+ k 18 9 18
29. For the function f (x, y) = 3x2 − xy + y, we have f (x + h) − f (x) = f (x + h1 , y + h2 ) − f (x, y) = 3(x + h1 )2 − (x + h1 )(y + h2 ) + (y + h2 ) − 3x2 − xy + y = [(6x − y) i + (1 − x) j] · (h1 i + h2 j) + 3h21 − h1 h2 = [(6x − y) i + (1 − x) j] · h + 3h21 − h1 h2 The remainder g(h) = 3h21 − h1 h2 = (3h1 i − h1 j) · (h1 i + h2 j) ,
and
|g(h)| 3h1 i − h1 j · h · cos θ = ≤ 3h1 i − h1 j h h Since 3h1 i − h1 j → 0 as h → 0 it follows that ∇f = (6x − y) i + (1 − x) j 30. f (x + h) − f (x) = [(x + 2y) i + (2x + 2y) j] · [h1 i + h2 j] + 12 h21 + 2h1 h2 + h22 ; g(h) = 12 h21 + 2h1 h2 + h22 is o(h).
789
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SECTION 16.1
31. For the function f (x, y, z) = x2 y + y 2 z + z 2 x, we have f (x + h) − f (x) = f (x + h1 , y + h2 , z + h3 ) − f (x, y, z)
= (x + h1 )2 (y + h2 ) + (y + h2 )2 (z + h3 ) + (z + h3 )2 (x + h1 ) − x2 y + y 2 z + z 2 x = 2xy + z 2 h1 + 2yz + x2 h2 + 2xz + y 2 h3 + (2xh2 + yh1 + h1 h2 ) h1 +
(2yh3 + zh2 + h2 h3 ) h2 + (2zh1 + xh3 + h1 h3 ) h3 = 2xy + z 2 i + 2yz + x2 j + 2xz + y 2 k · h + g(h) · h, where g(h) = (2xh2 + yh1 + h1 h2 ) i + (2yh3 + zh2 + h2 h3 ) j + (2zh1 + xh3 + h1 h3 ) k Since
|g(h)| → 0 as h → 0 h
it follows that
∇f = 2xy + z 2 i + 2yz + x2 j + 2xz + y 2 k
32. f (x + h) − f (x) = (2xy + 2h2 x + h1 y) i + 2x2 j + g(h) = h21 h2 is o(h) and ∇f = 4xy i = 2x2 j + 33. ∇f = F(x, y) = 2xy i + 1 + x2 j
⇒
∂f = x2 + g (y) = 1 + x2 ∂y Thus, f (x, y) = x2 y + y + C
k.
∂f = 2xy ∂x g (y) = 1
⇒
Now,
1 z2
1 k · (h1 i + h2 j + h3 k) + h21 ; z(z + h3 )
⇒ ⇒
f (x, y) = x2 y + g(y) for some function g. g(y) = y + C, C a constant.
1 34. ∇f = (2xy + x)i + (x2 + y)j =⇒ fx = 2xy + x =⇒ f (x, y) = x2 y + x2 + g(y) 2 Now, fy = x2 + g (y) = x2 + y =⇒ g (y) = y =⇒ g(y) = 12 y 2 + C Thus,
f (x, y) = x2 y + 12 x2 + 12 y 2 + C
35. ∇f = F(x, y) = (x + sin y) i + (x cos y − 2y) j ⇒ for some function g. ∂f Now, = x cos y + g (y) = x cos y − 2y ∂y Thus, f (x, y) = 12 x2 + x sin y − y 2 + C.
⇒
∂f = x + sin y ⇒ f (x, y) = ∂x g (y) = −2y
⇒
1 2
x2 + x sin y + g(y)
g(y) = −y 2 + C, C a constant.
36. ∇f = yzi + (xz + 2yz)j + (xy + y 2 )k =⇒ fx = yz =⇒ f (x, y, z) = xyz + g(y, z). fy = xz + gy = xz + 2yz =⇒ gy = 2yz =⇒ g(y, z) = y 2 z + h(z) =⇒ f (x, y, z) = xyz + y 2 z + h(z). fx = xy + y 2 + h (z) = xy + y 2 =⇒ h (z) = 0 =⇒ h(z) = C. Thus, f (x, y, z) = xyz + y 2 z + C. 37. With r = (x2 + y 2 + z 2 )1/2 we have ∂r x = , ∂x r (a)
∇(ln r) =
∂r y = , ∂y r
∂r z = . ∂z r
∂ ∂ ∂ (ln r) i + (ln r) j + (ln r)k ∂x ∂y ∂z
1 ∂r 1 ∂r 1 ∂r i+ j+ k r ∂x r ∂y r ∂z x y z r = 2 i+ 2 j+ 2 k= 2 r r r r
=
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SECTION 16.1 (b)
∇(sin r) =
791
∂ ∂ ∂ (sin r) i + (sin r) j + (sin r)k ∂x ∂y ∂z
∂r ∂r ∂r i + cos r j + cos r k ∂x ∂y ∂z x y z = (cos r) i + (cos r) j + (cos r) k r r r
cos r r = r
= cos r
(c) ∇er =
er r
r
[ same method as in (a) and (b) ]
38. With rn = (x2 + y 2 + z 2 )n/2 we have n ∂rn = (x2 + y 2 + z 2 )(n/2)−1 (2x) = n(x2 + y 2 + z 2 )(n−2)/2 x = nrn−2 x. ∂x 2 Similarly ∂rn = nrn−2 y ∂y
and
∂rn = nrn−2 z. ∂z
Therefore ∇rn = nrn−2 xi + nrn−2 yj + nrn−2 zk = nrn−2 (xi + yj + zk) = nrn−2 r
39. (a) ∇f = 2x i + 2y j = 0
=⇒
x = y = 0;
∇f = 0 at (0, 0).
(b)
40. (a) (c)
(c) f has an absolute minimum at (0, 0)
∇f =
−1 4 − x2 − y 2
(xi + yj) = 0 at (0, 0)
f has a maximum at (0, 0)
(b)
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SECTION 16.2
41. (a) Let c = c1 i + c2 j + c3 k. First, we take h = hi. Since c · h is o(h), c·h c1 h = lim = c1 . h→0 h h
0 = lim
h→0
Similarly, c2 = 0 and c3 = 0. (b) (y − z) · h = [f (x + h) − f (x) − z · h] + [y · h − f (x + h) + f (x) ] = o(h) + o(h) = o(h), so that, by part (a), y − z = 0. 42.
lim g(h) = lim
h→0
h→0
g(h) h h
=
lim h
h→0
g(h) lim h→0 h
= (0)(0) = (0).
43. (a) In Section 15.6 we showed that f was not continuous at (0, 0). It is therefore not differentiable at (0, 0). (b) For (x, y) = (0, 0), ∂f 2 2y 3 = 4 = ∂x y y
2y(y 2 − x2 ) ∂f = . As (x, y) tends to (0, 0) along the positive y-axis, ∂x (x2 + y 2 )2
tends to ∞.
SECTION 16.2 1. ∇f = 2xi + 6yj,
∇f (1, 1) = 2i + 6j,
2. ∇f = [1 + cos(x + y)]i + cos(x + y)j, √ fu (0, 0) = ∇f (0, 0) · u = 5
u=
1 2
√
∇f (0, 0) = 2i + j,
3. ∇f = (ey − yex ) i + (xey − ex ) j, ∇f (1, 0) = i + (1 − e)j, 1 fu (1, 0) = ∇f (1, 0) · u = (7 − 4e) 5 1 (−2yi + 2xj), ∇f (1, 0) = 2j, (x − y)2 √ fu (1, 0) = ∇f (1, 0) · u = − 3
4. ∇f =
u=
1 u = √ (2i + j), 5
u=
1 (3i + 4j), 5
√ 1 (i − 3j), 2
(b − a)x a−b (a − b)y (i − j), i+ j, ∇f (1, 1) = (x + y)2 (x + y)2 4 1√ 2 (a − b) fu (1, 1) = ∇f (1, 1) · u = 4
5. ∇f =
√ fu (1, 1) = ∇f (1, 1) · u = −2 2
2 (i − j),
u=
1√ 2 (i − j), 2
d−c 1 [(d − c)yi + (c − d)xj] , ∇f (1, 1) = (i − j), 2 (cx + dy) (c + d)2 d−c √ fu (1, 1) = ∇f (1, 1) · u = (c + d) c2 + d2
6. ∇f =
2x 2y i+ 2 j, ∇f (0, 1) = 2 j, 2 +y x + y2 2 fu (0, 1) = ∇f (0, 1) · u = √ 65
7. ∇f =
x2
1 u = √ (8 i + j), 65
u= √
c2
1 (ci − dj), + d2
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SECTION 16.2 π π 8. ∇f = 2xyi + (x2 + sec2 y)j, ∇f (−1, ) = − i + 3j, 4 2
π 1 π π = ∇f −1, · u = −√ +6 fu −1, 4 4 5 2
1 u = √ (i − 2j) 5
9. ∇f = (y + z)i + (x + z)j + (y + x)k, ∇f (1, −1, 1) = 2j, 2√ 6 fu (1, −1, 1) = ∇f (1, −1, 1) · u = 3 10. ∇f = (z 2 + 2xy)i + (x2 + 2yz)j + (y 2 + 2zx)k, √ 10 fu (1, 0, 1) = ∇f (1, 0, 1) · u = 10
u=
1 6
√
6 (i + 2j + k),
∇f (1, 0, 1) = i + j + 2k,
11. ∇f = 2 x + y 2 + z 3 i + 2yj + 3z 2 k , ∇f (1, −1, 1) = 6(i − 2j + 3k), √ fu (1, −1, 1) = ∇f (1, −1, 1) · u = −3 2 12. ∇f = (2Ax + Byz)i + (Bxz + 2Cy)j + Bxyk, u= √
1 (Ai + Bj + Ck); 2 A + B2 + C 2
13. ∇f = tan−1 (y + z) i + 1 u = √ (i + j − k), 3
793
1 u = √ (3j − k) 10
u=
1 2
√
2 (i + j),
∇f (1, 2, 1) = 2(A + B)i + (B + 4C)j + 2Bk
fu (1, 2, 1) = ∇f (1, 2, 1) · u =
2A2 + B 2 + 2AB + 6BC √ A2 + B 2 + C 2
π 1 1 ∇f (1, 0, 1) = i + j + k, 4 2 2 √ 3 π π fu (1, 0, 1) = ∇f (1, 0, 1) · u = √ = 12 4 3
x x j+ k, 1 + (y + z)2 1 + (y + z)2
14. ∇f = (y 2 cos z − 2πyz 2 cos πx + 6zx)i + (2xy cos z − 2z 2 sin πx)j + (−xy 2 sin z − 4yz sin πx + 3x2 )k ∇f (0, −1, π) = (2π 3 − 1)i; 15. ∇f =
x2
y x i+ 2 j, 2 +y x + y2
u=
1 (2i − j + 2k), 3 1
u=
x2
+
y2
(−xi − yj) ,
17. ∇f = (2Ax + 2By) i + (2Bx + 2Cy) j, ∇f (a, b) = (2aA + 2bB)i + (2aB + 2bC) j √ √ (a) u = 12 2 (−i + j), fu (a, b) = ∇f (a, b) · u = 2 [a(B − A) + b(C − B)] √ √ (b) u = 12 2 (i − j), fu (a, b) = ∇f (a, b) · u = 2 [a(A − B) + b(B − C)] z z i − j + ln x y
1 u = √ (i + j − k); 3
x k, y
∇f (1, 1, 2) = 2i − 2j
fu (1, 1, 2) = ∇f (1, 1, 2) · u = 0
1
fu (x, y) = ∇f · u = −
16. ∇f = exy (y 2 + xy 3 − y 3 )i + (x − 1)(2y + xy 2 )j , ∇f (0, 1) = −2j 1 4√ u = √ (−i + 2j), fu (0, 1) = ∇f (0, 1) · u = − 5 5 5
18. ∇f =
2 (2π 3 − 1). 3
fu (0, −1, π) = ∇f (0, −1, π) · u =
x2
+ y2
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SECTION 16.2
19. ∇f = ey
2
−z 2
∇f (1, 2, −2) = i + 4j + 4k, r (t) = i − 2 sin (t − 1) j − 2et−1 k, √ 7√ r (1) = i − 2k, u = 15 5 (i − 2k), fu (1, 2, −2) = ∇f (1, 2, −2) · u = − 5 5
(i + 2xyj − 2xzk),
at (1, 2, −2)
t = 1,
20. ∇f = 2xi + zj + yk,
∇f (1, −3, 2) = 2i + 2j − 3k 1 1√ Direction: r (−1) = −2i + 3j − 3k, u = √ (−2i + 3j − 3k), fu (1, −3, 2) = ∇f (1, −3, 2) · u = 22 2 22 21. ∇f = (2x + 2yz) i + 2xz − z 2 j + (2xy − 2yz) k, ∇f (1, 1, 2) = 6 i − 2 k
1 The vectors v = ±(2 i + j − 3 k) are direction vectors for the given line; u = ± √ [2 i + j − 3 k] 14 18 √ are corresponding unit vectors; fu (1, 1, 2) = ∇f (1, 1, 2) · (±u) = ± 14 22. ∇f = ex (cos πyzi − πz sin πyzj − πy sin πyzk),
∇f (0, 1, 12 ) = −
π j−πk 2
The vectors v = ±(2 i + 3 j + 5 k) are direction vectors for the line; u = ± 13π fu (0, 1, 12 ) = ∇f (0, 1, 12 ) · (±u) = ∓ √ 2 38
are corresponding unit vectors;
1 √ [2 i + 3 j + 5 k] 38
√ ∇f = 2 2,
∇f 1 = √ (i + j) ∇f 2 √ 1 f increases most rapidly in the direction u = √ (i + j); the rate of change is 2 2. 2 √ 1 f decreases most rapidly in the direction v = − √ (i + j); the rate of change is −2 2. 2
23. ∇f = 2y 2 e2x i + 2ye2x j,
∇f (0, 1) = 2 i + 2 j,
24. ∇f = [1 + cos(x + 2y)]i + 2 cos(x + 2y)j, ∇f (0, 0) = 2i + 2j √ 1 Fastest increase in direction u = √ (i + j), rate of change ∇f (0, 0) = 2 2 2 √ 1 Fastest decrease in direction v = − √ (i + j), rate of change −2 2 2 x
25. ∇f =
x2
+
y2
z2
y
i+
x2
y2
z2
j+
+ + + 1 ∇f (1, −2, 1) = √ (i − 2 j + k), ∇f = 1 6
z x2
+ y2 + z2
k,
1 f increases most rapidly in the direction u = √ (i − 2 j + k); the rate of change is 1. 6 1 f decreases most rapidly in the direction v = − √ (i − 2 j + k); the rate of change is −1. 6 26. ∇f = (2xzey + z 2 )i + x2 zey j + (x2 ey + 2xz)k, ∇f (1, ln 2, 2) = 12i + 4j + 6k 1 Fastest increase in direction u = (6i + 2j + 3k), rate of change ∇f (1, ln 2, 2) = 14 7 1 Fastest decrease in direction v = − (6i + 2j + 3k), rate of change −14 7 27. ∇f = f (x0 ) i. If f (x0 ) = 0, the gradient points in the direction in which f increases: to the right if f (x0 ) > 0, to the left if f (x0 ) < 0.
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SECTION 16.2 28. 0;
the vector c =
795
∂f ∂f (x0 , y0 )i − (x0 , y0 )j is perpendicular to the gradient ∇f (x0 , y0 ) and ∂y ∂x
points along the level curve of f at (x0 , y0 ). 29. (a) (b) 30. (a) (b)
√ f (h, 0) − f (0, 0) h2 |h| = lim = lim does not exist lim h→0 h→0 h→0 h h h no; by Theorem 16.2.5 f cannot be differentiable at (0, 0) o(h) g(x + h)o(h) = g(x + h) → g(x)(0) = 0 h h |[g(x + h) − g(x)]∇f (x) · h| [g(x + h) − g(x)]∇f (x)h ≤ h h
by Schwarz’s inequality = |g(x + h) − g(x)| · ∇f (x) → 0 31. ∇λ(x, y) = − 83 xi − 6yj 8 (a) ∇λ(1, −1) = − i = 6j, 3
8 i − 6j −∇λ(1, −1) 2√ = 32 √ , λu (1, −1) = ∇λ(1, −1) · u = − 97 ∇λ(1, −1) 3 3 97 (b) u = i, λu (1, 2) = ∇λ(1, 2) · u = − 83 i − 12j · i = − 83 √ √ 1√ 26 (c) u = 12 2 (i + j), λu (2, 2) = ∇λ(2, 2) · u = − 16 2 3 i − 12 j · 2 2 (i + j) = − 3
u=
32. ∇I = −4xi − 2yj. We want the curve r(t) = x(t)i + y(t)j which begins at (−2, 1) and has tangent vector r (t) in the direction ∇I. We can satisfy these conditions by setting x (t) = −4x(t), x(0) = −2;
y (t) = −2y(t), y(0) = 1.
These equations imply that x(t) = −2e−4t ,
y(t) = e−2t .
Eliminating the parameter, we get x = −2y 2 ; the particle will follow the parabolic path x = −2y 2 toward the origin. 33. (a) The projection of the path onto the xy-plane is the curve C : r(t) = x(t)i + y(t)j which begins at (1, 1) and at each point has its tangent vector in the direction of −∇f. Since ∇f = 2xi + 6yj, we have the initial-value problems x (t) = −2x(t),
x(0) = 1
and
y (t) = −6y(t),
x(t) = e−2t
and
y(t) = e−6t .
y(0) = 1.
From Theorem 7.6.1 we find that
Eliminating the parameter t, we find that C is the curve y = x3 from (1, 1) to (0, 0).
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SECTION 16.2 (b) Here x (t) = −2x(t),
x(0) = 1
y (t) = −6y(t),
and
y(0) = −2
so that x(t) = e−2t
and
y(t) = −2e−6t .
Eliminating the parameter t, we find that the projection of the path onto the xy-plane is the curve y = −2x3 from (1, −2) to (0, 0). 1 2 x − y 2 ; ∇f = xi − 2yj, so we choose the projection r(t) = x(t)i + y(t)j of the path 2 onto the xy-plane such that x (t) = x(t), y (t) = −2y(t) 1 (a) With initial point (−1, 1, − 12 ), we get x(t) = −et , y(t) = e−2t , or y = 2 x from (−1, 1), in the direction of decreasing x.
34. z = f (x, y) =
(b) With initial point (1, 0, 12 ), we get x(t) = et , y(t) = 0, or the x-axis from (1, 0), in the direction of increasing x. 35. The projection of the path onto the xy-plane is the curve C : r(t) = x(t)i + y(t)j which begins at (a, b) and at each point has its tangent vector in the direction of −∇f = − 2a2 xi + 2b2 yj . We can satisfy these conditions by setting x (t) = −2a2 x(t),
x(0) = a2
and
y (t) = −2b2 y(t),
y(0) = b
so that x(t) = ae−2a
2
Since
x b2 a a2
C is the curve (b)
b2
x
b2
= (a) y
a2
t
and
y(t) = be−2b t . 2
b2 y a2 2 = e−2a t = , b
from (a, b) to (0, 0).
36. The particle must go in he direction −∇T = −ey cos xi − ey sin xj, so we set x (t) = −ey(t) cos x(t), y (t) = −ey(t) sin x(t). Dividing, we have
sin x(t) dy y (t) = , or = tan x. x (t) cos x(t) dx
With initial point
(0, 0), we get y = ln | sec x|, in the direction of decreasing x (since x (0) < 0). 37. We want the curve C : r(t) = x(t)i + y(t)j which begins at (π/4, 0) and at each point has its tangent vector in the direction of √ √ ∇T = − 2 e−y sin x i − 2 e−y cos x j. From √ x (t) = − 2 e−y sin x
and
√ y (t) = − 2 e−y cos x
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SECTION 16.2
797
we obtain dy y (t) = = cot x dx x (t) so that y = ln | sin x| + C. √ √ Since y = 0 when x = π/4, we get C = ln 2 and y = ln | 2 sin x|. As ∇T (π/4, 0) = −i − j, the curve √ y = ln | 2 sin x| is followed in the direction of decreasing x. 38. ∇z = (1 − 2x)i + (2 − 6y)j,
so the projection of the path onto the xy-plane satisfies
2 − 6y dy = . With initial point dx 1 − 2x 3y = (2x − 1)3 + 1, in the direction of increasing x. 1 − 2x(t), y (t) = 2 − 6y(t),
39.
or
(0, 0),
x (t) =
this gives the curve
f 2 + h, (2 + h)2 − f (2, 4) 3(2 + h)2 + (2 + h)2 − 16 = lim lim h→0 h→0 h h 2 4h + h = lim 4 = lim 4(4 + h) = 16 h→0 h→0 h
2 h+8 h+8 , 4 + h − f (2, 4) f 3 + (4 + h) − 16 4 4 = lim lim h→0 h→0 h h
(a)
(b)
+ 3h + 12 + 4 + h − 16 h→0 h 3 = lim 16 h + 4 = 4 = lim
3 2 16 h
h→0
(c) u =
1 17
√
17 (i + 4j),
∇f (2, 4) = 12i + j;
fu (2, 4) = ∇f (2, 4) · u =
16 17
√
17
(d) The limits computed in (a) and (b) are not directional derivatives. In (a) and (b) we have, in essence, computed ∇f (2, 4) · r0 taking r0 = i + 4j in (a) and r0 = 14 i + j in (b). In neither case is r0 a unit vector. 40. ∇f =
−GM m −GM m (xi + yj + zk) = r r3 (x2 + y 2 + z 2 )3/2
∂f ∂f i+ j; ∂x ∂y
∂f ∂f ∂f ∂f i+ j · (cos θ i + sin θ j) = cos θ + sin θ fu (x, y) = ∇f · u = ∂x ∂y ∂x ∂y (b) ∇f = 3x2 + 2y − y 2 i + (2x − 2xy) j, ∇f (−1, 2) = 3 i + 2 j √ 2 3−3 fu (−1, 2) = 3 cos(2π/3) + 2 sin(2π/3) = 2
41. (a) u = cos θ i + sin θ j,
∇f (x, y) =
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SECTION 16.3
√ √ √ 2 2 ∂f 5π ∂f 5π 2y 2 2y − − 42. = cos + sin = 2xe + 2x e = − 2xe2y (1 + x) ∂x 4 ∂y 4 2 2 √ √ fu (2, ln 2) = − 2 · 2 · e2 ln 2 (1 + 2) = −24 2
∂(f g) ∂(f g) ∂(f g) ∂g ∂f ∂g ∂f ∂g ∂f i+ j+ k= f +g i+ f +g j+ f +g k 43. ∇(f g) = ∂x ∂y ∂z ∂x ∂x ∂y ∂y ∂z ∂z
∂g ∂f ∂g ∂g ∂f ∂f =f i+ j+ k +g i+ j+ k ∂x ∂y ∂z ∂x ∂y ∂z fu (x, y)
= f ∇g + g ∇f
f ∂ f ∂ f ∂ f ∇ = i+ j+ k g ∂x g ∂y g ∂z g
44.
∂f ∂g ∂f ∂f ∂g ∂g g−f g−f g−f ∂y ∂y ∂x ∂x ∂z ∂z k = i+ j+ g2 g2 g2 = 45. ∇f n =
g(x)∇f (x) − f (x)∇g(x) g 2 (x)
∂f ∂f ∂f ∂f n ∂f n ∂f n i+ j+ k = nf n−1 i + nf n−1 j + nf n−1 k = nf n−1 ∇f ∂x ∂y ∂z ∂x ∂y ∂z
SECTION 16.3 1. f (b) = f (1, 3) = −2; f (a) = f (0, 1) = 0; f (b) − f (a) = −2 ∇f = 3x2 − y i − x j; b − a = i + 2 j and ∇f · (b − a) = 3x2 − y − 2x The line segment joining a and b is parametrized by x = t,
y = 1 + 2t,
0≤t≤1
Thus, we need to solve the equation 3t2 − (1 + 2t) − 2t = −2,
which is the same as
The solutions are: t = 13 , t = 1. Thus, c = ( 13 , 53 )
3t2 − 4t + 1 = 0, 0 ≤ t ≤ 1
satisfies the equation.
Note that the endpoint b also satisfies the equation. 2. ∇f = 4zi − 2yj + (4x + 2z)k, b − a = i + 2j + k,
f (a) = f (0, 1, 1) = 0,
f (b) = f (1, 3, 2) = 3
so we want (x, y, z) such that
∇f · (b − a) = 4z − 4y + 4x + 2z = 6z − 4y + 4x = f (b) − f (a) = 3 Parametrizing the line segment from a to b by x(t) = t, y(t) = 1 + 2t, z(t) = 1 + t, we get t = 12 , or c = ( 12 , 2, 32 ) 3. (a) f (x, y, z) = a1 x + a2 y + a3 z + C
(b)
f (x, y, z) = g(x, y, z) + a1 x + a2 y + a3 z + C
4. Using the mean-value theorem 16.3.1, there exists c such that ∇f (c) · (b − a) = 0
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SECTION 16.3
799
5. (a) U is not connected (b)
(i) g(x) = f (x) − 1
(ii) g(x) = −f (x)
6. By the mean-value theorem f (x1 ) − f (x2 ) = ∇f (c) · (x1 − x2 ) for some point c on the line segment x1 x2 .
Since Ω is convex, c is in Ω.
Thus
|f (x1 ) − f (x2 )| = |∇f (c) · (x1 − x2 )| ≤ ∇f (c)x1 − x2 ≤ M x1 − x2 . ∧ by Schwarz’s inequality 7. ∇f = 2xyi + x2 j;
∇f (r(t)) · r (t) = 2i + e2t j · (et i − e−t j) = et
8. ∇f = i − j;
∇f (r(t)) · r (t) = (i − j) · (ai − ab sin atj) = a(1 + b sin at)
2 cos t −2x 2y −2 sin t i+ j ∇f (r(t)) = 2 i+ 2 j, 2 2t 2 2 2 2 1 + cos 1 + cos2 2t 1 + (y − x ) 1 + (y − x )
−2 sin t 2 cos t −4 sin t cos t −2 sin 2t i+ j · (cos t i − sin t j) = = ∇f (r(t)) · r (t) = 1 + cos2 2t 1 + cos2 2t 1 + cos2 2t 1 + cos2 2t
9. ∇f =
10. ∇f =
1 (4xi + 3y 2 j) 2x2 + y 3
1 −2/3 8e4t + 1 2t 2/3 2t (4e t i + 3t j) · (2e i + j) = 2e4t + t 3 2e4t + t
1 11. ∇f = (ey − ye−x ) i + (xey + e−x ) j; ∇f (r(t)) = (tt − ln t) i + tt ln t + j t
1 1 1 1 ∇f (r(t)) · r (t) = (tt − ln t) i + tt ln t + j · i + [1 + ln t] j = tt + ln t + [ln t]2 + t t t t 1
∇f (r(t)) · r (t) =
12. ∇f =
2 (xi + yj + zk) x2 + y 2 + z 2
∇f (r(t)) · r (t) =
2 4e4t (sin ti + cos tj + e2t k) · (cos ti − sin tj + 2e2t k) = 4t 1+e 1 + e4t
13. ∇f = yi + (x − z)j − yk; ∇f (r(t)) · r (t) = t2 i + t − t3 j − t2 k · i + 2tj + 3t2 k = 3t2 − 5t4 14. ∇f = 2x i + 2y j ∇f (r(t)) · r (t) = (2a cos ωti + 2b sin ωtj) · (−ωa sin ωti + ωb cos ωtj + bωk) = 2ω(b2 − a2 ) sin ωt cos ωt 15. ∇f = 2xi + 2yj + k; ∇f (r(t)) · r (t) = (2a cos ωt i + 2b sin ωt j + k) · (−aω sin ωt i + bω cos ωt j + bωk) = 2ω b2 − a2 sin ωt cos ωt + bω
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SECTION 16.3
16. ∇f = y 2 cos(x + z)i + 2y sin(x + z)j + y 2 cos(x + z)k ∇f (r(t)) · r (t) = [cos2 t cos(2t + t3 )i + 2 cos t sin(2t + t3 )j + cos2 t cos(2t + t3 )k] · (2i − sin tj + 3t2 k) = cos t[(2 + 3t2 ) cos t cos(2t + t3 ) − 2 sin t sin(2t + t3 )] 17.
∂u dx ∂u dy du = + = (2x − 3y)(− sin t) + (4y − 3x)(cos t) dt ∂x dt ∂y dt = 2 cos t sin t + 3 sin2 t − 3 cos2 t = sin 2t − 3 cos 2t
du y x ∂u dx ∂u dy 1 2 = · + · = 1+2 3t + 2 −3 − 2 dt ∂x dt ∂y dt x y t
1 3 2 = 1 + 2 3t2 + (2t2 − 3) − 2 = 3t2 + 4 + 2 t t t
18.
19.
∂u dx ∂u dy du = + dt ∂x dt ∂y dt
= (ex sin y + ey cos x) 12 + (ex cos y + ey sin x) (2) = et/2 12 sin 2t + 2 cos 2t + e2t 12 cos 12 t + 2 sin 12 t
20.
21.
∂u dx ∂u dy du = · + · = (4x − y)(−2 sin 2t) + (2y − x) cos t dt ∂x dt ∂y dt = 2 sin 2t(sin t − 4 cos 2t) + cos t(2 sin t − cos 2t) du ∂u dx ∂u dy = + = (ex sin y) (2t) + (ex cos y) (π) dt ∂x dt ∂y dt 2
= et [2t sin(πt) + π cos(πt)] 22.
23.
y ∂u dx ∂u dy ∂u dz z z 1 du √ + ln = · + · + · = − 2t + et (1 + t) dt ∂x dt ∂y dt ∂z dt x y2 t x √
t 2t2 et et =− 2 + + ln 2 et (1 + t) t +1 2 t +1 du ∂u dx ∂u dy ∂u dz = + + dt ∂x dt ∂y dt ∂z dt = (y + z)(2t) + (x + z)(1 − 2t) + (y + x)(2t − 2) = (1 − t)(2t) + (2t2 − 2t + 1)(1 − 2t) + t(2t − 2) = 1 − 4t + 6t2 − 4t3
24.
du ∂u dx ∂u dy ∂u dz = · + · + · = (sin πy + πz sin πx)2t + πx cos πy(−1) − cos πx(−2t) dt ∂x dt ∂y dt ∂z dt = 2t sin[π(1 − t)] + π(1 − t2 ) sin(πt2 ) − πt2 cos[π(1 − t)] + 2t cos(πt2 )
25. V =
1 2 πr h, 3
dV ∂V dr ∂V dh = + = dt ∂r dt ∂h dt
2 πrh 3
dr + dt
1 2 πr 3
dh . dt
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SECTION 16.3
801
At the given instant, dV 2 1 1288 = π(280)(3) + π(196)(−2) = π. dt 3 3 3 1288 π in.3 / sec . The volume is increasing at the rate of 3 26. v = πr2 h, dr = −2, dt
dh dv ∂v dr ∂v dh dr = · + · = 2πrh + πr2 dt ∂r dt ∂h dt dt dt dh dv = 3, r = 13, h = 18 =⇒ = −429π : dt dt
decreasing at the rate of
429π cm3 /sec. 27. A =
1 2
xy sin θ;
dA ∂A dx ∂A dy ∂A dθ = + + = dt ∂x dt ∂y dt ∂θ dt
1 2
(y sin θ)
dx dy dθ + (x sin θ) + (xy cos θ) . dt dt dt
At the given instant 1 dA = [(2 sin 1) (0.25) + (1.5 sin 1) (0.25) − (2(1.5) cos 1) (0.1)] ∼ = 0.2871 ft2 /s ∼ = 41.34 in2 /s dt 2 28.
dx y dy dz = 2x + . dt dt 2 dt
But x2 + y 2 = 13 =⇒ 2x
dy dy x dx dx + 2y = 0 =⇒ =− dt dt dt y dt
dx y x dx 3x dx dz = 2x + − = = 15. z is increasing 15 centimeters per second dt dt 2 y dt 2 dt ∂u ∂u ∂x ∂u ∂y = + = (2x − y)(cos t) + (−x)(t cos s) ∂s ∂x ∂s ∂y ∂s
=⇒ 29.
= 2s cos2 t − t sin s cos t − st cos s cos t ∂u ∂x ∂u ∂y ∂u = + = (2x − y)(−s sin t) + (−x)(sin s) ∂t ∂x ∂t ∂y ∂t = −2s2 cos t sin t + st sin s sin t − s cos t sin s 30.
∂u ∂x ∂u ∂y ∂u = · + · ∂s ∂x ∂s ∂y ∂s = [cos(x − y) − sin(x + y)]t + [− cos(x − y) − sin(x + y)]2s = (t − 2s) cos(st − s2 + t2 ) − (t + 2s) sin(st + s2 − t2 ) ∂u ∂u ∂x ∂u ∂y = · + · ∂t ∂x ∂t ∂y ∂t = [cos(x − y) − sin(x + y)]s + [− cos(x − y) − sin(x + y)](−2t) = (s + 2t) cos(st − s2 + t2 ) − (s − 2t) sin(st + s2 − t2 )
31.
∂u ∂u ∂x ∂u ∂y = + = (2x tan y)(2st) + x2 sec2 y (1) ∂s ∂x ∂s ∂y ∂s 3 2 = 4s t tan s + t2 + s4 t2 sec2 s + t2 ∂u ∂x ∂u ∂y ∂u = + = (2x tan y) s2 + x2 sec2 y (2t) ∂t ∂x ∂t ∂y ∂t 4 = 2s t tan s + t2 + 2s4 t3 sec2 s + t2
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SECTION 16.3 ∂u ∂u ∂x ∂u ∂y ∂u ∂z = · + · + · ∂s ∂x ∂s ∂y ∂s ∂z ∂s = z 2 y sec xy tan xy(2t) + z 2 x sec xy tan xy + 2z sec xy(2st) = sec[2st(s − t2 )] 2s4 t3 (s − t2 ) tan[2st(s − t2 )] + 2s3 t2 tan[2st(s − t2 )] + 4s3 t2 ∂u = z 2 y sec xy tan xy(2s) + z 2 x sec xy tan xy(−2t) + 2z sec xy(s2 ) ∂t = sec[2st(s − t2 )] 2s5 t2 (s − t2 ) tan[2st(s − t2 )] − 4s5 t4 tan[2st(s − t2 )] + 2s4 t
33.
∂u ∂u ∂x ∂u ∂y ∂u ∂z = + + ∂s ∂x ∂s ∂y ∂s ∂z ∂s = (2x − y)(cos t) + (−x)(− cos (t − s)) + 2z(t cos s) = 2s cos2 t − sin (t − s) cos t + s cos t cos (t − s) + 2t2 sin s cos s ∂u ∂x ∂u ∂y ∂u ∂z ∂u = + + ∂t ∂x ∂t ∂y ∂t ∂z ∂t = (2x − y)(−s sin t) + (−x)(cos (t − s)) + 2z(sin s) = −2s2 cos t sin t + s sin (t − s) sin t − s cos t cos (t − s) + 2t sin2 s
34.
∂u ∂u ∂x ∂u ∂y ∂u ∂z = · + · + · ∂s ∂x ∂s ∂y ∂s ∂z ∂s = eyz
2
1 2 2 + xz 2 eyz · 0 + 2xyzeyz 2s s
1 t3 (s2 +t2 )2 3 2 2 2 + 4st3 (s2 + t2 ) ln(st)et (s +t ) e s ∂u ∂x ∂u ∂y ∂u ∂z ∂u = · + · + · ∂t ∂x ∂t ∂y ∂t ∂z ∂t =
= eyz
2
1 2 2 + xz 2 eyz 3t2 + 2xyzeyz 2t t
1 t3 (s2 +t2 )2 3 2 2 2 + t2 (s2 + t2 )(3s2 + 7t2 ) ln(st)et (s +t ) e t d r (t) [f (r(t) ) ] = ∇f (r(t) ) · r (t) dt r (t) =
35.
= fu(t) (r(t)) r (t)
36.
where
u(t) =
∂ d ∂r ∂r x [f (r)] = [f (r)] = f (r) = f (r) ; ∂x dr ∂x ∂x r ∂ y [f (r)] = f (r) ∂y r
similarly
∂ z [f (r)] = f (r) . ∂z r x y z r ∇f (r) = f (r) i + f (r) j + f (r) k = f (r) . r r r r
Therefore 37. (a) (cos r)
r (t) r (t)
and
r r
38. (a) ∇(r ln r) = (1 + ln r)
(b) (r cos r + sin r) r r
2
r r
(b) ∇(e1−r ) = −2re1−r
2
r r
2
= −2e1−r r
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SECTION 16.3
r 39. (a) (r cos r − sin r) 3 r
(b)
40. (a)
(b)
sin r − r cos r sin2 r
r r
∂u dx ds ∂u dy ds du = + dt ∂x ds dt ∂y ds dt
41. (a)
∂u ∂u = (b) ∂r ∂x
∂x ∂w ∂x ∂t + ∂w ∂r ∂t ∂r
∂u + ∂y
∂y ∂w ∂y ∂t + ∂w ∂r ∂t ∂r
To obtain ∂u/∂s, replace each r by s. 42. (a)
(b)
∂u ∂x ∂u ∂z ∂u ∂w ∂u = + + , ∂r ∂x ∂r ∂z ∂r ∂w ∂r
∂u ∂u ∂y ∂u ∂w = + ∂v ∂y ∂v ∂w ∂v
∂u + ∂z
∂z ∂w ∂z ∂t + ∂w ∂r ∂t ∂r
.
803
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SECTION 16.3 ∂u dx ∂u dy du = + dt ∂x dt ∂y dt 2 2 d u ∂ 2 u dy ∂u d2 y dy ∂ 2 u dx ∂ 2 u dy ∂u d x dx ∂ 2 u dx + + + 2 = + + dt2 ∂x dt2 dt ∂x2 dt ∂y∂x dt ∂y dt2 dt ∂x ∂y dt ∂y dt and the result follows.
44.
∂u ∂u ∂x ∂u ∂y = + ∂s ∂x ∂s ∂y ∂s ∂2u ∂u ∂ 2 x ∂x = + ∂s2 ∂x ∂s2 ∂s =
∂2u ∂x2
∂x ∂s
∂ 2 u ∂x ∂ 2 u ∂y + ∂x2 ∂s ∂y∂x ∂s
2 +2
∂ 2 u ∂x ∂y ∂ 2 u + 2 ∂x∂y ∂s ∂s ∂y
∂u ∂ 2 y ∂y + + ∂y ∂s2 ∂s
∂y ∂s
2 +
∂ 2 u ∂x ∂ 2 u ∂y + 2 ∂x∂y ∂s ∂y ∂s
∂u ∂ 2 x ∂u ∂ 2 y + ∂x ∂s2 ∂y ∂s2
∂u ∂x ∂u ∂y ∂u ∂u ∂u = + = cos θ + sin θ ∂r ∂x ∂r ∂y ∂r ∂x ∂y
45. (a)
∂u ∂u ∂x ∂u ∂y ∂u ∂u = + = (−r sin θ) + (r cos θ) ∂θ ∂x ∂θ ∂y ∂θ ∂x ∂y 2 2 2 ∂u ∂u ∂u ∂u ∂u 2 (b) cos θ sin θ + = cos θ + 2 sin2 θ, ∂r ∂x ∂x ∂y ∂y 2 2 2 ∂u ∂u 1 ∂u ∂u ∂u 2 = sin θ − 2 cos2 θ, cos θ sin θ + 2 r ∂θ ∂x ∂x ∂y ∂y 2 2 2 2 2 2 1 ∂u ∂u 2 ∂u 2 ∂u ∂u ∂u + 2 = + cos θ + sin2 θ + sin θ + cos2 θ = ∂r r ∂θ ∂x ∂y ∂x ∂y
46. (a) By Exercise 45 (a) ∂w ∂w ∂w = cos θ + sin θ, ∂r ∂x ∂y ∂w ∂x (b) To obtain the first pair of equations set w = r; Solve these equations simultaneously for
∂w ∂w ∂w =− r sin θ + r cos θ. ∂θ ∂x ∂y and
∂w . ∂y
to obtain the second pair of equations set w = θ. (c) θ is not independent of x; r = x2 + y 2 gives ∂r r cos θ x = = = cos θ 2 2 ∂x r x +y 47. Solve the equations in Exercise 45 (a) for
∂u ∂u and : ∂x ∂y
∂u 1 ∂u ∂u = cos θ − sin θ, ∂x ∂r r ∂θ Then
∇u =
∂u ∂u 1 ∂u = sin θ + cos θ ∂y ∂r r ∂θ
∂u ∂u 1 ∂u ∂u i+ j= (cos θ i + sin θ j) + (− sin θ i + cos θ j) ∂x ∂y ∂r r ∂θ
48. u(r, θ) = r2 =⇒ ∇u = 2rer
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SECTION 16.3 49. u(x, y) = x2 − xy + y 2 = r2 − r2 cos θ sin θ = r2 1 −
1 2
∂u = r(2 − sin 2θ), ∂r
sin 2θ
805
∂u = −r2 cos 2θ ∂θ
∂u 1 ∂u er + eθ = r(2 − sin 2θ)er − r cos 2θ eθ ∂r r ∂θ ∂u ∂u ∂x ∂u ∂y ∂u ∂u = + = −r sin θ + r cos θ ∂θ ∂x ∂θ ∂y ∂θ ∂x ∂y
∇u = 50.
∂2u ∂u = − sin θ − r sin θ ∂r∂θ ∂x = − sin θ
∂ 2 u ∂x ∂ 2 u ∂y + ∂x2 ∂r ∂y∂x ∂r
∂u ∂u + cos θ + r sin θ cos θ ∂x ∂y
∂u + cos θ + r cos θ ∂y
∂2u ∂2u − ∂y 2 ∂x2
∂ 2 u ∂x ∂ 2 u ∂y + 2 ∂x∂y ∂r ∂y ∂r
+ r(cos2 θ − sin2 θ)
∂2u ∂x∂y
51. From Exercise 45 (a), ∂2u ∂2u ∂2u ∂2u sin θ cos θ + 2 sin2 θ = cos2 θ + 2 2 2 ∂r ∂x ∂y ∂x ∂y ∂2u ∂2u 2 2 ∂2u 2 ∂2u 2 r = r sin θ − 2 sin θ cos θ + r cos2 θ − r ∂θ2 ∂x2 ∂y ∂x ∂y 2 The term in parentheses is The result follows.
∂u = 2x − 2y, ∂x
∂u = −2x + 4y 3 ∂y
∂u dy 2y − 2x y−x = − ∂x = 3 = 3 ∂u dx 4y − 2x 2y − x ∂y 53. Set u = xey + yex − 2x2 y.
Then ∂u = xey + ex − 2x2 ∂y
dy ∂u/∂x ey + yex − 4xy =− =− y . dx ∂u/∂y xe + ex − 2x2 54. u(x, y) = x2/3 + y 2/3 ,
∂u ∂u cos θ + sin θ . ∂x ∂y
∂u . Now divide the second equation by r2 and add the two equations. ∂r
52. u(x, y) = x2 − 2xy + y 4 − 4,
∂u = ey + yex − 4xy, ∂x
2 ∂u = x−1/3 , ∂x 3
2 ∂u = y −1/3 ∂y 3
∂u
y 1/3 dy = − ∂x = − =⇒ ∂u dx x ∂y 55. Set u = x cos xy + y cos x − 2. ∂u = cos xy − xy sin xy − y sin x, ∂x
Then ∂u = − x2 sin xy + cos x ∂y
∂u/∂x cos xy − xy sin xy − y sin x dy =− = . dx ∂u/∂y x2 sin xy − cos x
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SECTION 16.4
56. Set u(x, y, z) = z 4 + x2 z 3 + y 2 + xy − 2. Then
∂u = 2xz 3 + y, ∂x
∂u = 2y + x, ∂y
∂u 2y + x ∂z ∂y =− =− 3 ∂u ∂y 4z + 3x2 z 2 ∂z
∂u ∂z 2xz 3 + y = − ∂x = − 3 , ∂u ∂x 4z + 3x2 z 2 ∂z
57. Set u = cos xyz + ln x2 + y 2 + z 2 . ∂u 2x , = −yz sin xyz + 2 ∂x x + y2 + z2
Then ∂u 2y , = −xz sin xyz + 2 ∂y x + y2 + z2
and
∂u 2z = −xy sin xyz + 2 . ∂z x + y2 + z2
2x − yz x2 + y 2 + z 2 sin xyz ∂z ∂u/∂x =− =− , ∂x ∂u/∂z 2z − xy (x2 + y 2 + z 2 ) sin xyz 2y − xz x2 + y 2 + z 2 sin xyz ∂z ∂u/∂y =− =− . ∂y ∂u/∂z 2z − xy (x2 + y 2 + z 2 ) sin xyz
58. (a)
Use
(b) (i)
du1 du2 du = i+ j and apply the chain rule to u1 , u2 . dt dt dt du = t(ex cos yi + ex sin yj) + π(−ex sin yi + ex cos yj) dt 2
= tet
/2
2
(ii) u(t) = et
2
(cos πti + sin πtj) + πet
/2
2
cos πti + et
/2
/2
(− sin πti + cos πtj)
sin πtj
du 2 2 2 2 = (−πet /2 sin πt + tet /2 cos πt)i + (πet /2 cos πt + tet /2 sin πt)j dt
59.
∂u ∂x ∂u ∂y ∂u = + , ∂s ∂x ∂s ∂y ∂s
∂u ∂u ∂x ∂u ∂y = + ∂t ∂x ∂t ∂y ∂t
60.
du ∂u dx ∂u dy ∂u dz = + + dt ∂x dt ∂y dt ∂z dt ∂u ∂u1 ∂u2 ∂u3 = i+ j+ k, ∂y ∂y ∂y ∂y
where
∂u ∂u1 ∂u2 ∂u3 = i+ j+ k, ∂x ∂x ∂x ∂x
∂u ∂u1 ∂u2 ∂u3 = i+ j+ k. ∂z ∂z ∂z ∂z
SECTION 16.4 1. Set f (x, y) = x2 + xy + y 2 .
Then,
∇f = (2x + y)i + (x + 2y)j, normal vector i + j; tangent vector i − j tangent line x + y + 2 = 0; normal line x − y = 0
∇f (−1, −1) = −3i − 3j.
∂u = 4z 3 + 3x2 z 2 ∂z
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SECTION 16.4 2. Set f (x, y) = (y − x)2 − 2x,
∇f = −2(y − x + 1)i + 2(y − x)j,
807
∇f (2, 4) = −6i + 4j
normal vector −3i + 2j; tangent vector 2i + 3j tangent line 3x − 2y + 2 = 0; normal line 2x + 3y − 16 = 0 2 3. Set f (x, y) = x2 + y 2 − 9 x2 − y 2 . Then,
√ √ ∇f = [4x(x2 + y 2 ) − 18x]i + 4y x2 + y 2 + 18y j, ∇f 2, 1 = −6 2 i + 30j. √ √ normal vector 2 i − 5 j; tangent vector 5i + 2 j √ √ √ tangent line 2x − 5y + 3 = 0; normal line 5x + 2 y − 6 2 = 0 ∇f = 3x2 i + 3y 2 j,
4. Set f (x, y) = x3 + y 3 ,
∇f (1, 2) = 3i + 12j
tangent vector 4i − j
normal vector i + 4j;
tangent line x + 4y − 9 = 0;
normal line 4x − y − 2 = 0
5. Set f (x, y) = xy 2 − 2x2 + y + 5x.
Then,
∇f = (y 2 − 4x + 5) i + (2xy + 1) j,
∇f (4, 2) = −7i + 17j.
normal vector 7i − 17j; tangent vector 17i + 7j tangent line 7x − 17y + 6 = 0; normal line 17x + 7y − 82 = 0 6. Set f (x, y) = x5 + y 5 − 2x3 . normal vector i − 5j;
∇f = (5x4 − 6x2 )i + 5y 4 j,
∇f (1, 1) = −i + 5j
tangent vector 5i + j
tangent line x − 5y + 4 = 0;
normal line 5x + y − 6 = 0
7. Set f (x, y) = 2x3 − x2 y 2 − 3x + y.
Then,
∇f = (6x2 − 2xy 2 − 3) i + (−2x2 y + 1) j,
∇f (1, −2) = −5i + 5j.
normal vector i − j; tangent vector i + j tangent line x − y − 3 = 0; normal line x + y + 1 = 0 8. Set f (x, y) = x3 + y 2 + 2x. ∇f = (3x2 + 2)i + 2yj, normal vector 5i + 6j;
tangent vector 6i − 5j
tangent line 5x + 6y − 13 = 0; 9. Set f (x, y) = x2 y + a2 y.
∇f (−1, 3) = 5i + 6j
normal line 6x − 5y + 21 = 0
By (15.4.4) m=−
2xy ∂f /∂x =− 2 . ∂f /∂y x + a2
At (0, a) the slope is 0. 10. Set f (x, y, z) = (x2 + y 2 )2 − z. Tangent plane: Normal:
∇f = 4x(x2 + y 2 )i + 4y(x2 + y 2 )j − k,
8x + 8y − z − 12 = 0
x = 1 + 8t,
y = 1 + 8t,
z =4−t
∇f (1, 1, 4) = 8i + 8j − k
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SECTION 16.4
11. Set f (x, y, z) = x3 + y 3 − 3xyz.
Then,
∇f = (3x2 − 3yz) i + (3y 2 − 3xz) j − 3xyk, ∇f 1, 2, 32 = −6i + 15 2 j − 6k; 3 tangent plane at 1, 2, 32 : −6(x − 1) + 15 2 (y − 2) − 6 z − 2 = 0, which reduces to 4x − 5y + 4z = 0. Normal:
y = 2 − 5t,
x = 1 + 4t,
3 2
+ 4t
∇f = y 2 i + 2xyj + 4zk,
12. Set f (x, y, z) = xy 2 + 2z 2 .
∇f (1, 2, 2) = 4i + 4j + 8k
x + y + 2z − 7 = 0
Tangent plane: Normal:
z=
x = 1 + t,
y = 2 + t,
z = 2 + 2t
1 = i + aj. 1,
a 1 , which reduces to x + ay − z − 1 = 0 a
13. Set z = g(x, y) = axy. Then, ∇g = ayi + axj, ∇g
1 tangent plane at 1, , 1 : z − 1 = 1(x − 1) + a y − a Normal: x = 1 + t, y = a1 + at, z = 1 − t 14. Set f (x, y, z) =
√
Tangent plane: Normal:
√
√
1 1 1 ∇f = √ i + √ j + √ k, 2 y 2 x 2 z 2x + y + 2z − 8 = 0
x+
x = 1 + 2t,
y+
z.
y = 4 + t,
Then,
∇g = [cos x + cos (x + y)] i + [cos y + cos (x + y)] j, tangent plane at (0, 0, 0) : z − 0 = 2(x − 0) + 2(y − 0), x = 2t,
y = 2t,
∇g(0, 0) = 2i + 2j;
2x + 2y − z = 0.
z = −t
16. Set f (x, y, z) = x2 + xy + y 2 − 6x + 2 − z.
∇f = (2x + y − 6)i + (x + 2y)j − k,
−k Tangent plane: Normal:
x = 4,
z = −10 y = −2,
z = −10 + t
17. Set f (x, y, z) = b2 c2 x2 − a2 c2 y 2 − a2 b2 z 2 .
Then,
∇f (x0 , y0 , z0 ) = 2b2 c2 x0 i − 2a2 c2 y0 j − 2a2 b2 z0 k; tangent plane at (x0 , y0 , z0 ) : 2b2 c2 x0 (x − x0 ) − 2a2 c2 y0 (y − y0 ) − 2a2 b2 z0 (z − z0 ) = 0, which can be rewritten as follows: b2 c2 x0 x − a2 c2 y0 y − a2 b2 z0 z = b2 c2 x0 2 − a2 c2 y0 2 − a2 b2 z0 2 = f (x0 , y0 , z0 ) = a2 b2 c2 . Normal: x = x0 + 2b2 c2 x0 t,
1 1 1 i+ j+ k 2 4 2
z = 1 + 2t
15. Set z = g(x, y) = sin x + sin y + sin (x + y).
Normal:
∇f (1, 4, 1) =
y = y0 − 2a2 c2 y0 t,
z = z0 − 2a2 b2 z0 t
∇f (4, −2, −10) =
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SECTION 16.4 18. Set f (x, y, z) = sin(x cos y) − z. ∇f = cos y cos(x cos y)i − x sin y cos(x cos y)j − k, ∇f (1, π2 , 0) = j − k Tangent plane: Normal:
x = 1,
y+z = y=
π 2
π + t, 2
z=t
19. Set z = g(x, y) = xy + a3 x−1 + b3 y −1 . ∇g = y − a3 x−2 i + x − b3 y −2 j,
∇g = 0 =⇒ y = a3 x−2 and x = b3 y −2 .
Thus, y = a3 b−6 y 4 , y 3 = b6 a−3 , y = b2 /a, x = b3 y −2 = a2 /b The tangent plane is horizontal at a2 /b, b2 /a, 3ab .
and g a2 /b, b2 /a = 3ab.
20. z = g(x, y) = 4x + 2y − x2 + xy − y 2 .
∇g = (4 − 2x + y)i + (2 + x − 2y)j 10 8 ∇g = 0 =⇒ 4 − 2x + y = 0, 2 + x − 2y = 0 =⇒ x = , y= 3 3 8 28 The tangent plane is horizontal at ( 10 , , ). 3 3 3
21. Set z = g(x, y) = xy.
Then, ∇g = yi + xj. ∇g = 0
=⇒
x = y = 0.
The tangent plane is horizontal at (0, 0, 0). 22. z = g(x, y) = x2 + y 2 − x − y − xy. ∇g = (2x − 1 − y)i + (2y − 1 − x)j ∇g = 0 =⇒ 2x − 1 − y = 0 = 2y − 1 − x = 0 =⇒ x = 1, y = 1 The tangent plane is horizontal at (1, 1, −1). 23. Set z = g(x, y) = 2x2 + 2xy − y 2 − 5x + 3y − 2.
Then,
∇g = (4x + 2y − 5) i + (2x − 2y + 3) j. ∇g = 0
4x + 2y − 5 = 0 = 2x − 2y + 3 1 The tangent plane is horizontal at 13 , 11 6 , − 12 . =⇒
=⇒
x = 13 ,
y=
24. (a) Set f (x, y, z) = xy − z.√ ∇f = yi + xj − k, ∇f (1, 1, 1) = i + j − k 3 (−i − j + k) upper unit normal = 3 1 1 1 1 (b) Set f (x, y, z) = − − z. ∇f = − 2 i + 2 j − k, ∇f (1, 1, 0) = −i + j − k x y√ x y 3 lower unit normal: = (−i + j − k) 3 25.
y − y0 z − z0 x − x0 = = (∂f /∂x)(x0 , y0 , z0 ) (∂f /∂y)(x0 , y0 , z0 ) (∂f /∂z)(x0 , y0 , z0 )
11 6 .
809
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SECTION 16.4
26. All the tangent planes pass through the origin. To see this, write the equation of the surface as xf (x/y) − z = 0.
The tangent plane at (x0 , y0 , z0 ) has equation
2 x0 x0 x0 x0 x0 +f − (y − y0 ) − (z − z0 ) = 0. f f (x − x0 ) y0 y0 y0 y0 2 y0
The plane passes through the origin:
x0 x0 x0 2 x0 x0 2 x0 − x0 f + + z0 = z0 − x0 f = 0. f f − y0 y0 y0 y0 y0 y0 27. Since the tangent planes meet at right angles, the normals ∇F and ∇G meet at right angles: ∂F ∂G ∂F ∂G ∂F ∂G + + = 0. ∂x ∂x ∂y ∂y ∂z ∂z 28. The sum of the intercepts is a. To see this, note that the equation of the tangent plane at (x0 , y0 , z0 ) can be written x − x0 y − y0 z − z0 + √ + √ = 0. √ x0 y0 z0 Setting y = z = 0 we have √ x − x0 √ = y 0 + z0 . √ x0 Therefore the x-intercept is given by √ √ √ √ √ √ √ x0 ( y0 + z0 ) = x0 + x0 ( a − x0 ) = x0 a. √ √ √ √ Similarly the y-intercept is y0 a and the z-intercept is z0 a. The sum of the intercepts is √ √ √ √ √ √ ( x0 + y0 + z0 ) a = a a = a. x = x0 +
√
29. The tangent plane at an arbitrary point (x0 , y0 , z0 ) has equation y0 z0 (x − x0 ) + x0 z0 (y − y0 ) + x0 y0 (z − z0 ) = 0, which simplifies to y0 z0 x + x0 z0 y + x0 y0 z = 3x0 y0 z0
and thus to
The volume of the pyramid is
x y z + + = 1. 3x0 3y0 3z0
1 1 (3x0 ) (3y0 ) 9 9 V = Bh = (3z0 ) = x0 y0 z0 = a3 . 3 3 2 2 2
30. The equation of the tangent plane at (x0 , y0 , z0 ) can be written x0 −1/3 (x − x0 ) + y0 −1/3 (y − y0 ) + z0 −1/3 (z − z0 ) = 0 Setting y = z = 0, we get the x-intercept x = x0 + x0 1/3 (y0 2/3 + z0 2/3 ) = x0 + x0 1/3 (a2/3 − x0 2/3 ) =⇒ x = x0 1/3 a2/3 Similarly, the y-intercept is y0 1/3 a2/3 and the z-intercept is z0 1/3 a2/3 . The sum of the squares of the intercepts is (x0 2/3 + y0 2/3 + z0 2/3 )a4/3 = a2/3 a4/3 = a2 .
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SECTION 16.4
811
31. The point (2, 3, −2) is the tip of r(1). 3 Since r (t) = 2i − 2 j − 4tk, we have r (1) = 2i − 3j − 4k. t Now set f (x, y, z) = x2 + y 2 + 3z 2 − 25. The function has gradient 2xi + 2yj + 6zk. At the point (2, 3, −2), ∇f = 2(2i + 3j − 6k). The angle θ between r (1) and the gradient gives 19 (2i − 3j − 4k) (2i + 3j − 6k) √ = √ ∼ · = 0.504. 7 29 7 29 Therefore θ ∼ = 1.043 radians. The angle between the curve and the plane is π −θ ∼ = 1.571 − 1.043 ∼ = 0.528 radians. 2 cos θ =
32. The curve passes through the point (3, 2, 1) at t = 1, and its tangent vector is r (1) = 3i + 4j + 3k. ∇f = 2xi + 4yj + 6zk,
For the ellipsoid, set f (x, y, z) = x2 + 2y 2 + 3z 2 . ∇f (3, 2, 1) = 6i + 8j + 6k, 33. Set
which is parallel to r (1).
f (x, y, z) = x2 y 2 + 2x + z 3 .
Then,
∇f = (2xy 2 + 2) i + 2x2 yj + 3z 2 k,
∇f (2, 1, 2) = 6i + 8j + 12k.
The plane tangent to f (x, y, z) = 16 at (2, 1, 2) has equation 6(x − 2) + 8(y − 1) + 12(z − 2) = 0, or 3x + 4y + 6z = 22. Next, set
g(x, y, z) = 3x2 + y 2 − 2z.
Then,
∇g = 6xi + 2yj − 2k,
∇g(2, 1, 2) = 12i + 2j − 2k.
The plane tangent to g(x, y, z) = 9 at (2, 1, 2) is 12(x − 2) + 2(y − 1) − 2(z − 2) = 0, or 6x + y − z = 11. 34. Sphere: f (x, y, z) = x2 + y 2 + z 2 − 8x − 8y − 6z + 24,
∇f = (2x − 8)i + (2y − 8)j + (2z − 6)k
∇f (2, 1, 1) = −4i − 6j − 4k Ellipsoid:
g(x, y, z) = x2 + 3y 2 + 2z 2 ,
∇g = 2xi + 6yj + 4zk
∇g(2, 1, 1) = 4i + 6j + 4k Since their normal vectors are parallel, the surfaces are tangent. 35. A normal vector to the sphere at (1, 1, 2) is 2xi + (2y − 4) j + (2z − 2)k = 2i − 2j + 2k. A normal vector to the paraboloid at (1, 1, 2) is 6xi + 4yj − 2k = 6i + 4j − 2k.
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SECTION 16.4 Since (2i − 2j + 2k) · (6i + 4j − 2k) = 0, the surfaces intersect at right angles.
36. Surface A: Set f (x, y, z) = xy − az 2 ,
∇f = yi + xj − 2azk
Surface B: Set g(x, y, z) = x2 + y 2 + z 2 ,
∇g = 2xi + 2yj + 2zk
Surface C: Set h(x, y, z) = z 2 + 2x2 − c(z 2 + 2y 2 ).
∇h = 4xi − 4cyj + 2(1 − c)zk
Where surface A and surface B intersect,
∇f · ∇g = 4(xy − az 2 ) = 0
Where surface A and surface C intersect,
∇f · ∇h = 4(1 − c)(xy − az 2 ) = 0
Where surface B and surface C intersect,
∇g · ∇h = 4[2x2 − 2cy 2 + (1 − c)z 2 ] = 0
37. (a) 3x + 4y + 6 = 0 since plane p is vertical. (b) y = − 14 (3x + 6) = − 14 [3(4t − 2) + 6] = −3t z = x2 + 3y 2 + 2 = (4t − 2)2 + 3(−3t)2 + 2 = 43t2 − 16t + 6 r(t) = (4t − 2)i − 3tj + (43t2 − 16t + 6)k (c) From part (b) the tip of r(1) is (2, −3, 33).
We take
r (1) = 4i − 3j + 70j as d to write R(s) = (2i − 3j + 33k) + s(4i − 3j + 70k). (d) Set g(x, y) = x2 + 3y 2 + 2.
Then,
∇g = 2xi + 6yj
and ∇g(2, −3) = 4i − 18j.
An equation for the plane tangent to z = g(x, y) at (2, −3, 33) is z − 33 = 4(x − 2) − 18(y + 3)
which reduces to
4x − 18y − z = 29.
(e) Substituting t for x in the equations for p and p1 , we obtain 3t + 4y + 6 = 0
and
4t − 18y − z = 29.
From the first equation y = − 34 (t + 2) and then from the second equation z = 4t − 18 − 34 (t + 2) − 29 =
35 2 t
− 2.
Thus, (∗)
r(t) = ti − ( 34 t + 32 )j +
35 2
t − 2 k.
Lines l and l are the same. To see this, consider how l and l are formed; to assure yourself, replace t in (∗) by 4s + 2 to obtain R(s) found in part (c).
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SECTION 16.4 38. (a) normal vector: (b) tangent line: (c) 2
12 25
i−
x=2+
14 25 14 25
j;
normal line:
t, y = 1 +
12 25
x=2+
12 25
t, y = 1 −
14 25
t
t
y
1
1
39. (a) normal vector: (b) tangent plane:
2
3
x
2 i + 2 j + 4 k;
normal line:
x = 1 + 2t, y = 2 + 2t, z = 2 + 4t
2(x − 1) + 2(y − 2) + 4(z − 2) = 0
or x + y + 2z − 7 = 0
(c)
40. (a)
(b)
1.5 1 0.5
2 1
0 0
0 -0.5
-1 -1
0
-1
1
(c) ∇f = 0 at (±1, 0),
-1.5
0, ±
3/2
-1.5 -1
-0.5
0
0.5
1
1.5
813
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SECTION 16.5
41. (a)
(b)
1.5 1 0.5 0 -0.5
1
-1
3 2
-1.5 -1.5 -1 -0.5
0
0
0.5
1
1.5
1 0 -1
-1 0 1
(c) ∇f = 4x3 − 4x i − 4y 3 − 4y j; ∇f = 0 :
4x3 − 4x = 0
x = 0, ±1;
=⇒
4y 3 − 4y = 0
y = 0, ±1
=⇒
∇f = 0 at (0, 0), (±1, 0), (0, ±1), (±1, ±1)
42. (a)
(b)
2
1
1
2
0 1
-1 -2
0
0 -1 -1
0
-1
1 2 -2
-2 -2
-1
0
1
√ √ (c) ∇f = 0 at (0, 0), ± 2/2, ± 2/2
SECTION 16.5 1. ∇f = (2 − 2x) i − 2y j = 0 only at (1, 0). The difference f (1 + h, k) − f (1, 0) = 2(1 + h) − (1 + h)2 − k 2 − 1 = −h2 − k 2 ≤ 0 for all small h and k; there is a local maximum of 1 at (1, 0). 2. ∇f = (2 − 2x) i + (2 + 2y) j = 0 only at (1, −1). The difference f (1 + h, −1 + k) − f (1, −1) = [2(1 + h) + 2(−1 + k) − (1 + h)2 + (−1 + k)2 + 5] − 5 = −h2 + k 2 does not keep a constant sign for small h and k; (1, −1) is a saddle point.
2
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SECTION 16.5
815
3. ∇f = (2x + y + 3) i + (x + 2y) j = 0 only at (−2, 1). The difference f (−2 + h, 1 + k) − f (−2, 1) = [(−2 + h)2 + (−2 + h)(1 + k) + (1 + k)2 + 3(−2 + h) + 1] − (−2) = h2 + hk + k 2 is nonnegative for all small h and k. To see this, note that h2 + hk + k 2 ≥ h2 − 2|h||k| + k 2 = (|h| − |k|)2 ≥ 0; there is a local minimum of −2 at (−2, 1). 4. ∇f = (3x2 − 3) i + j is never 0 ;
there are no stationary points and no local extreme values.
5. ∇f = (2x + y − 6) i + (x + 2y) j = 0 only at (4, −2). fxx = 2,
fxy = 1,
fyy = 2.
At (4, −2), D = 3 > 0 and A = 2 > 0 so we have a local min; the value is −10. 6. ∇f = (2x + 2y + 2) i + (2x + 6y + 10) j = 0 ∂2f = 2, ∂x2
∂2f = 2, ∂y∂x
∂2f = 6; ∂y 2
only at (1, −2).
D = 2 · 6 − 22 > 0, A = 2 =⇒ local min; the value is −8.
7. ∇f = (3x2 − 6y) i + 3y 2 − 6x j = 0 at (2, 2) and (0, 0). fxx = 6x,
fxy = −6,
D = 36xy − 36.
fyy = 6y,
At (2, 2), D = 108 > 0 and A = 12 > 0 so we have a local min; the value is −8. At (0, 0), D = −36 < 0 so we have a saddle point. 8. ∇f = (6x + y + 5) i + (x − 2y − 5) j = 0 ∂2f = 6, ∂x2
∂2f = 1, ∂y∂x
∂2f = −2; ∂y 2
at
5 35 − ,− . 13 13
D = 6 · (−2) − 12 < 0;
9. ∇f = (3x2 − 6y + 6) i + (2y − 6x + 3) j = 0 at (5, fxx = 6x,
fxy = −6,
fyy = 2,
27 2 )
(−5/13, −35/13) is a saddle point.
and (1, 32 ).
D = 12x − 36.
117 At (5, 27 2 ), D = 24 > 0 and A = 30 > 0 so we have a local min; the value is − 4 .
At (1, 32 ), D = −24 < 0 so we have a saddle point. 10. ∇f = (2x − 2y − 3) i + (−2x + 4y + 5) j = 0 ∂2f ∂2f = 2, = −2, ∂x2 ∂y∂x the value is − 13 4 .
∂2f = 4; ∂y 2
11. ∇f = sin y i + x cos y j = 0 fxx = 0,
fxy = cos y,
at
( 12 , −1).
D = 2 · 4 − (−2)2 > 0, A = 2 =⇒ local minimum;
at (0, nπ) for all integral n.
fyy = −x sin y.
Since D = − cos nπ = −1 < 0, each stationary point is a saddle point. 2
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SECTION 16.5
12. ∇f = sin y i + (1 + x cos y) j = 0 at (−1, 2nπ) and (1, (2n + 1)π) for all integral n. ∂2f = 0, ∂x2
∂2f = cos y, ∂y∂x
∂2f = −x sin y; ∂y 2
D = 0 · (−x sin y) − cos2 y < 0
at the above points
so they are all saddle points 13. ∇f = (2xy + 1 + y 2 ) i + x2 + 2xy + 1 j = 0 at fxx = 2y,
fxy = 2x + 2y,
(1, −1) and (−1, 1).
D = 4xy − 4(x + y)2 .
fyy = 2x,
At both (1, −1) and (−1, 1) we have saddle points since D = −4 < 0. 14. ∇f =
1 y + 2 y x
x 1 x2 + y 2 x2 + y 2 i+ − 2 − j j= i − y x x2 y xy 2
is never 0;
no stationary points, no local extreme values. 1 15. ∇f = (y − x−2 ) i + x − 8y −2 j = 0 only at 2, 4 . fxx = 2x−3 , fxy = 1, fyy = 16y −3 , D = 32x−3 y −3 − 1. At 12 , 4 , D = 3 > 0 and A = 16 > 0 so we have a local min; the value is 6. 16. ∇f = (2x − 2y) i + (−2x − 2y) j = 0 ∂2f = 2, ∂x2
∂2f = −2, ∂y∂x
only at (0, 0)
∂2f = −2; ∂y 2
D = 2 · (−2) − (−2)2 < 0;
17. ∇f = (y − x−2 ) i + x − y −2 j = 0 only at fxx = 2x−3 ,
fxy = 1,
fyy = 2y −3 ,
(0, 0) is a saddle point.
(1, 1).
D = 4x−3 y −3 − 1.
At (1, 1), D = 3 > 0 and A = 2 > 0 so we have a local min; the value is 3. 18. ∇f = (2xy − y 2 − 1) i + (x2 − 2xy + 1) j = 0 at (1, 1), (−1, −1) ∂2f = 2y, ∂x2
∂2f = 2(x − y), ∂y∂x
∂2f = −2x; ∂y 2
D = −4xy − 4(x − y)2 < 0
at the above points;
(1, 1) and (−1, −1) are saddle points. 19. ∇f =
fxx =
2 x2 − y 2 − 1 (x2
+
y2
2
+ 1)
i+
(x2
−4x3 + 12xy 2 + 12x , (x2 + y 2 + 1)3
f (1, 0) = −1;
4xy j = 0 at + y 2 + 1)2 fxy =
(1, 0) and (−1, 0).
4y 3 + 4y − 12x2 y , (x2 + y 2 + 1)3
fyy =
4x3 + 4x − 12xy 2 . (x2 + y 2 + 1)3
point
A
B
C
D
(1, 0)
1
0
1
1
loc. min.
(−1, 0)
−1
0
−1
1
loc. max.
f (−1, 0) = 1
result
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SECTION 16.5
x−3 3 i+ j=0 20. ∇f = ln xy + 1 − x y ∂2f 1 3 = + 2, 2 ∂x x x
∂2f 1 = , ∂y∂x y
∂2f 2 = , ∂x2 3
At (3, 1/3),
3−x ∂2f = 2 ∂y y2
∂2f = 3, ∂y∂x
21. ∇f = 4x3 − 4x i + 2y j = 0 fxx = 12x2 − 4,
at (3, 1/3)
fxy = 0,
at
∂2f =0 ∂y 2
and D = −9 < 0 =⇒ saddle point.
(0, 0), (1, 0), and (−1, 0).
fyy = 2. point
A
B
C
D
result
(0, 0)
−4
0
2
−8
saddle
(1, 0)
8
0
2
16
loc. min.
(−1, 0)
8
0
2
16
loc. min.
f (±1, 0) = −3. 2
22. ∇f = 2xex
−y 2
2
(1 + x2 + y 2 ) i + 2yex
−y 2
(1 − x2 − y 2 ) j = 0 at (0, 0), (0, 1), (0, −1)
A=
∂2f 2 2 2 2 = 2xex −y (2x + 2x3 + 2xy 2 ) + ex −y (2 + 6x2 + 2y 2 ) ∂x2
B=
∂2f 2 2 2 2 = −2yex −y (2x + 2x3 + 2xy 2 ) + ex −y (4xy) ∂y∂x
C=
∂2f 2 2 2 2 = −2yex −y (2y − 2yx2 − 2y 3 ) + ex −y (2 − 2x2 − 6y 2 ) ∂y 2
At (0, 0), At (0, ±1),
AC − B 2 = (2)(2) = 4 > 0, A > 0
local minimum; the value is 0.
AC − B 2 = (4e−1 )(−4e−1 ) = −8e−2 > 0, 1
23. ∇f = cos x sin y i + sin x cos y j = 0 at fxx = − sin x sin y,
1 2 π, 2 π
point 1 2 π, 2 π 1 3 2 π, 2 π
1 3 , 2 π, 2 π , (π, π), 32 π, 12 π , 32 π, 32 π .
fyy = − sin x sin y
fxy = cos x cos y, 1
saddle points
A
B
C
D
−1
0
−1
1
loc. max.
1
0
1
1
loc. min.
(π, π) 0 1 0 −1 1 1 0 1 1 2 π, 2 π 3 3 −1 0 −1 1 2 π, 2 π 1 1 3 3 1 3 3 1 f 2 π, 2 π = f 2 π, 2 π = 1; f 2 π, 2 π = f 2 π, 2 π = −1 3
result
saddle loc. min. loc. max.
24. ∇f = − sin x cosh y i + cos x sinh y j = 0 at (−π, 0) , (0, 0) , (π, 0). fxx = − cos x cosh y,
fxy = − sin x sinh y,
D = − cos2 x cosh y − sin2 x sinh y < 0; 2
2
fyy = cos x cosh y
(−π, 0) , (0, 0) , (π, 0) are saddle points.
817
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SECTION 16.5
25. (a) ∇f = (2x + ky) i + (2y + kx) j and ∇f (0, 0) = 0 independent of the value of k. (b) fxx = 2,
fxy = k,
fyy = 2,
D = 4 − k 2 . Thus, D < 0 for |k| > 2 and (0, 0) is a saddle point
(c) D = 4 − k2 > 0 for |k| < 2. Since A = fxx = 2 > 0, (0, 0) is a local minimum. (d) The test is inconclusive when D = 4 − k 2 = 0 i.e., for k = ±2. (If k = ±2, f (x, y) = (x ± y)2 and (0, 0) is a minimum.) 26. (a) ∇f = (2x + ky) i + (kx + 8y) j = 0 (b)
∂2f = 2, ∂x2
∂2f = k, ∂y∂x
∂2f = 8; ∂y 2
at (0, 0). we want
16 − k 2 < 0, or |k| > 4
(c) We want 16 − k2 > 0, or |k| < 4 (d) k = ±4. (If k = ±4, f (x, y) = (x ± 2y)2 and (0, 0) is a minimum.) 27. Let P (x, y, z) be a point in the plane. We want to find the minimum of f (x, y, z) =
x2 + y 2 + z 2 .
However, it is sufficient to minimize the square of the distance: F (x, y, z) = x2 + y 2 + z 2 . It is clear that F has a minimum value, but no maximum value. Since P lies in the plane, 2x − y + 2z = 16 which implies y = 2x + 2z − 16 = 2(x + z − 8). Thus, we want to find the minimum value of F (x, z) = x2 + 4(x + z − 8)2 + z 2 Now, ∇F = [2x + 8(x + z − 8)] i + [8(x + z − 8) + 2z] k The gradient is 0 when 2x + 8(x + z − 8) = 0
and
8(x + z − 8) + 2z = 0
32 16 , from which it follows that y = − . 9 9 16 32 The point in the plane that is closest to the origin is P 32 , − , . 9 9 9 The only solution to this pair of equations is: x = z =
The distance from the origin to the plane is: F (P ) = Check using (13.6.5): d(P, 0) =
16 3 .
|2 · 0 − 0 + 2 · 0 − 16| 16 . = 2 2 2 3 2 + (−1) + 2
28. We want to minimize (x + 1)2 + (y − 2)2 + (z − 4)2 on the plane. Since z = −16 − 32 x + 2y, we need to minimize f (x, y) = (x + 1)2 + (y − 2)2 + (−20 − 32 x + 2y)2 ; ∇f = 13 2 x − 6y + 62 i + (−84 − 6x + 10y) j = 0 at (−4, 6) √ Closest point (−4, 6, 2), distance= (−1 − (−4))2 + (2 − 6)2 + (4 − 2)2 = 29 29. f (x, y) = (x − 1)2 + (y − 2)2 + z 2 = (x − 1)2 + (y − 2)2 + x2 + 2y 2
since z =
x2 + 2y 2
2 1 , y= . 2 3 fxx = 4 > 0, fxy = 0, fyy = 6, D = 24 > 0. Thus, f has a local minimum at (1/2, 2/3). √ The shortest distance from (1, 2, 0) to the cone is f 12 , 23 = 16 114
∇f = [2(x − 1) + 2x] i + [2(y − 2) + 4y] j = 0
=⇒
x=
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SECTION 16.5 30. V = 8xyz,
x2 + y 2 + z 2 = a2 =⇒ V (x, y) = 8xy
a2 − x2 − y 2 , x > 0, y > 0, x2 + y 2 < a2
√ √ 8y(a2 − x2 − y 2 ) − 8x2 y 8x(a2 − x2 − y 2 ) − 8xy 2 i+ j = 0 at a/ 3, a/ 3 a2 − x2 − y 2 a2 − x2 − y 2 2a 2a 2a 8 3√ a 3 dimensions: √ × √ × √ , maximum volume: 9 3 3 3
∇V =
31. (a)
-2
(b)
-1 0
1
1 2 4
0.5
3
0
2
-0.5
1 1
-1
0 2
0 -1
-1
-0.5
0
0.5
1
-2
(c) ∇f = (4y − 4x3 ) i + (4x − 4y 3 ) j = 0 at (0, 0), (1, 1), (−1, −1). fxx = −12x2 ,
fxy = 4,
fyy = −12y 2 ,
D = 144x2 y 2 − 16
point
A
B
C
D
result
(0, 0)
0
4
0
−16
saddle
(1, 1)
−12
4
−12
128
loc. max.
(−1, −1)
−12
4
−12
128
loc. max.
f (1, 1) = f (−1, −1) = 3 32. (a)
(b)
1.5 1 0.5
3 2
0
1
1 0
-1 0
-0.5
-1
-1
1 -1.5 -1.5
-1
-0.5
(c) ∇f = (4x3 − 4x) i + 2y j = 0 at (0, 0), (1, 0), (−1, 0). fxx = 12x2 − 4,
fx,y = 0,
f (1, 0) = f (−1, 0) = 0
fyy = 2,
D = 24x2 − 8
point
A
B
C
D
result
(0, 0)
−8
0
2
−8
saddle
(1, 0)
8
0
2
16
loc. min.
(−1, 0)
8
0
2
16
loc. min.
0
0.5
1
1.5
819
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SECTION 16.5
33. (a)
(b)
4
1.5 1 0.5
2 0 -0.5 -1
0 -2
0
-1.5
2
-1.5 -1 -0.5 0
0.5
1
1.5
f (1, 1) = 3 is a local max.; f has a saddle at (0, 0).
34. (a)
(b)
2
1 0.6 0.4 0.2 0 -2
2 1
0
0 -1
-1
0 1 2
-1
-2
-2 -2
-1
0
1
2
f (0, 0) = 0 is a local min.; f (0, 1) = f (0, −1) = 2e−1 are local maxima; f has a saddle at (±1, 0).
35. (a)
2 0
-2
0
(b)
2
2
-2 1
1
0
0 -1
-1
f (1, 0) = −1 is a local min.; f (−1, 0) = 1 is a loc. max.
-2 -2
-1
0
1
2
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SECTION 16.6 36. (a)
(b)
821
10
8
3 2 1 0 -1 0
6
10 8
4
6 2
4 4
2
2
6 8
100 0 0
2
4
6
8
10
f (π/2, π/2) = f (π/2, 5π/2) = f (5π/2, π/2) = f (5π/2, 5π/2) = 3 are local maxima; (π/2, 3π/2), (3π/2, π/2), (3π/2, 3π/2), (3π/2, 5π/2), (5π/2, 3π/2) are saddle points of f ; f (7π/6, 7π/6) = f (11π/6, 11π/6) = − 32 are local minima.
SECTION 16.6 y 4
1. ∇f = (4x − 4) i + (2y − 2) j = 0 at (1, 1) in D; f (1, 1) = −1 Next we consider the boundary of D. We parametrize each side of the triangle: C1 : r1 (t) = t i,
1
2
x
t ∈ [ 0, 2 ],
C2 : r2 (t) = 2 i + t j, C3 : r3 (t) = t i + 2t j,
t ∈ [ 0, 4 ], t ∈ [ 0, 2 ],
Now, f1 (t) = f (r1 (t)) = 2(t − 1)2 ,
t ∈ [ 0, 2 ]; critical number: t = 1,
f2 (t) = f (r2 (t)) = (t − 1) + 1,
t ∈ [ 0, 4 ];
critical number: t = 1,
f3 (t) = f (r3 (t)) = 6t2 − 8t + 2,
t ∈ [ 0, 2 ];
critical number: t = 23 .
2
Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (0) = f3 (0) = f (0, 0) = 2; f2 (1) = f (2, 1) = 1;
f1 (1) = f (1, 0) = 0;
f2 (4) = f3 (2) = f (2, 4) = 10;
f1 (2) = f2 (0) = f (2, 0) = 2; f3 (2/3) = f (2/3, 4/3) = − 23 .
f takes on its absolute maximum of 10 at (2, 4) and its absolute minimum of −1 at (1, 1). 2. ∇f = −3 i + 2 j = 0; no stationary points in D; Next we consider the boundary of D. We parametrize each side of the triangle: t ∈ [ 0, 4 ], 3 C2 : r2 (t) = t i + − 2 t + 6 j, C1 : r1 (t) = t i,
C3 : r3 (t) = t j,
t ∈ [ 0, 6 ],
t ∈ [ 0, 4 ],
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SECTION 16.6 and evaluate f : f1 (t) = f (r1 (t)) = 2 − 3t,
t ∈ [ 0, 4 ]; no critical numbers,
f2 (t) = f (r2 (t)) = −6t + 14, f3 (t) = f (r3 (t)) = 2 + 2t,
t ∈ [ 0, 4 ];
t ∈ [ 0, 6 ];
no critical numbers, no critical numbers.
Evaluating these functions at the endpoints of their domains, we find that: f1 (4) = f2 (4) = f (4, 0) = −10;
f1 (0) = f3 (0) = f (0, 0) = 2;
f2 (0) = f3 (6) = f (0, 6) = 14;
f takes on its absolute maximum of 14 at (0, 6) and its absolute minimum of −10 at (4, 0). 3. ∇f = (2x + y − 6) i + (x + 2y) j = 0 at (4, −2) in D;
y 1
f (4, −2) = −13
2
3
4
5
x
-1
Next we consider the boundary of D. We -2
parametrize each side of the rectangle: C1 : r1 (t) = −t j,
t ∈ [ 0, 3 ]
-3
C2 : r2 (t) = t i − 3 j,
t ∈ [ 0, 5 ]
C3 : r3 (t) = 5 i − t j,
t ∈ [ 0, 3 ]
C4 : r4 (t) = t i,
t ∈ [ 0, 5 ]
Now, f1 (t) = f (r1 (t)) = t2 − 1,
t ∈ [ 0, 3 ];
no critical numbers
f2 (t) = f (r2 (t)) = t2 − 9t + 8,
t ∈ [ 0, 5 ];
critical number: t =
9 2
f3 (t) = f (r3 (t)) = t2 − 5t − 6,
t ∈ [ 0, 3 ];
critical number: t =
5 2
f4 (t) = f (r4 (t)) = t2 − 6t − 1,
t ∈ [ 0, 5 ];
critical number: t = 3
Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (0) = f4 (0) = f (0, 0) = −1;
f1 (−3) = f2 (0) = f (0, −3) = 8;
f2 (5) = f3 (3) = f (5, −3) = −12;
f3 (5/2) = f (5, −5/2) = − 49 4 ;
f2 (9/2) = f (9/2, −3) = − 49 4 ; f3 (0) = f4 (5) = f (5, 0) = −6.
f4 (3) = f (3, 0) = −10 f takes on its absolute maximum of 8 at (0, −3) and its absolute minimum of −13 at (4, −2). 4. ∇f = (2x + 2y) i + (2x + 6y) j = 0 at (0, 0) in D; Next we consider the boundary of D. C1 : r1 (t) = t i − 2 j,
t ∈ [ −2, 2 ],
C2 : r2 (t) = 2 i + t j,
t ∈ [ −2, 2 ],
C3 : r3 (t) = t i + 2 j,
t ∈ [ −2, 2 ],
C4 : r4 (t) = −2 i + t j,
f (0, 0) = 0
We parametrize each side of the square:
t ∈ [ −2, 2 ],
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SECTION 16.6
823
and evaluate f : f1 (t) = f (r1 (t)) = t2 − 4t + 12,
t ∈ [ −2, 2 ];
f2 (t) = f (r2 (t)) = 4 + 4t + 3t2 ,
t ∈ [ −2, 2 ];
critical number: t = − 23 ,
f3 (t) = f (r3 (t)) = t2 + 4t + 12,
t ∈ [ −2, 2 ];
no critical numbers,
f4 (t) = f (r4 (t)) = 4 − 4t + 3t2 ,
t ∈ [ −2, 2 ];
critical number: t = 23 .
no critical numbers,
Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (−2) = f4 (−2) = f (−2, −2) = 24; f2 (2) = f3 (2) = f (2, 2) = 24;
f1 (2) = f2 (−2) = f (2, −2) = 8;
f3 (−2) = f4 (2) = f (−2, 2) = 8;
f2 (−2/3) = f (2, −2/3) = 83 ;
f4 (2/3) = f (−2, 2/3) = 83 .
f takes on its absolute maximum of 24 at (−2, −2) and (2, 2) and its absolute minimum of 0 at (0, 0). Note that x2 + 2xy + 3y 2 = (x + y)2 + 2y 2 . The results follow immediately from this observation. 5. ∇f = (2x + 3y) i + (2y + 3x) j = 0 at (0, 0) in D; Next we consider the boundary of D. C : r(t) = 2 cos t i + 2 sin t j,
f (0, 0) = 2
We parametrize the circle by:
t ∈ [ 0, 2π ]
The values of f on the boundary are given by the function F (t) = f (r(t)) = 6 + 12 sin t cos t, F (t) = 12 cos2 t − 12 sin2 t :
F (t) = 0
=⇒
t ∈ [ 0, 2π ]
cos t = ± sin t
=⇒
t = 14 π,
3 5 7 4 π, 4 π, 4 π
Evaluating F at the endpoints and critical numbers, we have:
√ √
√ √ F (0) = F (2π) = f (2, 0) = 6; F 14 π = F 54 π = f 2, 2 = f ( − 2, − 2 = 12;
√ √
√ √ 3 7 2, − 2 = 0 4 π = f − 2, 2 = F 4 π = f √ √ √ √ f takes on its absolute maximum of 12 at 2, 2 and at − 2, − 2 ; f takes on its absolute √ √ √ √ 2, − 2 . minimum of 0 at − 2, 2 and at F
6. ∇f = yi + (x − 3)j = 0 at (3, 0), which is not in the interior of D. The boundary is r(t) = 3 cos t i + 3 sin t j.
f (r(t)) = 3 sin t(3 cos t − 3) = 9 sin t(cos t − 1),
t ∈ [0, 2π].
df df = 9(2 cos2 t − cos t − 1); = 0 =⇒ cos t = 1, − 12 which yields the points A (3, 0), dt dt √ √ √ √ B (− 32 , 3 2 3 ), C (− 32 , − 3 2 3 ) : f (A) = 0, f (B) = − 274 3 min, f (C) = 274 3 max 7. ∇f = 2(x − 1)i + 2(y − 1) j = 0
only at (1, 1) in D.
As the sum of two squares, f (x, y) ≥ 0.
Thus, f (1, 1) = 0 is a minimum. To examine the behavior of f on the boundary of D, we note that f represents the square of the distance between (x, y) and (1, 1). Thus, f is maximal at the point √ √ of the boundary furthest from (1, 1). This is the point − 2, − 2 ; the maximum value of f is √ √ √ f − 2, − 2 = 6 + 4 2.
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SECTION 16.6
8. ∇f = (y + 1) i + (x − 1) j = 0 at (1, −1) which is not in the interior of D. Next we consider the boundary of D. C1 : r1 (t) = t j + t2 j, C2 : r2 (t) = t i + 4 j,
We parametrize the boundary by:
t ∈ [ −2, 2 ], t ∈ [ −2, 2 ],
and evaluate f : f1 (t) = f (r1 (t)) = t3 − t2 + t + 3, f2 (t) = f (r2 (t)) = 5t − 1,
t ∈ [ −2, 2 ];
t ∈ [ −2, 2 ];
no critical numbers,
no critical numbers.
Evaluating these functions at the endpoints of their domains, we find that: f1 (−2) = f2 (−2) = f (−2, 4) = −11;
f1 (2) = f2 (2) = f (2, 4) = 9.
f takes on its absolute maximum of 9 at (2, 4) and its absolute minimum of −11 at (−2, 4). 9. ∇f =
4xy 2x2 − 2y 2 − 2 i+ 2 j = 0 at (1, 0) and (−1, 0) in D; f (1, 0) = −1, (x2 + y 2 + 1)2 (x + y 2 + 1)2
f (−1, 0) = 1.
Next we consider the boundary of D. We parametrize each side of the squre: C1 : r1 (t) = −2 i + t j,
t ∈ [ −2, 2 ]
C2 : r2 (t) = t i + 2 j,
t ∈ [ −2, 2 ]
C3 : r3 (t) = 2 i + t j,
t ∈ [ −2, 2 ]
C4 : r4 (t) = t i,
t ∈ [ −2, 2 ]
Now, 4 , +5 −2t f2 (t) = f (r2 (t)) = 2 , t +5 −4 f3 (t) = f (r3 (t)) = 2 , t +5 −2t f4 (t) = f (r4 (t)) = 2 , t +5
f1 (t) = f (r1 (t)) =
t2
t ∈ [ −2, 2 ]; critical number: t = 0 t ∈ [ −2, 2 ]; no critical numbers t ∈ [ −2, 2 ]; critical number: t = 0 t ∈ [ −2, 2 ]; no critical numbers
Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (−2) = f4 (−2) = f (−2, −2) = 49 ; f4 (2) = f3 (−2) = f (2, −2) = − 49 ;
f1 (0) = f (−2, 0) = 45 ; f3 (0) = f (2, 0) = − 45 ;
f1 (2) = f2 (−2) = f (−2, 2) = 49 ; f2 (2) = f3 (2) = f (2, 2) = − 49 .
f takes on its absolute maximum of 1 at (−1, 0) and its absolute minimum of −1 at (1, 0). 10. ∇f =
2x2 − 2y 2 − 2 4xy i+ 2 j = 0 at (1, 0) in D; f (1, 0) = −1. (x2 + y 2 + 1)2 (x + y 2 + 1)2
Next we consider the boundary of D. We parametrize each side of the triangle: C1 : r1 (t) = t i − t j,
t ∈ [ 0, 2 ]
C2 : r2 (t) = 2 i + t j,
t ∈ [ −2, 2 ]
C3 : r3 (t) = t i + t j,
t ∈ [ 0, 2 ],
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SECTION 16.6
825
and evaluate f : √ −2t , t ∈ [ 0, 2 ]; critical number: t = 1/ 2, 2 2t + 1 −4 f2 (t) = f (r2 (t)) = 2 , t ∈ [ −2, 2 ]; critical number: t = 0 t +5 √ −2t f3 (t) = f (r3 (t)) = 2 , t ∈ [ 0, 2 ]; critical number: t = 1/ 2. 2t + 1 f1 (t) = f (r1 (t)) =
Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: √ √ √ √ f1 (0) = f3 (0) = f (0, 0) = 0; f1 (1/ 2) = f (1/ 2, −1/ 2) = −1/ 2; f1 (2) = f2 (−2) = f (2, −2) = − 49 ; f2 (0) = f (2, 0) = − 45 ; √ √ √ √ f3 (1/ 2) = f (1/ 2, 1/ 2) = −1/ 2.
f2 (2) = f3 (2) = f (2, 2) = − 49 ;
f takes on its absolute maximum of 0 at (0, 0) and its absolute minimum of −1 at (1, 0). 11. ∇f = (4 − 4x) cos y i − (4x − 2x2 ) sin y j = 0 at (1, 0) in D: f (1, 0) = 2 Next we consider the boundary of D. We parametrize each side of the rectangle: C1 : r1 (t) = t j, t ∈ − 14 π, 14 π C2 : r2 (t) = t i − 14 π j, C3 : r3 (t) = 2 i + t j,
t ∈ [ 0, 2 ] t ∈ − 14 π, 14 π t ∈ [ 0, 2 ]
C4 : r4 (t) = t i + 14 π j, Now, f1 (t) = f (r1 (t)) = 0; √
2 (4t − 2t2 ), 2 f3 (t) = f (r3 (t)) = 0; √ 2 (4t − 2t2 ), f4 (t) = f (r4 (t)) = 2 f2 (t) = f (r2 (t)) =
t ∈ [ 0, 2 ];
t ∈ [ 0, 2 ];
f at the vertices of the rectangle has the value 0;
critical number: t = 1;
critical number: t = 1; √ f2 (1) = f4 (1) = f 1, − 14 π = f 1, 14 π = 2.
f takes on its absolute maximum of 2 at (1, 0) and its absolute minimum of 0 along the lines x = 0 and x = 2. 12. ∇f = 2(x − 3)i + 2yj = 0
at
(3, 0)
which is not in the interior of D. df = 2(x − 3) + 4x3 = 0 at x = 1 =⇒ (1, 1) Boundary: On y = x2 , f = (x − 3)2 + x4 , dx 3 3 12 df = 2(x − 3) + 32x = 0 at x = =⇒ ( , ). On y = 4x, f = (x − 3)2 + 16x2 , dx 17 17 17 So the maximum and minimum occur at one or more of the following points: (0, 0), (4, 16), (1, 1), (
3 12 , ). 17 17
Evaluating f at these points, we find that f (1, 1) = 5 is the absolute minimum of f ; f (4, 16) = 257 is the absolute maximum of f .
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SECTION 16.6
13. ∇f = (3x2 − 3y) i + (−3x − 3y 2 ) j = 0 at (−1, 1) in D;
y 2
f (−1, 1) = 1 Next we consider the boundary of D. We -2
parametrize each side of the triangle: C1 : r1 (t) = −2 i + t j,
-1
t ∈ [ −2, 2 ],
C2 : r2 (t) = t i + t j,
t ∈ [ −2, 2 ],
C3 : r3 (t) = t i + 2 j,
t ∈ [ −2, 2 ],
and evaluate f :
x
-2
t ∈ [ −2, 2 ];
f3 (t) = f (r3 (t)) = t3 − 6t − 8,
2
√ t ∈ [ −2, 2 ]; critical numbers: t = ± 2,
f1 (t) = f (r1 (t)) = −8 + 6t − t3 , f2 (t) = f (r2 (t)) = −3t2 ,
1
critical number: t = 0,
√ critical numbers: t = ± 2.
t ∈ [ −2, 2 ];
Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: √ √ √ f1 (−2) = f2 (−2) = f (−2, −2) = −12; f1 (− 2) = f (−2, − 2) = −8 − 4 2 ∼ = −13.66; √ √ √ f1 ( 2) = f (−2, 2) = −8 + 4 2 ∼ f1 (2) = f3 (−2) = f (−2, 2) = −4; = −2.34; f2 (0) = f (0, 0) = 0; f2 (2) = f3 (2) = f (2, 2) = −12; √ √ √ √ √ √ f3 ( 2) = f ( 2, 2) = −8 − 4 2 f3 (− 2) = f (− 2, 2) = −8 + 4 2;
√ √ f takes on its absolute maximum of 1 at (−1, 1) and its absolute minimum of −8 − 4 2 at ( 2, 2) √ and (−2, − 2). 14. ∇f = 2(x − 4)i + 2yj = 0 at (4, 0) which is not in the interior of D. Next we examine f on the boundary of D: C1 : r1 (t) = ti + 4tj,
t ∈ [ 0, 2, ],
C2 : r2 (t) = ti + t3 j,
t ∈ [ 0, 2 ].
Note that f1 (t) = f (r1 (t)) = 17t2 − 8t + 16, f2 (t) = f (r2 (t)) = (t − 4)2 + t6 . Next f1 (t) = 34t − 8 = 0
=⇒
t = 4/17
and gives x = 4/17, y = 16/17
and f2 (t) = 6t5 + 2t − 8 = 0
=⇒
t=1
and gives x = 1, y = 1.
The extreme values of f can be culled from the following list: 4 16 f (0, 0) = 16, f (2, 8) = 68, f 17 , 17 =
256 17 ,
f (1, 1) = 10.
We see that f (1, 1) = 10 is the absolute minimum and f (2, 8) = 68 is the absolute maximum.
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SECTION 16.6 15. ∇f =
827
4xy 2y 2 − 2x2 − 2 i + j = 0 at (0, 1) and (0, −1) in D; (x2 + y 2 + 1)2 (x2 + y 2 + 1)2
f (0, 1) = −1,
f (0, −1) = 1
Next we consider the boundary of D. C : r(t) = 2 cos t i + 2 sin t j,
We parametrize the circle by:
t ∈ [ 0, 2π ]
The values of f on the boundary are given by the function F (t) = f (r(t)) = − 45 sin t, F (t) = − 45 cos t :
F (t) = 0
=⇒
t ∈ [ 0, 2π ]
cos t = 0
Evaluating F at the endpoints and critical numbers, we have: F (0) = F (2π) = f (2, 0) = 0; F 12 π = f (0, 2) = − 45 ;
=⇒
F
t = 12 π,
3 2 π.
3 2 π = f (0, −2) =
4 5
f takes on its absolute maximum of 1 at (0, −1) and its absolute minimum of −1 at (0, 1). 16. ∇f = (2x + 1) i + (8y − 2) j = 0 at − 12 , 14 in D; Next we consider the boundary of D. C : r(t) = 2 cos t i + sin t j,
f − 12 , 14 = − 12
We parametrize the ellipse by:
t ∈ [ 0, 2π ]
The values of f on the boundary are given by the function F (t) = f (r(t)) = 4 cos2 t + 4 sin2 t + 2 cos t − 2 sin t = 4 + 2 cos t − 2 sin t, F (t) = −2 sin t − 2 cos t :
F (t) = 0
=⇒
cos t = − sin t
=⇒
t ∈ [ 0, 2π ]
t = 34 π, or 74 π
Evaluating F at the endpoints and critical numbers, we have: F (0) = F (2π) = f (2, 0) = 6;
√ √
√ √ √ √ 3 F 4 π = f − 2, 2/2 = 4 − 2 2; F 74 π = f 2, − 2/ = 4 + 2 2 √ √ √ f takes on its absolute maximum of 4 + 2 2 at 2, − 2/2 ; f takes on its absolute minimum of − 12 at − 12 , 14 . 17. ∇f = 2(x − y)i − 2(x − y) j = 0 at each point of the line segment y = x from (0, 0) to (4, 4). Since f (x, x) = 0 and f (x, y) ≥ 0, f takes on its minimum of 0 at each of these points. Next we consider the boundary of D. We parametrize each side of the triangle: C1 : r1 (t) = tj,
t ∈ [ 0, 12 ]
C2 : r2 (t) = ti,
t ∈ [ 0, 6 ]
C3 : r3 (t) = ti + (12 − 2t) j,
t ∈ [ 0, 6 ]
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SECTION 16.6 and observe from f (r1 (t)) = t2 , 2
f (r2 (t)) = t ,
t ∈ [ 0, 12 ] t ∈ [ 0, 6 ]
f (r3 (t)) = (3t − 12)2 ,
t ∈ [ 0, 6 ]
that f takes on its maximum of 144 at the point (0, 12). 18. ∇f =
1 (x2 + y 2 )3/2
(−xi − yj) = 0 in D. Note that f (x, y) is the reciprocal of the distance of (x, y)
√ from the origin. The point of D closest to the origin (draw a figure) is (1, 1). Therefore f (1, 1) = 1/ 2 is the maximum value of f. The point of D furthest from the origin is (3, 4). Therefore f (3, 4) = 1/5 is the least value taken on by f . 19. Using the hint, we want to find the maximum value of f (x, y) = 18xy − x2 y − xy 2 in the triangular region. The gradient of f is: ∇D = 18y − 2xy − y 2 i + 18x − x2 − 2xy j The gradient is 0 when 18y − 2xy − y 2 = 0
and
18x − x2 − 2xy = 0
The solution set of this pair of equations is: (0, 0), (18, 0), (0, 18), (6, 6). It is easy to verify that f is a maximum when x = y = 6. The three numbers that satisfy x + y + z = 18 and maximize the product xyz are: x = 6, y = 6, z = 6. 20. f (y, z) = 30yz − y z − yz , 2
2 2
3
∇f = (30z − 2yz − z )j + (60yz − 2y z − 3yz )k = 0 at 2
2
15 , 15 2
3
=
2
2
154 . 4
(other points are not in the interior);
f
On the line y + z = 30, f (y, z) = 0
so the maximum of xyz 2 occurs at x = y =
21. f (x, y) = xy(1 − x − y),
0 ≤ x ≤ 1,
15 , 15 2
15 , 2
z = 15
0 ≤ y ≤ 1 − x.
[ dom (f ) is the triangle with vertices (0, 0), (1, 0), (0, 1).] ∇f = (y − 2xy − y 2 )i + x − 2xy − x2 j = 0 =⇒ x = y = 0, x = 1, y = 0, x = 0, y = 1, x = y = 13 . (Note that [ 0, 0 ] is not an interior point of the domain of f .) fxx = −2y, fxy = 1 − 2x − 2y, fyy = −2x. 1 1 1 At 3 , 3 , D = 3 > 0 and A < 0 so we have a local max; the value is 1/27. Since f (x, y) = 0 at each point on the boundary of the domain, the local max of 1/27 is also the absolute max.
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SECTION 16.6
829
x y z x y x y + + = 1 =⇒ V (x, y) = xyc 1 − − , x > 0, y > 0, + 0.
V (x, y) = xy
S − 2xy 2(x + y)
and
2(x + y)(−2y) − (S − 2xy)(2) S − 2xy + xy i ∇V = y 2(x + y) 4(x + y)2 S − 2xy 2(x + y)(−2x) − (S − 2xy)(2) + x + xy j 2(x + y) 4(x + y)2 ∂V ∂V = = 0 and simplifying, we get the pair of equations Setting ∂x ∂y 2S − 4x2 − 8xy = 0 2S − 4y 2 − 8xy = 0 from which it follows that x = y = S/6. From practical considerations, we conclude that V has a maximum value at ( S/6, S/6). Substituting these values into the equation for z, we get z = S/6 and so the box of maximum volume is a cube.
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SECTION 16.6
26. V = xyz,
S = xy + 2xz + 2yz
=⇒
V (x, y) = xy
(S − xy) , x > 0, y > 0, xy < S. 2(x + y)
y 2 (S − x2 − 2xy) x2 (S − y 2 − 2xy) i + j 2(x + y)2 2(x + y)2 S S , y= ; dimensions for maximum volume: ∇V = 0 =⇒ x = 3 3
∇V =
27.
f (x, y) =
3
(x − xi )2 + (y − yi )2
S × 3
S 1 × 3 2
S 3
i=1
∇f (x, y) = 2 [(3x − x1 − x2 − x3 ) i + (3y − y1 − y2 − y3 ) j]
x1 + x2 + x3 y1 + y2 + y3 ∇f = 0 only at , = (x0 , y0 ) . 3 3 The difference
f (x0 + h, y0 + k) − f (x0 , y0 ) 3 = (x0 + h − xi )2 + (y0 + k − yi )2 − (x0 − xi )2 − (y0 − yi )2 i=1
=
3
2h (x0 − xi ) + h2 + 2k (y0 − yi ) + k 2
i=1
= 2h (3x0 − x1 − x2 − x3 ) + 2k (3y0 − y1 − y2 − y3 ) + 3h2 + 3k 2 = 3h2 + 3k 2 is nonnegative for all h and k. Thus, f has its absolute minimum at (x0 , y0 ) . 28. Profit P (x, y) = N1 (x − 50) + N2 (y − 60) = 250(y − x)(x − 50) + [32, 000 + 250(x − 2y)](y − 60) ∇P = 250(2y − 2x − 10)i + [32, 000 + 250(2x + 70 − 4y)]j = 0 =⇒ x = 89,
y = 94 1 x tan θ , A = xy + x 2 2
x P = x + 2y + 2 sec θ , 2 P 1 . 0 < θ < π, 0 < x < 2 1 + sec θ
29.
A(x, θ) = 12 x(P − x − x sec θ) + 14 x2 tan θ,
2
P x x2 x 2 ∇A = − x − x sec θ + tan θ i + sec θ − sec θ tan θ j, 2 2 4 2 (Here j is the unit vector in the direction of increasing θ.) ∇A =
x2 1 [P + x(tan θ − 2 sec θ − 2)] i + sec θ (sec θ − 2 tan θ) j. 2 4
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SECTION 16.6 From
831
∂A ∂A = 0 we get θ = 16 π and then from = 0 we get ∂θ ∂x √ √ √ P + x 13 3 − 43 3 − 2 = 0 so that x = (2 − 3)P.
Next, Axx = 12 (tan θ − 2 sec θ − 2), x sec θ (sec θ − 2 tan θ), 2 x2 = sec θ sec θ tan θ − sec2 θ − tan2 θ . 2
Axθ = Aθθ Apply the second-partials test:
√ √ C = − 13 P 2 3 (2 − 3 )2 , D < 0. √ √ Since, D > 0 and A < 0, the area is a maximum when θ = 16 π, x = (2 − 3 ) P and y = 16 (3 − 3 )P. A = − 12 (2 +
√
3 ),
B = 0,
30. (a) ∇f = (2ax + by)i + (bx + 2cy)j ∂2f = 2a, ∂x2
∂2f = b, ∂y∂x
∂2f = 2c; ∂y 2
D = 4ac − b2 .
(b) The point (0, 0) is the only stationary point. If D < 0, (0, 0) is a saddle point; if D > 0, (0, 0) is a local minimum if a > 0 and a local maximum if a < 0. √ √ √ √ (c) (i) if b > 0, f (x, y) = ( ax + cy)2 ; every point on the line ax + cy = 0 is a stationary point and at each such point f takes on a local and absolute min of 0 √ √ √ √ if b < 0, f (x, y) = ( ax − cy)2 ; every point on the line ax − cy = 0 is a stationary point and at each such point f takes on a local and absolute min of 0 |a|x − |c|y = 0 (ii) if b > 0, f (x, y) = −( |a|x − |c|y)2 ; every point on the line is a stationary point and at each such point f takes on a local and absolute max of 0 if b < 0, f (x, y) = −( |a|x + |c|y)2 ; every point on the line |a|x + |c|y = 0 is a stationary point and at each such point f takes on a local and absolute max of 0 31. From
x = 12 y = 13 z = t
and
x=y−2=z =s
(t, 2t, 3t)
and
(s, 2 + s, s)
we take
as arbitrary points on the lines. It suffices to minimize the square of the distance between these points: f (t, s) = (t − s)2 + (2t − 2 − s)2 + (3t − s)2 = 14t2 − 12ts + 3s2 − 8t + 4s + 4,
t, s real.
Let i and k be the unit vectors in the direction of increasing t and s, respectively. ∇f = (28t − 12s − 8)i + (−12t + 6s + 4) j; ftt = 28,
fts = −12,
fss = 6,
∇f = 0
=⇒
t = 0, s = −2/3.
D = 6(28) − (−12)2 = −24 < 0.
By the second-partials test, the distance is a minimum when t = 0, s = −2/3; √ problem tells us the minimum is absolute. The distance is f (0, −2/3) = 23 6.
the nature of the
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SECTION 16.6
32. We want to minimize S = 4πr2 + 2πrh given that V = 43 πr3 + πr2 h = 10, 000.
V 4 4 2V 2 − r = πr2 + S(r) = 4πr + 2πr πr2 3 3 r V 8πr3 − 6V 4 3 6V , h= 2 − r=0 = 0 =⇒ r = S (r) = 3r2 8π πr 3 The optimal container is a sphere of radius r = 3 7500/π meters. 33. (a) Let x and y be the cross-sectional measurements of the box, and let l be its length. Then V = xyl,
where
2x + 2y + l ≤ 108,
x, y > 0
To maximize V we will obviously take 2x+2y+l = 108. Therefore, V (x, y) = xy(108−2x−2y) and ∇V = [y(108 − 2x − 2y) − 2xy] i + [x(108 − 2x − 2y) − 2xy] j Setting
∂V ∂V = = 0, we get the pair of equations ∂x ∂y ∂V = 108y − 4xy − 2y 2 = 0 ∂x ∂V = 108x − 4xy − 2x2 = 0 ∂y
from which it follows that x = y = 18
=⇒
l = 36.
Now, at (18, 18), we have A = Vxx = −4y = −72 < 0,
B = Vxy = 108 − 4x − 4y = −36,
C = Vyy = −4x = −72,
and D = (36)2 − (72)2 < 0.
Thus, V is a maximum when x = y = 18 inches and l = 36 inches. (b) Let r be the radius of the tube and let l be its length. Then V = π r2 l,
where
2π r + l ≤ 108,
r>0
To maximize V we take 2π r + l = 108. Then V (r) = π r2 (108 − 2π r) = 108π r2 − 2π 2 r3 . Now dV = 216π r − 6π 2 r2 dr Setting
dV = 0, we get dr 216π r − 6π 2 r2 = 0
=⇒
r=
36 π
=⇒
Now, at r = 36/π, we have d2 V 36 = − 216π < 0 = 216π − 12π 2 2 dr π Thus, V is a maximum when r = 36/π inches and l = 36 inches.
l = 36
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SECTION 16.6 34. Let (x, y, z) be on the ellipsoid, x > 0, y > 0, z > 0.
833
Then
V = 2x · 2y · 2z = 8xyz. Note that V achieves its maximum ⇐⇒ Let s = x2 y 2 z 2 ,
=⇒
then
y2 2x2 + = 1, 9 4
x2 y 2 z 2 achieves its maximum.
x2 y2 − , s = x2 y 2 1 − 9 4
∂s y2 2x2 = 2xy 2 1 − − =0 ∂x 9 4
∂s x2 2y 2 2 = 2x y 1 − − =0 ∂y 9 4 x2 2y 2 + =1 9 4
Thus, Vmax
=⇒
3 x= √ , 3
2 y=√ , 3
1 z=√ 3
√ 3 2 1 16 3 = 8xyz = 8 · √ · √ · √ = . 3 3 3 3
35. Let S denote the cross-sectional area. Then S= where
1 1 (12 − 2x + 12 − 2x + 2x cos θ) x sin θ = 12x sin θ − 2x2 sin θ + x2 sin 2θ, 2 2
0 < x < 6, 0 < θ < π/2
Now, with j in the direction of increasing θ, ∇S = (12 sin θ − 4x sin θ + x sin 2θ) i + (12x cos θ − 2x2 cos θ + x2 cos 2θ) j Setting
∂S ∂S = = 0, we get the pair of equations ∂x ∂θ 12 sin θ − 4x sin θ + x sin 2θ = 0 12x cos θ − 2x2 cos θ + x2 cos 2θ = 0
from which it follows that x = 4, θ = π/3. Now, at (4, π/3), we have 3√ 3, B = Sxθ = 12 cos θ − 4x cos θ + 2x cos 2θ = −6, 2 √ = −12x sin θ + 2x2 sin θ − 2x2 sin 2θ = −24 3 and D = 108 − 36 > 0.
A = Sxx = −4 sin θ + sin 2θ = − C = Sθθ
Thus, S is a maximum when x = 4 inches and θ = π/3. 36.
96 = xyz, C = 30xy + 10(2xz + 2yz) 96 = 30xy + 20(x + y) . xy
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SECTION 16.7 64 64 + , C(x, y) = 30 xy + x y
∇C = 30(y − 64x−2 )i + 30(x − 64y −2 ) j = 0
=⇒
Cxx = 128x−3 ,
x = y = 4. Cyy = 128y −3 .
Cxy = 1,
When x = y = 4, we have D = 3 > 0 and A = 2 > 0 so the cost is minimized by making the dimensions of the crate 4 × 4 × 6 meters. 37. (a) f (m, b) = [2 − b]2 + [−5 − (m + b)]2 + [4 − (2m + b)]2 . fm = 10m + 6b − 6, fmm = 10,
fb = 6m + 6b − 2;
fmb = 6,
fbb = 6,
Answer: the line y = x −
fm = fb = 0
D = 24 > 0
=⇒
m = 1, b = − 23 .
=⇒
a minimum.
2 3.
(b) f (α, β) = [2 − β]2 + [−5 − (α + β)]2 + [4 − (4α + β)]2 . fα = 34α + 10β − 22, fαα = 34,
fβ = 10α + 6β − 2;
⎧ ⎨α =
fα = fβ = 0
fαβ = 10,
fββ = 6, D = 104 > 0 1 Answer: the parabola y = 13 14x2 − 19 .
=⇒
=⇒
⎩
14 13
β = − 19 13
⎤ ⎦.
a minimun.
38. (a) f (m, b) = [2 − (−m + b)]2 + [−1 − b]2 + [1 − (m + b)]2 fm = 4m + 2, fmm = 4,
fb = 6b − 4,
2 1 m=− , b= 2 3 =⇒ minimum
fm = fb = 0
=⇒
fmb = 0,
fbb = 6, D = 24 > 0 2 1 Answer: the line y = − x + 2 3 (b) f (α, β) = [2 − (α + β)]2 + [−1 − β]2 + [1 − (α + β)]2 fα = 4α + 4β − 6, fαα = 4,
fβ = 4α + 6β − 4;
fα = fβ = 0
fαβ = 4,
fββ = 6, D = 8 > 0 5 Answer: the parabola y = x2 − 1 2
=⇒
=⇒
α=
5 , β = −1 2
minimum
SECTION 16.7 g(x, y) = xy − 1
f (x, y) = x2 + y 2 ,
1.
∇f = 2xi + 2yj, ∇f = λ∇g
=⇒
∇g = yi + xj.
2x = λy and 2y = λx.
Multiplying the first equation by x and the second equation by y, we get 2x2 = λxy = 2y 2 . Thus, x = ±y. From g(x, y) = 0 we conclude that x = y = ±1. The points (1, 1) and (−1, −1) clearly give a minimum, since f represents the square of the distance of a point on the hyperbola from the origin. The minimum is 2.
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SECTION 16.7 2. f (x, y) = xy,
835
g(x, y) = b2 x2 + a2 y 2 − a2 b2 ∇f = yi + xj,
∇g = 2b2 xi + 2a2 yj
∇f = λ∇g =⇒ y = 2λb2 x,
x = 2λa2 y =⇒ a2 y 2 = b2 x2 a b From g(x, y) = 0 we get 2b2 x2 = a2 b2 =⇒ x = ± √ , y = ± √ 2 2 √ √ √ √ 1 The maximum value of xy is 2 ab, achieved at (a/ 2, b 2) and at (−a/ 2, −b 2). g(x, y) = b2 x2 + a2 y 2 − a2 b2
f (x, y) = xy,
3.
∇f = yi + xj, ∇f = λ∇g
∇g = 2b2 xi + 2a2 yj. =⇒
y = 2λb2 x and x = 2λa2 y.
Multiplying the first equation by a2 y and the second equation by b2 x, we get a2 y 2 = 2λa2 b2 xy = b2 x2 . √ √ Thus, ay = ±bx. From g(x, y) = 0 we conclude that x = ± 12 a 2 and y = ± 12 b 2. Since f is continuous and the ellipse is closed and bounded, the minimum exists. It occurs at √ 1 √ 1 √ 1 √ 1 1 2 a 2, − 2 b 2 and − 2 a 2, 2 b 2 ; the minimum is − 2 ab. 4. f (x, y) = xy 2 ,
g(x, y) = x2 + y 2 − 1
∇f = y 2 i + 2xyj,
∇g = 2xi + 2yj
∇f = λ∇g =⇒ y 2 = 2λx,
2xy = 2λy =⇒ y = 0
y = 0: From g(x, y) = 0 we get x = ±1;
or y 2 = 2x2
f (±1, 0) = 0
√ 2 y = ±√ 3
1 y = 2x : From g(x, y) = 0 we get 3x = 1, =⇒ x = ± √ , 3 √ √ √ √ 2 3 at (−1/ 3, ± 2 3) The Minimum of xy 2 is: − 9 2
2
2
5. Since f is continuous and the ellipse is closed and bounded, the maximum exists. g(x, y) = b2 x2 + a2 y 2 − a2 b2
f (x, y) = xy 2 , ∇f = y 2 i + 2xyj, ∇f = λ∇g
=⇒
∇g = 2b2 xi + 2a2 yj.
y 2 = 2λb2 x and
2xy = 2λa2 y.
Multiplying the first equation by a2 y and the second equation by b2 x, we get a2 y 3 = 2λa2 b2 xy = 2b2 x2 y. We can exclude y = 0; it clearly cannot produce the maximum. Thus, a2 y 2 = 2b2 x2 and, from g(x, y) = 0, 3b2 x2 = a2 b2 . √ √ √ √ This gives us x = ± 13 3 a and y = ± 13 6 b. The maximum occurs at x = 13 3 a, y = ± 13 6 b; the √ value there is 29 3 ab2 .
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SECTION 16.7
6. f (x, y) = x + y, ∇f = i + j,
g(x, y) = x4 + y 4 − 1
∇g = 4x3 i + 4y 3 j
∇f = λ∇g =⇒ 1 = 4λx3 ,
1 = 4λy 3 =⇒ x = y
From g(x, y) = 0 we get 2x4 = 1 =⇒ x = y = ±2−1/4 The maximum of x + y is:
2 · 2−1/4 = 23/4 , achieved at (2−1/4 , 2−1/4 ).
7. The given curve is closed and bounded. Since x2 + y 2 represents the square of the distance from points on this curve to the origin, the maximum exists. g(x, y) = x4 + 7x2 y 2 + y 4 − 1 ∇g = 4x3 + 14xy 2 i + 4y 3 + 14x2 y j.
f (x, y) = x2 + y 2 , ∇f = 2xi + 2yj,
We use the cross-product equation (16.7.4) : 2x(4y 3 + 14x2 y) − 2y(4x3 + 14xy 2 ) = 0, 20x3 y − 20xy 3 = 0, xy(x2 − y 2 ) = 0. Thus, x = 0, y = 0, or x = ±y. From g(x, y) = 0 we conclude that the points to examine are √ 1√ (0, ±1), (±1, 0), ± 3 3, ± 13 3 . The value of f at each of the first four points is 1;
the value at the last four points is 2/3. The
maximum is 1. 8. f (x, y, z) = xyz,
g(x, y, z) = x2 + y 2 + z 2 − 1
∇f = yzi + xzj + xyk,
∇g = 2xi + 2yj + 2zk
∇f = λ∇g =⇒ yz = 2λx,
xz = 2λy,
xy = 2λz =⇒ x2 = y 2 = z 2 or λ = 0.
λ = 0: In this case, at least two of x, y, z are 0 and f = 0. x2 = y 2 = z 2 From g(x, y, z) = 0 we get 3x2 = 1 =⇒ x = ± √13 , y = ± √13 , z = ± √13 √ √ √ √ √ √ 1√ 3 at (−1/ 3, −1/ 3, −1/ 3), (−1/ 3, 1/ 3, 1/ 3), 9 √ √ √ √ √ √ (1/ 3, −1/ 3, 1/ 3), (1/ 3, 1/ 3, −1/ 3). The minimum of xyz is:
−
9. The maximum exists since xyz is continuous and the ellipsoid is closed and bounded. y2 z2 x2 + + −1 a2 b2 c2 2y 2z 2x ∇f = yzi + xzj + xyk, ∇g = 2 i + 2 j + 2 k. a b c 2y 2z 2x ∇f = λ∇g =⇒ yz = 2 λ, xz = 2 λ, xy = 2 λ. a b c
f (x, y, z) = xyz,
g(x, y, z) =
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SECTION 16.7
837
We can assume x, y, z are non-zero, for otherwise f (x, y, z) = 0, which is clearly not a maximum. Then from the first two equations yza2 xzb2 = 2λ = x y
so that a2 y 2 = b2 x2
or
x2 y2 = . a2 b2
Similarly from the second and third equations we get b2 z 2 = c2 y 2
y2 z2 = 2. 2 b c
or
a b 3x2 From g(x, y, z) = 0, we get 2 = 1 =⇒ x ± √ , from which it follows that y = ± √ , a 3 3 √ c 1 z = ± √ . The maximum value is 9 3 abc. 3 g(x, y, z) = x2 + y 2 + z 2 − 7
10. f (x, y, z) = x + 2y + 4z, ∇f = i + 2j + 4k,
∇g = 2xi + 2yj + 2zk
∇f = λ∇g =⇒ 1 = 2λx, 2 = 2λy, 4 = 2λz =⇒ y = 2x, z = 4x 1 From g(x, y, z) = 0 we get 21x2 = 7 =⇒ x = ± √ 3 √ √ √ √ Minimum of x + 2y + 4z is: −7 3, achieved at (−1/ 3, −2/ 3, −4/ 3). 11. Since the sphere is closed and bounded and 2x + 3y + 5z is continuous, the maximum exists. g(x, y, z) = x2 + y 2 + z 2 − 19
f (x, y, z) = 2x + 3y + 5z, ∇f = 2i + 3j + 5k, ∇f = λ∇g
∇g = 2xi + 2yj + 2zk.
=⇒
2 = 2λx, 3 = 2λy, 5 = 2λz.
Since λ = 0 here, we solve the equations for x, y and z: x=
1 , λ
y=
3 , 2λ
z=
5 , 2λ
and substitute these results in g(x, y, z) = 0 to obtain 1 9 25 + 2 + 2 − 19 = 0, 2 λ 4λ 4λ
38 − 19 = 0, 4λ2
λ=±
1√ 2. 2
The positive value of λ will produce positive values for x, y, z and thus the maximum for f. We get √ √ √ √ x = 2, y = 32 2, z = 52 2, and 2x + 3y + 5z = 19 2. g(x, y, z) = x + y + z − 1
12. f (x, y, z) = x4 + y 4 + z 4 , ∇f = 4x i + 4y j + 4z k, 3
3
3
∇g = i + j + k
∇f = λ∇g =⇒ 4x = λ, 4y 3 = λ, 4z 3 = λ =⇒ x = y = z 3
From g(x, y, z) = 0 we get 3x = 1, =⇒ x = 1 Minimum is: 27 13.
1 3
=y=z
x y z + + −1 a b c 1 1 1 ∇f = yzi + xzj + xyk, ∇g = i + j + k. a b c λ λ λ ∇f = λ∇g =⇒ yz = , xz = , xy = . a b c f (x, y, z) = xyz,
g(x, y, z) =
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SECTION 16.7 Multiplying these equations by x, y, z respectively, we obtain xyz =
λx , a
xyz =
λy , b
xyz =
λz . c
Adding these equations and using the fact that g(x, y, z) = 0, we have
x y z 3xyz = λ + + = λ. a b c Since x, y, z are non-zero, yz = Similarly, y =
λ 3xyz = , a a
1=
3x , a
x=
a . 3
c 1 b and z = . The maximum is abc. 3 3 27
14. Maximize area A = xy given that the perimeter P = 2x + 2y f (x, y) = xy,
g(x, y) = 2x + 2y − P
∇f = yi + xj,
∇g = 2i + 2j;
∇f = λ∇g =⇒ y = 2λ,
x = 2λ =⇒ x = y.
The rectangle of maximum area is a square. 15. It suffices to minimize the square of the distance from (0, 1) to a point on the parabola. Clearly, the minimum exists. f (x, y) = x2 + (y − 1)2 ,
g(x, y) = x2 − 4y
∇f = 2xi + 2(y − 1)j,
∇g = 2xi − 4j.
We use the cross-product equation (16.7.4): 2x(−4) − 2x(2y − 2) = 0,
4x + 4xy = 0,
x(y + 1) = 0.
Since y ≥ 0, we have x = 0 and thus y = 0. The minimum is 1. 16. Minimize f (x, y) = (x − p)2 + (y − 4p)2 ∇f = 2(x − p)i + 2(y − 4p)j,
subject to
g(x, y) = 2px − y 2 = 0
∇g = 2pi − 2yj
∇f = λ∇g =⇒ 2(x − p) = 2λp,
2(y − 4p) = −2λy =⇒ x =
8p3 = y 2 =⇒ y = 2p, x = 2p y √ Distance to parabola is: f (x, y) = 5p
4p2 y
From g(x, y) = 0 we get
17. It suffices to maximize and minimize the square of the distance from (2, 1, 2) to a point on the sphere. Clearly, these extreme values exist. f (x, y, z) = (x − 2)2 + (y − 1)2 + (z − 2)2 , ∇f = 2(x − 2) i + 2(y − 1) j + 2(z − 2) k, ∇f = λ∇g
=⇒
g(x, y, z) = x2 + y 2 + z 2 − 1 ∇g = 2x i + 2y j + 2z k.
2(x − 2) = 2xλ, 2(y − 1) = 2yλ, 2(z − 2) = 2zλ
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SECTION 16.7
839
Thus, x=
2 , 1−λ
y=
1 , 1−λ
2 . 1−λ
z=
Using the fact that x2 + y 2 + z 2 = 1, we have
At λ = −2, At λ = 4,
2 1−λ
2
+
1 1−λ
(x, y, z) = (2/3, 1/3, 2/3)
2
+
2 1−λ
2 =1
=⇒
λ = −2, 4
and f (2/3, 1/3, 2/3) = 4
(x, y, z) = (−2/3, −1/3, −2/3)
and f (−2/3, −1/3, −2/3) = 16
Thus, (2/3, 1/3, 2/3) is the closest point and (−2/3, −1/3, −2/3) is the furthest point. 18. f (x, y, z) = sin x sin y sin z,
g(x, y, z) = x + y + z − π
∇f = cos x sin y sin zi + sin x cos y sin zj + sin x sin y cos zk,
∇g = i + j + k
∇f = λ∇g =⇒ cos x sin y sin z = λ = sin x cos y sin z = sin z sin y cos z =⇒ cos x = cos y = cos z π =⇒ x = y = z = 3 √ 3 3 Maximum of sin x sin y sin z is: 8 f (x, y, z) = 3x − 2y + z,
19.
g(x, y, z) = x2 + y 2 + z 2 − 14
∇f = 3 i − 2 j + k, ∇f = λ∇g
∇g = 2x i + 2y j + 2z k.
=⇒
3 = 2xλ,
3 , 2λ
y=−
−2 = 2yλ,
1 = 2zλ.
Thus, x=
1 , λ
z=
1 . 2λ
Using the fact that x2 + y 2 + z 2 = 14, we have
At λ =
1 , 2
3 2λ
2
2 2 1 1 + − + = 14 λ 2λ
=⇒
(x, y, z) = (3, −2, 1) and f (3, −2, 1) = 14
1 At λ = − , (x, y, z) = (−3, 2, −1) and f (−3, 2, −1) = −14 2 Thus, the maximum value of f on the sphere is 14. 20. f (x, y, z) = xyz,
g(x, y, z) = x2 + y 2 + z − 4
∇f = yzi + xzj + xyk,
∇g = 2xi + 2yj + k
∇f = λ∇g =⇒ yz = 2λx, xz = 2λy, xy = λ =⇒ x2 = y 2 = From g(x, y, z) = 0 we get 4x2 = 4 =⇒ x = 1, y = 1, z = 2. Maximum volume is 2
z 2
1 λ=± . 2
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SECTION 16.7
21. It’s easier to work with the square of the distance; the minimum certainly exists. f (x, y, z) = x2 + y 2 + z 2 ,
g(x, y, z) = Ax + By + Cz + D
∇f = 2xi + 2yj + 2zk, ∇f = λ∇g
=⇒
∇g = Ai + Bj + Ck.
2x = Aλ, 2y = Bλ, 2z = Cλ.
Substituting these equations in g(x, y, z) = 0, we have 1 2 −2D . λ A + B 2 + C 2 + D = 0, λ = 2 2 A + B2 + C 2 Thus, in turn, −DA −DB −DC , y= 2 , z= 2 A2 + B 2 + C 2 A + B2 + C 2 A + B2 + C 2 −1/2 so the minimum value of x2 + y 2 + z 2 is |D| A2 + B 2 + C 2 . x=
22. f (x, y, z) = xyz,
g(x, y, z) = 2xy + 2xz + 2yz − 6a2
∇f = yzi + xzj + xyk,
∇g = 2(y + z)i + 2(x + z)j + 2(x + y)k
∇f = λ∇g =⇒ yz = 2λ(y + z),
xz = 2λ(x + z),
xy = 2λ(x + y) =⇒ x = y = z
From g(x, y, z) = 0 we get 6x2 = 6a2 =⇒ x = y = z = a Maximum volume is a3 . 23.
area A = 12 ax + 12 by + 12 cz. The geometry suggests that x2 + y 2 + z 2 has a minimum.
g(x, y, z) = ax + by + cz − 2A
f (x, y, z) = x2 + y 2 + z 2 , ∇f = 2xi + 2yj + 2zk,
∇g = ai + bj + ck. ∇f = λ∇g
=⇒
2x = aλ, 2y = bλ, 2z = cλ.
Solving these equations for x, y, z and substituting the results in g(x, y, z) = 0, we have a2 λ b2 λ c2 λ + + − 2A = 0, 2 2 2
λ=
a2
4A + b2 + c2
and thus x=
a2
2aA , + b2 + c2
The minimum is 4A2 (a2 + b2 + c2 )−1 .
y=
a2
2bA , + b2 + c2
z=
a2
2cA . + b2 + c2
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SECTION 16.7
841
24. Use figure 16.7.6 and write the side condition as x + y + z = 2π. For (a) maximize f (x, y, z) = 8R3 sin 12 x sin 12 y sin 12 z. For (b) maximize f (x, y, z) = 4R2 (sin2 12 x + sin2 12 y + sin2 12 z). 2π . This gives an equilateral triangle. Each maximum occurs with x = y = z = 3 25. Since the curve is asymptotic to the line y = x as x → −∞ and as x → ∞, the maximum exists. The distance between the point (x, y) and the line y − x = 0 is given by |y − x| 1√ √ 2 |y − x|. (see Section 1.4) = 2 1+1 Since the points on the curve are below the line y = x, we can replace |y − x| by x − y. To simplify √ the work we drop the constant factor 12 2. f (x, y) = x − y, ∇f = i − j,
g(x, y) = x3 − y 3 − 1 ∇g = 3x2 i − 3y 2 j.
We use the cross-product equation (16.7.4): 1 −3y 2 − 3x2 (−1) = 0,
3x2 − 3y 2 = 0,
x = −y (x = y).
Now g(x, y) = 0 gives us x3 − (−x)3 − 1 = 0,
2x3 = 1,
x = 2−1/3 .
The point is 2−1/3 , −2−1/3 . 26. Let r, s, t be the intercepts. We wish to minimize the volume 1 1 V = rst [volume of pyramid = base × height] 6 3 a b c subject to the side condition + + = 1. The minimum occurs when all the intercepts are: r s t r = 3a, s = 3b, t = 3c 27. It suffices to show that the square of the area is a maximum when a = b = c. f (a, b, c) = s(s − a)(s − b)(s − c),
g(a, b, c) = a + b + c − 2s
∇f = −s(s − b)(s − c)i − s(s − a)(s − c) j − s(s − a)(s − b)k,
∇g = i + j + k.
(Here i, j, k are the unit vectors in the directions of increasing a, b, c.) ∇f = λ∇g
=⇒
−s(s − b)(s − c) = −s(s − a)(s − c) = −s(s − a)(s − b) = λ.
Thus, s − b = s − a = s − c so that a = b = c. This gives us the maximum, as no minimum exists. [The area can be made arbitrarily small by taking a close to s.] 28. f (x, y, z) = 8xyz,
g(x, y, z) = a2 − x2 − y 2 − z 2 , x > 0, y > 0, z > 0.
∇f = 8yzi + 8xzj + 8xyk,
∇g = −2x i − 2y j − 2z k
∇f = λ∇g =⇒ 8yz = −2λx,
8xz = −2λy,
8xy = −2λz =⇒ x = y = z
The rectangular box of maximum volume inscribed in the sphere is a cube.
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SECTION 16.7 g(x, y) = x + y − k, (x, y ≥ 0, k a nonnegative constant)
29. (a) f (x, y) = (xy)1/2 , ∇f =
y 1/2 x1/2 i + j, 2x1/2 2y 1/2
∇g = i + j.
∇f = λ∇g
=⇒
x1/2 y 1/2 = λ = 2x1/2 2y 1/2
Thus, the maximum value of f is: f (k/2, k/2) =
=⇒
x=y=
k . 2
k . 2
(b) For all x, y (x, y ≥ 0) we have (xy)1/2 = f (x, y) ≤ f (k/2, k/2) =
x+y k = . 2 2
k k , where (xyz)1/3 = . 3 3 x+y+z k 1/3 . (xyz) ≤ = 3 3
30. (a) The maximum occurs when x = y = z = (b) If x + y + z = k,
then, by (a),
31. Simply extend the arguments used in Exercises 29 and 30. 32. T (x, y, z) = xy 2 z,
g(x, y, z) = x2 + y 2 + z 2 − 1
∇T = y 2 zi + 2xyzj + xy 2 k,
∇g = 2xi + 2yj + 2zk
∇T = λ∇g =⇒ y 2 z = 2λx,
2xyz = 2λy, xy 2 = 2λz =⇒ x2 =
y2 = z2 2
From g(x, y, z) = 0 we get 4x2 = 1 =⇒ x = ± 12 , y = ± √12 , z = ± 12 Maximum
1 8
at ( 12 , ± √12 , 12 ),
(− 12 , ± √12 , − 12 )
Minimum − 18 at ( 12 , ± √12 , − 12 ),
(− 12 , ± √12 , 12 ) g(r, h) = πr2 h − V, (V constant)
S(r, h) = 2πr2 + 2πrh,
33.
∇S = (4πr + 2πh) i + 2πr j, ∇S = λ∇g
∇g = 2πrh i + πr2 j.
4πr + 2πh = 2πrhλ, 2πr = πr2 λ =⇒ 4V 3 V 3 16π 2 =⇒ r = , h= 3 . Now πr h = V, =⇒ λ = V 2π π 4V 3 V , and h = 3 . To minimize the surface area, take r = 2π π 34. f (x, y, z) = xyz,
=⇒
r=
g(x, y, z) = x + y + z − 18
∇f = yzi + xzj + xyk, ∇f = λ∇g
=⇒
∇g = i + j + k
yz = xz = xy
35. Same as Exercise 13.
=⇒
x=y=z
=⇒
x=y=z=6
2 , λ
h=
4 . λ
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SECTION 16.7 36. f (x, y, z) = xyz 2 ,
g(x, y, z) = x + y + z − 30
∇f = yz 2 i + xz 2 j + 2xyzk, ∇f = λ∇g
=⇒
843
∇g = i + j + k
yz 2 = λ = xz 2 = 2xyz
=⇒
x=y=
z 2
=⇒
x=y=
15 , 2
z = 15
37. Let x, y, z denote the length, width and height of the box. We want to maximize the volume V of the box given that the surface area S is constant. That is: maximize V (x, y, z) = xyz
subject to
S(x, y, z) = 2xy + 2xz + 2yz = S constant
Let g(x, y, z) = 2xy + 2xz + 2yz − S. Then ∇V = yz i + xz j + xy k,
∇g = (2y + 2z) i + (2x + 2z) j + (2x + 2y) k
∇V = λ ∇g and the side condition yield the system of equations: yz = λ (2y + 2z) xz = λ (2x + 2z) xy = λ (2x + 2y) xy + 2xz + 2yz = S. Multiply the first equation by x, the second by y and subtract. This gives 0 = 2λ z(x − y)
=⇒
x=y
since z = 0
=⇒
V = 0.
Multiply the second equation by y, the third by z and subtract. This gives 0 = 2λ x(y − z)
=⇒
y=z
since x = 0
=⇒
V = 0.
Thus the closed rectangular box of maximum volume is a cube. The cube has side length x =
S/6.
38. Let x, y, z denote the length, width and height of the box. We want to maximize the volume V of the box given that the surface area S is constant. That is: maximize V (x, y, z) = xyz
subject to
S(x, y, z) = 2xy + 2xz + 2yz = S constant
Let g(x, y, z) = xy + 2xz + 2yz − S. Then ∇V = yz i + xz j + xy k,
∇g = (y + 2z) i + (x + 2z) j + (2x + 2y) k
∇V = λ ∇g and the side condition yield the system of equations: yz = λ (y + 2z) xz = λ (x + 2z) xy = λ (2x + 2y) xy + 2xz + 2yz = S. Multiplying the first equation by x, the second by y and subtracting, we get 0 = 2λ z(x − y)
=⇒
x=y
since z = 0
=⇒
V = 0.
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SECTION 16.7 Now put y = x in the third equation. This gives x2 = 4λ x
x(x − 4λ) = 0
=⇒
=⇒
x = 4λ since x = 0
=⇒
V = 0.
Thus, x = y = 4λ. Substituting x = 4λ in the second equation gives z = 2λ. Finally, substituting these values for x, y, z in the fourth equation, we get 1 S S 2 2 =⇒ λ = 48λ = S =⇒ λ = 48 4 3 S 1 S To maximize the volume, take x = y = and z = . 3 2 3 39. S(r, h) = 4πr2 + 2πrh,
4 3 πr + πr2 h − 10, 000 3 ∇g = (4πr2 + 2πrh)i + πr2 j
g(r, h) =
∇S = (8πr + 2πh)i + 2πrj,
(Here i, j are the unit vectors in the directions of increasing r and h.) ∇S = λ∇g
2π(4r + h) = 2πrλ(2r + h), 2πr = λπr2 Maximum volume for sphere of radius r = 3 7500/π meters.
40. (a)
=⇒
=⇒
h=0
g(x, y, l) = 2x + 2y + l − 108,
f (x, y, l) = xyl, ∇f = yl i + xl j + xy k, ∇f = λ∇g
∇g = 2 i + 2 j + k. =⇒
Now 2x + 2y + l = 108,
yl = 2λ,
=⇒
xl = 2λ,
xy = λ
=⇒
y = x and l = 2x.
x = 18 and l = 36.
To maximize the volume, take x = y = 18 in. and l = 36 in. g(r, l) = 2πr + l − 108,
f (r, l) = πr2 l,
(b)
∇g = 2π i + j.
∇f = 2πrl i + πr j, 2
∇f = λ∇g
2πrl = 2πλ, πr2 = λ, l = πr. 36 and l = 36. Now 2πr + l = 108, =⇒ r = π To maximize the volume, take r = 36/π in. and l = 36 in. 41. f (x, y, z) = 8xyz,
=⇒
g(x, y, z) = 4x2 + 9y 2 + 36z 2 − 36.
∇f (x, y, z) = 8yzi + 8xzj + 8xyk, ∇f = λ∇g
∇g(x, y, z) = 8xi + 18yj + 72zk.
gives yz = λx, 4
xyz = 4x2 , λ
4xz = 9λy, 4
xyz = 9y 2 , λ
xy = 9λz. 4
xyz = 36z 2 . λ
Also notice 4x2 + 9y 2 + 36z 2 − 36 = 0 We have 12
xyz = 36 λ
=⇒
x=
√ 3,
2 y=√ , 3
1 z=√ . 3
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SECTION 16.7 Thus, V = 8xyz = 8 ·
√
2 1 16 3· √ · √ = √ . 3 3 3
42. Let x, y, z denote the length, width and height of the crate, and let C be the cost. Then C(x, y, z) = 30xy + 20xz + 20yz
g(x, y, z) = xyz − 96 = 0
subject to
∇C = (30y + 20z) i + (30x + 20z) j + (20x + 20y) k,
∇g = yz i + xz j + xy k
∇C = λ ∇g implies 30y + 20z = λ yz 30x + 20z = λ xz 20x + 20y = λ xy xyz = 96 Multiplying the first equation by x, the second by y and subtracting, we get 20z(x − y) = 0
=⇒
x=y
since z = 0
Now put y = x in the third equation. This gives 40x = λ x2
=⇒
x(λ x − 40) = 0
=⇒
x=
40 λ
since x = 0
Thus, x = y = 40/λ. Substituting x = 40/λ in the second equation gives z = 60/λ. Finally, substituting these values for x, y, z in the fourth equation, we get 40 40 60 = 96 λ λ λ
=⇒
96λ3 = 96, 000
=⇒
λ3 = 1000
=⇒
λ = 10
To minimize the cost, take x = y = 4 meters and z = 6 meters. 43. To simplify notation we set x = Q1 , f (x, y, z) = 2x + 8y + 24z,
y = Q2 ,
g(x, y, z) = x2 + 2y 2 + 4z 2 − 4, 500, 000, 000
∇f = 2i + 8j + 24k, ∇f = λ∇g
z = Q3 .
∇g = 2xi + 4yj + 8zk. =⇒
2 = 2λx,
8 = 4λy,
24 = 8λz.
Since λ = 0 here, we solve the equations for x, y, z: x=
1 , λ
y=
2 , λ
z=
and substitute these results in g(x, y, z) = 0 to obtain
9 4 1 + 4 − 45 × 108 = 0, + 2 λ2 λ2 λ2
3 , λ
45 = 45 × 108 , λ2
λ = ±10−4 .
Since x, y, z are non-negative, λ = 10−4 and x = 104 = Q1 ,
y = 2 × 104 = Q2 ,
z = 3 × 104 = Q3 .
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SECTION 16.7
44. f (x, y, z) = 8xyz,
g(x, y, z) =
x2 y2 z2 + + − 1. We take a, b, c, x, y, z > 0 a2 b2 c2
∇f (x, y, z) = 8yzi + 8xzj + 8xyk;
∇g(x, y, z) =
2y 2z 2x i + 2 j + 2 k. 2 a b c
∇f = λ ∇g and the side condition yield the system of equations: 8yz =
2xλ a2
8xz =
2yλ b2
8xy =
2zλ c2
x2 y2 z2 + + =1 a2 b2 c2 Multiply the first equation by x, the second by y and subtract. This gives 0=
2λy 2 2λx2 − 2 2 a b
=⇒
y=
b x since λ = 0 a
=⇒
V = 0.
Multiply the second equation by y, the third by z and subtract. This gives 0=
2λy 2 2λz 2 − 2 2 b c
c c z = y = x. b a
=⇒
Substituting these results into the side condition, we get: 3x2 =1 a2
a x= √ 3
c and z = √ . 3 √
b c a 8 3 √ √ abc. = The volume of the largest rectangular box is: V = 8 √ 9 3 3 3 =⇒
=⇒
b y=√ 3
PROJECT 16.7 1. f (x, y, z) = xy + z 2 ,
g(x, y, z) = x2 + y 2 + z 2 − 4,
∇f = y i + x j + 2z k,
∇g = 2x i + 2y j + 2z k,
∇f = λ∇g + μ∇h
=⇒
2z = 2λz
=⇒
λ=0
y = 2λx − μ,
h(x, y, z) = y − x
∇h = −i + j.
x = 2λy − μ,
2z = 2λz
or z = 1.
λ=0
=⇒
y = −x which contradicts y = x.
z=1
=⇒
x2 + y 2 = 3, which, with y = x implies x = ± 3/2;
± 3/2, ± 3/2
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SECTION 16.7 Adding the first two equations gives x + y = 2λ(x + y) x=y=0
z = ±2;
=⇒
(x + y)[2λ − 1] = 0
=⇒
=⇒
λ=
1 c
or
x = y = 0.
(0, 0, ±2).
1 =⇒ z = 0 and y = x =⇒ 2 5
f ± 3/2, ± 3/2, 1 = ; f (0, 0, ±2) = 4; 2
2x4 = 4;
λ=
√ x = ± 2;
√ √ (± 2, ± 2, 0).
√ √ f (± 2, ± 2, 0) = 2.
The maximum value of f is 4; the minimum value is 2.
2. D(x, y, z) = x2 + y 2 + z 2 , ∇D = 2xi + 2yj + 2zk, ∇D = λ∇g + μ∇h
g(x, y, z) = x + 2y + 3z, ∇g = i + 2j + 3k,
=⇒
∇h = 2i + 3j + k
2x = λ + 2μ,
2y = 2λ + 3μ,
Then g(x, y, z) = 0 and h(x, y, z) = 0 give x = Closest point
68 16 76 , ,− 75 15 75
3. f (x, y, z) = x2 + y 2 + z 2 , ∇f = 2x i + 2y j + 2z k, ∇f = λ∇g + μ∇h
h(x, y, z) = 2x + 3y + z − 4
68 , 75
y=
g(x, y, z) = x + y − z + 1, ∇g = i + j − k, =⇒
2z = 3λ + μ 16 , 15
z=−
=⇒
z = 5y − 7x
76 75
h(x, y, z) = x2 + y 2 − z 2
∇h = 2x i + 2y j − 2z k.
2x = λ + 2xμ,
2y = λ + 2yμ,
2z = − λ − 2zμ
Multiplying the first equation by y, the second equation by x and subtracting, yields λ(y − x) = 0. Now λ = 0
=⇒
μ=1
x = y = z = 0. This is impossible since x + y − z = −1. √ =⇒ z = ± 2 x.
=⇒
Therefore, we must have y = x √ Substituting y = x, z = 2 x into the equation x + y − z + 1 = 0, we get √ √ √ 2 2 x = −1 − =⇒ y = −1 − , z = −1 − 2 2 2 √ Substituting y = x, z = − 2 x into the equation x + y − z + 1 = 0, we get √ √ √ 2 2 x = −1 + =⇒ y = −1 + , z = −1 + 2 2 2
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SECTION 16.8 Since
√ √ √ √ 2 2 −1 − , −1 − , −1 − 2 = 6 + 4 2 and 2 2
f
f
√ √ √ √ 2 2 −1 + , −1 + , −1 + 2 = 6 − 4 2, 2 2
it follows that
√ √ √ 2 2 −1 + , −1 + , −1 + 2 is closest to the origin and 2 2
√ √ 2 2 −1 − , −1 − , −1 − 2 2 2 √
is furthest from the origin.
SECTION 16.8 1. df = 3x2 y − 2xy 2 Δx + x3 − 2x2 y Δy 2. df =
∂f ∂f ∂f Δx + Δy + Δz = (y + z)Δx + (x + z)Δy + (x + y)Δz ∂x ∂y ∂z
3. df = (cos y + y sin x) Δx − (x sin y + cos x) Δy 4. df = 2xye2z Δx + x2 e2z Δy + 2x2 ye2z Δz 5. df = Δx − (tan z) Δy − y sec2 z Δz x−y x−y + ln(x + y) Δx + − ln(x + y) Δy 6. df = x+y x+y
7. df =
y(y 2 + z 2 − x2 ) x(x2 + z 2 − y 2 ) 2xyz Δx + Δz 2 2 Δy − 2 2 2 2 2 2 2 (x + y + z ) (x + y + z ) (x + y 2 + z 2 )2
8. df =
2y 2x xy 2 xy Δy + e (1 + xy) Δx + + x e x2 + y 2 x2 + y 2
9. df = [cos(x + y) + cos(x − y)] Δx + [cos(x + y) − cos(x − y)] Δy 10. df = ln
1+y 1−y
Δx +
2x Δy 1 − y2
x Δz 11. df = y 2 zexz + ln z Δx + 2yexz Δy + xy 2 exz + z 12. df = y(1 − 2x2 )e−(x
2
+y 2 )
Δx + x(1 − 2y 2 )e−(x
2
+y 2 )
Δy
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SECTION 16.8 Δu = (x + Δx)2 − 3(x + Δx)(y + Δy) + 2(y + Δy)2 − x2 − 3xy + 2y 2 = (1.7)2 − 3(1.7)(−2.8) + 2(−2.8)2 − 22 − 3(2)(−3) + 2(−3)2
13.
= (2.89 + 14.28 + 15.68) − 40 = −7.15 du = (2x − 3y) Δx + (−3x + 4y) Δy = (4 + 9)(−0.3) + (−6 − 12)(0.2) = −7.50 14. du =
√ x+y x−y+ √ 2 x−y
Δx +
√
x+y x−y− √ 2 x−y
Δy = 1
15. Δu = (x + Δx)2 (z + Δz) − 2(y + Δy)(z + Δz)2 + 3(x + Δx)(y + Δy)(z + Δz) − x2 z − 2yz 2 + 3xyz = (2.1)2 (2.8) − 2(1.3)(2.8)2 + 3(2.1)(1.3)(2.8) − (2)2 3 − 2(1)(3)2 + 3(2)(1)(3) = 2.896 du = (2xz + 3yz) Δx + −2z 2 + 3xz Δy + x2 − 4yz + 3xy Δz = [2(2)(3) + 3(1)(3)](0.1) + [−2(3)2 + 3(2)(3)](0.3) + [22 − 4(1)(3) + 3(2)(1)](−0.2) = 2.5 16. du =
x3 + xz 2 xyz 77 y 3 + yz 2 Δx + Δy − 2 Δz = 2 2 2 3/2 2 2 2 3/2 2 2 3/2 (x + y + z ) (x + y + z ) (x + y + z ) 4(14)3/2
17. f (x, y) = x1/2 y 1/4 ;
x = 121, y = 16, Δx = 4, Δy = 1
f (x + Δx, y + Δy) ∼ = f (x, y) + df = x1/2 y 1/4 + 12 x−1/2 y 1/4 Δx + 14 x1/2 y −3/4 Δy √ √ √ √ 4 4 125 17 ∼ = 121 16 + 12 (121)−1/2 (16)1/4 (4) + 14 (121)1/2 (16)−3/4 (1) 1 (2)(4) + 14 (11) 18 = 11(2) + 12 11 = 22 +
4 11
+
11 32
∼ = 22 249 352 = 22.71
√ √ 18. f (x, y) = (1 − x)(1 + y), x = 9, y = 25, Δx = 1, Δy = −1 √ √ 1+ y 4 1− x df = − √ Δx + √ Δy = − 2 y 5 2 x 4 4 f (10, 24) ∼ = f (9, 25) − = −12 5 5 π π π f (x, y) = sin x cos y; x = π, y = , Δx = − , Δy = − 19. 4 7 20 df = cos x cos y Δx − sin x sin y Δy f (x + Δx, y + Δy) ∼ = f (x, y) + df 6 1 π π π π π sin π cos π ∼ − − sin π sin − = sin π cos + cos π cos 7 5 4 4 7 4 20 √
π 1√ π 2∼ 2 +0= = 0+ = 0.32 2 7 14
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SECTION 16.8 √
1 π , Δx = −1, Δy = π 4 16 √ 1 1 3π df = √ tan yΔx + x sec2 yΔy = − + 6 8 2 x
π 1 3π 17 3 ∼ 5 − + = + π = 4.01 f 8, π ∼ = f 9, 16 4 6 8 6 8
20. f (x, y) =
x tan y,
x = 9, y =
21. f (2.9, 0.01) ∼ = f (3, 0) + df, where df is to be evaluated at x = 3, y = 0, Δx = −0.1, Δy = 0.01. df = 2xexy + x2 yexy Δx + x3 exy Δy = 2(3)e0 + (3)2 (0)e0 (−0.1) + 33 e0 (0.01) = − 0.33 Thus, f (2.9, .01) ∼ = 32 e0 − 0.33 = 8.67. 22. x = 2, y = 3, z = 3, Δx = 0.12, Δy = −0.08, Δz = 0.02 df = 2xy cos πzΔx + x2 cos πzΔy − πx2 y sin πzΔz = −12(0.12) + 4(0.08) = −1.12 f (2.12, 2.92, 3.02) ∼ = f (2, 3, 3) − 1.12 = −13.12 23. f (2.94, 1.1, 0.92) ∼ = f (3, 1, 1) + df,
where df is to be evaluated at x = 3, y = 1, z = 1,
Δx = −0.06, Δy = 0.1, Δz = −0.08 xz xy π Δy + Δz = (−0.06) + (1.5)(0.1) + (1.5)(−0.08) ∼ = −0.0171 2 2 2 2 1+y z 1+y z 4 Thus, f (2.94, 1.1, 0.92) ∼ = 34 π − 0.0171 ∼ = 2.3391 df = tan−1 yz Δx +
24. x = 3, y = 4, Δx = 0.06, Δy = −0.12 x y 3 4 df = Δx + Δy = (0.06) + (−0.12) = −0.06 2 2 2 2 5 5 x +y x +y ∼ f (3.06, 3.88) = f (3, 4) − 0.06 = 4.94 ∂z 2y ∂z 2x Δx + Δy = Δx − Δy 2 ∂x ∂y (x + y) (x + y)2 With x = 4, y = 2, Δx = 0.1, Δy = 0.1, we get
25. df =
4 8 1 df = 36 (0.1) − 36 (0.1) = − 90 . 2 1 1 4.1 − 2.1 4 − 2 − = − =− . The exact change is 4.1 + 2.1 4 + 2 6.2 3 93
26. V (r, h) = πr2 h,
r = 8,
h = 12,
Δr = −0.3,
Δh = 0.2
dV = 2πrhΔr + πr Δh = 192π(−0.3) + 64π(0.2) = −44.8π 2
decreases by approximately 44.8π cubic inches. 27. S = 2πr2 + 2πrh;
r = 8, h = 12, Δr = −0.3, Δh = 0.2 dS =
∂S ∂S Δr + Δh = (4πr + 2πh) Δr + (2πr) Δh ∂r ∂h = 56π(−0.3) + 16π(0.2) = −13.6π.
The area decreases about 13.6π in.2 .
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SECTION 16.8 28. dT = 2x cos πzΔx − 2y sin πzΔy − (πx2 sin πz + πy 2 cos πz)Δz 2 4 = 4(0.1) − 4π(0.2) = − π 5 5 2 4 T decreases by about π − ∼ = 2.11 5 5 29. S(9.98, 5.88, 4.08) ∼ = S(10, 6, 4) + dS = 248 + dS, where dS = (2w + 2h) Δl + (2l + 2h) Δw + (2l + 2w) Δh = 20(−0.02) + 28(−0.12) + 32(0.08) = −1.20 Thus, S(9.98, 5.88, 4.08) ∼ = 248 − 1.20 = 246.80. πr 2 h , r = 7, h = 10, Δr = 0.2, Δh = 0.15 3 πr2 140 49 π 2πrh Δr + Δh = π(0.2) + π(0.15) = (35.35) df = 3 3 3 3 3 π π ∼ ∼ f (7.2, 10.15) = f (7, 10) + (35.35) = 525.35 = 550.15 3 3
30. f (r, h) =
31. (a) dV = yz Δx + xz Δy + xy Δz = (8)(6)(0.02) + (12)(6)(−0.05) + (12)(8)(0.03) = 0.24 (b) ΔV = (12.02)(7.95)(6.03) − (12)(8)(6) = 0.22077 32. (a) S(x, y, z) = 2(xy + xz + yz),
x = 12, y = 8, z = 6, Δx = 0.02, Δy = −0.05, Δz = 0.03
dS = 2(y + z)Δx + 2(x + z)Δy + 2(x + y)Δz = 28(0.02) + 36(−0.05) + 40(0.03) = −0.04 (b) ΔS = S(12.02, 7.95, 6.03) − S(12, 8, 6) = −0.0438 33. T (P ) − T (Q) ∼ = dT = (−2x + 2yz) Δx + (−2y + 2xz) Δy + (−2z + 2xy) Δz Letting x = 1, y = 3, z = 4, Δx = 0.15, Δy = −0.10, Δz = 0.10, we have dT = (22)(0.15) + (2)(−0.10) + (−2)(0.10) = 2.9 34. Amount of paint is increase in volume. f (x, y, z) = xyz, x = 48 in, y = 24 in, z = 36 in, Δx = Δy = Δz =
2 16
2 Δf ∼ ) = 468 = df = yzΔx + xzΔy + xyΔz = 3774( 16
in.
The amount of paint is approximately 468 cubic inches. 35. (a) πr2 h = π(r + Δr)2 (h + Δh)
=⇒
Δh =
df = (2πrh) Δr + πr2 Δh,
r2 h (2r + Δr)h −h=− Δr. 2 (r + Δr) (r + Δr)2 df = 0
=⇒
Δh =
−2h Δr. r
(b) 2πr2 + 2πrh = 2π(r + Δr)2 + 2π(r + Δr)(h + Δh). Solving for Δh, 2r + h + Δr r2 + rh − (r + Δr)2 −h=− Δr. r + Δr r + Δr
2r + h df = (4πr + 2πh) Δr + 2πr Δh, df = 0 =⇒ Δh = − Δr. r Δh =
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SECTION 16.9
36. The area is given by A =
1 2
x2 tan θ.
(a) The change in area is approximated by: dA = x tan θ Δx +
1 2
x2 sec2 θ Δθ = 3Δx +
25 2 Δθ.
(b) The actual change in area is 1 1 1 (x + Δx)2 tan(θ + Δθ) − x2 tan θ = [4 + Δx]2 tan[arctan (3/4) + Δθ] − 6. 2 2 2 (c) The area is more sensitive to a change in θ. x2 1 2 x sin θ; ΔA ∼ cos θΔθ = dA = x sin θΔx + 2 2 (b) The area is more sensitive to changes in θ if x > 2 tan θ, otherwise it is more sensitive to changes
37. (a) A =
in x. 38. (a) dV ∼ = yzΔx + xzΔy + xyΔz,
x = 60 in, y = 36 in, z = 42 in
1 Maximum possible error = 6192( 12 ) = 516 cubic inches.
(b) dS ∼ = 2(y + z)Δx + 2(x + z)Δy + 2(x + y)Δz 1 Maximum possible error = 552( 12 ) = 46 square inches
39. s =
A ; A = 9, W = 5, ΔA = ±0.01, ΔW = ±0.02 A−W ∂s ∂s −W A ds = ΔA + ΔW = ΔA + ΔW 2 ∂A ∂W (A − W ) (A − W )2 =−
9 5 (±0.01) + (±0.02) ∼ = ±0.014 16 16
The maximum possible error in the value of s is 0.014 lbs; 2.23 ≤ s + Δs ≤ 2.27 40. Assuming A > W , s is more sensitive to change in A. SECTION 16.9 1.
∂f = xy 2 , ∂x Thus,
2. 3.
4.
∂f = x, ∂x
f (x, y) = 12 x2 y 2 + φ(y),
φ (y) = 0, φ(y) = C,
and
∂f = x2 y + φ (y) = x2 y. ∂y f (x, y) = 12 x2 y 2 + C.
∂f 1 = y =⇒ f (x, y) = (x2 + y 2 ) + C ∂y 2
∂f ∂f = y, f (x, y) = xy + φ(y), = x + φ (y) = x. ∂x ∂y Thus, φ (y) = 0, φ(y) = C, and f (x, y) = xy + C. x3 ∂f = x2 + y =⇒ f (x, y) = + xy + φ(y); ∂x 3
∂f 1 1 = x + φ (y) = y 3 + x =⇒ f (x, y) = x3 + y 4 + xy + C ∂y 3 4 ∂ ∂ 5. No; y 3 + x = 3y 2 whereas x2 + y = 2x. ∂y ∂x
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SECTION 16.9 6.
7.
8.
9.
∂f = y 2 ex − y =⇒ f (x, y) = y 2 ex − xy + φ(y); ∂x ∂f = 2yex − x + φ (y) = 2yex − x =⇒ f (x, y) = y 2 ex − xy + C ∂y ∂f ∂f = cos x − y sin x, f (x, y) = sin x + y cos x + φ(y), = cos x + φ (y) = cos x. ∂x ∂y Thus, φ (y) = 0, φ(y) = C, and f (x, y) = sin x + y cos x + C. ∂f = 1 + ey =⇒ f (x, y) = x + xey + φ(y); ∂x ∂f y3 = xey + φ (y) = xey + y 2 =⇒ f (x, y) = x + xey + +C ∂y 3 ∂f = ex cos y 2 , ∂x Thus,
10.
11.
12.
15.
17.
φ (y) = 0, φ(y) = C,
and
∂f = −2yex sin y 2 + φ (y) = −2yex sin y 2 . ∂y
f (x, y) = ex cos y 2 + C.
∂2f ∂2f = ex sin y = ; ∂x∂y ∂y∂x
not a gradient.
∂f ∂f = xex − e−y , f (x, y) = xyex + e−y + φ(x), = yex + xyex + φ (x) = yex (1 + x). ∂y ∂x Thus, φ (x) = 0, φ(x) = C, and f (x, y) = xyex + e−y + C. ∂f = ex + 2xy =⇒ f (x, y) = ex + x2 y + φ(y); ∂x =⇒ f (x, y) = ex + x2 y − cos y + C ∂ xy xe + x2 = x2 exy ∂y
∂f = x2 + φ (y) = x2 + sin y ∂y
∂ (yexy − 2y) = y 2 exy ∂x
whereas
∂f = x sin x + 2y + 1 =⇒ f (x, y) = xy sin x + y 2 + y + φ(x) ∂y ∂f = y sin x + xy cos x + φ (x) = y sin x + xy cos x =⇒ f (x, y) = xy sin x + y 2 + y + C ∂x ∂f = 1 + y 2 + xy 2 , f (x, y) = x + xy 2 + ∂x Thus,
16.
f (x, y) = ex cos y 2 + φ(y),
∂2f = −ex sin y, ∂y∂x
13. No;
14.
853
φ (y) = y + 1,
φ(y) =
1 2
1 2
x2 y 2 + φ(y),
y2 + y + C
∂f = 2xy + x2 y + φ (y) = x2 y + y + 2xy + 1. ∂y
and f (x, y) = x + xy 2 +
1 ∂f = 2 ln 3y + =⇒ f (x, y) = 2x ln 3y + ln |x| + φ(y); ∂x x y3 +C f (x, y) = 2x ln 3y + ln |x| + 3 x ∂f = , 2 ∂x x + y2 Thus,
f (x, y) =
φ (y) = 0, φ(y) = C,
x2 + y 2 + φ(y), and
1 2
x2 y 2 +
1 2
y 2 + y + C.
∂f 2x 2x = + φ (y) = + y2 ∂y y y
y ∂f y = + φ (y) = . 2 2 2 ∂y x +y x + y2 f (x, y) = x2 + y 2 + C.
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SECTION 16.9 ∂f x2 = x tan y + sec2 x =⇒ f (x, y) = tan y + tan x + φ(y); ∂x 2 ∂f x2 π x2 x2 = sec2 y + φ (y) = sec2 y + πy; =⇒ f (x, y) = tan y + tan x + y 2 + C. ∂y 2 2 2 2 ∂f = x2 sin−1 y, ∂x Thus,
f (x, y) = 13 x3 sin−1 y + φ(y),
φ (y) = − ln y,
=⇒
x3 ∂f x3 = + φ (y) = − ln y. ∂y 3 1 − y2 3 1 − y2
φ(y) = y − y ln y + C,
and
1 f (x, y) = x3 sin−1 y + y − y ln y + C. 3 20.
∂f x x2 tan−1 y + =√ =⇒ f = sin−1 x tan−1 y + + φ(y); ∂x y 2y 1 − x2 ∂f x2 sin−1 x x2 x2 sin−1 x − + φ (y) = − 2 + 1 =⇒ f (x, y) = sin−1 x tan−1 y + = + y + C. 2 2 2 ∂y 1+y 2y 1+y 2y 2y
21. (a) Yes
(b) Yes
22. (a) f (x, y) = (x − y)e−x
2
y
(c) No
+C
(b) f (x, y) = sin(x + y) − cos(x − y) + C;
f (π/3, π/4) = 6
=⇒
C=6
f (x, y) = sin(x + y) − cos(x − y) + 6. 23.
∂f = f (x, y), ∂x Thus,
24.
∂f /∂x = 1, f (x, y)
ln f (x, y) = x + φ(y),
φ (y) = 1, φ(y) = y + K,
and
∂f /∂y = 0 + φ (y), f (x, y)
∂f = f (x, y). ∂y
f (x, y) = ex+y+K = Cex+y .
∂f = eg(x,y) gx (x, y) =⇒ f (x, y) = eg(x,y) + φ(y); ∂x ∂f = eg(x,y) gy (x, y) + φ (y) = eg(x,y) gy (x, y) =⇒ f (x, y) = eg(x,y) + C. ∂y
25. (a) P = 2x, Q = z, R = y;
∂P ∂Q =0= , ∂y ∂x
∂P ∂R =0= , ∂z ∂x
∂Q ∂R =1= ∂z ∂y
(b), (c), and (d) ∂f = 2x, f (x, y, z) = x2 + g(y, z). ∂x ∂g ∂f ∂g ∂f =0+ with = z =⇒ = z. ∂y ∂y ∂y ∂y Then, g(y, z) = yz + h(z)
=⇒
∂f = 0 + y + h (z) and ∂z Thus, h(z) = C 26.
and
f (x, y, z) = x2 + yz + h(z), ∂f =y ∂z
=⇒
h (z) = 0.
f (x, y, z) = x2 + yz + C.
∂f ∂g ∂f = yz =⇒ f (x, y, z) = xyz + g(y, z); = xz + = xz =⇒ f = xyz + h(z) ∂x ∂y ∂y ∂f = xy + h (z) = xy =⇒ f (x, y, z) = xyz + C ∂z
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SECTION 16.9 27. The function is a gradient by the test stated before Exercise 25. Take
P = 2x + y,
R = y − 2z.
Q = 2y + x + z, ∂P ∂Q =1= , ∂y ∂x
Then
∂P ∂R =0= , ∂z ∂x
∂Q ∂R =1= . ∂z ∂y
∇f = P i + Qj + Rk.
Next, we find f where
∂f = 2x + y ∂x
=⇒
∂f ∂g = x+ ∂y ∂y
with
f (x, y, z) = x2 + xy + g(y, z). ∂f = 2y + x + z ∂y
=⇒
∂g = 2y + z. ∂y
Then, g(y, z) = y 2 + yz + h(z), f (x, y, z) = x2 + xy + y 2 + yz + h(z). ∂f = y + h (z) = y − 2z ∂z Thus, 28.
h(z) = −z 2 + C
h (z) = −2z.
=⇒
f (x, y, z) = x2 + xy + y 2 + yz − z 2 + C.
and
∂f = 2x sin 2y cos z =⇒ f (x, y, z) = x2 sin 2y cos z + g(y, z); ∂x ∂f ∂g = 2x2 cos 2y cos z + = 2x2 cos 2y cos z =⇒ f (x, y, z) = x2 sin 2y cos z + h(z) ∂y ∂y ∂f = −x2 sin 2y sin z + h (z) = −x2 sin 2y sin z =⇒ f (x, y, z) = x2 sin 2y cos z + C ∂z
29. The function is a gradient by the test stated before Exercise 25. Take
P = y 2 z 3 + 1,
Q = 2xyz 3 + y,
∂P ∂Q = 2yz 3 = , ∂y ∂x Next, we find f where
R = 3xy 2 z 2 + 1. ∂P ∂R = 3y 2 z 2 = , ∂z ∂x
Then ∂Q ∂R = 6xyz 2 = . ∂z ∂y
∇f = P i + Qj + Rk. ∂f = y 2 z 3 + 1, ∂x
f (x, y, z) = xy 2 z 3 + x + g(y, z). ∂f ∂g = 2xyz 3 + ∂y ∂y
with
∂f = 2xyz 3 + y ∂y
=⇒
∂g = y. ∂y
Then, g(y, z) =
1 2
y 2 + h(z),
f (x, y, z) = xy 2 z 3 + x +
1 2
y 2 + h(z).
∂f = 3xy 2 z 2 + h (z) = 3xy 2 z 2 + 1 ∂z Thus,
h(z) = z + C
and
f (x, y, z) = xy 2 z 3 + x +
1 2
=⇒
y 2 + z + C.
h (z) = 1.
855
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December 7, 2006
REVIEW EXERCISES y ∂f xy = − ez =⇒ f (x, y, z) = − xez + g(y, z) ∂x z z ∂f x ∂g x xy = + = + 1 =⇒ f (x, y, z) = + y − xez + h(z) ∂y z ∂y z z ∂f xy xy xy = − 2 − xez + h (z) = −xez − 2 =⇒ f (x, y, z) = − xez + y + C ∂z z z z
31. F(r) = ∇
GmM r
⎧ ⎪ ⎪ ⎨∇
32.
h(r) =
⎪ ⎪ ⎩
k n+2 , r n+2
n = 2
∇ (k ln r) ,
n = −2.
REVIEW EXERCISES 1.
∇f (x, y) = (4x − 4y)i + (3y 2 − 4x)j
2.
∇f (x, y) =
y 3 − x2 y x3 − xy 2 i + j (x2 + y 2 )2 (x2 + y 2 )2
3. ∇f (x, y) = (yexy tan 2x + 2exy sec2 2x) i + xexy tan 2x j 4. ∇f =
1 (x i + y j + z k) x2 + y 2 + z 2
5. ∇f (x, y) = 2xe−yz sec z i, −zx2 e−yz sec zj − (x2 ye−yz sec z − x2 e−yz sec z tan z)k 6. ∇f (x, y) = ye−3z cos xy i + e−3z (x cos xy + sin y) j, −3e−3z (sin xy − cos y) k 7. ∇f (x, y) = (2x − 2y) i − 2x j,
∇f (1, −2) = 6 i − 2 j;
1 2 ua = √ i + √ j; 5 5
2 fu a (1, −2) = ∇f (1, −2) · ua = √ . 5 8. ∇f (x, y) = (e
xy
xy
2 xy
+ xye ) i + x e
fu a (2, 0) = ∇f (2, 0) · ua =
j,
∇f (2, 0) = i + 4j;
√ 1 + 2 3. 2
9. ∇f (x, y, z) = (y 2 + 6xz) i + (2xy + 2z) j + (2y + 3x2 ) k, ua =
√ 3 1 j ua = i + 2 2
∇f (1, −2, 3) = 22 i + 2 j − k;
2 2 1 16 i − j + k; fu a (1, −2, 3) = ∇f (1, −2, 3) · ua = . 3 3 3 3
10. ∇f (x, y, z) =
x2
2x 2y 2z i+ 2 j+ 2 k, 2 2 2 2 +y +z x +y +z x + y2 + z2
1 1 1 ua = √ i − √ j + √ k; 3 3 3
∇f (1, 2, 3) =
2 fu a (1, 2, 3) = ∇f (1, 2, 3) · ua = √ . 7 3
1 (i + 2 j + 3 k); 7
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REVIEW EXERCISES 11. ∇f (x, y) = (6x − 2y 2 ) i − 4xy j,
857
∇f (3, −2) = 10 i + 24 j;
a = (0, 0) − (3, −2) = (−3, 2) = −3 i + 2 j,
−3 2 ua = √ i + √ j; 13 13
18 fu a (3, −2) = ∇f (3, −2) · ua = √ . 13 12. ∇f (x, y, z) = (y 2 z − 3yz) i + (2xyz − 3xz) j + (xy 2 − 3xy) k, r (t) = i − π sin πt j + 2et−1 k,
a = r (1) = i + 2k,
∇f (1, −1, 2) = 8 i − 10 j + 4 k;
1 2 ua = √ i + √ k; 5 5
16 fu a (1, −1, 2) = ∇f (1, −1, 2) · ua = √ . 5 13. ∇f (x, y, z) =
1 x2 + y 2 + z 2
(x i + y j + z k),
1 ∇f (3, −1, 4) = √ (3 i − j + 4 k); 26
1 ua = ± √ (4 i − 3 j + k); 26
a = ±(4 i − 3 j + k),
fu a (3, −1, 4) = ∇f (3, −1, 4) · ua = ±
14. ∇f (x, y) = 2e2x (cos y − sin y) i − e2x (sin y + cos y) j, ∇f √ maximum directional derivative: ∇f 12 , − 12 π = e 5. 15. ∇f (x, y, z) = cos xyz(yz i + xz j + xy k),
1
1 2, −2π
√ √ π 3 π 3 6 i+ 4 √ 2 +3 − 39π 12
∇f ( 12 , 13 , π) =
minimum directional derivative: fu = −∇f ( 12 , 13 , π) =
19 . 26
= 2e i + e j;
√
3 12
j+
k;
16. Let r(t) = x(t) i + y(t) j be the path of the particle. ∇I(x, y) = −2x i − 6y j. Then x (t) = −2x(t),
y (t) = −6y(t)
=⇒
x(t) = C1 e−2t ,
r(0) = (4, 3)
=⇒
C1 = 4,
y(t) = C2 e−6t .
C2 = 3.
Therefore the path of the particle is: r(t) = 4e−2t i + 3e−6t j, t ≥ 0,
or,
y=
00
result saddle loc. min.
f (6, 6) = −216 42. ∇f (x, y) = (3x2 − 12x) i + (2y + 1) j = 0 at (0, − 12 ), (4, − 12 ). fxx = 6x − 12, fxy = 0, fyy = 2.
f (4, − 12 ) = − 145 4
point
A
B
C
D
result
(0, − 12 ) (4, − 12 )
−12
0
2
−24
saddle
12
0
2
24
loc. min.
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REVIEW EXERCISES
861
43. ∇f (x, y) = (1 − 2xy + y 2 ) i + (−1 − x2 + 2xy) j = 0 at (1, 1), (−1, −1). fxx = −2y, fxy = −2x + 2y, fyy = 2x.
44. ∇f (x, y) = e−(x
2
+y 2 )/2
fxx = e−(x
2
point
A
B
C
D
result
(1, 1)
−2
0
2
−4
saddle
(−1, −1)
2
0
−2
−4
saddle
√ (y 2 − x2 y 2 ) i + (2xy − xy 3 ) j = 0 at (±1, ± 2), (x, 0), x any real number.
+y 2 )/2
(−3xy 2 + x3 y 2 ), fyy = e−(x
2
fxy = e−(x
+y 2 )/2
point A B √ (1, 2) −4e−3/2 0 √ (1, − 2) −4e−3/2 0 √ (−1, 2) 4e−3/2 0 √ (−1, − 2) 4e−3/2 0 √ √ local maxima: f (1, 2) = f (1, − 2) = 2e−3/2 ;
2
+y 2 )/2
(2y − y 3 − 2x2 y + x2 y 3 ),
(2x − 5xy 2 + xy 4 ). C
D
−4e−3/2
16e−3
loc. max
−4e−3/2
16e−3
loc. max
4e−3/2
16e−3
loc. min
4e−3/2
16e−3
loc. min
local minima: f (−1,
result
√
√ 2) = f (−1, − 2) = −2e−3/2 .
At (x, 0), D = 0 and f (x, 0) ≡ 0. For x < 0, f (x, y) < f (x, 0) for all y = 0; for x > 0, f (x, y) > f (x, 0) for all y > 0; (0, 0) is a saddle point. Here is a graph of the surface.
45. ∇f = (2x − 2) i + (2y + 2) j = 0 at (1, −1) in D; Next we consider the boundary of D. C : r(t) = 2 cos t i + 2 sin t j,
f (1, −1) = 0
We parametrize the circle by:
t ∈ [ 0, 2π ]
The values of f on the boundary are given by the function F (t) = f (r(t)) = 6 − 4 cos t + 4 sin t, F (t) = 4 sin t + 4 cos t :
F (t) = 0
=⇒
t ∈ [ 0, 2π ]
sin t = − cos t
=⇒
t=
7 3 π, π 4 4
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REVIEW EXERCISES Evaluating F at the endpoints and critical numbers, we have:
√ √ √ F (0) = F (2π) = f (2, 0) = 2; F 34 π = f − 2, 2 = 6 + 4 2;
√ √ √ 7 π = f 2, − 2 = 6 − 4 2. 4 √ √ √ f takes on its absolute maximum of 6 + 4 2 at − 2, 2 ; f takes on its absolute minimum of 0 at F
(1, −1). 46. ∇f (x, y) = (4x − 4) i + (2y − 4) j = 0 at (1, 2) on the boundry of D; no critical points in D. Next we consider the boundary of D. We
y 3
parametrize each side of the triangle: C1 : r1 (t) = t i + 2t j,
t ∈ [ 0, 1 ]
C2 : r2 (t) = (1 − t) i + 2 j, C3 : r3 (t) = (2 − t) j,
2
t ∈ [ 0, 1 ]
t ∈ [ 0, 2 ]
1
Now, f1 (t) = f (r1 (t)) = 6t2 − 8t + 3, f2 (t) = f (r2 (t)) = 2t2 − 3+, f3 (t) = f (r3 (t)) = t2 − 1,
t ∈ [ 0, 1 ];
t ∈ [ 0, 1 ];
t ∈ [ 0, 2 ];
critical number: t =
1
2 3
2
3
x
critical number critical number
Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (0) = f3 (2) = f (0, 0) = 3;
f1 (2/3) = f (2/3, 4/3) = − 73 ;
f1 (1) = f2 (0) = f (1, 2) = −3;
f2 (1) = f3 (0) = f (0, 2) = −1. f takes on its absolute maximum of 3 at (0, 0) and its absolute minimum of −3 at (1, 2). 47. ∇f (x, y) = (8x − y) i + (−x + 2y + 1) j = 0 at (−1/15, −8/15) in D;
f (−1/15, −8/15) = −4/15.
On the boundary of D : x = cos t, y = 2 sin t. Set F (t) = f (cos t, 2 sin t) = 4 + 2 sin t − 2 sin t cos t,
0 ≤ t ≤ 2π.
Then F (t) = 2 cos t − 4 cos2 t + 2 = −2(2 cos t + 1)(cos t − 1);
F (t) = 0
=⇒
t=
2π 4π , . 3 3
Evaluating F at the endpoints of the interval and at the critical points, we get √ √ 3 3 F (0) = F (2π) = f (1, 0) = 4, F (2π/3) = f (−1/2, 3) = 4 + , 2 √ √ 4 3 3 >− F (4π/3) = f (−1/2, − 3) = 4 − 2 15 f takes on its absolute maximum of 2 at (0, 1); (−1/15, −8/15).
f takes on its absolute minimum of −4/15 at
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REVIEW EXERCISES 48. ∇f (x, y) = 4x3 i + 6y 2 j = 0 at (0, 0) in D;
863
f (0, 0) = 0.
On the boundary of D : x = cos t, y = sin t. Set 0 ≤ t ≤ 2π.
F (t) = f (cos t, sin t) = cos4 t + 2 sin3 t, Then
F (t) = 4 cos3 t sin t + 6 sin2 t cos t = 2 sin t cos t(2 sin t − 1)(sin t + 2); F (t) = 0
=⇒
t = π/6, π/2, 5π/6, π, 3π/2
Evaluating F at the endpoints of the interval and at the critical points, we get √ F (0) = F (2π) = f (1, 0) = 1, F (π/6) = f ( 3/2, 1/2) = 13/16, F (π/2) = f (0, 1) = 2, √ F (5π/6) = f (− 3/2, 1/2) = 13/16, F (π) = f (−1, 0) = 1, F (3π/2) = f (0, −1) = −2. f takes on its absolute maximum of 2 at (0, 1); f takes on its absolute minimum of −2 at (0, −1). 49. Set f (x, y, z) = D2 = (x − 1)2 + (y + 2)2 + (z − 3)2 , ∇f = 2(x − 1) i + 2(y + 2) j + 2(z − 3) k,
g(x, y) = 3x + 2y − z − 5.
∇g = 3 i + 2 j − k.
Set ∇f = λ∇g : 2(x − 1) = 3λ
=⇒
x = 32 λ + 1,
2(y + 2) = 2λ
=⇒
y = λ − 2,
2(z − 3) = −λ
=⇒
z = − 12 λ + 3.
41 5 33 9 =⇒ x = , y=− , z= . 7 14 7 14 The point on the plane that is closest to (1, −2, 3) is (41/14, −5/7, 33/14). The distance from the 9 point to the plane is √ . 14 Substituting these values in 3x + 2y − z = 5 gives λ =
50. Set f (x, y, z) = 3x − 2y + z, ∇f = 3 i − 2 j + k,
g(x, y, z) = x2 + y 2 + z 2 − 14,
∇g = 2x i + 2y j + 2z k.
Set ∇f = λ∇g : 3 = 2λx
=⇒
x = 3/2λ,
−2 = 2λy
=⇒
y = −1/λ,
Substituting these values in x2 + y 2 + z 2 = 14 gives λ = ± 12
1 = 2λz
=⇒
z = 1/2λ.
x = 3, y = −2, z = 1 or
=⇒
x = −3, y = 2, z = −1. Evaluating f : f (3, −2, 1) = 14, f (−3, 2, −1) = −14. The maximum value of f on the sphere is 14. 51. Set f (x, y, z) = x + y − z, ∇f = i + j − k,
g(x, y, z) = x2 + y 2 + 4z 2 − 4,
∇g = 2x i + 2y j + 8z k.
Set ∇f = λ∇g : 1 = 2λx
=⇒
x = 1/2λ,
1 = 2λy
=⇒
y = 1/2λ,
−1 = 8λz
=⇒
Substituting these values in x2 + y 2 + 4z 2 = 4 gives λ = ± 38
=⇒
or x = −4/3, y = −4/3, z = 1/3. Evaluating f :
f (− 43 , − 43 , 13 )
value of f is 3, the minimum value is −3.
f ( 43 , 43 , − 13 )
= 3,
z = −1/8λ.
x = 4/3, y = 4/3, z = −1/3 = −3. The maximum
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REVIEW EXERCISES
52. Let the length, width and height be x, y, z respectively. Then the total cost is f (x, y, z) =
1 2
xy +
1 2
xz +
1 2
yz +
1 10
xy =
3 5
xy +
1 2
xz +
1 2
yz
with the condition g(x, y, z) = xyz − 16 = 0. xyz = 16 =⇒ x = 0, y = 0 z = 0. ∇f = 35 y + 12 z i + 35 x + 12 z j + 12 y + 12 x k,
Note first that
∇f = λ ∇g
=⇒
3 5
y+
1 2
z = λyz,
3 5
x+
1 2
∇g = yz i + xz j + xy k
z = λxz,
1 2
x+
1 2
y = λxy
Multiply the first equation by x, the second equation by y and subtract. This gives: 1 2 (xz
− yz) = 0
=⇒
z(x − y) = 0
=⇒
Substituting y = x in the third equation yields x = λx2 =⇒ 1 6 Substituting x = y = in the first equation yields z = . λ 5λ
x=
y=x 1 . λ
√ 3 3 . Finally, substituting these values for x, y and z into the equation xyz = 16, we get λ = √ 3 2 5 √ 3 12 ∼ 10 ∼ 6 2 5 = √ Therefore, x = y = √ = 2.37 and z = x = √ = 2.85. 3 3 3 5 3 75 75 53. df = (9x2 − 10xy 2 + 2) dx + (−10x2 y − 1) dy 54. df = (2xy sec2 x2 − 2y 2 ) dx + (tan x2 − 4xy) dy 55. df =
xz 2 + zx2 x2 y + y 2 x y2 z + z2 y dx + dy + dz (x + y + z)2 (x + y + z)2 (x + y + z)2
z 2y x yz yz dx + ze − 2 dy + ye − 2 dz 56. df = − 2 y + xz y + xz y + xz 57. Set f (x, y, z) = ex
y + z 3 . Then df = ex
1 ex ex 3z 2 y + z 3 Δx + Δy + Δz. 2 y + z3 2 y + z3
With x = 0, y = 15, z = 1, Δx = 0.02, Δy = 0.2, Δz = 0.01, df = 4 Δx + √ Therefore, e0.02 15.2 + (1.01)3 ∼ = e0 15 + 1 + 0.1088 = 4.1088.
1 8
Δy +
3 8
Δz ∼ = 0.1088.
58. Set f (x, y) = x1/3 cos2 y. Then df =
1 3
x−2/3 cos2 y Δx − 2x1/3 cos y sin y Δy.
√ 1 π , df = Δx − 2 3 Δy ∼ = 0.1287. 90 64 Therefore, (64.5)1/3 cos2 (28◦ ) ∼ = 641/3 cos2 (30◦ ) + 0.1287 = 3.1287. With x = 64, y = 30◦ = π/6, Δx = 0.5, Δy = −2◦ = −
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REVIEW EXERCISES 59. V = πr2 h;
r = 5 ft.,
h = 22 ft.,
Δr = 0.01 in. =
1 1200
ft.,
Δh = 0.01 =
865
1 1200
dV = 2πrh Δr + πr2 Δh Using the values given above, dV = 2π(5)(22)
1 1 ∼ + π(25) = 0.6414 cu. ft. ∼ = 1108.35 cu. in.; 1200 1200
1108.35 ∼ = 4.80. 231
Approximately 4.80 gallons will be needed. 60.
∂P ∂Q = 12x2 y − 8x = ; the vector function is a gradient. ∂y ∂x ∂f = 6x2 y 2 − 8xy + 2x, ∂x
f (x, y) = 2x3 y 2 − 4x2 y + x2 + φ(y),
∂f = 4x3 y − 4x2 + φ (y) = 4x3 y − 4x2 − 8. ∂y Thus, 61.
φ (y) = −8, φ(y) = −8y + C,
f (x, y) = 2x3 y 2 − 4x2 y + x2 − 8y + C.
and
∂Q ∂P = 2x − sin x = ; the vector function is a gradient. ∂y ∂x ∂f = 2xy + 3 − y sin x, ∂x
f (x, y) = x2 y + 3x + y cos x + φ(y),
∂f = x2 + cos x + φ (y) = x2 + 2y + 1 + cos x. ∂y Thus, 62.
63.
φ (y) = 2y + 1, φ(y) = y 2 + y + C,
∂P = 2xy + 4y; ∂y
∂Q = −2xy + 2; ∂x
and
∂P ∂Q = ; ∂y ∂x
f (x, y) = x2 y + 3x + y cos x + y 2 + y + C. the vector function is not a gradient.
∂P ∂P ∂Q ∂Q ∂R ∂R = ey sin z = , = ey cos z = , = xey cos z = ; ∂y ∂x ∂z ∂x ∂z ∂y the vector function is a gradient. f (x, y, z) = (ey sin z + 2x) dx = xey sin z + x2 + φ(y, z), ∂φ ∂φ 1 = xey sin z − y 2 =⇒ = −y 2 =⇒ φ = − y 3 + ψ(z), ∂y ∂y 3 1 f (x, y, z) = xey sin z + x2 − y 3 + ψ(z), fz = xey cos z + ψ (z) = xey cos z =⇒ ψ (x) = 0 =⇒ 3 ψ(x) = C fy = xey sin z +
Therefore f (x, y, z) = xey sin z + x2 − 13 y 3 + C.
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SECTION 17.1
CHAPTER 17 SECTION 17.1 1.
3 3
2i−1 3j+1 =
⎞ ⎛ 3 2i−1 ⎝ 3j+1 ⎠ = (1 + 2 + 4)(9 + 27 + 81) = 819
3
i=1 j=1
i=1
j=1
2. 2 + 22 + 3 + 32 + 4 + 42 + 5 + 52 = 68 3.
4 3
(i2 + 3i)(j − 2) =
i=1 j=1
4.
5.
4
⎡ (i2 + 3i) ⎣
i=1
3
⎤ (j − 2)⎦ = (4 + 10 + 18 + 28)(−1 + 0 + 1) = 0
j=1
4 2 2 2 2 2 2 4 4 4 4 4 4 6 6 6 6 6 6 + + + + + + + + + + + + + + + + + = 19 . 2 3 4 5 6 7 2 3 4 5 6 7 2 3 4 5 6 7 35 m
Δxi = Δx1 + Δx2 + · · · + Δxm = (x1 − x0 ) + (x2 − x1 ) + · · · + (xm − xm−1 )
i=1
= xm − x0 = a2 − a1
6. (y1 − y0 ) + (y2 − y1 ) + · · · + (yn − yn−1 ) = yn − y0 = b2 − b1
7.
m n
Δxi Δyj =
i=1 j=1
8.
9.
⎞ ⎛ n Δxi ⎝ Δyj ⎠ = (a2 − a1 ) (b2 − b1 )
m i=1
q n
⎛ Δyj Δzk = ⎝
j=1
n
j=1 k=1
j=1
m
m
(xi + xi−1 )Δxi =
i=1
⎞ Δyj ⎠
q
Δzk
= (b2 − b1 ) (c2 − c1 )
k=1 m
(xi + xi−1 )(xi − xi−1 ) =
i=1
(xi 2 − x2i−1 )
i=1
= xm 2 − x0 2 = a2 2 − a1 2 10.
n 1 j=1
11.
2
m n
(xi + xi−1 )Δxi Δyj =
i=1 j=1
∧ (Exercise 9)
12.
1 3 1 (yj − yj−1 3 ) = (b2 3 − b1 3 ) 2 j=1 2 n
(yj 2 + yj yj−1 + yj−1 2 )Δyj =
m n i=1 j=1
m
(xi + xi−1 )Δxi
i=1
n
Δyj
j=1
= a2 2 − a1 2 (b2 − b1 )
(yi + yj−1 )Δxi Δyj =
m i=1
⎤ ⎡ n Δxi ⎣ (yj 2 − yj−1 2 )⎦ = (a2 − a1 )(b2 2 − b1 2 ) j=1
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SECTION 17.2 13.
m n
m
(2Δxi − 3Δyj ) = 2
i=1 j=1
Δxi
867
m n n 1 −3 1 Δyj
i=1
j=1
i=1
j=1
= 2n(a2 − a1 ) − 3m(b2 − b1 ) 14.
m n
(3Δxi − 2Δyj ) = 3
i=1 j=1
15.
m n
Δxi − 2
i=1 j=1
q m n
Δxi Δyj Δzk =
i=1 j=1 k=1
m
m n
Δyj = 3n(a2 − a1 ) − 2m(b2 − b1 ).
i=1 j=1
⎞ ⎛ n q Δxi ⎝ Δyj ⎠ Δzk
i=1
j=1
k=1
= (a2 − a1 )(b2 − b1 )(c2 − c1 ) 16.
q m n
(xi + xi−1 )Δxi Δyj Δzk =
i=1 j=1 k=1
m
⎞
⎛ n q 2 2 (xi − xi−1 ) ⎝ Δyj ⎠ Δzk
i=1
j=1
k=1
= (a2 2 − a1 2 )(b2 − b1 )(c2 − c1 ) 17.
n n n
δijk aijk = a111 + a222 + · · · + annn =
appp
p=1
i=1 j=1 k=1
18. Start with
n
m n
aij .
Take all the aij
(there are only a finite number of them) and order them
i=1 j=1
in any order you chose. Call the first one b1 , the second b2 , and so on. Then m r n aij = bp where r = m × n. p=1
i=1 j=1
SECTION 17.2 1. Lf (P ) = 2 14 , 3. (a) Lf (P ) =
Uf (P ) = 5 34 m n
2. Lf (P ) = 3,
(xi−1 + 2yj−1 ) Δxi Δ yj ,
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
m n
Uf (P ) = 5
(xi + 2yj ) Δxi Δyj
i=1 j=1
m n yj−1 + yj xi−1 + xi +2 Δxi Δyj ≤ Uf (P ). 2 2 i=1 j=1
The middle expression can be written m m n n 2 1 2 2 xi − x2i−1 Δyj + yj − yj−1 Δxi . 2 i=1 j=1 i=1 j=1
The first double sum reduces to m n 1 i=1 j=1
2
xi − 2
x2i−1
1 Δyj = 2
m
xi − 2
x2i−1
i=1
In like manner the second double sum also reduces to 2. bounded above by the plane z = x + 2y and below by R.
n j=1
Δyj
=
1 (4 − 0) (1 − 0) = 2. 2
Thus, I = 4; the volume of the prism
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SECTION 17.2
4. Lf (P ) = −7/16, 6. (a) Lf (p) =
5. Lf (P ) = −7/24,
Uf (P ) = 7/16
m n
(xi−1 − yj )Δxi Δyj ,
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
m n
Uf (P ) = 7/24
(xi − yj−1 )Δxi Δyj
i=1 j=1
m n xi + xi−1
2
i=1 j=1
−
yj + yj−1 2
Δxi Δyj ≤ Uf (P )
The middle expression can be written m n 1 i=1 j=1
2
(xi 2 − xi−1 2 )Δyj −
m n 1
2
i=1 j=1
(yj 2 − yj−1 2 )Δxi .
The first sum reduces to ⎞ m ⎛ n 1 1 1 (xi 2 − xi−1 2 ) ⎝ Δyj ⎠ = (1 − 0)(1 − 0) = . 2 i=1 2 2 j=1 In like manner the second sum also reduces to 12 . 7. (a) Lf (P ) =
m n
(4xi−1 yj−1 ) Δxi Δyj ,
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
Thus I = m n
1 2
−
1 2
= 0.
(4xi yj ) Δxi Δyj
i=1 j=1
m n
(xi + xi−1 ) (yj + yj−1 ) Δx1 Δyj ≤ Uf (P ).
i=1 j=1
The middle expression can be written m n
xi − 2
x2i−1
yj − 2
2 yj−1
i=1 j=1
= ∧
m
xi − 2
x2i−1
i=1
n
yj − 2
2 yj−1
j=1
by (17.1.5) = b2 − 02 d2 − 02 = b2 d2 . It follows that I = b2 d2 . 8. (a) Lf (P ) =
m n
3(xi−1 2 + yj−1 2 )Δxi Δyj ,
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
m n
3(xi 2 + yj 2 )Δxi Δyj
i=1 j=1
m n
(xi 2 + xi xi−1 + xi−1 2 ) + (yj 2 + yj yj−1 + yj−1 2 ) Δxi Δyj ≤ Uf (P )
i=1 j=1
Since in general (A2 + AB + B 2 )(A − B) = A3 − B 3 , m n
(xi 3 − xi−1 3 )Δyj +
i=1 j=1
which reduces to m
m n
the middle expression can be written
(yj 3 − yj−1 3 )Δxi ,
i=1 j=1
⎞ ⎞ ⎛ n ⎛ n m xi 3 − xi−1 3 ⎝ Δyj ⎠ + Δxi ⎝ yj 3 − yj−1 3 ⎠ .
i=1
j=1 3
3
i=1 2
2
This can be evaluated as b d + bd = bd(b + d ).
j=1
It follows that I = bd(b2 + d2 ).
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SECTION 17.2 9. (a) Lf (P ) =
m n
3 x2i−1 − yj 2 Δxi Δyj ,
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
m
n
m n
869
2 Δxi Δyj 3 xi 2 − yj−1
i=1 j=1
2 xi 2 + xi xi−1 + x2i−1 − yj 2 + yj yj−1 + yj−1 Δxi Δyj ≤ Uf (P ).
i=1 j=1
Since in general A2 + AB + B 2 (A − B) = A3 − B 3 ,
the middle expression can be written
m n
m n 3 3 xi 3 − x3i−1 Δyj − yj − yj−1 Δxi ,
i=1 j=1
which reduces to m
i=1 j=1
⎞ ⎞ ⎛ n ⎛ n m 3 ⎠ . xi 3 − x3i−1 ⎝ Δyj ⎠ − Δxi ⎝ yj 3 − yj−1
i=1
j=1
i=1
This can be evaluated as b3 d − bd3 = bd b2 − d
2
j=1
. It follows that I = bd (b2 − d2 ).
10. On each subrectangle, the minimum and the maximum of f are equal, so f is constant on each subrectangle and therefore (since f is continuous) on the entire rectangle R. Then f (x, y) dxdy = f (a, c)(b − a)(d − c). R
11.
dxdy =
Ω
b
φ(x) dx a
12. Suppose that there is a point (x0 , y0 ) on the boundary of Ω at which f is not zero. As (x, y) tends to (x0 , y0 ) through that part of R which is outside Ω, f (x, y), being zero, tends to zero. Since f (x0 , y0 ) is not zero, f (x, y) does not tend to f (x0 , y0 ). Thus the extended function f can not be continuous at (x0 , y0 ). 13. Suppose f (x0 , y0 ) = 0. Assume f (x0 , y0 ) > 0. Since f is continuous, there exists a disc Ω with radius centered at (x0 , y0 ) such that f (x, y) > 0 on Ω . Let R be a rectangle contained in Ω . Then f (x, y) dxdy > 0, which contradicts the hypothesis. R
14.
(x + 2y) dx dy = 4;
area(R) = (2)(1) = 2,
so average value =
4 =2 2
R
4xy dxdy = 22 32 = 36. Thus
15. By Exercise 7, Section 17.2, R
favg =
1 area (R)
4xy dxdy =
1 (36) = 6 6
R
(x2 + y 2 ) dxdy =
16. R
bd(b2 + d2 ) ; 3
area(R) = bd,
so average value =
b 2 + d2 3
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SECTION 17.3
17. By Theorem 16.2.10, there exists a point (x1 , y1 ) ∈ Dr such that
dxdy = f (x1 , y1 )πr2
f (x, y) dxdy = f (x1 , y1 ) Dr
=⇒
f (x1 , y1 ) =
1 πr2
R
f (x, y) dxdy Dr
As r → 0, (x1 , y1 ) → (x0 , y0 ) and f (x1 , y1 ) → f (x0 , y0 ) since f is continuous. The result follows. 18. 0 ≤ sin(x + y) ≤ 1 for all (x, y) ∈ R. Thus, 0 ≤
sin(x + y) dxdy ≤
R
19. z =
4 − x2 − y 2 dxdy is the volume V of one
Ω
quarter of a hemisphere; V = 43 π.
dxdy = 1
R
4 − x2 − y 2 on Ω : x2 + y 2 ≤ 4, x ≥ 0, y ≥ 0;
20. 8 − 4
x2 + y 2 on Ω is a cone with height h = 8 and radius r = 2;
V =
32π 3
x y z + + = 1; the solid is the tetrahedron bounded by the coordinate planes 3 2 6 x y z and the plane: + + = 1; V = 16 (3)(2)(6) = 6 3 2 6 2 ∼ ∼ 22. (a) Lf (P ) = 35.4603; Uf (P ) = 36.5403 (c) 3y − 2x dxdy = 36 21. z = 6 − 2x − 3y ⇒
R
SECTION 17.3 1 3 2 1. x dy dx = 0
0 3
1
x+y
e
3x2 dx = 1
0
2. 0
1
0 1
3
dx dy = 0
3
1
2
3.
xy dy dx = 0
0 1
0
x
3
4.
1
x y dy dx = 0
0 1
x
xy 3 dy dx =
0 1
x3
x
x2 y 2 dy dx = 0
7.
1
3
0
1
dx =
9x dx = 0
0
1 4 y 4
x2
x
dx = 0
0
9 2
1
1 5 1 x dx = 4 24
x3 1 dx = 3 18
π/2
sin (x + y) dy dx = 0
x2 1 dx = 2 12
1
x
0 π/2
1 x y3 3
0
6. 0
0
5. 0
3 (e1+y − ey ) dy = e1+y − ey 0 = e4 − e3 − e + 1
0
π/2
π/2
[− cos (x + y)]0
dx = 0
π/2
π dx = 2 cos x − cos x + 2
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SECTION 17.3
π/2
π/2
8. 0
sin
0
π/2
π/2
π/2
10.
0
1 y + xy 2 2
√
−
−1
1
y
11.
1
0
13.
(x + 3y ) dx dy =
−2
2 3/2 x 3
y
y(ey
4−y
1 2 2y
2
dx dy =
x2
2
4x7 x8 x10 = − − 21 8 30
1
0
1 1 1 1 2 π + π x dx = π 2 + π 4 2 8 4 64
(integrand is odd)
2 2 2 y − y 7/2 dy = 3 27
2
1 = 0
1 (e − 2) 2
1 4 − y2 2
−
1 2 y 2
dy
512 16 − 8y 2 + y 4 dy = 15
x4 y +
0
15. 0
1
(x4 + y 2 ) dy dx =
x3
0
1
0
y2
0 1
dy =
2
π/2
dx = 0
1 − y 2 dy = 0
4−y
−2
y
2
=2
14. I =
π/2
y2 ey − − 1) dy = 2 2
2
0 4− 12 y 2
6y 3
1
yex dx dy =
0 2
√
1
xy dx dy = 0
y2
1
−1
1−y 2
√
12.
3
y2
0
0
√1−y2
1
π π/2 + y − sin y dy = cos y − cos +y =0 2 2 0
(1 + xy) dy dx = 0
π
0
9.
y3 3
x2
1
dx = 0
x3
4x6 x9 − x7 − 3 3
dx
9 280
= 0
by symmetry (integrand odd in y, Ω symmetric about x-axis)
1
2y
−y 2 /2
16.
e 0
2
0
x/2
2ye
2
2
0
0
−1
−y 2 /2
dy = −2e
1 0
1
x5 +
x8 x6 − x4 − 2 2
x6 x9 x5 x7 + − − = 6 18 5 14
x3
(x + y) dydx = x4
0
1 =2 1− √ e
2 1 x2 1 x2 1 4 = xe dx = e e −1 2 4 4 0
x4
x3
ex dy dx =
(x + y) dy dx +
−1
=
−y 2 /2
0
0
0
1
dx dy =
0
17.
18.
π/2
cos(x + y) dx dy =
0
1
dx + 0
x4 +
0
−1
xy +
y2 2
x6 x8 − x5 − 2 2
x5 x7 x6 x9 + + − − 5 14 6 18 −1
1 =− 0
x4 x3
1 3
dx
1
xy +
dx + 0
y2 2
x3 dx x4
871
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SECTION 17.3 20.
19.
1
y 1/4
f (x, y) dx dy
21.
1
√
f (x, y) dydx x
22.
0
−1
1
−x
1
1
f (x, y) dy dx +
f (x, y) dy dx 0
1/2
√ 3
x
23.
y 1/3
1
f (x, y) dxdy + 1/2
1/2
y 1/3
f (x, y) dxdy y
1/2
24.
2
y
4
y
f (x, y) dx dy + 1
f (x, y) dx dy
1
2
y/2 8
4
−2
3
1 2 x+2
dy dx =
4y−y 2
26.
3
dx dy = 0
y
27.
dx dy = 0
2y 3/2
0
−y
f (x, y) dx dy +
9 2
1 y − 2y 3/2 dy = 160
1
9
+
3
f (x, y) dx dy
−1
1
1 1 2 x + 2 − x dx = 9 2 4
(3y − y 2 ) dy =
1/4
3
y/2
0
y 1/4
4
−2
1/4x2
−3
f (x, y) dx dy 4
−1
4
+
25.
0
y 1/2
0
1
1
3
√
f (x, y) dx dy y
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SECTION 17.3
3
5−x
28.
3
dy dx = 2
6/x
(5 − x − 6/x) dx =
2
2 5 + 6 ln 2 3
29.
1
y2
sin 0
0
30.
1
y3 + 1 2
x2 dy dx = =
ln 2
32.
1
0
2
2/y
dx dy =
33. 1
y−1 1
34.
1
2
0
0
√
y
1−x
0
1
x(1 − x) +
y3 + 1 y sin dx dy = dy 2 0 3 1 y +1 2 = − cos 3 2 0 1 2 cos − cos 1 = 3 2
1
−1
1
2
x2 (1 − x2 ) dx
4 15
e−x dy dx =
ln 2
e−x (2 − ex ) dx
0
ln 2 = −2e−x − x 0 = 1 − ln 2
x3 x4 + y 2
2 1 − (y − 1) dy = ln 4 − y 2
(x + y) dy dx = 0
2
ex
0
0
x2 −1
−1
31.
873
1 (1 − x)2 dx = 2 3
1 √ ( 2 − 1)y dy 0 2 1 √ = ( 2 − 1) 4
dx dy =
1
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December 7, 2006
SECTION 17.3
2
3− 32 x
35. 0
3 2− 2 y 3 4 4 4 − 2x − y dy dx = 4 − 2x − y dx dy = 4 3 3 0 0
0 1
1
36.
1
(2x + 3y) dy dx = 0
0 2
1− 12 x
37. 0
38.
2
39.
0 √
1−x2
√ − 1−x2
−1
√
2−2y
x3 y dx dy =
0
(x2 + y 2 ) dy dx = 2
(2x + 1) dy dx =
−1
=
40.
√ 1− 1−x2
−1
1
√ 1+ 1−x2
41.
1
2
42.
0 1
√
−1
1
1−x2
− 1−x2
2
2
0
2
5−y
(1 + xy) dx dy = 1
2y−1 a
√
a2 −x2
45. a
0
b(1−x/a)
0 1
47. 0
1
1
0
1
1 6
dx =
11 2x3 − x4 − x6 dx = 70
2
3 55 (6 + 9y − 3y 2 − y 3 ) dy = 2 8 a
dy dx =
1
0 cos−1 y
(a2 − x2 ) dx =
2 3 a 3
a bc x 2 x y abc dy dx = 1− c 1− − dx = a b a 6 0 2
ey/x dx dy =
y
48. 0
−
x2
= 3π
a2
2
43 1 2 2 2 3/2 2 2 6 1 − x − x 1 − x − (1 − x ) dx = π. 2 3 16 −1
0
46.
π
(2 1 − x2 − 2x 1 − x2 ) dx = π
0
0
1
1 − y 2 dy = 6
4 1 2x − x3 − x + 3 3
1
0
44.
1−y 2
1
(x + 3y ) dy dx = x2
(2x + 1) dx dy 1−y 2
2
−1
x
43.
(1 − x) dy dx =
√
1−y 2
0 √
1
−1
(x + y ) dy dx = 0
√
1−
1 (4 − y − x2 ) dy dx = 4
2
1 π 1 − x2 + (1 − x2 )3/2 dx = 3 2
1+√1−y2 x2 + x √ dy
2
1−x
2 15
−1
=6
1+
1−
1
1
x2
1
1
−1
2x−x2
√ − 2x−x2
0
2
x3 y dy dx =
0
1
0
5 3 (2x + ) dx = 2 2
x
ey/x dy dx =
0
1
xey/x
0
esin x dx dy = 0
π/2
0
cos x
x
dx =
0
0
esin x dy dx = 0
π/2
1
x(e − 1) dx =
1 (e − 1) 2
cos xesin x dx = e − 1
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SECTION 17.3
1
1
49.
2 y4
1
y
x e dy dx = 0
y
50.
0
y2
e 0
51. favg
1
−1
4
2
yey dy =
1
−1
1
2
8x dx =
1
1−x2
2 π
0
0
ln 2
Ω=1
√
xy dy dx =
2 ln 2
1 dy = 3 0
1
4
y 3 ey dy =
0
1 (e − 1) 12
−1
x2 dx =
2 3
2 ln 2
1
1 1 dy dx = xy (ln 2)2
ln 2
x(1 − x2 ) dx =
0 2 ln 2
1 2π
1 ln 2 dx = 1 x
ln 2
(parallelogram)
1
x+1
x+y
e
Average value = 0
55.
y
(quarter circle of radius 1)
4 Average value = π 1 (ln 2)2
1 3 y4 x e 3
1 (e − 1) 2
1 x y dy dx = 8
0
0
2
52. area of Ω = 14 π
54. area of
0
0
1 = 8
53. favg =
1
dx dy =
0
1
x e dx dy =
x
1
2 y4
x−1
d
1 e2x+1 − e2x−1 dx = (e3 − 2e + e−1 ). 2
0
f (x)g(y) dxdy =
b
d
=
a
f (x)g(y) dx c
d
g(y) c
b
f (x) dx
b
f (x)g(y) dx dy = c
R
1
dy dx =
dy
a
a
b
dy =
d
f (x) dx
g(y) dy
a
c
56. Symmetry about the origin [ we want Ω to contain (−x, −y) whenever it contains (x, y)]. Ω = { (x, y) : 0 ≤ x ≤ y,
57. Note that Set
0 ≤ y ≤ 1}.
Ω = { (x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 1}. 1 y f (x)f (y) dxdy = f (x)f (y) dx dy 0
Ω
0 1
x
=
f (y)f (x) dy dx 0
x and y are dummy variables
0
1
=
x
f (x)f (y) dy dx = 0
0
f (x)f (y) dxdy. Ω
Note that Ω and Ω don’t overlap and their union is the unit square R = { (x, y) : 0 ≤ x ≤ 1,
0 ≤ y ≤ 1}.
875
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December 7, 2006
SECTION 17.3 If 0
1
f (x) dx = 0, then 1 0= f (x) dx 0
1
f (y) dy
=
f (x)f (y) dxdy
0
R
by Exercise 55
f (x)f (y) dxdy +
=
f (x)f (y) dxdy
=2 and therefore
f (x)f (y) dxdy Ω
Ω
Ω
f (x)f (y) dxdy = 0. Ω
x
b
58. 0
a
∂f (t, y) dy dt = ∂x
b
a
x
0
b
=
∂f (t, y) dt dy ∂x
[f (x, y) − f (0, y)] dy
a
b
=
f (0, y) dy.
a
Thus
b
x
b
a
and d dx
a
b
0
d f (x, y) dy = dx =
d dx
a
x
0
b
f (0, y) dy a
b ∂f d f (0, y) dy (t, y) dy dt + ∂x dx a
b
a
x
a
∂f (t, y) dy dt + ∂x
f (x, y) dy =
b
f (x, y) dy −
H(t) dt + 0 = H(x) =
0
b
∂f (x, y) dy. ∂x
a
59. Let M be the maximum value of | f (x, y) | on Ω.
φ2 (x+h) φ1 (x+h)
φ1 (x)
= φ1 (x+h)
| F (x + h) − F (x) | =
φ2 (x+h)
+ φ1 (x)
φ2 (x+h)
φ2 (x)
+
φ2 (x)
f (x, y) dy −
φ1 (x+h)
=
φ2 (x)
φ1 (x)
φ1 (x)
f (x, y) dy
φ2 (x+h)
f (x, y) dy + φ1 (x+h)
≤
φ1 (x)
φ1 (x+h)
φ2 (x)
f (x, y) dy +
f (x, y) dy
φ2 (x+h)
φ2 (x)
f (x, y) dy
≤ φ1 (x) − φ1 (x + h) M + | φ2 (x + h) − φ2 (x) M. The expression on the right tends to 0 as h tends to 0 since φ1 and φ2 are continuous.
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SECTION 17.4 60. (a)
3
2
3
x2 −2x+2
1 dy dx =
4
xy dy dx ∼ = 4.5720 x2 + y 2
1
1 3
2
(b)
2x + y dx dy ∼ = 2.8133
√ 1+ y−1
y 2 −2y+2
1
3−y
62. (a) 0
7
(b)
√ 1+ x−1
1 1
dy dx ∼ = −25.9893
2
61. (a)
−xy
xe
−1
5
2
x/2
(b)
2y
0
877
1 dx dy =
1 3
3
3−x
2x + y dy dx +
0
2
2x + y dy dx
0
∼ = 2.8133 SECTION 17.4 π/2 sin θ r cos θ dr dθ = 1. 0
0 π/4
0
cos 2θ
π/4
0
0 π/2
0
3 sin θ
3.
2
0 2π/3
−π/3
π/2
r dr dθ = 0
π/2 1 1 1 = sin2 θ cos θ dθ = sin3 θ 2 6 6 0
cos2 2θ π dθ = 2 16
r dr dθ =
2.
4.
π/2
1 (1 − cos θ) sin θ dθ = 9 − cos θ + cos3 θ 3
0
2 cos θ
r sin θ dr dθ = 0
2π/3
2 cos2 θ sin θ dθ =
−π/3
5. (a) Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1 cos r2 r drdθ =
2π
0
Γ
Γ
2
cos r2 r dr dθ = 2π
Γ
cos r2 r dr dθ = 2π
1
Γ
1
r cos r2 dr = π sin 1
2
r cos r2 dr = π(sin 4 − sin 1)
1
0
1
r sin r dr = 2π(sin 1 − cos 1)
0
(b) Γ : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2. 2π 2 (sin r)r dr dθ = (sin r)r dr dθ = 2π 0
0
6. (a) Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1. 2π 1 (sin r)r dr dθ = (sin r)r dr dθ = 2π 0
1
2
r sin r dr = 2π[cos 1 − 2 cos 2 + sin 2 − sin 1]
1
7. (a) Γ : 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 1 (r cos θ + r sin θ)r drdθ =
π/2
0
1
r2 (cos θ + sin θ) dr dθ
0
Γ
=6 0
1 6
0
(b) Γ : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2 2π cos r2 r drdθ = 0
1
π/2
2
9 sin θ dθ = 9 0
π/2
3
π/2
=
1
(cos θ + sin θ) dθ 0
2
r dr 0
1 2 = =2 3 3
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SECTION 17.4 (b) Γ : 0 ≤ θ ≤ π/2, 1 ≤ r ≤ 2 (r cos θ + r sin θ)r drdθ =
π/2
0
r2 (cos θ + sin θ) dr dθ
1
Γ
2
π/2
0
8. Γ : 0 ≤ θ ≤
0≤r≤
π 3,
+
y2
π/2
1
−π/2
0
√
1
1−x2
1/2 sec θ
12. 0
π/3
= 0
1
√
1−x2
π/3
1
sec θ
π/2
2π
π/3
+
1
y2
e−r r dr dθ = 2π 2
1
0
√
2x−x2
15.
π/2
π/3
sec3 θ dθ
0
2
π/2
1
r2 dr dθ =
0
1 1 − sec2 θ 2 8
dθ =
π 2
2
r2 dr =
0
4 π 3
√ 3 1 π− 6 8
r4 cos θ sin θ dr dθ
0
1
π/2
0
(sin 1 − cos 1) dθ =
0
π (sin 1 − cos 1) 2
1 2 re−r dr = π(1 − ) e
0
0
sin(r) r dr dθ =
2 cos θ
8 r cos θ r dr dθ = 3
x dy dx = 0
π/2
dy dx =
0
2
0
0
1 7 1 19 cos θ sin θ dθ = + = 5 480 40 480
0
0
1 2
x2
π/3
r dr dθ =
r4 cos θ sin θ dr dθ +
π/2
0
sin θ dθ + 5(32) cos4 θ
sin
14.
0
π/3
13. 0
dθ 1 = 3 cos3 θ 3
10.
0
r · r dr dθ =
1 π 3
r2 dr dθ =
0
π/3
π/3 √ 1√ 1 1 1 1 = 3 + ln(2 + 3) sec θ tan θ + ln | sec θ + tan θ| = 3 2 2 3 6 0
0
π/3
1/ cos θ
0
dy dx =
π/3
0
11. 1/2
7 14 = 3 3
=2
1
dx dy =
Γ
9.
r2 dr
1 cos θ
x2
2
(cos θ + sin θ) dθ
=
0
π/2
cos4 θ dθ =
0
∧
8 3 1 π π · · · = 3 4 2 2 2
(See Exercise 62, Section 8.3) 16. The region is the inside of the circle (x − 1/2)2 + y 2 = 1/4, which has polar equation r = cos θ. So the integral becomes
π/2
−π/2
π/3
17. A =
3 sin 3θ
r dr dθ = 0
0
9 2
0
π/3
cos θ
r3 dr dθ =
0
sin2 3θ dθ =
9 4
0
π/2
−π/2
π/3
1 3π cos4 θ dθ = 4 32
(1 − 6 cos θ) dθ =
3π 4
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SECTION 17.4
2π
2(1−cos θ)
18. A =
2π
2(1 − cos θ)2 dθ = 6π
r dr dθ = 0
0
0
19. First we find the points of intersection: r = 4 cos θ = 2
=⇒
π θ=± . 3
=⇒
A=
π/3
r dr dθ =
−π/3
1+2 cos θ
0
b
23.
3 cos θ
1+cos θ
π/4
−π/3
r2 sin θ + br dr dθ =
2π
1
24. V = 0 π/2
π/3
π/3 1 3θ 2 2 9 cos θ − (1 + cos θ) dθ = + sin 2θ − sin θ =π 2 2 −π/3
b 1 3 b r sin θ + r2 dθ 3 2 0 0 2π 1 1 sin θ + dθ = b3 π = b3 3 2 0
0
2π
1
(1 − r )r dr dθ = 2π 2
0
0
0
(r − r3 ) dr =
0
r 12 − 3r2 dr dθ = 8 2
2
25. 8
π/2
π/2
=8 0
2π
26. V = 0
2π
r 0
−π/2
3/2 1 12 − 3r2 18
2 dθ 0
4√ 16 √ 3 dθ = 3π 3 3 √
5 − r2 r dr dθ = 2π
5
r
6
5 − r2 dr =
30 1/6 (5) π 7
3/2 1 1 4 − r2 dθ 3 0 0 2π √ 8 √ 2 − 3 dθ = (8 − 3 3 )π = 3 3 0
4 − r2 dr dθ =
π/2
−
π 2
0
0
28. V =
6
0 1
27.
5
0
√
0
√ 4π +2 3 3
0
r dr dθ =
−π/3
−π/3
(2 + 4 cos 2θ) dθ =
cos 2θ dθ = 4
0
π/3
π/3
1 (1 + 2 cos θ)2 dθ = 3π 2
r dr dθ = 8
2π
√ 2 cos 2θ
π/4
0
0
(8 cos2 θ − 2) dθ =
0
21. A = 4
2π
π/3
r dr dθ = 0
22. A =
1 2
−π/3
2
2π
4 cos θ
20. A =
cos θ =
cos θ
2π
−
(1 − r2 )r dr dθ =
π/2
−π/2
cos2 θ cos4 θ − 2 4
dθ =
5π 32
879
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SECTION 17.5
29.
December 7, 2006
π/2
−π/2
2 cos θ
2r2 cos θ dr dθ =
π/2
−π/2
0
=
π/2
2 3 r cos θ 3
2 cos θ dθ 0
16 32 cos4 θ dθ = 3 3
−π/2
π/2
cos4 θ dθ =
0
∧
32 3
3 π 16
= 2π
Ex. 46, Sect. 8.3 30.
31.
π/2
−π/2
b a
2a cos θ
r2 dr dθ =
0
a sin θ 1 2 b π 2 3/2 2 2 − a −r r a − r dr dθ = dθ a 0 3 0 π 1 1 = a2 b 1 − cos3 θ dθ = πa2 b 3 3 0
a sin θ
0
2π
r
32. (a) V = 2 0
(b)
R2
−
s2
s ds dθ = 4π
0
r
S 0
4π 3 R − (R2 − r2 )3/2 . R2 − S 2 dS = 3
4 3 4 πR − V = π(R2 − r2 )3/2 3 3
π/4
2
ex
0
1 2
0
+y 2
2π
dxdy = 0
Ω
4
π/4
r dr dθ = 2 0
2 cos 2θ
33. A = 2
34.
8a3 32 3 cos3 θdθ = a 3 9
−π/2
0
π
π/2
π/4
(2 cos 2θ)2 dθ = 2
(1 + cos 4θ) dθ = 0
2 rer dr dθ = π e16 − e4
2
SECTION 17.5 1 1 2 1. M = x2 dy dx = 3 −1 0 xM M =
1
−1
yM M =
1
√
x3 dy dx = 0
=⇒
1
1
−1
1 2 1 x dx = 2 3
2
x y dy dx = 0 x
2. M =
1
(x + y) dy dx = 0
0 √
x
xM M =
1
5/2
x(x + y) dy dx = 0
1
x
0
yM M =
√
0
x
y(x + y) dy dx = 0
0
0
1
yM =
=⇒
1/3 1 = 2/3 2
x 13 dx = 2 20
x3/2 +
0
1
xM = 0
0
−1
1
1
x2 + 2
x2 x3/2 + 2 3
dx =
19 190 =⇒ xM = 42 273
dx =
3 6 =⇒ yM = 10 13
π 2
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SECTION 17.5
1
1
π
sin x
1 1 1 3. M = xy dy dx = (x − x5 ) dx = 2 6 0 x2 0 1 1 4 1 1 2 2 2/21 2 xM M = =⇒ xM = = x y dy dx = (x − x6 ) dx = 2 21 1/6 7 2 0 x 0 1 1 1 3 1 1 1/8 =⇒ yM = = yM M = xy 2 dy dx = (x − x7 ) dx = 3 8 1/6 4 2 0 x 0
π
sin2 x π dx = 2 4 0 0 0 π sin x π 2 π2 sin x π dx = =⇒ xM = xy dy dx = x xM M = 2 8 2 0 0 0 π sin x π 3 sin x 4 16 dx = =⇒ yM = y 2 dy dx = yM M = 3 9 9π 0 0 0 y dy dx =
4. M =
8
x1/3
5. M = 0
0
y 2 dy dx =
1 3
8
x dx = 0
32 3
1 8 2 512 xM M = xy dy dx = x dx = 3 9 0 0 0 8 x1/3 1 8 4/3 96 y 3 dy dx = x dx = yM M = 4 0 7 0 0
a
8
√
x1/3
2
a2 −x2
=⇒ =⇒
xM = yM =
16 512/9 = 32/3 3
9 96/7 = 32/3 7
a
x 2 a4 (a − x2 ) dx = 8 0 0 0 2 a √a2 −x2 a 2 x 2 a5 8 xM M = (a − x2 ) dx = =⇒ xM = a x2 y dy dx = 2 15 15 0 0 0 a √a2 −x2 a x 2 a5 8 2 yM M = (a − x2 )3/2 dx = =⇒ yM = a xy dy dx = 3 15 15 0 0 0 xy dy dx =
6. M =
1
3x
7. M =
xy dy dx = 0
2x
1
3x
xM M =
0 1
2x 3x
yM M = 0
2x
5 2
1
x3 dx =
0
5 x y dy dx = 2
1
2
xy 2 dy dx =
x4 dx =
0
19 3
5 8
1
1 =⇒ 2
x4 dx =
0
=⇒
1/2 4 = 5/8 5
yM =
152 19/15 = 5/8 75
1 3 3 x(3 − x) + (3 − x)2 dx = 5 2 2 2 0 0 0 2 3− 3 x 2 2 x 3 2 7 3 7 2 =⇒ xM = x (3 − x) + (3 − x) dx = x(x + y) dy dx = xM M = 2 2 2 2 10 0 0 0 2 3− 3 x 2 2 (3 − 32 x)3 x 3 6 (3 − x)2 + dx = 6 =⇒ yM = y(x + y) dy dx = yM M = 2 2 3 5 0 0 0
8. M =
2
3− 32 x
(x + y) dy dx =
2
19 15
xM =
881
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December 7, 2006
SECTION 17.5
2π
1+cos θ
9. M = 0
0
2π
1 r dr dθ = 3
1+cos θ
xM M = 0
2π
2
0
(1 + 3 cos θ + 3 cos2 θ + cos3 θ) dθ =
0
1 r cos θ dr dθ = 4
2π
3
1 = 4
5π 3
(1 + cos θ)4 cos θ dθ
0 2π
cos θ + 4 cos2 θ + 6 cos3 θ + 4 cos4 θ + cos5 θ dθ
0
7π = 4 21 7π/4 = . 5π/3 20
Therefore, xM =
2π
1+cos θ
yM M = 0
0
1 r sin θ dr dθ = 4
2π
3
0
2π 1 1 5 (1 + cos θ) (1 + cos θ) sin θ dθ = =0 4 5 0 4
Therefore, yM = 0.
10. M =
5π/6
y dx dy =
5π/6
r sin θ r dr dθ = π/6
Ω
2 sin θ
1
π/6
by symmetry 5π/6 2 sin θ yM M = r2 sin2 θ r dr dθ =
8 1 sin4 θ − sin θ 3 3
√ 3 3 + 8π dθ = 12
xM = 0
π/6
1
Therefore, yM
5π/6
1 4 sin θ − sin2 θ 4
6
π/6
√ 11 3 + 12π dθ = 16
√ 33 3 + 36π = √ 12 3 + 32π
11. Ω : −L/2 ≤ x ≤ L/2, −W/2 ≤ y ≤ W/2 M 2 4M W/2 L/2 2 1 y dx dy = y dxdy = MW 2 Ix = LW LW 0 12 0 Ω
∧ symmetry
M 2 1 x dxdy = ML2 , LW 12
Iy =
Iz =
Ω
M 2 1 x + y 2 dxdy = M L2 + W 2 LW 12
Ω
√ √ W 3 L 3 , Ky = Iy /M = Kx = Ix /M = 6 6 √ √ 3 L2 + W 2 Kz = Iz /M = 6 L 12. λ(x, y) = k x + 2 Ix =
−W/2
=
W/2
W3 12
L/2
M=
L k x+ 2 −L/2
M W
=
W/2
−W/2
L/2
L k x+ 2 −L/2
y 2 dx dy =
1 MW2 12
W/2
−W/2
dx dy =
y 2 dy
L/2
kW L2 2
L k x+ 2 −L/2
dx
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SECTION 17.5 Iy =
L/2
L/2
L k x+ 2 −L/2
−W/2
Iz =
W/2
W/2
L k x+ 2 −L/2
−W/2
x2 dx dy =
1 kL4 W = M L2 24 12
(x2 + y 2 ) dx dy = Ix + Iy =
1 M (L2 + W 2 ). 12
1 1 L 1 dxdy = kL dxdy = kL( area of Ω) = kL2 W k x+ 2 2 2 2
13. M = Ω
Ω
∧ symmetry L 1 kx2 + Lx dxdy x k x+ dxdy = 2 2
xM M = Ω
kx2 dxdy = 4k
=
0
Ω
Ω
W/2
L/2
x2 dx dy =
0
∧
∧
symmetry =
1
1 6
2 2 kL W
yM M =
1 kWL3 12
symmetry
L=
1 6
ML;
xM =
L dxdy = 0; y k x+ 2
1 6
L
yM = 0
Ω
∧ by symmetry
λ(x, y)[x2 + y 2 ] dx dy =
14. Iz = Ω
Ω
Since Iz = Ix + Iy , 15. Ix = Ω
4M = πR2
λ(x, y)x2 dx dy +
Iy = 14 MR2 ,
π/2
π/2
0
2
sin θ dθ 0
Ω
M Kz 2 = M Kx 2 + M Ky 2
we have
4M 2 4M y dxdy = 2 πR πR2
λ(x, y)y 2 dx dy = Ix + Iy .
R
Kz 2 = Kx 2 + Ky 2 .
r3 sin2 θ dr dθ
0 R
3
r dr 0
therefore
1 4M π 1 4 R = MR2 = 2 πR 4 4 4
Iz = 12 MR2
Kx = Ky = 12 R, 16. Iz = IM + d2 M.
√ Kz = R/ 2 Rotation doesn’t change d,
doesn’t change M , and doesn’t change IM .
883
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SECTION 17.5
17. IM , the moment of inertia about the vertical line through the center of mass, is M 2 x + y 2 dxdy πR2 Ω
where Ω is the disc of radius R centered at the origin. Therefore 2π R M 1 r3 dr dθ = MR2 . IM = πR2 0 2 0 1 2
MR2 + d2 M where d is the distance from the center of the disc to the origin. Solving √ M. this equation for d, we have d = I0 − 12 MR2 We need I0 =
b
f (x)
18. Ix = a
0
b
f (x)
Iy =
λ λy dy dx = 3
b
2
b
2
λx dy dx = λ a
[f (x)]3 dx
a
0
x2 f (x) dx.
a
19. Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ b b√ 2 2 4M 2 4M a a a −x 2 1 Ix = y dxdy = y dy dx = M b2 πab πab 0 0 4 Ω
Iy =
4M 2 4M x dxdy = πab πab
a
0
Ω
b a
√
a2 −x2
x2 dy dx =
0
1 M a2 4
Iz = 14 M a2 + b2
1
√
x
20. Ix =
1
2
(x + y)y dy dx =
x5/2 x2 + 3 4
0
0 1
0 √
x
Iy =
1
2
0 1
0 1
21. Ix = x2
0
1
1
Iy = 1
1
Iz = 8
0
1
(x − x9 ) dx =
0
1
√ 3
x
(x3 − x7 ) dx =
y 2 · y 2 dy dx =
8
0
x
y 2 · x2 dy dx = 0
5 28
dx =
25 ; 72
Iz = Ix + Iy .
1 16
13 80
8
x5/3 96 dx = 5 5
8
x3 1024 dx = ; 3 3
0 √ 3
dx =
1 10
0
0
Iy = 0
1 2
x3 + 2
xy(x2 + y 2 ) dy dx = Ix + Iy =
22. Ix =
1 4
x2
0
x3 y dy dx =
x2
0
xy 3 dy dx =
7/2
x
(x + y)x dy dx = 0
Iz = Ix + Iy
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SECTION 17.5
2π
1+cos θ
23. Ix =
0
2π
0 1+cos θ
Iy =
0 2π
0
1+cos θ
(1 + cos θ)5 sin2 θ dθ =
0
1 5
r4 cos2 θ dr dθ =
2π
(1 + cos θ)5 cos2 θ dθ =
0
33π 40 93π 40
63π 20
r4 dr dθ = Ix + Iy =
0
x1 M1 + x2 M2 , M 1 + M2
24. xM =
2π
2
0
Iz =
1 r sin θ dr dθ = 5 4
y1 M1 + y2 M2 M1 + M 2 A = π r22 − r12
yM =
25. Ω : r12 ≤ x2 + y 2 ≤ r22 ,
(a) Place the diameter on the x-axis. M 2 M 2π r2 2 2 1 y dxdy = Ix = r sin θ r dr dθ = M r22 + r12 A A 0 4 r1 Ω (b) 14 M r22 + r12 + M r12 = 14 M r22 + 5r12 (parallel axis theorem) (c) 14 M r22 + r12 + M r22 = 14 M 5r22 + r12 26. Set r1 = r2 = r in the proceeding problem. Then the required moments of inertia are 1 2 2Mr
(a)
(b)
3 2 2Mr .
27. Ω : r12 ≤ x2 + y 2 ≤ r22 , A = π r22 − r12 M 2 1 M 2π r2 3 2 x + y dxdy = I= r dr dθ = M (r22 + r12 ) A A 0 2 r1 Ω
28. Let l be the x-axis and let the plane of the plate be the xy-plane. Then 2 λ(x, y)y dx dy − λ(x, y)(y − yM )2 dx dy I − IM = Ω
Ω
2 λ(x, y)[2yM y − yM ] dx dy
= Ω
yλ(x, y) dx dy −
= 2yM
2 yM
λ(x, y) dx dy
Ω
= 29. M =
2 2yM M
Ω
−
2 yM M
k R − x2 + y 2 dxdy = k 0
Ω
π
=
2 yM M
R
1 Rr − r2 dr dθ = kπR3 6
0
by symmetry y k R − x2 + y 2 dxdy = k yM M = yM = R/π
= d M.
xM = 0
Ω
2
0
π
0
R
1 Rr2 − r3 sin θ dr dθ = kR4 6
885
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SECTION 17.5 k(R −
30. Ix =
x2
+
y 2 )y 2
π
0
Ω
π
R
Iy = k 0
R
dx dy = k
(R − r) r2 sin2 θ r dr dθ =
0
3M R2 kπR5 = . 40 20
3M R2 kπR5 = 40 20
(R − r)r2 cos2 θr dr dθ =
0
3M R2 . 10
Iz = Ix + Iy =
31. Place P at the origin. M= k x2 + y 2 dxdy Ω
π
2R sin θ
=k 0
r2 dr dθ =
0
32 3 kR 9
xM = 0 by symmetry yM M = y k x2 + y 2 dxdy = k 0
Ω
π
2R sin θ
r3 sin θ dr dθ =
0
64 4 kR 15
yM = 6R/5 Answer: the center of mass lies on the diameter through P at a distance 6R/5 from P . 32. Putting the right angle at the origin, we have λ(x, y) = k(x2 + y 2 ). b h− h x b 1 k(x2 + y 2 ) dy dx = kbh(b2 + h2 ) M= 12 0 0 b h− h x b kb2 h(3b2 + h2 ) b(3b2 + h2 ) kx(x2 + y 2 ) dy dx = =⇒ xM = xM M = 60 5(b2 + h2 ) 0 0
b
yM M = 0
h− h bx
ky(x2 + y 2 ) dy dx =
0
kbh2 (b2 + 3h2 ) h(b2 + 3h2 ) =⇒ yM = 60 5(b2 + h2 )
33. Suppose Ω, a basic region of area A, is broken up into n basic regions Ω1 , · · · , Ωn with areas A1 , · · · , An . Then
xA =
x dxdy = Ω
n i=1
⎛ ⎝
Ωi
⎞ x dxdy ⎠ =
n
xi Ai = x1 A1 + · · · + xn An .
i=1
The second formula can be derived in a similar manner. 2 1 4 (x + y) dy dx = 34. (a) M = 3 0 x/2 2 1 7 7 xM M = xM = x(x + y) dy dx = ; 6 8 0 x/2 2 1 3 yM M = y(x + y) dy dx = 1; yM = 4 0 x/2 2 1 2 1 4 4 2 (b) Ix = y (x + y) dy dx = x2 (x + y) dy dx = Iy = 5 3 0 x/2 0 x/2
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SECTION 17.6
887
SECTION 17.6 1. They are equal; they both give the volume of T . 2. (a) Lf (P ) =
q m n
xi−1 yj−1 zk−1 Δxi Δyj Δzk ,
Uf (P ) =
i=1 j=1 k=1
(b) xi−1 yj−1 zk−1 ≤
xi yj zk Δxi Δyj Δzk
i=1 j=1 k=1
xi + xi−1 2
xi−1 yj−1 zk−1 Δxi Δyj Δzk ≤
yj + yj−1 2
zk + zk−1 2
≤ xi yj zk
1 2 xi − xi−1 2 yj 2 − yj−1 2 zk 2 − zk−1 2 ≤ xi yj zk Δxi Δyj Δzk 8
1 2 xi − xi−1 2 yj 2 − yj−1 2 zk 2 − zk−1 2 ≤ Uf (P ). 8 i=1 j=1 m
Lf (P ) ≤
q m n
q
n
k=1
The middle term can be written ⎞ ⎛ n m q 1 2 1 1 xi − xi−1 2 ⎝ yj 2 − yj−1 2 ⎠ zk 2 − zk−1 2 = (1)(1)(1) = . 8 i=1 8 8 j=1 k=1
Therefore
I=
3.
1 . 8 dx dy dz = α (volume of Π) = α(a2 − a1 )(b2 − b1 )(c2 − c1 )
α dx dy dz = α Π
Π
4. Since the volume is 1, the average value is
xyz dx dy dz =
1 . 8
Ω
5. Let P1 = {x0 , · · · , xm },
P2 = {y0 , · · · , yn },
P3 = {z0 , · · · , zq } be partitions of [ 0, a ], [ 0, b ], [ 0, c ]
respectively and let P = P1 × P2 × P3 . Note that yj + yj−1 xi + xi−1 ≤ xi yj xi−1 yj−1 ≤ 2 2 and therefore xi−1 yj−1 Δxi Δyj Δzk ≤
1 4
xi 2 − x2i−1
2 yj 2 − yj−1 Δzk ≤ xi yj Δxi Δyj Δzk .
It follows that Lf (P ) ≤
q m n 1 2 2 xi − x2i−1 yj 2 − yj−1 Δzk ≤ Uf (P ). 4 i=1 j=1 k=1
The middle term can be written ⎞ m ⎛ n q 1 2 1 2 2 2 xi − xi−1 ⎝ yj − yj−1 ⎠ Δzk = a2 b2 c. 4 i=1 4 j=1 k=1
6. Ix = Ixy + Ixz ,
Iy = Ixy + Iyz ,
Iz = Ixz + Iyz
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SECTION 17.6
7. x1 = a,
y 1 = b,
z 1 = c;
x0 = A,
x1 V1 + xV = x0 V0
y 0 = B,
z0 = C
=⇒
a2 bc + (ABC − abc) x = A2 BC
=⇒
x=
A2 BC − a2 bc ABC − abc
similarly y= 8. Encase T in a box Π.
AB 2 C − ab2 c , ABC − abc
z=
ABC 2 − abc2 ABC − abc
A partition P of Π breaks up II into little boxes Πijk .
Since f is
nonnegative on Π, all the mijk are nonnegative. Therefore 0 ≤ Lf (P ) ≤ f (x, y, z) dx dy dz. T
9. M =
Kz dxdydz Π
Let P1 = {x0 , · · · , xm }, P = P1 × P2 × P3 .
P2 = {y0 , · · · , yn },
P3 = {z0 , · · · , zq } be partitions of [ 0, a ] and let
Note that zk−1 ≤
1 2
(zk + zk−1 ) ≤ zk
and therefore 2 Kzk−1 Δxi Δyj Δzk ≤ 12 K Δxi Δyj zk 2 − zk−1 ≤ Kzk Δxi Δyj Δzk . It follows that Lf (P ) ≤
q m n 1 2 K Δxi Δyj zk 2 − zk−1 ≤ Uf (P ). 2 i=1 j=1 k=1
The middle term can be written ⎞ ⎛ n m q 1 1 1 2 2 Δxi ⎝ Δyj ⎠ zk − zk−1 = K(a) (a) (a2 ) = Ka4 . K 2 2 2 i=1 j=1 k=1
M = 12 Ka4 where K is the constant of proportionality for the density function. 10. xM M =
Kzx dx dy dz =
Ka5 1 =⇒ xM = a 4 2
Kzy dx dy dz =
Ka5 1 =⇒ yM = a 4 2
Kz 2 dx dy dz =
Ka5 2 =⇒ zM = a. 3 3
Π
yM M = Π
zM M = Π
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SECTION 17.7 11.
Kz x2 + y 2 dxdydz
Iz = Π
Kx2 z dxdydz +
= !
Π
"#
Ky 2 z dxdydz . Π
!
$
"#
I1
$
I2
We will calculate I1 using the partitions we used in doing Exercise 9. Note that 2 xi + xi xi−1 + x2i−1 zk + zk−1 x2i−1 zk−1 ≤ ≤ xi 2 zk 3 2 and therefore 2 ≤ Kxi 2 zk 2 Δxi Δyj Δzk . Kx2i−1 zk−1 Δxi Δyj Δzk ≤ 16 K xi 3 − x3i−1 Δyj zk 2 − zk−1 It follows that Lf (P ) ≤
q m n 1 3 2 K xi − x3i−1 Δyj zk 2 − zk−1 ≤ Uf (P ). 6 i=1 j=1 k=1
The middle term can be written ⎞ ⎛ n m q 1 1 1 3 2 3 2 ⎠ ⎝ K xi − xi−1 Δyj zk − zk−1 = Ka3 (a)(a2 ) = Ka6 . 6 6 6 i=1 j=1 k=1 Similarly I2 = 16 Ka6 and therefore Iz = 13 Ka6 = 23 12 Ka4 a2 = 23 M a2 . ∧ by Exercise 9 2 12. (a) Lf (P ) ∼ 3y − 2x dxdy = 57 (c) = 56.4803 Uf (P ) ∼ = 57.5603 R
SECTION 17.7 a b c 1. dx dy dz = 0
0
1
0 x
a
0
y
1
0
0
0
1
2y
3.
bc dz = abc 0
x
y dz dy dx = 0
a
c dy dz =
0
2.
b
1
y 2 dy dx =
0
0
x
1
2y
(x + 2z) dz dx dy = 0
1
0
0
1
1
0 1
1+x
4.
1
xy
0
1−x
2
5. 0
= 0
1
3
−1 0 2 1 −1
2 3 x 3
2y
0
2
(z − xy) dz dy dx =
(4 − 2xy) dy dx = 0
0 2
1
2 2
0 1
−1
1 2 z − xyz 2
2y − xy 2
1 −1
1
0
1
dy dx 1
2
8 dy = 16 0
2y
0
1
16 3 2 y − 3 3
2x2 dx dy
dy =
2 3
2x2 11 (1 + x)3 − (1 − x)3 dx = 3 9
3
dx =
1
dx dy =
1−x
dy =
2x y dy dx =
0
x z2 0
1+x
4z dz dy dx =
xz +
=
x3 1 dx = . 3 12
889
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December 7, 2006
SECTION 17.7
2
6.
−1
0
1
√
0
1
= 0
−1
√
1
y−2
e2
e
y2
x+y dz dx dy = z
ln x
−1
2
z
y
2
0
y2
y2
y(x − 1) dx dy =
= y
1
π/2
2
1
π/2
10.
1 2 x y − xy 2
y2
0
0
1
0
2
dy = y
π/2
z
π/2
e cos x sin y dz dy dx = 0
1 dz = 0
1 3
π/2
cos z dz = 0
1 3
(y − 2)2 − 1 3 + y(y − 3) dy = 2 2
x [yez ]ln dx dy 0
1
2
−1
1
y
1
y−2
(x + y) dx dy =
ye dz dx dy =
9. 1
1 − (1 − x2 )3/2 cos z 3
2
2
[xy cos z]0 1−x dx dz
0
π/2
0
1
2
x 1 − x2 cos z dx dz =
0
2
4z dz = 8 0
π/2
0
0
π/2
2
(2z − 4x) dx dz =
x cos z dy dx dz = 0
8.
1−x2
7.
1
−1
0
1
2
(z − xy) dy dx dz =
1
π/2
3
1 5 3 3 47 2 y − y + y dy = 2 2 24
(e − 1) cos x sin y dy dx
0 π/2
=
(e − 1) cos x dx = e − 1
0
11.
c2
b2
a2
f (x)g(y)h(z) dxdydz = c1
Π
b1
c2
f (x)g(y)h(z) dx dy dz
a1
b2
= c1
c2
b1
=
a1
a2
x3 dx
0
2
1
13.
x2 dx
0
2
3
y 2 dy
0
0
kxyz dx dy dz = k
a
=
x dx
b
c
y dy
z dz
0
2
a
kx yz dxdydz = k
2
x dx
=k 1 8
ka2 b2 c2 .
1 3
a3
1 2 2 2 ka b c 8
b
z dz
0
1
Therefore x =
2 2 3
b2
1 2
c2 =
c
y dy
0
Π
=
0
By Exercise 14, M =
c1
8 27 1 =8 3 3 3
0
xM M =
h(z) dz
8 9 1 =3 4 3 2
z 2 dz
Π
15.
=
0
14. M =
c2
g(y) dy b1
z dz
0
b2
f (x) dx 3
y 2 dy
g(y) dy dz
b1
a1 1
b2
f (x) dx a1
=
a2
h(z) c1
12.
f (x) dx dy dz
a2
g(y)h(z)
0 1 12
ka3 b2 c2 .
a. Similarly, y =
2 3
b and z =
2 3
c.
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SECTION 17.7
891
16. (a) I= kxyz (x − a)2 + (y − b)2 dx dy dz Π
a
x(x − a) dx
=k
c
y dy
0
=
b
2
z dz
0
x dx
a
+k
0
0
b
y(y − b) dy 2
0
M = 18 ka2 b2 c2 ,
Since IM −
(b)
1 2 (a + b2 ) 18
I = 16 M (a2 + b2 ).
by parallel axis theorem
17.
18. V =
1
1
1−y
dx dy dz =
dz dy dx = 0
T
0
0
1 2
19. center of mass is the centroid 1 2
x=
by symmetry yV = y dxdydz =
1
1
1−y
0
0
1
= V =
1 2
1 2
0
1
0
1
0
0
1 1 − 2y + y 2 dy dx = 2 y = 13 ,
z=
1
y − y 2 dy dx
0
0
(by Exercise 18 );
20. Ix =
1
0
T
0
1 1 1 2 1 3 1 1 y − y dx = dx = = 2 3 6 0 0 6 0 1 1 1−y zV = z dxdydz = z dz dy dx =
1
y dz dy dx =
0
T
0
1
1 y − y2 y3 3
1
0
1 (1 − y)2 dy dx 2
1 dx = 0
1 2
0
1
1 1 dx = 3 6
1 3
M 2 1 (y + z 2 ) dx dy dz = M V 3
T
Iy = T
M 2 1 (x + z 2 ) dx dy dz = M V 2
Iz = T
z dz 0
1 2 2 2 2 ka b c (a + b2 ). 48
M 2 1 (x + y 2 ) dx dy dz = M V 2
c
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SECTION 17.7
21.
r
−r
22.
−φ(x)
√
23.
√
25.
1
√
x−x2
√
−
−1 √
k
2
dz dy dx 1−x2 /2
2+y 2
k 2−y 2
1 k z − 3x2 − y 2 dz dy dx 4
4−x2 −y 2 /4
4−z 2
x2 + y 2 + z 2 dz dy dx
4−x2 −y 2
√2−y2 −
x2 + y 2 dx dz dy
z 2 +2y 2
2 x z + y dx dy dz =
27.
k the constant of proportionality
dz dy dx
3x2 +y 2 /4
√
2
3
= 0
2
1
1
y
1 z+y 3
28.
1
2
0
0 1
x
x2 y 2 z 2 dx dy dz =
T
=
=
=
=
0
−1
1 3 1 3 1 9
y+1
0 0
−1
0
0
−1
1 3 2 2 x y z 3
y+1
y+1
0
2
dz = 0
1
3
1
0
0
1
1
x2 y 2 z 2 dx dz dy +
1 dy dz 0
2 28 z + 4 dz = 3 3 1
(4y 2 ey − 2yey + 2y − 2y 2 ) dy = 4e −
1
1
0
1−y
1 3 2 2 x y z 3
0
1−y
y2 z2
1
0
1
1 3 x z + xy 3
0
0
1 3
0
0
dy + 0
0
1 3
3
x
dz dy +
y+1
2
0
y
0
1
y 2 z 2 dz dy +
1 2 3 y z 3
0
−1
0
−1
2
2y(x + y)e dx dy =
0
29.
1 1 zy + y 2 3 2
2ye dz dx dy = 0
1
0
dy dz =
x+y
3
x z + y dx dy dz =
0
T
r2 − x2 ,
1−y 2
√ −2 2−2x2
√ − 2
with φ(x) =
x2 +y 2
−2x−3y−10
√ 2 2−2x2
1
1−x2
√1−x2 /2
2
√ − 2
r2 − (x2 + y 2 ) ,
√ − x−x2
0
24.
−ψ(x,y)
− 1−x2
1
k r − x2 + y 2 + z 2 dz dy dx
ψ(x,y)
√
−1
√
1
φ(x)
ψ(x, y) =
26.
QC: PBU/OVY
JWDD027-Salas-v1
1 2 3 y z 3
1 y 5 + 3y 4 + 3y 3 + y 2 dy + 9
0
0
1−y
1
x2 y 2 z 2 dx dz dy
0
1 dz dy 0
dz dy
1−y
0
dy 0 1
1 y 2 − 3y 3 + 3y 4 − y 5 dy = 270
29 3
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December 7, 2006
SECTION 17.7
2
√
√4−x2 −y2
4−x2
30.
√
2
4−x2
xy dz dy dx = 0
0
xy 0
0
0
π/2
2
= 0
1 2
=
3
y 2 dx dy dz =
31.
0
T
3
0
r2 cos θ sin θ
2−2x/3
4 − r2 r dr dθ
0
2
r3
0
6−2x−3y
1 4 √ 32 4 − r2 dr = (4 u − u3/2 ) du = 4 0 15
3
y 2 dz dy dx =
0
= 0
2−2x/3
4 − x2 − y 2 dy dx
0
2−2x/3
y2 z
0
6−2x−3y 0
dy dx
(6y 2 − 2xy 2 − 3y 3 ) dy dx
2−2x/3 2 3 3 4 xy − y dx 3 4 0 0 2 1 3 12 2 − x dx = = 4 0 3 5
3
=
1
1−x2
√
0
2y 3 −
1−y
1
2
32.
y dz dy dx = 0
0
0
2
0 2
x+2
x
V = x2
0 2
x
xV = x2
0 2
2
x2
2
2
x+2
x
x2
11 , 10
x=
0
x2 2
1
0
9 , 4
y=
1
z=
kz dx dy dz = 0
0
1
0
1
M=
2
−1
3
0
0
4−x2
1
x+2
2−x
x2
44 x3 + 2x2 − x4 dx = 15
2
1 3 x + 4x2 + 4x − x5 dx = 6 2
0
1 2 x dy dx = 2
2
0
1 k 2
k(x2 + y 2 + z 2 ) dz dy dx = k
0
dz dy dx = 0
11 20
1
M=
2
0
x+2
z dz dy dx = 0
2
xy dy dx = 0
zV =
8 x2 + 2x − x3 dx = 3
x2
0
0
x+2
x dy dx =
x
2
0
y dz dy dx = 0
35. V =
x2
0
x+2
yV =
(b)
0
x dz dy dx =
34. (a)
x dy dx =
0
x+2
1 − y dy dx
x+2
dz dy dx =
1 2 4 2 16 dx = − x3 + x5 − x7 + 3 5 7 105 12
1
y2
0
0
=
33.
1−x2
27 2 ;
(x, y, z) =
1
3 12 2, 2, 5
1 3 22 x + 2x2 − x4 dx = 2 15
893
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894
December 7, 2006
SECTION 17.7
(x − x) dx dy dz =
36.
x dx dy dz − x
T
T
dx dy dz = xV − xV = 0. T
Similarly, the other two integrals are zero.
a
φ(x)
x 1 abc with φ(x) = b 1 − , 6 a
ψ(x,y)
dz dy dx =
37. V = 0
0
0
1
(x, y, z) =
1 4
a,
4
b,
1 4
c
M 2 1 (x + y 2 ) dx dy dz = V 30
38. Iz =
x y ψ(x, y) = c 1 − − a b
M V
=
1 M 5
T
39. Π : 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c a b c M 2 1 (a) Iz = x + y 2 dz dy dx = M a2 + b2 3 0 0 0 abc 1 (b) IM = Iz − d2 M = 13 M a2 + b2 − 14 a2 + b2 M = 12 M a2 + b2 ∧ parallel axis theorem (17.5.7) 1 1 (c) I = IM + d2 M = 12 M a2 + b2 + 14 a2 M = 13 M a2 + 12 M b2 ∧ parallel axis theorem (17.5.7)
2
2
40. V = 1
2
2
1
2
2
zV = 1
1
1
y
41. M =
2
0
z dz dy dx =
73 73 =⇒ z = . 12 72
2
0
(xM , yM , zM ) =
42. T is symmetric
1
109 109 =⇒ x = =y 12 72
k x +y +z 0
2
x dz dy dx =
1+x+y
−2
1
1+x+y
−2
1
(3 + x + y) dy dx = 6 1
xV =
2
dz dy dx =
−2
1
1+x+y
7 34 37 12 , 45 , 90
2
1 3 2 3 k x y + y + y dy dx dz dy dx = 3 0 0 1 1 2 1 1 = x + dx = k k 2 3 2 0
(a) about the yz-plane,
(d) about the origin.
by symmetry
1
1
(b) about the xz-plane,
(c) about the xy-plane,
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December 7, 2006
SECTION 17.7 43. (a)
0 by symmetry (b) (a1 x + a2 y + a3 z + a4 ) dxdydz = a4 dxdydz = a4 (volume of ball) = T
4 3
T
∧ by symmetry
2
2
4−y 2
0
352 45
x2 y 2 dz dy dx =
44. 2−y
0
a
√
a2 −x2
√a2 −x2 −y2
45. V = 8
a
√
a2 −x2
dz dy dx = 8 0
0
0
∧
π/2
a
a2 − r2 r dr dθ
=8 0
0
π/2
= −4 =
a
8 3
46. 8 0
47.
M=
2
√
4−x2 /2
√ − 4−x2 /2
−2
2
√
0
4x − x − 4xy 3
2π
V =2 0
49. M =
2
−1
3
0
2
4−x2
2
2−z
2
√
1
2
2−x
(xM , yM , zM ) =
dy dz dx 0
5
0
0
2
(c) V =
2−x
9
√
9−y
dz dx dy + 0
0
0
2−x
dz dx dy 5
3/2 128 k x 4 − x2 dx = 15
9−x2
(b) V =
0
2
(r − r3 ) dr dθ = π
0
0
4−y 2
0
dy dx dz 0
x2 +3y 2
9−x2
50. (a) V =
4−x2 /2
0
4 dy dx = k 3
135 k; k(1 + y) dz dy dx = 4
2−x
4 π a3 3
kx dz dy dx 0
dθ 0
k|x| dz dy dx = 4 x2 +3y 2
0
48. using polar coordinates
a
4 πabc. 3
4−y 2
4−x2 /2
= 4k
π/2
0
0
2 2 (a − r2 )3/2 3
dθ =
dz dy dx = 0
0
b√1−x2 /a2 c√1−x2 /a2 −y2 /b2
a2 − x2 − y 2 dy dx
0
0
polar coordinates
0
0
1 9 12 , , 2 5 5
πa4
895
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December 7, 2006
SECTION 17.8
51. (a)
6
3
6−x
V =
dy dx dz 0
z/2
3
2x
x
6−x
(b) V =
dy dz dx 0
(c)
0
6
x
3
y
V =
6
(12−z)/2
6−y
dx dy dz + 0
z/2
Ωxy
(c) V =
dx dy dz 0
3
z/2
2 4 − y dx dy
52. (a) V =
z/2
4
−4
4
√
V =
dz dy dx
(d)
Ωxy
4
(c) V =
√
(b) V =
4−y
√ − 4−y
0
dx dz dy
−y
(d) V =
(4 − z − |x|) dx dz
54. (a) V =
−2
(d) V =
4
0
5
√
3
4+x
√ − 4+x 2
55. (a) 2
z 2 −4
−2
4−z
2
1
4−z
2
4−z 2
2
dy dz dx +
|x|
0
√
4−x
√ − 4−x
√ − 4−y
dx
dydz
y
−y
dx dy dz
4−z 2
|x|
dz dx dy
dy
dx dz
3−(3/2)x
4−z 2
dy dz dx
|x|
ln xy dz dy dx ∼ = 6.80703 z
4−y
y
4−z 2
dx dy
dy dx dz
56. Let f (x, z) = 36 − 9x2 − 4z 2 and g(x, z) = 1 −
dz
2
|x|
√
0
Ωxy
2
−2
(b) V =
Ωxy
(c) V =
2
2
−y
Ωxy y
y
−y
0
2y dydz
4
V =
53. (a) V =
4−y
√ − 4−y
Ωxy 4−y
√ − 4−y
|x|
(b)
√
4
2
(b) 0 1 2
x−
1 3
1
0
3
√ 16 3 √ √ 4 2−2 x yz dz dy dx = 3
z. Then
f (x,z)
1 dy dz dx = 71
V = 0
0
g(x,z)
SECTION 17.8 1. r2 + z 2 = 9
2. r = 2
3. z = 2r
4. r cos θ = 4z
5. 4r2 = z 2
6. r2 sin2 θ + z 2 = 8
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JWDD027-Salas-v1
December 7, 2006
SECTION 17.8
π/2
2
4−r 2
7.
r dz dr dθ 0
0
π/2
0
2
= 0
2
1
0
4
4r − r2 dr dθ
0 π/2
2
4 dθ = 2π
= 0
0 1
0 2
π/4
1
√
1−r 2
8.
r dz dr dθ = 0
0
0
2π
2
π 12
r2
9.
r dz dr dθ 0
0
0
2π
2
= 0
r3 dr dθ
4
0 2π
=
-2 -
4 dθ = 8π 0
2 0 0 2 0
2
3
2π
3
10.
r dz dθ dr = 9π 0
0
r
1
√
1−x2
√4−(x2 +y2 )
π/2
1
√
4−r 2
dz dy dx =
11. 0
0
0
π/2
1
r
= 0
= 0
π
1
12. 0
0
r
1
z 3 r dz dr dθ = 0
r dz dr dθ 0
π
0
1
0
0
4 − r2 dr dθ
0 π/2
8 √ − 3 3
1 π (1 − r4 ) r dr dθ = 4 12
dθ =
√ 1 8−3 3 π 6
897
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898
December 7, 2006
SECTION 17.8
3
√9−y2 √9−x2 −y2
13. 0
0
x2
0
1 +
π/2
3
√
9−r 2
dz dx dy =
y2
0
0 π/2
0 3
= 0
1 · r dz dr dθ r
9 − r2 dr dθ
0
3 π/2 r 9 −1 r 2 9 − r + sin dθ = 2 2 3 0 0
9π = 4
π/2
1
√
1−r 2
14.
π/2
1
0
0 1
√
1−x2
0
0
0
0
2
π/2
sin(x2 + y 2 ) dz dy dx =
15. 0
0
1
0
2π
1
2−r 2
0
0
r2 dz dr dθ = 4π
r2
1
dθ = 0
0
π/2
9 2 π 8
0
sin(r2 )r dz dr dθ
2r sin(r2 ) dr dθ = 12 π(1 − cos 1) ∼ = 0.7221
(1 − r2 )r2 dr =
0
2
0 1
=
16.
π/2
r π (1 − r2 ) dr dθ = 2 16
zr dz dr dθ = 0
8π 15
17. (0, 1, 2) → (1, 12 π, 2)
18. (0, 1, −2) → (1, 12 π, −2)
19. (0, −1, 2) → (1, 32 π, 2)
20. (0, 0, 0) → (0, arbitrary, 0)
21. V =
π/2
π/2
−π/2
22. V =
π/2
π/2
−π/2
π/2
−π/2
0
2a cos θ
r 2 /a
r dz dr dθ = 0
0
a cos θ
2a cos θ
r2 dr dθ
0
0
0
3
a
0
1 1 cos2 θ − cos3 θ 2 3
2 cos θ
0
3 3 πa 2
a−r
r dz dr dθ =
π/2
−π/2
r
8 3 32 3 a cos3 θ dθ = a 3 9
−π/2
0
−π/2
23. V =
24. V =
π/2
2a cos θ
r dz dr dθ =
−π/2
=
=
2+ 12 r cos θ
π/2
−π/2
dθ =
r dz dr dθ =
a cos θ
r(a − r) dr dθ
0
1 3 a (9π − 16) 36
5 π 2
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JWDD027-Salas-v1
December 7, 2006
SECTION 17.8 25. V =
π/2
3
√
25−r 2
0
2π
r+1 √
1/2
27. V = 0
a
√
√
0
2π
3
1
2π
2r
9−r 2
√ √ 1 r 1 − r2 − r2 3 dr dθ = π 2 − 3 3
2
0
2π
0
1 2
1
√
1/2
3
r
0
30. V = 0
r2 cos θ − r3 dr dθ
2π 3 √ a ( 2 − 1) 3
a2 +r 2
r dz dr dθ =
√
2π
r dz dr dθ =
0
0
29. V = 0
cos θ
−π/2
r dz dr dθ =
r 3
28. V = 0
1−r 2
√
0
2π
41 π 3
r dz dr dθ =
26. V = 0
π/2
1 1 cos4 θ dθ = π 12 32
−π/2
2π
r2
0
π/2
r cos θ
r dz dr dθ =
−π/2
=
cos θ
899
36−r 2
r dz dr dθ =
0
9 − r2 dr dθ =
1
32 3 π
√
2
√ √ 1 π(35 35 − 128 2). 3
31. Set the lower base of the cylinder on the xy-plane so that the axis of the cylinder coincides with the z-axis. Assume that the density varies directly as the distance from the lower base.
2π
R
M=
h
kzr dz dr dθ = 0
0
0
1 kπR2 h2 2
32. xM = yM = 0 by symmetry 2π R h 1 zM M = kz 2 r dz dr dθ = kπR2 h3 3 0 0 0 M=
1 kπR2 h2 , 2
zM =
2 h 3
The center of mass lies on the axis of the cylinder at a distance density.
2π
R
h
33. I = Iz = k =
1 4
0
0
4 2
1 2
kπR h =
1 2
zr3 dr dθ dz
0
kπR2 h2 R2 =
1 2
∧ from Exercise 31
MR2
2 3h
from the base of zero mass
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900
SECTION 17.9
34. (a)
M I= πR2 h
(b)
0
R
0 2π
M πR2 h
I=
2π
0
h
r3 dz dr dθ =
0 R
0
h
1 M R2 2
0
I = 14 M R2 + 13 M h2 − M ( 12 h)2 = 14 M R2 +
(c)
1 1 M R 2 + M h2 4 3
(r2 sin2 θ + z 2 )r dz dr dθ = 1 2 12 M h
35. Inverting the cone and placing the vertex at the origin, we have
h
2π
(R/h)z
r dr dθ dz =
V = 0
36. xM = yM = 0
2π
(R/h)z
zM M = 0
0
0
M V
zr dr dθ dz =
On the axis of the cone at a distance
M 37. I = V
h
0
M 38. I = V
0
2π
2π
0
h
1
(R/h)z
r3 dr dθ dz =
0
2π
0
(R/h)z
1−r 2
0
2π
0
1
1−r 2
41. M = 0
0
0
M V
πR2 h2 3 πR2 h2 =⇒ zM = = h 4 4V 4
from the vertex.
3 MR2 10
z 2 r dr dθ dz =
r dz dr dθ =
3 4h
0
39. V = 0
0
by symmetry
h
0
1 2 πR h. 3
3 M h2 . 5
1 π 2
40.
2π
1
1−r 2
kzr dz dr dθ =
M= 0
0
0
1 k r2 + z 2 r dz dr dθ = kπ 4
SECTION 17.9 1. 3. 5.
√
3,
( 34 ,
3 4
1 4
√
π, cos−1 3,
3 2
√
1 √ 3 3
3)
√ √ 4 6 2 2 2 ρ = 2 + 2 + (2 6/3) = 3 √ 2 6/3 π √ φ = cos−1 = cos−1 (1/2) = 3 4 6/3 π θ = tan−1 (1) = 4 √ 4 6 π π , , (ρ, θ, φ) = 3 4 3
2. 4.
1√ 2
1 2
√ (2 10,
6.
6,
√
√ 2
2,
2 3 π,
√ 3 cos−1 [ 10 10])
π 5π 8, − , 4 6
1 πk 6
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SECTION 17.9
901
8. (a) (5, 12 π, arccos 45 )
7. x = ρ sin φ cos θ = 3 sin 0 cos(π/2) = 0
(b) (5, 32 π, arccos 45 )
z = ρ cos φ = 3 cos 0 = 3 y = ρ sin φ sin θ = 3 sin 0 sin(π/2) = 0 (x, y, z) = (0, 0, 3) 9. The circular cylinder x2 + y 2 = 1;
the radius of the cylinder is 1 and the axis is the z-axis.
10. The xy-plane. 11. The lower nappe of the circular cone z 2 = x2 + y 2 . 12. Vertical plane which bisects the first and third quadrants of the xy-plane. 13. Horizontal plane one unit above the xy-plane. 14. sphere
x2 + y 2 + (z − 12 )2 =
1 4
of radius
1 2
and center (0, 0, 12 )
15. Sphere of radius 2 centered at the origin: 2π π 2 8 2π π 16 2π 32π ρ2 sin φ dρ dφ dθ = sin φ dφ dθ = dθ = 3 0 3 0 3 0 0 0 0 16. That part of the sphere of radius 1 that lies in the first quadrant between the x, z-plane and the plane y=x
π/4
0
π/2
0
1
π 12
ρ2 sin φ dρ dφ dθ =
0
17. The first quadrant portion of the sphere that lies between the x, y-plane and the plane z =
π/2
π/2
3
π/2
2
π/2
ρ sin φ dρ dθ dφ = 9 π/6
0
sin φ dθ dφ
0
π/6
π/4
2π
π
=
9 2
√ π/2 π [− cos φ]π/6 = 94 π 3
sec φ
1
19. 0
0
√
1−x2
√2−x2 −y2 √
0 π/4
0
π/2
√
2
=
2 3
√
2 3
0
π/4
π/2
2
ρ2 sin φ dρ dθ dφ
sin φ dθ dφ
√
=
0
ρ2 sin φ dρ dθ dφ =
0
dz dy dx = x2 +y 2
sin φ dφ π/6
18. A cone of radius 1 and height 1; 0
π/2
9 2
=
0
0
π
0
π/4
sin φ dφ = 0
√
√ 2 π (2 − 2) 6
1 π 3
3 2
√
3.
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902
December 7, 2006
SECTION 17.9
π/4
π/2
2
0
21.
0 3
16π 5
ρ4 sin φ dρ dθ dφ =
20. 0
π/4
sin φ dφ = 0
√ 8π (2 − 2) 5
√9−y2 √9−x2 −y2 z x2 + y 2 + x2 dz dx dy 0
0
0
π/2
π/2
3
= 0
0 π/2
ρ cos φ · ρ · ρ2 sin φ dρ dθ dφ
0
1 sin 2φ dφ 2
= 0
π/2
3 π/2 π 1 5 1 ρ dρ = − cos 2φ ρ 4 2 5 0 0
3
4
dθ 0
0
243π = 20
π/2
π/2
1
1 2 π ρ sin φ dρ dθ dφ = 2 ρ 2
22. 0
0
0
2π
π
R
23. V = 0
0
0
24. r = ρ sin φ,
α
θ = θ,
π
R
25. V = 0
0
2π
ρ2 sin φ dρ dφ dθ =
0
0
π
0 2π
z = ρ cos φ
ρ2 sin φ dρ dφ dθ = R
26. M =
2π
4
= =
kh 4
tan−1 (r/h)
0
h sec φ
kh4 tan φ sec3 φ dφ dθ 4
0 2π
1 3 tan−1 (r/h) kh4 sec φ 0 dθ = 3 4
2π
0
⎡ √ ⎤ 3 2 2 1⎣ r +h − 1⎦ dθ 3 h
3/2 1 − h3 kπh r2 + h2 6
2π
tan−1 r/h
h sec φ
28. V = 0
=
kρ3 sin φ dρ dφ dθ
0
tan−1 (r/h)
0
1 kπR4 3
0
= 0
2 αR3 3
k(R − ρ)ρ2 sin φ dρ dφ dθ =
27. M = 0
4 πR3 3
0
ρ2 sin φ dρ dφ dθ =
0
r 1 π 3 h tan2 (tan−1 = πr2 h 3 h 3
29. center ball at origin; density = (a) I =
3M 4πR3
(b) I =
2 5
2π
0 2
0 2
π
R
7 5
tan−1 (r/h)
0
3M M = V 4πR3
ρ4 sin3 φ dρ dφ dθ =
0
MR + R M =
2π 3
MR2
2 MR2 5
(parallel axis theorem)
h3 tan φ sec2 φ dφ
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SECTION 17.9 30. The center of mass is the centroid;
2π
π/2
R
zV = 0
0
3 z = R; 8
2 3 πR 3
(ρ cos φ)ρ2 sin φ dρ dφ dθ =
0
3 (x, y, z) = 0, 0, R 8
31. center balls at origin;
(a) I =
V =
3M 4π R2 3 − R1 3
density
2π
0
=
π
1 4 πR 4
3M M = V 4π R2 3 − R1 3
R2
ρ4 sin3 φ dρ dφ dθ =
R1
0
903
5 R 2 − R1 5 2 M 5 R2 3 − R1 3
This result can be derived from Exercise 29 without further integration. View the solid as a ball of mass M2 from which is cut out a core of mass M1 . 4 MR2 3 M 3M 3 3 V2 = πR = ; M2 = 2 V 3 4π R2 − R1 3 R2 3 − R1 3 Then I = I2 − I1 =
2 5
M2 R 2 2 −
2 5
similarly
M1 =
MR1 3 . R2 3 − R1 3
MR2 3 MR1 3 2 2 − R R1 2 2 5 R2 3 − R1 3 R2 3 − R1 3 5 R2 − R1 5 2 . = M 5 R2 3 − R1 3
M1 R 1 2 =
2 5
(b) Outer radius R and inner radius R1 gives
5 R − R1 5 2 moment of inertia = M . 5 R 3 − R1 3
[part (a) ]
As R1 → R, R 5 − R1 5 R4 + R3 R1 + R2 R1 2 + RR1 3 + R1 4 5R4 5 = −→ = R2 . 3 2 3R2 3 R 3 − R1 R2 + RR1 + R1 Thus the moment of inertia of spherical shell of radius R is 2 5 2 2 M R = MR2 . 5 3 3 (c) I = 23 MR2 + R2 M =
5 3
MR2
(parallel axis theorem)
32. (a) The center of mass is the centroid;
using the result of Exercise 30,
x=y=0
3 4 3 4 R2 πR2 3 − R1 πR1 3 z 2 V2 − z 1 V1 3(R2 2 + R1 2 )(R2 + R1 ) 8 3 8 3 z= = = 4 V 8(R2 2 + R2 R1 + R1 2 ) π(R2 3 − R1 3 ) 3 R1 = R2 = R
(b) Setting
2π
α
33. V = 0
0
0
a
in (a), we get
ρ2 sin φ dρ dφ dθ =
x = y = 0,
2 π (1 − cos α) a3 3
z = 12 R
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SECTION 17.9
2π
π/4
1
34. 0
0
3
eρ ρ2 sin φ dρ dφ dθ =
0
35. (a) Substituting
x = ρ sin φ cos θ,
z = ρ cos φ
ρ2 sin2 φ + (ρ cos φ − R)2 = R2 ,
we have
which simplifies to (b) 0 ≤ θ ≤ 2 π,
2π
ρ = 2R cos φ.
0 ≤ φ ≤ π/4,
π/2
2R cos φ
0
0 2π
0 π/2
2R cos φ
(b) M = 0
0
2π
2R cos φ
0 2π
π/4
π
38. V = 0
2
2
2π
π/2
ρ sin φ dρ dφ dθ + 0
2π
1 2 4 kπ R 8
kρ3 cos2 θ sin2 φ dρ dφ dθ =
V =
=
1 2 4 kπ R 4
0
37.
8 kπR4 5
kρ3 sin2 φ dρ dφ dθ =
0
π/2
(c) M = 0
R sec φ ≤ ρ ≤ 2R cos φ
kρ3 sin φ dρ dφ dθ =
36. (a) M =
0
y = ρ sin φ sin θ,
x2 + y 2 + (z − R)2 = R2
into
√ 1 π(e − 1) 2 − 2 3
0
0
0
π/4
√ 2 2 cos φ
ρ2 sin φ dρ dφ dθ
0
√ 1 16 − 6 2 π 3
1−cos φ
ρ2 sin φ dρ dφ dθ =
0
39. Encase T in a spherical wedge W . f to be zero outside of T .
8 π 3
W has spherical coordinates in a box Π that contains S.
Then F (ρ, θ, φ) = f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)
is zero outside of S and
f (x, y, z) dxdydz =
T
f (x, y, z) dxdydz W
F (ρ, θ, φ) ρ2 sin φ dρdθdφ
= Π
F (ρ, θ, φ) ρ2 sin φ dρdθdφ.
= S
Define
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SECTION 17.9 40. Break up T into little basic solids T1 , . . . , TN . all the mass as concentrated there.
905
Choose a point (x∗i , yi∗ , zi∗ ) from each Ti and view
Now Ti attracts m with a force
Gmλ(x∗i , yi∗ , zi∗ )(Volume of Ti ) Fi ∼ ri =− ri 3 where ri is the vector from (x∗i , yi∗ , zi∗ ) to (a, b, c).
We therefore have
Gmλ(x∗i , yi∗ , zi∗ )[(x∗i − a)i + (yi∗ − b)j + (zi∗ − c)k] Fi ∼ = [(x∗i − a)2 + (yi∗ − b)2 + (zi∗ − c)2 ]3/2
(Volume of Ti ).
The sum of these approximations is a Riemann sum for the triple integral given and tends to that triple integral as the maximum diameter of the Ti tends to zero. 41. T is the set of all (x, y, z) with spherical coordinates (ρ, θ, φ) in the set 0 ≤ θ ≤ 2π,
S: T has volume V =
2 3
0 ≤ φ ≤ π/4,
R sec φ ≤ ρ ≤ 2R cos φ.
πR3 . By symmetry the i, j components of force are zero and ⎧ ⎫ ⎨ 3GmM ⎬ z F= dxdydz k ⎩ 2πR3 ⎭ (x2 + y 2 + z 2 )3/2 T
⎧ ⎫ ⎨ 3GmM ρ cos φ ⎬ = ρ2 sin φ dρdθdφ k 3 3 ⎩ 2πR ⎭ ρ + = =
S
3GmM 2πR3
cos φ sin φ dρ dφ dθ 0
0
R sec φ
GmM √ 2 − 1 k. R2
42. With the coordinate system shown in the figure, T is the set of all points (x, y, z) with cylindrical coordinates (r, θ, z) in the set S:
0 ≤ r ≤ R,
0 ≤ θ ≤ 2π,
α ≤ z ≤ α + h.
The gravitational force is ⎡
F =⎣ ⎡
GmM V
T
GmM =⎣ πR2 h
⎤ z dxdydz ⎦ k (x2 + y 2 + z 2 )3/2
⎤ zr dr dθ dz ⎦ k (r2 + z 2 )3/2
S
=
GmM πR2 h
0
2π R α+h 0
α
,
2π π/4 2R cos φ
zr dzdrdθ k (r2 + z 2 )3/2
2GmM 2 2− 2 + (α + h)2 + h k = R + α R R2 h
k
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SECTION 17.10
SECTION 17.10 1.
ad − bc
2.
1
3.
2 v 2 − u2
4.
u ln v − u
5.
−3u2 v 2
6.
1+
7.
abc
8.
2
9.
ρ2 sin φ
cos θ 10. |J(r, θ, z)| = −r sin θ 0
0 0 = r. 1
sin θ r cos θ 0
sin φ cos θ 11. J(ρ, θ, φ) = −ρ sin φ sin θ ρ cos φ cos θ
sin φ sin θ
= −ρ2 sin φ; 0 −ρ sin φ cos θ
ρ sin φ cos θ ρ cos φ sin θ
dx − by = (ad − bc)u0
12. (a)
(b)
v = x − y.
13. Set u = x + y,
1 uv
|J(ρ, θ, φ)| = ρ2 sin φ.
cx − ay = (bc − ad)v0
Then
x=
u+v , 2
y=
u−v 2
1 and J (u, v) = − . 2
Ω is the set of all (x, y) with uv-coordinates in 0 ≤ u ≤ 1,
Γ: Then
x2 − y 2 dxdy =
Ω
1 = 2
1
u du 0
1 1 uv dudv = 2 2
Γ
2
v dv 0
1 = 2
1
0 ≤ v ≤ 2. 2
uv dv du 0
0
1 1 (2) = . 2 2
14. Using the changes of variables from Exercise 13, 1 2 2 u − v2 1 1 1 2 2 dv du = 4xy dx dy = 4 u − v 2 dv du = −1 4 2 2 0 0 0 0 Ω
15.
1 2
1
2
u cos (πv) dv du = 0
0
16. Set u = x − y,
1 2
v = x + 2y.
1
u du 0
2
cos (πv) dv 0
=
1 2
1 (0) = 0 2
Then
2u + v v−u 1 , y= , and J(u, v) = 3 3 3 Ω is the set of all (x, y) with uv-coordinates in the set x=
Γ:
0 ≤ u ≤ π,
0 ≤ v ≤ π/2.
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SECTION 17.10 Therefore
1 1 (u + 2v) du dv = 9 9
(x + y) dxdy = Ω
Γ
17. Set u = x − y,
π/2
(u + 2v) dv du = 0
0
1 3 π . 18
v = x + 2y. Then 2u + v , 3
x= Ω
π
y=
v−u , 3
and J(u, v) =
1 . 3
is the set of all (x, y) with uv-coordinates in the set Γ : 0 ≤ u ≤ π,
0 ≤ v ≤ π/2.
Therefore 1 1 π π/2 sin u cos v dudv = sin (x − y) cos (x + 2y) dxdy = sin u cos v dv du 3 3 0 0 Ω Γ π π/2 2 1 1 sin u du cos v dv = (2)(1) = . = 3 3 3 0 0 18. Using the change of variables from Exercise 16, π π/2 1 sin 3x dx dy = sin(2u + v) du dv = 0. 3 0 0 Ω
19. Set u = xy,
v = y.
xy = 1,
xy = 4
y = x,
y = 4x
Then x = u/v,
y=v
u = 1,
u=4
=⇒ =⇒
and
J (u, v) = 1/v.
4u/v = v =⇒ v 2 = u,
u/v = v,
v 2 = 4u
Ω is the set of all (x, y) with uv-coordinates in the set
(a)
A=
1 dudv = v
Γ
4 (b) xA = 1
√ 2 u
√
4 yA = 1
u
4
√
1
u
1 dv du = v
u 7 dv du = ; 2 v 3
√ 2 u
√
√ 2 u
dv du = u
√
1 ≤ u ≤ 4,
Γ:
14 ; 3
x=
y=
√ u ≤ v ≤ 2 u.
4
ln 2 du = 3 ln 2 1
7 9 ln 2
14 9 ln 2
Γ : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π 2π 1 abr dr dθ = ab r dr dθ = πab A=
20. J(r, θ) = abr,
0
Γ
21. Set u = x + y,
v = 3x − 2y. x=
0
Then
2u + v , 5
y=
3u − v 5
and
1 J (u, v) = − . 5
907
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SECTION 17.10 0 ≤ u ≤ 1,
0≤v≤2 1 2 1 2 M= λ dv du = λ where 5 5 0 0
With Γ :
Then
1
2
2
Ix = 0
0 1
Iy = 0
0
3u − v 5 2u + v 5
Iz = Ix + Iy =
2
2
λ is the density.
1 8λ 4 λ dv du = = 5 375 75 1 28λ 14 λ dv du = = 5 375 75
2 4 λ = M, 5 75
2 14 λ = M, 5 75
18 M. 75
v−u 1 u+v , y= , J(u, v) = Γ: 2 2 2 2 1 1 2 −u 16 du dv = A= dv du = 2 2 −2 −4 3
−2 ≤ u ≤ 2,
22. x =
−4 ≤ v ≤ −u2
Γ
23. Set u = x − 2y,
v = 2x + y. x=
Then
u + 2v , 5
v − 2u 5
y=
and
J (u, v) =
1 . 5
Γ is the region between the parabola v = u2 − 1 and the line v = 2u + 2. that Γ:
−1 ≤ u ≤ 3,
Then A=
24. xA = A=
1 2
2
−2
−u2
−4
1 1 (area of Γ) = 5 5
u+v 32 dv du = − 2 5
3
u2 − 1 ≤ v ≤ 2u + 2.
−1
yA =
A sketch of the curves shows
1 2
32 . (2u + 2) − u2 − 1 du = 15
2
−2
−u2
−4
v−u 32 dv du = − 2 5
16 6 =⇒ x = y = − 3 5
25. The choice θ = π/6 reduces the equation to 13u2 + 5v 2 = 1. This is an ellipse in the uv-plane √ √ with area πab = π/ 65. Since J(u, v) = 1, the area of Ω is also π/ 65. 26.
e−(x−y) 1 dx dy = 2 1 + (x + y) 2 2
Sa
e−u du dv 1 + v2 2
Γ
where Γ is the square in the uv-plane with vertices (−2a, 0), (0, −2a), (2a, 0), (0, 2a). Γ contains the square −a ≤ u ≤ a, −a ≤ v ≤ a and is contained in the square −2a ≤ u ≤ 2a, −2a ≤ v ≤ 2a. Therefore 1 2
a
−a
a
−a
e−u 1 du dv ≤ 2 1+v 2 2
Γ
e−u 1 dudv ≤ 2 1+v 2 2
2a
−2a
2a
−2a
e−u du dv. 1 + v2 2
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SECTION 17.10 The two extremes can be written a a 1 1 2 e−u du dv 2 2 −a −a 1 + v As a → ∞ both expressions tend to ∞ −∞
27. J = abcρ2 sin φ;
2π 0
28. x = y = 0 2π zV = 0
π/2
π
V =
0
1
0
1 2 ∞
−∞ 1
1 2
and
√
( π) (π) = 12 π 3/2 .
2a
−2a
e−u du 2
2a
−2a
1 dv . 1 + v2
It follows that
e−(x−y) 1 dx dy = π 3/2 . 1 + (x + y)2 2 2
abcρ2 sin φ dρ dφ dθ =
0
(cρ cos φ)abcρ2 sin φ dρ dφ dθ =
0
4 πabc 3
πabc2 3 =⇒ z = c . 4 8
2 M 3M V = πabc, λ = = 3 V 2πabc 2π π/2 1 2 2 3M b ρ sin2 φ sin2 θ + c2 ρ2 cos2 φ abcρ2 sin φ dρ dφ dθ Ix = 2πabc 0 0 0 2 1 2 = 5M b +c Iy = 15 M a2 + c2 , Iz = 15 M a2 + b2
29.
2π
1
π
30. I = 0
0
0
PROJECT 17.10 1. (a)
ρ2 (abcρ2 sin φ) dφ dρ dθ =
−1
θ = tan
ay 1/α , bx
α/2 x 2/α y 2/α r= + a b ⎫ ar1 (cos θ1 )α = ar2 (cos θ2 )α ⎪ ⎪ ⎬ α α =⇒ r1 = r2 , br1 (sin θ1 ) = br2 (sin θ2 ) ⎪ ⎭ 1 ⎪ r > 0, 0 < θ < π
(b)
1
2. J = abαr cos 3. (a)
α−1
4 πabc 5
α−1
θ sin
2
θ
y a
a x
θ 1 = θ2
909
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REVIEW EXERCISES 2
2
2
(b) x = ar cos3 θ, y = ar sin3 θ; x 3 + y 3 = a 3 =⇒ r = 1 and x = a cos3 θ, y = a sin3 θ 0 0 A= y(θ)x (θ) dθ = a sin3 θ(3a cos2 θ[− sin θ]) dθ π 2
π 2
π 2
= 3a2 2
= 3a =
sin4 θ cos2 θ dθ = 3a2
0
3·1 π 5·3·1 π − 4·2 2 6·4·2 2
3a2 π 32
(sin4 θ − sin6 θ) dθ
0
(See Exercise 62(b) in 8.3)
3a2 π 3a2 π = 32 8
4·
(c) Entire area enclosed:
π 2
4. (a)
a = 3, b = 2
a = 2, b = 3
(b) From Problem 2, Jacobian J = 8abr cos7 θ sin7 θ π π 1 2 2 ab 8abr cos7 θ sin7 θ dθ = 4ab cos7 θ sin7 θ dθ = A= 70 0 0 0
REVIEW EXERCISES
1
√
y
1.
xy dx dy = 0
y 1
2.
1
y
x+y
e
3x
1
dx dy =
3
−y
3
y 2 ex
y
2
0
y
xe
2 sin θ
5. 0
0 2
−1
4
1
xyz dx dy dz = 0
0
2
−1
0
1
(e2y − 1) dy =
1
dx = 0
2
dx =
2
dy =
4
2
r cos θ
1 2
x2 yz
2 sin θ
0
8
y4 −
1 0
=
1 5 1 1 = y 0 10 40 e2 3 − 2 2
3 1 3 1 8 8 3 3 (9x2 ex − x2 ex ) dx = 3ex − ex = e− 0 3 3 3 1 2 5 x3 − x2 = 1 3 2 6
1
x(x − 1) dx =
1
1
e2y − y
2
0
1
0
1
1 3 1 4 y − y 2 2
1
π/4
r cos θ dr dθ =
dy =
dy =
x
0
1
0
3x
ln x
1
y
0
ln x
π/4
√y y
1
xe dy dx =
x+y
e
2yex dy dx =
4. 1
x y
2
x 2
2 2
0
3. 0
1
0
−y
0
6.
1
2
0
π/4
dθ =
2 sin θ cos θ dθ = 0
dy dz =
2
2
−1
0
4
1 yz dy dz = 2
2 3
3
sin θ
2
−1
4z dz = 6
π/4 0
√ =
2 6
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REVIEW EXERCISES
2
2−3x
x+y
7.
2
2−3x
xz
x dz dy dx = 0
0
0
0
1 x2 y + xy 2 2
= 0
8.
π 2
0
π 2
sin z
π 2
2
3x sin y dx dy dz =
z
0
0
2
x+y
0
0
2−3x
π 2
0
−π 2
−π 6
π 2
r2
r2 cos θ dz dr dθ =
1
0
3
x sin y
−π 6
=
1
0
1
1
1
1
0
−π 6
2
2y
π2 1 1 sin4 z = 0 4 4
1
1
xy dy dx =
15. 0
16.
3
√ − 3
2
3x−x2
17. 0
4−y 2
y 2 /3
1
0
0 √
1
x
2
(x − y) dx dy =
1
1 x2 1 e−1 e = 0 2 2
2
xex dx =
0
2y x cos y 2 dy = 0
0
1−y
1
1
0
0
xy √
2
√1−x2 0
0
1 2y cos y 2 dy = sin y 2 = sin 1 0
x2
√ − 3
2
(x − xy) dy dx = 2
0
2
dx =
− xy)
y(sin 1 − sin y) dy
0
0
3
1
dy =
1 2 = y sin 1 + y cos y − sin y 2
−π 2
1−y
y sin(x + y)
1−x2
32 16 sin5 θ cos θ dθ = − 5 15
y cos(x + y) dx dy 0
√
0
dy = 1
1−x
sin3 z sin y dy dz
z
1
=
1
π 2
0
0
0
r4 cos θ dr dθ =
0
dxdy = 1 − y2
π 2
dy dz =
0
0
sin z
2 x ex y dx =
0
1
2
1 sin ϕ cos ϕ dθ dϕ 4
cos y 2 dx dy =
y cos(x + y) dy dx = 0
1
ex dy dx =
(x2 + xy) dy dx
0
π π2 π 3π sin 2ϕ dϕ = − cos 2ϕ π = −6 16 32 64
0
0
13.
x
1
cos y 2 dy dx =
√1−y2
0
π 2
0
π 2
0
x 2
0
14.
1
0
12.
2
ex dx dy =
y 2
π 2
3
0
11.
2 sin θ
0
ρ sin ϕ cos ϕ dρ dθ dϕ =
0
0
2−3x
3 3 x − 4x2 + 2x dx 2 0 2 3 4 2 = x4 − x3 + x2 = − 0 8 3 3
sin3 z cos z dz =
−π 2
0
0
π 2
10.
2 sin θ
0
0
9.
2
dx =
z π 2
=
dy dx =
911
4−y2 y 2 /3
1
1 = cos 1 − 0
1 1 x − x3 2 2
dy =
√
3
√ − 3
3x−x2 1 x y − xy 2 dx = x 2
2
0
1 sin 1 2
2
dx =
1 8
√ 4 4 4 3 48 3 2 y + y − 4y − 4y + 8 dy = 9 3 5
1 2x − 2x − x5 2 4
3
dx = −
8 15
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REVIEW EXERCISES
2
2−x
0
2
x(x − 1)exy dy dx =
18. 0
(x − 1)exy
2−x
0
0
2
dx = 0
1
2
=−
19.
√ 2 y 4−y
2
2xyz dz dx dy = 0
0
0
0
0
1
x
√
2
4−x2
21.
1
0
0
0
√4−x2 −y2
√
2
4−x2
0
2
0 2
0
(x + 2z) dx dy dz =
22.
−2
T
=
2
23.
√4−y2 √ 2 2 e x +y dx dy =
0
√
π/2
0
= 0
3
2π
0 1
0
√
x
26. 0
x2
3
er r dr dθ =
1
1
(2 − x − y ) dy dx = − 2
1−x 0
(x2 + y 2 ) dy dx = 0
1 12
dy dx
dx =
2
rer dr =
0
θ dθ + 2
π
π/2
π
π/2
210 21
1
2 π r π re − er = (e2 + 1) 0 2 2
(θ − π)r dr dθ
0
1 (θ − π) dθ = 0 2 81π 2
52 2x1/2 − x5/2 − 13 x3/2 − 2x2 + x4 + 13 x6 dx = 105
0 1
π 2
1
(9 − r2 )r dr =
0
2
27. V = 0
(9 − r2 )r dr dθ = 2π
(1 − x)2 x dx =
(4x2 − x2 y + 16 − 8y + y 2 ) dy dx
0 π/2
1
(x2 + 2z) dz dy dx
rθ dr dθ +
0
25. V =
4−y
0
0
√
1 6 64 x − 8x2 + 6 3
2
xy(4 − y 2 ) dx dy
0
arctan (y/x) dy dx =
−1
4
x2
π/2
1−x2
24.
x2
−2
0
0
1
2
4
2
−2
=
2
2
0
0
3 32 1 x(4 − x2 ) 2 dx = 3 15
=
√4−x2 −y2
(x − 1) dx
0
2 x2 − x = 0
0
2
4−x2 1 − x 4 − x2 − y 2 dx 0 3
=
2
xyz
2
1
(1 − x)2 dy dx =
0
0
0
2
−
0
xy dz dy dx = 0
0
x
z dz dy dx = 0
T
1−x
z dxdydz = 2
2
0
1 3 8 y (4 − y 2 ) dy = 2 3
=
20.
0
x
2
e2x−x
2 √ 2 2 4−y xyz 2 dx dy =
2
2
(x − 1)e2x−x dx −
1 1 3 1 4 1 1 1 x − x − (1 − x)4 = x2 − x3 + (1 − x)3 dx = 3 3 4 12 6 0
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REVIEW EXERCISES
3
π/2
29. M =
0
0
π/2
cos x
y dy dx =
π/2
xy dy dx = 0 by symmetry 0
−π/2
1
1 π cos2 x dx = 2 4
cos x
cos x
y 2 dy dx =
0
y
π/2
1 4 cos3 x dx = 3 9
−π/2
16 9π )
The center of mass is: (0,
π/2
−π/2
0 π/2
−π/2
yM M =
9 sin θ dθ = 9 0
−π/2
xM M =
π/2
r2 sin θ dr dθ =
28. V =
913
1
2 15 0 y2 0 1 y 1 2 3 2 6 1 2 xM M = y − y dy = 2x dx dy = 3 3 14 0 y2 0 1 y 1 1 yM M = 2xy dx dy = (y 3 − y 5 ) dy = 12 0 y2 0 2x dx dy =
30. M =
(y 2 − y 4 ) dy =
The center of mass is: (15/28, 5/8) π/2 R π 31. M = u3 du dθ = (R4 − r4 ); 8 0 r
(polar coordinates [u, θ])
By symmetry, x = y. π/2 R 1 xM M = u4 cos θ du dθ = (R5 − r5 ); 5 0 r
π
2(1+cos θ)
xM =
8(R5 − r5 ) 5π(R4 − r4 )
π
8 20 (1 + cos θ)3 dθ = π 3 0 0 0 3 π 2(1+cos θ) π xM M = r3 cos θ dr dθ = 4(1 + cos θ)4 cos θ dθ = 14π
32. M =
0
π
r2 dr dθ =
0 2(1+cos θ)
yM M = 0
0 π
r3 sin θ dr dθ =
0
0
The center of mass is:
4 128 4(1 + cos θ)4 sin θ dθ = − (1 + cos θ)5 |π0 = 5 5
96 ( 21 10 , 25π )
y
33. Introduce a coordinate system as shown in the figure. (a) A = 12 bh; yA = = (b) I =
0
−b/2
bh2 6 0
−b/2
h
by symmetry, x = 0 2h b b (x+ 2 )
0
2h b b (x+ 2 )
b 2
0
0
=⇒
y dy dx +
y=
b − 2h b (x− 2 )
y dy dx
0
h 3
−b 2
λy 2 dy dx + 0
b/2
0
b − 2h b (x− 2 )
λy 2 dy dx =
λbh3 = 16 M h2 12
b 2
x
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REVIEW EXERCISES
b/2
b − 2h b (x− 2 )
(c) I = 2 0
34. Let λ =
1 λhb3 = 48
λx2 dx dy =
0
1 2 24 M b
k x2
+ y2 π R (a) M = k dr dθ = kπ(R − r) 0
r
By symmetry, xM = 0;
π
R
yM M = 0
kr sin θ dr dθ = k(R2 − r2 )
r
R+r π )
The center of mass is: (0, π R π k (b) Ix = kr2 sin2 θ dr dθ = (R3 − r3 ) sin2 θ dθ = 3 0 r 0 π R π k (c) Iy = kr2 cos2 θ dr dθ = (R3 − r3 ) cos2 θ dθ = 3 0 r 0
2
x
2x+2y+1
35. V =
2
0
1
36. V =
0
4(x2 +y 2 )
0
1
x
dz dy dx =
−1
x2
0
(2x + 2y + 1) dy dx = 10
0
x
(4x2 + 4y 2 + 1) dy dx
x2
0 1
16 3 107 4 6 2 4 x + x − x − 4x − x dx = 3 3 210
= 0
x2 + y 2 = 4, x = 3
37. The curve of intersection of the two surfaces is the circle: V =
2
√
4−x2
√
− 4−x2
−2
12−x2 −2y 2
dz dy dx =
4−x2
√
− 4−x2
−2
2x2 +y 2
√
2
2π
2π
2
=3
0
38.
1
0
√1−y2
(2−y−z)/2
V =
4 − r2 r dr dθ
2r2 −
1
π/2
dx dz dy = 0
0
0
3 4 − x2 − y 2 dy dx
0
=3
kπ 3 (R − r3 ) 6
x
dz dy dx = 0
kπ 3 (R − r3 ) 6
0
0
1 4
r4
2 0
2π
dθ = 12
= 24π 0
2 − r cos θ − r sin θ r dθ dr 2
1 3π − 4 (π − 2r)r dr = 2 12 0 2π π/3 2 1 3 ρ2 sin φ dρ dφ dθ = ρ dφ dθ 3 1
=
39.
2π
π/3
2
V = 0
0
sec φ
0
=
1 3
1 = 3 1 = 3
sec φ
0
2π
0
8 − sec3 φ sin φ dφ dθ
0 2π
0
π/3
− 8 cos φ −
5 5π (2π) = 2 3
1 2
π/3
sec2 φ
0
dθ
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REVIEW EXERCISES
4
(12−3x)/4
16−x2
40. V =
4
12−3x 4
dz dy dx = 0
0
0
2π
π/2
41. V =
0
0
1
2π
2
π/2
ρ sin φ dρ dφ dθ = 0
π/4
2π
0
π/4
0
1
42. V = 0
0
0
1
43. (a) V =
x
π/4
1 sin φ dφ dθ = 3
0
1−x2
1
1
x
0
0
0
0
(64 − 16x − 4x2 + x3 ) dx = 80
0
√
2π 3
√1−y2
y
0
1
0
0
2 x 1 − x2 dx = 3
y
1
1 − x2 dy dx = 2
By symmetry, x = y. 1 x √1−x2 xV = x dz dy dx + 0
4
dz dx dy 0
0
=2
√ 2− 2 π ρ sin φ dρ dφ dθ = 3
dz dy dx + 0
3 4
2
√
(16 − x2 ) dy dx =
√1−y2 x dz dx dy
0
0
0
For the first integral:
1
x
√
1−x2
1
x 1 − x2 dy dx
x
x dz dy dx = 0
0
0
0
0
1
2
x
=
1−
x2
π/2
dx =
0
∧
sin2 u cos2 u du =
0
x = sin u For the second integral: 1 y √1−y2 x dz dx dy = 0
0
0
3π 32
=⇒
Now calculate z: 1 zV = 0 1
x
0
Thus, x V =
1
√
0
x
0
x=y=
√
1
0
z dz dy dx = 0
0
0
Therefore,
zV =
1 4
y
√1−y2
=⇒
z dz dx dy; 0
1−x2
1−y 2
y
1
z dz dy dx +
and similarly, √ 1
1 2 π y 1 − y 2 dy = 2 32
9π 64
0
0
0
1
0
1−x2
z dz dy dx = 0
x 1 − y 2 dx dy =
y
0
x
0
1 1 (1 − x2 )dy dx = 2 2
1 . 8 z=
0
3 8
0
1
915
1 x − x3 dx = 8
π 16
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REVIEW EXERCISES
1
√
x
1−x2
(b) Iz =
λ 0
0
1
0 √
x
1−x2
0
0
x2 + y 2
λ( 0
2π
1
44. (a) V = 0
1
λ 0
π/4
1
(r − r )r dr dθ = 2π
0
(r2 − r3 ) dr =
0
x2 + y 2
sec θ
r sin θ 0
3 λ 20
=⇒
Iz =
3 λ 10
π 6
Here, K is the density of the mass.
0
0
0
r
u3 du =
0
πhr4 2
(b) By symmetry, xM = yM = 0 2π r h πh2 r4 =⇒ zM = h/2 (c) zM M = u3 z dz du dθ = 4 0 0 0 x2 + y 2 2π (a) M =
46. λ =
0
π/2
0
r
ρ3 sin2 φ dρ dφ dθ =
0
r4 π 2
π 2
sin2 φ dφ =
0
r4 π 2 8
(b) By symmetry, xM = yM = 0 2π π/2 r 2r5 π 16r =⇒ zM = ρ4 sin2 φ cos φ dρ dφ dθ = zM M = 15 15π 0 0 0
1
2π
1
r2 dz dθ dr = 2π
47. (a) M = 0
0
r
1
0
0 1
0 2π
r
1
(c) Iz = 0
48.
0
2u J(u, v) = −2v
r
2v 2u
1 0
r4 dz dθ dr =
π 15
= 4u2 + 4v 2
r2 dz dr = 2π
r
(b) By symmetry, xM = yM = 0 1 2π 1 zM M = r2 z dz dθ dr = π
1
dz dx dy;
λr3 dz dr dθ =
By symmetry, x = y = 0 2π 1 r π and hence z = 12 zV = zr dz dr dθ = 12 2 0 0 r 2π 1 r 1 πK (b) Iz = K r2 dz dr dθ = 2πK r2 (r − r2 )dr = 10 0 0 r2 0
45. Denote polar coordinates by [u, θ]. 2π r h (a) M = u3 dz du d θ = 2πh
2
0
0
x2 + y 2 )2 dz dx dy =
2
dz dy dx =
0
0
√1−y2
dz dy dx +
2
y
0
√ 2 y 1−y
1
and
x2 + y 2
2
0
λ 0
1
r2 (1 − r) dr =
0
r2 (1 − r2 ) dr =
2π 4 =⇒ zM = 15 5
π 6
3 λ 20
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REVIEW EXERCISES eu cos v 49. J(u, v) = −eu sin v
50.
2u Ju, v, w) = 2w 2v
eu sin v eu cos v
= e2u
vw uw = (vw − u2 )(4uw − 4v 2 ) uv
2w 2v 2u
v+u 1 v−u , y= =⇒ u = y − x, v = y + x, 1 ≤ v ≤ 2, J = − 2 2 2 at x = 0, y = u, y = v =⇒ u = v
51. Set
x=
at y = 0, −x = u, x = v =⇒ u = −v 2 v 2 u y−x 3 1 dx dy = cos du dv = cos v sin 1 dv = sin 1 y+x v 2 1 −v 2 i Ω
52.
cos θ u r cos θ By the hint, J = − u2 r sin θ − u
2π
2
sin θ u r sin θ − 2 u r cos θ u
dx dy dz = T
0
1
1
2
2r 2r3 0 =− 3 u 0
2r3 45π dr du dθ = 3 u 8
917
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January 4, 2007
SECTION 18.1
CHAPTER 18 SECTION 18.1 1. (a) h(x, y) = y i + x j;
r(u) = u i + u2 j, u ∈ [ 0, 1 ]
x(u) = u, y(u) = u2 ;
x (u) = 1, y (u) = 2u
h(r(u)) · r (u) = y(u) x (u) + x(u) y (u) = u2 (1) + u(2u) = 3u2 1 h(r) · dr = 3u2 du = 1 0
C
(b) h(x, y) = y i + x j;
r(u) = u3 i − 2u j, u ∈ [ 0, 1 ]
x(u) = u3 , y(u) = −2u;
x (u) = 3u2 , y (u) = −2
h(r(u)) · r (u) = y(u) x (u) + x(u) y (u) = (−2u)(3u2 ) + u3 (−2) = −8u3 1 h(r) · dr = −8u3 du = −2 C
0
1
h · dr =
2. (a) C
(u i + u j) · (i + 2u j) du =
0
C
1
0 1
h · dr =
(b)
2
(u + 2u3 ) du = 1
1
(u3 i − 2u j) · (3u2 i − 2 j) du =
0
(3u5 + 4u) du =
0
5 2
r(u) = cos u i − sin u j, u ∈ [ 0, 2π ]
3. h(x, y) = y i + x j;
x(u) = cos u, y(u) = − sin u;
x (u) = − sin u, y (u) = − cos u
h(r(u)) · r (u) = y(u) x (u) + x(u) y (u) = sin2 u − cos2 u 2π h(r) · dr = (sin2 u − cos2 u) du = 0 0
C
1
h · dr =
4. (a)
0
C
h · dr =
(b) C
(e−u i + 2 j)·(eu i − e−u j) du =
0
2
2 j·(1 − u) i du =
0
2
0 du = 0 0
5. (a) r(u) = (2 − u) i + (3 − u) j, u ∈ [ 0, 1 ] 1 17 h(r) · dr = (−5 + 5u − u2 ) du = − 6 C 0 (b) r(u) = (1 + u) i + (2 + u) j, u ∈ [ 0, 1 ] 1 17 h(r) · dr = (1 + 3u + u2 ) du = 6 C 0
1
(1 − 2e−u ) du = 2e−1 − 1
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SECTION 18.1
4
h · dr =
6. (a) C
1
1
h · dr =
(b) C
0
7. C = C1 ∪ C2 ∪ C3
4 1 1 1 8 1 1 √ √ i+ √ du = ln i+ √ j · j du = u(1 + u) 5 u(1 + u) 2 u u 1+u 2 1+u 1 1 1 2 3 (i + j) · (i + j) du = du = 3 (1 + u)3 (1 + u) 4 0
where,
C1 : r(u) = (1 − u)(−2 i) + u(2 i) = (4u − 2) i, u ∈ [ 0, 1 ] C2 : r(u) = (1 − u)(2 i) + u(2 j) = (2 − 2u) i + 2u j,
u ∈ [ 0, 1 ]
C3 : r(u) = (1 − u)(2 j) + u(−2 i) = −2u i + (2 − 2u) j, = + + = 0 + (−4) + (−4) = −8 C
C1
C2
u ∈ [ 0, 1 ]
C3
8. r(u) = (−1 + 2u) i + (1 + u) j, u ∈ [0, 1] 1 1 e5 − e2 + 6e − 6 h · dr = (e−2+u i + e3u j) · (2i + j) du = (2e−2+u + e3u ) du = 3e2 0 0 C
C1 : r(u) = (−1 + 2u) i,
9.
u ∈ [ 0, 1 ]
C2 : r(u) = cos u i + sin u j,
= C
= 0 + (−π) = −π
+ C1
C2
10. Bottom:
1
r(u) = u i;
0
r(u) = (1 − u)i + j; r(u) = (1 − u)j;
1
3(1 − u) i · (−i) du = 2
0 1
2(1 − u)j·(−j) du =
0
h · dr = sum of the above = 0 C
1
1
[3ui + (1 + 2u)j]·j du =
Left:
0 du = 0 0
r(u) = i + uj;
Top:
1
u3 j · i du =
0
Right side:
919
0
(1 + 2u) du = 2 0 1
−3(1 − u)2 du = −1
0 1
−2(1 − u) du = −1
u ∈ [ 0, π ]
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920
SECTION 18.1
11. (a)
r(u) = u i + u j + uk, u ∈ [ 0, 1 ] 1 h(r) · dr = 3u2 du = 1
C
0 1
h(r) · dr =
(b) C
0
C
1
eu (i + j + k) · (i + j + k) du =
0
3eu du = 3(e − 1)
0
1
h · dr =
(b)
23 21
1
h · dr =
12. (a)
(2u3 + u5 + 3u6 ) du =
u2
u
(e i + e
u3
j+e
k)·( i + 2u j + 3u k) du=
0
C
1
2
2
3
(eu + 2ueu + 3u2 eu ) du = 3(e − 1).
0
13. (a) r(u) = 2u i + 3u j − u k, u ∈ [ 0, 1 ] 1 1 h(r) · dr = (2 cos 2u + 3 sin 3u + 3u2 ) du = sin 2u − cos 3u + u3 0 = 2 + sin 2 − cos 3 C
0
1
h(r) · dr =
(b) C
0
1
h · dr =
14. (a) C
h · dr =
C
1
(i + ue2u j + uk) · (eu i − e−u j + k) du =
0
C
0
r(u) = u i; 0
C2 :
1
u i · i du =
r(u) = (1 − u)i + j; F · dr = C
1
u du = 0
1
0
W =
0 1
2
17 6
(eu − ueu + u) du = e −
3 2
2u3 + 6u2 + u du = 26
0
r(u) = i + uj;
C3 :
(−4u2 − 6u3 ) du = −
0
15. r(u) = u i + u2 j, u ∈ [ 0, 2 ] 2 (u + 2u2 ) + (2u + u2 )2u du = F(r) · dr =
16. C1 :
1
(−2u2 i + 4u3 j − 2u3 k)·(2 i − j + k) du =
0
(b)
1 1 4 2u cos u2 + 3u2 sin u3 − u4 du = sin u2 − cos u3 − u5 = + sin 1 − cos 1 5 5 0
1 2
1
1 −u sin 1 du = − sin 1 2 1 1 1 [(1 − u) cos 1i − sin(1 − u)j] ·(−i) du = (u − 1) cos 1 du = − cos 1 2 0 0
(cos ui − u sin 1j) · j du =
0
1 1 1 1 − sin 1 − cos 1 = (1 − sin 1 − cos 1) 2 2 2 2
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SECTION 18.1
921
17. r(u) = (1 − u)(j + 4 k) + u(i − 4 k) = ui + (1 − u)j + (4 − 8u)k, u ∈ [ 0, 1 ] 1 1 F(r) · dr = (−32u + 97u2 − 64u3 ) du = 3 C 0
18. C1 :
r(u) = u i;
C2 :
C1 1
r(u) = i + uj; 0
C3 :
F · dr = 0
F(r(u)) = 0,
1
uk · j du =
r(u) = i + j + u k;
0 du = 0 0
1
(ui + uj + k) · k du =
0
1
du = 1 0
W =1 19. r(u) = cos u i + sin u j + u k, u ∈ [ 0, 2π ] 2π 2 2 2 F(r) · dr = − cos u sin u + cos u sin u + u du = 0
C
2π
0
u2 du =
8π 3 3
20. Place the origin at the center of the circular path C and use the time parameter t. Motion along C at constant speed is uniform circular motion r(t) = r(cos ωt i + sin ωt j). Differentiation gives r (t) = rω(− sin ωti + cos ωt j),
r (t) = −rω 2 (cos ωti + sin ωt j).
The force on the object is F(r(t)) = mr (t). Note that F(r(t)) · r (t) = 0 for all t, and therefore W is 0 on every time integral. P hysical explanation :
At each instant the force on the object is perpendicular to the path of the
object. Thus the component of force in the direction of the motion is always zero.
b
q · dr =
21.
[q · r (u)] du =
a
C
a
b
d [q · r(u)] du du
= [q · r (b)] − [q · r (a)]
= q · [r (b) − r (a)]
b
r · dr = C
a
[r (u) · r (u)] du
1 b = r d r (see Exercise 57, Section 14.1) 2 a 1 = r(b)2 − r(a)2 2
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SECTION 18.1
r(u) = (1 − 2u) i;
22. (a)
h · dr = C1
1
h · dr =
(b) C2
(1 − 2u) i · (−2 i) du =
u du +
1
−2(1 − 2u)2 du = −
[(1 − 2u)2 i + j] · (−2i) du +
0
1
1
0
1
(i + u j) · j du +
0
(c)
2
0
0
=
1
1
2 3
[i + (1 − u)j] · (−j) du
0
−2(1 − 2u)2 du +
0
1
−(1 − u) du = −
0
2 3
r(u) = cos u i + sin u j, u ∈ [0, π] π π 2 2 h · dr = (cos u i + sin uj) · (− sin ui + cos u j) du = (− sin u cos2 u + sin u cos u) du = − 3 C3 0 0
f(r) · dr =
23. C
b
[f(r (u)) · r (u)] du =
a
b
[f (u) i · i] du =
a
b
f (u) du a
24. Follows from the linearity of the dot product and of ordinary integrals.
25. E : r(u) = a cos u i + b sin u j,
2π
W = 0
u ∈ [ 0, 2π ]
1 1 1 2π − b sin u (−a sin u) + a cos u (b cos u) du = ab du = πab 2 2 2 0
If the ellipse is traversed in the opposite direction, then W = −πab. In both cases |W | = πab = area of the ellipse. 26. force at time t:
mr (t) = 2mβ j
work during time interval:
1
W =
4mβ 2 t dt = 2mβ 2
0
27. r(t) = αt i + βt2 j + γt3 k r (t) = α i + 2βt j + 3γt2 k force at time t = mr (t) = m(2βj + 6γtk)
1
[m(2β j + 6γt k) · (α i + 2βtj + 3γt2 k)] dt 0 1 9 2 2 2 3 2 =m (4β t + 18γ t ) dt = 2β + γ m 2 0
W =
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SECTION 18.1 28. (a) (b)
v ⊥ k,
v ⊥ r,
v = ω
and ωk, r, v, form a right-handed triple
We can parametrize C counterclockwise by r(t) = a cos t i + a sin t j, Then
0 ≤ t ≤ 2π.
r (t) = −a sin t i + a cos t j
and
(ω k × r) · dr =
C
2π
(ω k × r(t)) · r (t) dt.
0
Now k × r(t) = a cos t j − a sin t i. So (k × r(t)) · r (t) = a2 (cos2 t + sin2 t) = a2 . Thus
2π
(ωk × r) · dr + C
ωa2 dt = ωa2 (2π) = 2ω(πa2 ) = 2ωA.
0
If C is parametrized clockwise, the circulation is −2ωA.
29. Take C : r(t) = r cos t i + r sin t j, t ∈ [ 0, 2π ] 2π v(r) · dr = [v(r(t)) · r (t)] dt C
0 2π
=
[f (x(t), y(t)) r(t) · r (t)] dt
0
=
2π
f (x(t), y(t)) [r (t) · r (t)] dt = 0
0
since for the circle r(t) · r (t) = 0 identically. The circulation is zero. 30. (a)
r(u) = i + u j;
W =
(b)
r(u) = ui + j;
2
0
W = 0
1
2 k ku k (i + u j) · j du = du = ln 5 2 2 1+u 2 0 1+u 1 k ku k (ui + j) · i du = du = ln 2. 2+1 u2 + 1 u 2 0
31. (a) r(u) = (1 − u)(i + 2 k) + u(i + 3 j + 2 k) = i + 3u j + 2 k, u ∈ [ 0, 1 ].
1 1 −k 9uk k k √ F(r) · dr = du = =√ −√ 2 )3/2 2 (5 + 9u 5 14 5 + 9u 0 C 0
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SECTION 18.2 (b) Let C be an arc on the sphere r = r = 5.
F(r) · dr =
C
C2
k r · dr = 3 r d r 5 C2 C2
(0, 4, 3) k 1 =0 r 2 = 3 5 2 (3,4,0) k = 3 5
kr · dr r 3
(see Exercise 57, Section 14.1)
32. Let (x0 , y0 , z0 ) and (x1 , y1 , z1 ) be the coordinates of a and b, respectively. Then k
W =
x21
y12
+
+
z12
k
−
x20
+ y02 + z02
33. r(u) = u i + α u(1 − u) j, r (u) = i + α(1 − 2u) j, u ∈ [ 0, 1 ] 1 2 2 (α u (1 − u)2 + 1 + [u + α u(1 − u)]α(1 − 2u)] dx F(r) · dr = W (α) = C
0 1
1 1 1 + (α + α2 )u − (2α + 2α2 )u2 + α2 u4 du = 1 − α + α2 6 30 0 1 15 5 1 α =⇒ α = = W (α) = − + 6 15 6 2 The work done by F is a minimum when α = 5/2. =
34. Suppose that C is the curve r(u), a ≤ u ≤ b.
b
∇f · dr = C
∇f (r(u)) · r (u) du =
a
a
b
b df du = f (r(u)) = f (r(b)) − f (r(a)). a du
SECTION 18.2 1. h(x, y) = ∇f (x, y) C is closed
where f (x, y) = 12 (x2 + y 2 ) =⇒ h(r) · dr = 0 C
2. x i + y j
is a gradient (Exercise 1);
h(r) · dr =
C
2π
we need integrate only y i.
y(t)x (t) dt = 0
2π
(b sin t)(−a sin t) dt = −πab
0
3. h(x, y) = ∇f (x, y) where f (x, y) = x cos πy; r(0) = 0, r(1) = i − j h(r) · dr = ∇f (r) · dr = f (r(1)) − f (r(0)) = f (1, −1) − f (0, 0) = −1 C
C
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SECTION 18.2 4. h = ∇f
x3 y3 with f (x, y) = + − xy, 3 3
925
h · dr = 0
and C is closed, so C
5. h(x, y) = ∇f (x, y) where f (x, y) = 12 x2 y 2 ; r(0) = j, r(1) = −j h(r) · dr = ∇f (r) · dr = f (r(1)) − f (r(0)) = f (0, −1) − f (0, 1) = 0 − 0 = 0 C
C
we need integrate only i − x j
6. ey i + xey j is a gradient; C = C1 ∪ C2 ∪ C3 ∪ C4
where
C1 : r(u) = (2u − 1)i − j,
u ∈ [0, 1]
C2 : r(u) = i + (2u − 1)j,
u ∈ [0, 1]
C3 : r(u) = (1 − 2u)i + j,
u ∈ [0, 1]
C4 : r(u) = −i + (1 − 2u)j, u ∈ [0, 1] = + + + = 2 + (−2) + (−2) + (−2) = −4 C
C1
C2
C3
C4
7. h(x, y) = ∇f (x, y) where f (x, y) = x2 y − xy 2 ; r(0) = i, r(π) = −i h(r) · dr = ∇f (r) · dr = f (r(π)) − f (r(0)) = f (−1, 0) − f (1, 0) = 0 − 0 = 0 C
C
8. h(x, y) = ∇f (x, y) where f (x, y) = (x2 + y 4 )3/2 h(r) · dr = ∇f (r) · dr = f (−1, 0) − f (1, 0) = 1 − 1 = 0 C
C
9. h(x, y) = ∇f (x, y) where f (x, y) = (x2 + y 4 )3/2 h(r) · dr = ∇f (r) · dr = f (1, 0) − f (−1, 0) = 1 − 1 = 0 C
C
10. h = ∇f
2
with f (x, y) = cosh x y;
h · dr = 0
and C is closed, so C
11. h(x, y) is not a gradient, but part of it, 2x cosh y i + (x2 sinh y − y)j, is a gradient. Since we are integrating over a closed curve, the contribution of the gradient part is 0. Thus
h(r) · dr =
C
(−yi) · dr. C
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SECTION 18.2 u ∈ [ 0, 1 ]
C1 : r(u) = i + (−1 + 2u)j, C2 : r(u) = (1 − 2u)i + j,
h(r) · dr = C
C4 : r(u) = (−1 + 2u)i − j,
u ∈ [ 0, 1 ]
(−y i) · dr + C2
0
1
+
0
(−y i) · dr +
C3
(−y i) · dr C4
−i · (−2 i) du +
0
=
u ∈ [ 0, 1 ]
(−y i) · dr +
=
C3 : r(u) = −i + (1 − 2u)j,
C1
u ∈ [ 0, 1 ]
+
0
1
+ 0
1
2 du
+
0
+
12. h(x, y) = ∇
2
x2 y 2 2
(u5 i + u4 j) · (i + 2u j) du =
(b)
f (2, 4) − f (0, 0) = 32 − 0 = 32
0
2
3u5 du = 32
0
13. h(x, y) = (3x2 y 3 + 2x) i + (3x3 y 2 − 4y) j;
C
(b)
0
4
(a)
(a)
∂P ∂Q = 9x2 y 2 = . ∂y ∂x
Thus h is a gradient.
r(u) = u i + eu j, r (u) = i + eu j, u ∈ [ 0, 1 ] 1 2 3u 1 h(r) · dr = (3u e + 2u) + 3u3 e3u − 4e2u ) du = u3 e3u + u2 − 2e2u 0 = e3 − 2e2 + 3 0
∂f = 3x2 y 3 + 2x =⇒ f (x, y) = x3 y 3 + x2 + g(y); ∂x ∂f = 3x3 y 2 + g (y) = 3x3 − 4y =⇒ g (y) = −4y ∂y
=⇒
g(y) = −2y 2
Therefore, f (x, y) = x3 y 3 + x2 − 2y 2 . Now, at u = 0, r(0) = 0 i + j = (0, 1); at u = 1, r(1) = i + e j = (1, e) and (1,e) h(r) · dr = x3 y 3 + x2 − 2y 2 (0,1) = e3 − 2e2 + 3 C
14. h(x, y) = ∇(x2 sin y − ex ) π (a) (2 cos u sin u − ecos u ) i + (cos2 u cos u)j · (− sin u i + j) du = e − e−1 0
(b)
1
2 du
0
=
i · (2 i) du
f (−1, π) − f (1, 0) = e − e−1
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SECTION 18.2 ∂P ∂Q = 2e2y − 2x = . ∂y ∂x
15. h(x, y) = (e2y − 2xy) i + (2xe2y − x2 + 1) j;
r (u) = (1 + u)eu i + j,
(a) r(u) = ueu i + (1 + u) j,
h(r) · dr =
C
(b)
1
Thus h is a gradient.
u ∈ [ 0, 1 ]
e2 (3ue3u + e3u − 2u3 e2u − 5u2 e2u − 2ue2u + 1 du
0
1 = e2 ue3u − u3 e2u − u2 e2u + u 0 = e5 − 2e2 + 1
∂f = e2y − 2xy =⇒ f (x, y) = xe2y − x2 y + g(y). ∂x ∂f = 2xe2y − x2 + g (y) = 3x3 − 4y =⇒ g (y) = 1 ∂y
=⇒
g(y) = y
Therefore, f (x, y) = xe2y − x2 y + y. Now, at u = 0, r(0) = 0 i + j = (0, 1); at u = 1, r(1) = e i + 2 j = (e, 2) and (e,2) h(r) · dr = xe2y − x2 y + y (0,1) = e5 − 2e2 + 1 C
16. h(x, y, z) = ∇f
with
h · dr = f (1, 1, 1) − f (0, 0, 0) = 1
f (x, y, z) = xy 2 z 3 C
17. h(x, y, z) = (2xz + sin y) i + x cos y j + x2 k; ∂Q ∂P = cos y = , ∂y ∂x
∂P ∂R = 2x = , ∂z ∂x
∂Q ∂R =0= . ∂z ∂y
∂f = 2xz + sin y, ∂x ∂f ∂g = x cos y + = x cos y, ∂y ∂y
=⇒
=⇒
g(y, z) = h(z)
∂f = x2 + h (z) = x2 ∂z Therefore, f (x, y, z) = x2 z + x sin y
h(r) · dr =
C
C
Thus h is a gradient.
f (x, y, z) = x2 z + x sin y + g(y, z) =⇒ f (x, y, z) = x2 z + x sin y + h(z) =⇒
h (z) = 0
h(z) = C
(take C = 0)
r(2π) (1,0,2π) ∇f · dr = x2 z + x sin y r(0) = x2 z + x sin y (1,0,0) = 2π
18. h(x, y, z) = ∇f with f (x, y, z) = yz sin πx √ 1 √ h · dr = f 12 , 23 , π3 − f (1, 0, 0) = π 3 6 C 19. h(x, y, z) = (2xy + z 2 ) i + x2 j + 2xz k; ∂P ∂Q = 2x = ∂y ∂x,
=⇒
∂P ∂R = 2z = ∂z ∂x,
∂Q ∂R =0= ∂z ∂y.
Thus h is a gradient.
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SECTION 18.2 ∂f = 2xy + z 2 ∂x ∂f ∂g = x2 + = x2 ∂y ∂y
=⇒
Therefore, f (x, y, z) = x2 y + xz 2 h(r) · dr = C
=⇒ f (x, y, z) = x2 y + xz 2 + h(z)
g(y, z) = h(z)
∂f = 2xz + h (z) = 2xz ∂z
f (x, y, z) = x2 y + xz 2 + g(y, z)
=⇒
h (z) = 0
=⇒
=⇒
h(z) = C
(take C = 0)
r(1)
(2,3,−1) ∇f · dr = x2 y + xz 2 = x2 y + xz 2 = 14 r(0)
C
(0,2,0)
20. h(x, y, z) = ∇f with f (x, y, z) = z 3 − e−x ln y h · dr = f (2, e2 , 2) − f (1, 1, 1) = 7 − 2e−2 C
∂P ∂Q = 2e2y = . ∂y ∂x
21. F(x, y) = (x + e2y ) i + (2y + 2xe2y ) j;
∂f = x + e2y ∂x ∂f = 2xe2y + g (y) = 2y + 2xe2y ∂y Therefore, f (x, y) =
with f (x, y, z) = x2 ln y − xyz
23. Set f (x, y, z) = g(x)
f (x, y) =
g (y) = 2y
1 2 x + xe2y + y 2 F(r) · dr = ∇f · dr = 2 C C
22. F = ∇f
=⇒
1 2 x + xe2y + g(y); 2
=⇒ g(y) = y 2 (take C = 0)
1 2 x + xe2y + y 2 . 2
=⇒
Thus F is a gradient.
and
r(2π) r(0)
1 2 x + xe2y + y 2 = 2
(3,0) =0 (3,0)
W = f (3, 2, 2) − f (1, 2, 1) = 8 ln 2 − 10
C : r(u) = u i,
u ∈ [ a, b ].
In this case ∇f (r(u)) = g (x(u))i = g (u)i so that
∇f (r) · dr =
C
Since
b
[ ∇f (r(u)) · r(u)] du =
a
gives
b
g (u) du = g(b) − g(a).
k (x i + y j + z j) = ∇f (x2 + y 2 + z 2 )n/2 f (r) = k ln r + C
g (u) du.
a
C
(a) n = 2 :
b
a
f (r(b)) − f (r(a)) = g(b) − g(a), ∇f (r) · dr = f (r(b)) − f (r(a))
24. F(x, y, z) =
and r(u) = i,
(b) n = 2 :
f (r) = −
k n−2
1 rn−2
+C
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SECTION 18.2 25. F(r) = kr r = k
x2 + y 2 + z 2 (x i + y j + z k),
∂P kxy ∂Q = = 2 2 2 ∂y ∂x, x +y +z
929
k > 0 constant.
∂P kxz ∂R = = 2 2 2 ∂z ∂x x +y +z
∂Q kyz ∂R = = 2 2 2 ∂z ∂y x +y +z
Therefore, F is a gradient field. 3/2 ∂f k 2 = kx x2 + y 2 + z 2 =⇒ f (x, y, z) = x + y2 + z2 + g(y, z). ∂x 3 3/2 ∂f ∂g k 2 = ky x2 + y 2 + z 2 + = ky x2 + y 2 + z 2 =⇒ f (x, y, z) = x + y2 + z2 + h(z) ∂y ∂y 3 ∂f = kz x2 + y 2 + z 2 + h (z) = kz x2 + y 2 + z 2 =⇒ h(z) = C, constant ∂z 3/2 + C. Therefore, f (x, y, z) = k3 x2 + y 2 + z 2 26. Set f (x, y, z) =
1 2
x2 +y 2 +z 2
g(u) du. Then 0
∇f = g(x2 + y 2 + z 2 ) [x i + y j + x k] = F(r) 27. F(r) = ∇ 28. (a)
mG r
;
F(r) · dr = mG
W = C
1 1 − r2 r1
Since the denominator is never 0 in Ω, P and Q are continuously differentiable on Ω. x2 − y 2 ∂Q ∂P = 2 . = ∂y (x + y 2 )2 ∂x
(b)
1 2
Take r(u) =
h · dr =
C
cos u i +
2π
1 2
0
1 2
sin u j.
sin u i− 1/4
1 2
2π cos u 1 1 j · − sin u i + cos u j du = − du = −2π 1/4 2 2 0
Therefore h is not a gradient since the integral over C (a closed curve) is not zero. (c)
Ω : 0 < x2 + y 2 < 1
29. F(x, y, z) = 0 i + 0 j +
is an open plane region but is not simply connected.
−mGr02 k; (r0 + z)2
∂Q ∂P =0= , ∂y ∂x
∂P ∂R =0= , ∂z ∂x
∂Q ∂R =0= . ∂z ∂y
Therefore, F(x, y, z) is a gradient. ∂f =0 ∂x
=⇒
f (x, y, z) = g(y, z);
∂f ∂g = =0 ∂y ∂y
=⇒
g(y, z) = h(z).
Therefore f (x, y, z) = h(z). Now
∂f −mGr02 = h (z) = ∂z (r0 + z)2
=⇒
f (x, y, z) = h(z) =
30. W = f (x, y, 0) − f (x, y, 300) = mGr0 −
mGr02 r0 + z
mGr0 2 = 279.07 mG. r0 + 300
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SECTION 18.3
SECTION 18.3 1. If f is continuous, then −f is continuous and has antiderivatives u. The scalar fields U (x, y, z) = u(x) are potential functions for F: ∂U ∂U ∂U du i+ j+ k= i = −f i = −F. ∂x ∂y ∂z dx
d 1 d 1 mv 2 = m(v · v) = m(v · a) = v · m a dt 2 dt 2 e = v · F = v · [v × B] = 0 c
∇U = 2.
3. The scalar field U (x, y, z) = αz + d is a potential energy function for F. We know that the total mechanical energy remains constant. Thus, for any times t1 and t2 , 1 2 2 m[v (t1 )]
+ U (r(t1 )) = 12 m[v (t2 )]2 + U (r(t2 )).
This gives 1 2 2 m[v (t1 )]
+ αz(t1 ) + d = 12 m[v (t2 )]2 + αz(t2 ) + d.
Solve this equation for v (t2 ) and you have the desired formula. 4. Throughout the motion, the total mechanical energy of the object remains constant:
At firing v = v0 ,
As r → ∞,
1 GmM mv 2 − = E. 2 r r = Re = the radius of the earth and we have
1 GmM = E. mv0 2 − 2 Re v → 0 (by assumption) and also −GmM/r → 0.
Thus E = 0 and we have 1 GmM mv0 2 = 2 Re
and v0 =
2GM . Re
(Note that v0 is independent of the mass of the projectile.) 5. (a) We know that −∇U points in the direction of maximum decrease of U . Thus F = −∇U attempts to drive objects toward a region where U has lower values. (b) At a point where u has a minimum, ∇U = 0 and therefore F = 0. 6. We have x(0) = 2, x (0) = v(0) = 1. Inserting these values in the formula for E we have 1 E = m + 2λ. 2 Since E = 12 mv2 + 12 λx2 is constant, the maximum value of v comes when x = 0. Then 1 1 E = mv2 = m + 2λ and v = 1 + 4λ/m. 2 2
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SECTION 18.4
931
The maximum value of x comes when v = 0 (at the endpoints of the oscillation). Then E=
1 1 2 λx = m + 2λ 2 2
and x =
m/λ + 4.
7. (a) By conservation of energy 12 mv2 + U = E. Since E is constant and U is constant, v is constant. (b) ∇U is perpendicular to any surface where U is constant. Obviously so is F = −∇U . 8. F(r) =
k r = ∇f r2
9. f (x, y, z) = −
where
f (r) = k ln r
k x2
+ y2 + z2
along C is:
is a potential function for F. The work done by F moving an object
F(r) · dr =
W =
b
∇f · dr = f [r(b)] − f [r(a)].
a
C
Since r(a) = (x0 , y0 , z0 ) and r(b) = (x1 , y1 , z1 ) are points on the unit sphere, f [r(b)] = f [r(a)] = −k
and so W = 0
SECTION 18.4 1. r(u) = u i + 2u j, u ∈ [ 0, 1 ] 1 (x − 2y) dx + 2x dy = {[x(u) − 2y(u)]x (u) + 2x(u) y (u)} du = C
2. r(u) = ui + 2u2 j,
0
u ∈ [0, 1]
C
1
{[x(u) − 2y(u)] x (u) + 2x(u)y (u)} du
0
=
1
(u + 4u2 ) du =
0
11 6
3. C = C1 ∪ C2 C1 : r(u) = u i, u ∈ [ 0, 1 ]; (x − 2y) dx + 2x dy =
C2 : r(u) = i + 2u j, u ∈ [ 0, 1 ] 1 1 1 x dx = x(u) x (u) du = u du = 2 C1 C1 0 0 1 (x − 2y) dx + 2x dy = 2x dy = 4 du = 4 C C2 0 2 9 = + = 2 C C1 C2
4. C = C1 ∪ C2 C1 : r(u) = 2u j, u ∈ [0, 1]; C2 : r(u) = u i + 2 j, (x − 2y) dx + 2x dy = 0 dy = 0 C1
u du =
0
(x − 2y) dx + 2x dy =
C1
1
u ∈ [0, 1]
1 2
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SECTION 18.4
(x − 2y) dx + 2x dy = C2
=
C
C2
0
1
y dx + xy dy = C
1
7. C = C1 ∪ C2 C1 : r(u) = u j, y dx + xy dy = 0 y dx + xy dy =
C
y dx =
C2
+
11 6
[y(u)x (u) + x(u)y(u)y (u)] du
1
(2u + 2u2 ) du =
0
(4u2 + 2u3 ) du =
0
=
=
7 2
0
C
C1
1
[y(u) x (u) + x(u) y(u) y (u)] du =
y dx + xy dy =
u ∈ [0, 1]
(u − 4) du = −
0
0
6. r(u) = 2u i + u j,
1
(x(u) − 4)x (u) du =
u ∈ [ 0, 1 ]
5. r(u) = 2u2 i + u j,
C2
7 2
=−
+ C1
(x − 4) dx = C2
1
u ∈ [ 0, 1 ];
1
5 3
C2 : r(u) = 2u i + j,
y(u) x (u) du = 0
u ∈ [ 0, 1 ]
1
2 du = 2 0
=2
C1
C2
8. r(u) = 2u3 i + u j,
u ∈ [0, 1] 1 y dx + xy dy = [y(u)x (u) + x(u)y(u)y (u)] du C
0
1
=
(6u3 + 2u4 ) du =
0
9. r(u) = 2u i + 4u j, u ∈ [ 0, 1 ] 2 2 y dx + (xy − x ) dy = C
19 10
y 2 (u)x (u) + x(u)y(u) − x2 (u) y (u) du
1
1
0
=
(4u)2 (2) + (8u2 − 4u2 )(4) du =
0
1
48u2 du = 16
0
u ∈ [0, 2]
10. r(u) = u i + u2 j,
y 2 dx + (xy − x2 ) dy = C
2
[y 2 (u)x (u) + (x(u)y(u) − x2 (u))y (u)] du
0
=
0
2
(3u4 − 2u3 ) du =
56 5
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SECTION 18.4 1 2 u i + u j, u ∈ [ 0, 4 ] 8 y 2 dx + (xy − x2 ) dy =
11. r(u) =
C
4
0
y 2 (u)x (u) + x(u)y(u) − x2 (u) y (u) du
4
2
12. C = C1 ∪ C2 C1 : r(u) = 2u i, u ∈ [0, 1]; y 2 dx + (xy − x2 ) dy = 0 dx = 0 C1
C1
y 2 dx + (xy − x2 ) dy = C2
+
C
C1
u
u ∈ [0, 1]
C2 : r(u) = 2 i + 4u j,
1
(2y − 4) dy = C2
=
2 2 u u2 = + (u) − u (1) du 4 8 8 0
4 3 3 104 1 4 u − u du = = 8 64 5 0
[2y(u) − 4]y (u) du =
0
1
16(2u − 1) du = 0
0
=0 C2
u ∈ [ 0, 1 ]
13. r(u) = u i + u j,
(y 2 + 2x + 1) dx + (2xy + 4y − 1) dy C 1 2 = [y (u) + 2x(u) + 1]x (u) + [2x(u)y(u) + 4y(u) − 1]y (u) du 0 1 1 2 2 2 (u + 2u + 1) + (2u + 4u − 1) du = 3u + 6u du = 4 0
0
u ∈ [0, 1]. (y 2 + 2x + 1) dx + (2xy + 4y − 1) dy
14. r(u) = ui + u2 j,
C 1
=
(y 2 (u) + 2x(u) + 1)x (u) + (2x(u)y(u) + 4y(u) − 1)y (u) du
0
=
1
(5u4 + 8u3 + 1) du = 4
0 3 15. r(u) = u i + u j, u ∈ [ 0, 1 ] (y 2 + 2x + 1) dx + (2xy + 4y − 1) dy C
[y 2 (u) + 2x(u) + 1]x (u) + [2x(u)y(u) + 4y(u) − 1]y (u) du 0 1 1 6 6 = (u + 2u + 1) + (2u4 + 4u3 − 1)3u2 du = 7u + 12u5 − 3u2 + 2u + 1 du = 4 =
1
0
0
16. C = C1 ∪ C2 ∪ C3 C1 : r(u) = 4u i, u ∈ [0, 1];
C2 : r(u) = 4 i + 2u j, u ∈ [0, 1];
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SECTION 18.4 C3 : r(u) = (4 − 3u)i + (2 − u)j, u ∈ [0, 1] 1 2 = (y + 2x + 1) dx = 4(8u + 1) du = 20 C1
C1
0
(8y + 4y − 1) dy =
= C2
C2
1
2(24u − 1) du = 22
0
−3 (2 − u)2 + 2(4 − 3u) + 1 − [2(4 − 3u)(2 − u) + 4(2 − u) − 1] du
1
= 0
C3
1
=
(−9u2 + 54u − 62) du = −38.
0
=
+
C
C1
= 20 + 22 − 38 = 4.
+ C2
C3
17. r(u) = u i + u j + u k, u ∈ [ 0, 1 ] 1 y dx + 2z dy + x dz = [y(u) x (u) + 2z(u) y (u) + x(u) z (u)] du = C
0
4u du = 2
0
1
y dx + 2z dy + x dz =
18.
1
C
[y(u)x (u) + 2z(u)y (u) + x(u)z (u)] du
0
=
1
(u2 + 3u3 + 4u4 ) du =
0
113 60
19. C = C1 ∪ C2 ∪ C3 C : r(u) = u k, u ∈ [ 0, 1 ]; 1 y dx + 2z dy + x dz = 0
C2 : r(u) = u j + k, u ∈ [ 0, 1 ];
C3 : r (u) = u i + j + k, u ∈ [ 0, 1 ]
C1
y dx + 2z dy + x dz =
C2
C2
y dx + 2z dy + x dz =
C3
=
C
+ C2
2z(u) y (u) du =
0 1
y dx =
y(u) x (u) du = 0
C3
+ C1
1
2z dy =
1
2 du = 2 0 1
du = 1 0
=3 C3
20. C = C1 ∪ C2 ∪ C3 C : r(u) = u i, u ∈ [0, 1]; 1 y dx + 2z dy + x dz = 0
C2 : r(u) = i + u j,
C1
y dx + 2z dy + x dz = 0 C2
y dx + 2z dy + x dz =
C3
x dz =
C3
du = 1 0
y dx + 2z dy + x dz = 0 + 0 + 1 = 1 C
1
u ∈ [0, 1];
C3 : r(u) = i + j + u k,
u ∈ [0, 1]
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SECTION 18.4 21. r(u) = 2u i + 2u j + 8u k, u ∈ [ 0, 1 ] xy dx + 2z dy + (y + z) dz C
1
=
{x(u)y(u)x (u) + 2z(u)y (u) + [y(u) + z(u)]z (u)} du
0
1
=
[(2u)(2u)(2) + 2(8u)(2) + (2u + 8u)(8)] du 0
1
=
176 8u2 + 112u du = 3
0
22. C = C1 ∪ C2 ∪ C3 C1 : r(u) = 2u i,
u ∈ [0, 1];
u ∈ [0, 1];
C2 : r(u) = 2 i + 2u j,
C3 : r(u) = 2 i + 2 j + 8u k, u ∈ [0, 1] xy dx + 2z dy + (y + z) dz = 0 C1
xy dx + 2z dy + (y + z) dz = 0 C2
xy dx + 2z dy + (y + z) dz =
C3
C2
23. r(u) = u i + u j + 2u2 k,
1
(y + z) dz =
8(2 + 8u) du = 48 0
u ∈ [ 0, 2 ]
xy dx + 2z dy + (y + z) dz C
2
=
{x(u)y(u)x (u) + 2z(u)y (u) + [y(u) + z(u)]z (u)} du
0
(u)(u)(1) + 2(2u2 )(1) + (u + 2u2 )(4u) du
2
2
= 0
=
8u3 + 9u2 du = 56
0
24. C = C1 ∪ C2 C1 : r(u) = 2u i + 2u j + 2u k, u ∈ [0, 1]; C2 : r(u) = 2 i + 2 j + (2 + 6u) k, u ∈ [0, 1]. 1 32 xy dx + 2z dy + (y + z) dz = 8(u2 + 2u) du = 3 C1 0 1 xy dx + 2z dy + (y + z) dz = (y + z) dz = 6(4 + 6u) du = 42 C2
=
C
C2
+
C1
C2
158 = 3
25. r(u) = (u − 1) i + (1 + 2u2 ) j + u k, u ∈ [ 1, 2 ] x2 y dx + y dy + xz dz C
0
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SECTION 18.4
x2 (u)y(u)x (u) + y(u)y (u) + x(u)z(u)z (u) du
2
2
2
= 1
=
(u − 1)2 (1 + 2u2 )(1) + (1 + 2u2 )(4u) + (u − 1)u du
1
=
1177 2u4 + 4u3 + 4u2 + u + 1 du = 30
1
u2 2
u2 j + u k, u ∈ [0, 2] 4 2 u2 π 8 u2 u2 2 2 + (2 − u ) + u(1 − ) du = − − −u 1 − y dx + yz dy + z(x − 1) dz = 4 2 2 2 15 C 0
26. r(u) =
27. (a)
2−
i+u
1−
∂P ∂Q = 6x − 4y = ∂y ∂x ∂f 1 = x2 + 6xy − 2y 2 =⇒ f (x, y) = x3 + 3x2 y − 2xy 2 + g(y) ∂x 3 ∂f = 3x2 − 4xy + g (y) = 3x2 − 4xy + 2y =⇒ g (y) = 2y =⇒ g(y) = y 2 + C ∂y 1 Therefore, f (x, y) = x3 + 3x2 y − 2xy 2 + y 2 (take C = 0) 3 (0,4)
(b) (i) C
(x2 + 6xy − 2y 2 ) dx + (3x2 − 4xy + 2y) dy = [f (x, y)](3,0) = 7
(ii) C
(0,3)
(x2 + 6xy − 2y 2 ) dx + (3x2 − 4xy + 2y) dy = [f (x, y)](4,0) = −
37 3
28. (a) F = ∇f where f (x, y, z) = x2 y + xz 2 − y 2 z (b) (i) (2xy + z 2 ) dx + (x2 − 2yz) dy + (2xz − y 2 ) dz = f (3, 2, −1) − f (1, 0, 1) = 25 − 1 = 24 C (ii) (2xy + z 2 ) dx + (x2 − 2yz) dy + (2xz − y 2 ) dz = f (1, 0, 1) − f (3, 2 − 1) = −24 C
[x (u)]2 + [y (u)]2 = a π/2 2 k(x + y) ds = k [x(u) + y(u)] s (u) du = ka (a) M =
29. s (u) =
C
0
xM M =
0
(cos u + sin u) du = 2ka2
0 π/2
kx(x + y) ds = k C
π/2
3
x(u) [x(u) + y(u)] s (u) du
π/2
= ka
(cos2 u + cos u sin u) du =
0
yM M =
π/2
ky(x + y) ds = k C
0
3
= ka
0
xM = yM =
1 8 a(π
+ 2)
1 3 ka (π + 2) 4
y(u) [x(u) + y(u)] s (u) du
π/2
(sin u cos u + sin2 u) du =
1 3 ka (π + 2) 4
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SECTION 18.4
(b)
π/2
k(x + y)y 2 ds = k
I= C
x(u)y 2 (u) + y 3 (u) s (u) du
0
= ka4
sin2 u cos u + sin3 u du
π/2
π/2
0
= ka4 = ka4
sin2 u cos u + (1 − cos2 u) sin u du
0
sin3 u − cos u +
1 3
1 3
π/2
cos3 u
0
= ka4
I = 12 a2 M . M 2 M a2 2π 1 cos2 u du = M a2 x ds = 30. (a) I = 2π 0 2 C L M 2 Ma a ds = (b) I = ds = M a2 L 2π C C
k(x + y)a2 ds = a2
31. (a) Iz =
k(x + y) ds = a2M = Ma2
C
C ∗
√ (b) The distance from a point (x , y ∗ ) to the line y = x is |y ∗ − x∗ |/ 2. Therefore
π/2 1 1 2 k(x + y) (a cos u + a sin u)(a sin u − a cos u)2 a du (y − x) ds = k I= 2 2 0 C π/2 d 1 (sin u − cos u) du (sin u − cos u)2 = ka4 2 du 0
π/2 1 1 1 = ka4 . = ka4 (sin u − cos u)3 2 3 3 0 M = 2ka2 .
From Exercise 29,
Therefore I = 16 (2ka2 )a2 = 16 Ma2 .
k ds =
32. (a) M = C
(b)
2π
k
sin u + (1 − 2
cos u)2
2π
du =
0
0
xM M =
1 (1 − cos u)(2k sin u) du 2
2π
kx ds = 0
C
2π
= 4k 0
yM M =
2π 0
= 2k 0
= 8πk;
1 32 k; sin3 u du = 2 3
xM =
4 3
1 (u − sin u)(2k sin u) du 2
ky ds = C
1 2k sin u du = 8k 2
2π
1 1 1 (u sin u − 2 sin2 u cos u) du 2 2 2 yM = π
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January 4, 2007
SECTION 18.4
√ 33. (a) s (u) = a2 + b2 2π L= ds = a2 + b2 du = 2π a2 + b2 C
(b) xM = 0,
0
(by symmetry) 2π 1 1 √ zM = z ds = bu a2 + b2 du = bπ L C 2π a2 + b2 0 M 2 M 2π 2 2 1 (a sin u + b2 u2 ) du = M (3a2 + 8b2 π 2 ) (y + z 2 ) ds = (c) Ix = L 2π 6 0 C
yM = 0
Iy = 16 M (3a2 + 8b2 π 2 ) Iz = Ma2
similarly
(all the mass is at distance a from the z-axis)
34. (a) s (u) = 2u2 + 1 a 2 a(2a2 + 3) ds = (2u2 + 1) du = a3 + a = L= 3 3 C 0 a 1 3 3a(a2 + 1) 3 (b) xM = x ds = (2u + u) du = L C a(2a2 + 3) 0 2(2a2 + 3) a 1 3 a2 (6a2 + 5) y ds = (2u4 + u2 ) du = yM = 2 L C a(2a + 3) 0 5(2a2 + 3) a 4 5 2 3 1 3 a3 (4a2 + 3) z ds = + du = u u zM = L C a(2a3 + 3) 0 3 3 6(2a3 + 3) (c)
M Iz = L
3M (x + y ) ds = a(2a3 + 3) C 2
= 35.
2
a
[(u2 + u4 )(2u2 + 1)] du
0
M a2 (30a4 + 63a2 + 35) 35(2a2 + 3)
k(x2 + y 2 + z 2 ) ds
M = C
=k
a2 + b2
2π
(a2 + b2 u2 ) du =
0
36. C : r = r(u),
2 2 πk a + b2 (3a2 + 4π 2 b2 ) 3
u ∈ [a, b]
h(r) · dr =
b
a
C
b
= a
b
=
[h(r(u)) · r (u)] du
r (u) r (u) du h(r(u)) · r (u)
[h(r(u)) · T(r(u))]s (u) du
a
[h(r) · T(r)] ds
= C
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SECTION 18.5 SECTION 18.5 1. (a) xy dx + x2 dy = C
xy dx + x2 dy +
xy dx + x2 dy +
C1
xy dx + x2 dy, where
C2
C1 : r(u) = u i + u j, u ∈ [ 0, 1 ];
C3
C2 : r(u) = (1 − u) i + j, u ∈ [ 0, 1 ]
C3 : r(u) = (1 − u) j, u ∈ [ 0, 1 ]. 1 2 2 xy dx + x dy = (u2 + u2 ) du = 3 C1 0 1 1 xy dx + x2 dy = −(1 − u) du = − 2 C2 0 1 xy dx + x2 dy = 02 (−1) du = 0 C3
0
1 2 1 Therefore, xy dx + x2 dy = − = . 3 2 6 C 1 y 2 (b) xy dx + x dy = x dx dy = x dx dy = C
Ω
0
0
1
0
1 2 x 2
y 0
1 du = 2
1
y 2 dy =
0
1 6
2. (a) C = C1 ∪ C3 ∪ C3 ∪ C4 C1 : r(u) = u i, u ∈ [0, 1];
C2 : r(u) = i + u j, u ∈ [0, 1]
C3 : r(u) = (1 − u)i + j, u ∈ [0, 1]; x2 y dx + 2y 2 dy = 0 C1
x2 y dx + 2y 2 dy = C2
x2 y dx + 2y 2 dy =
C3
C4
= C1
2y 2 dy =
C4
+
=−
+
C2
C3
1
C4
u ∈ [0, 1]
2 3
−(1 − u)2 du = −
1 3
−2(1 − u)2 du = −
0
+
1
0
x2 y dx + 2y 2 dy =
2u2 du =
0
x2 dx = C3
1
2y 2 dy = C2
C
C4 : r(u) = (1 − u)j,
2 3
1 3
1 1 ∂ ∂ 2 1 x2 y dx + 2y 2 dy = (b) (2y 2 ) − (x y) dx dy = −x2 dx dy = − ∂x ∂y 3 C 0 0 Ω
3. (a) C : r(u) = 2 cos u i + 3 sin u j, u ∈ [ 0, 2π ] (3x2 + y) dx + (2x + y 3 ) dy C
(12 cos2 u + 3 sin u)(− 2 sin u) + (4 cos u + 27 sin3 u)3 cos u du
2π
2π
= 0
=
−24 cos2 u sin u − 6 sin2 u + 12 cos2 u + 81 sin3 u cos u du
0
= 8 cos3 u − 3u +
2π 3 81 = 6π sin 2u + 6u + 3 sin 2u + sin4 u 0 2 4
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January 4, 2007
SECTION 18.5 (3x2 + y) dx + (2x + y 3 ) dy = (b) 1 dx dy = area of ellipse Ω = 6π C
Ω
4. (a) C = C1 ∪ C2 C1 : r(u) = u i + u2 j, u ∈ [0, 1]; C2 : r(u) = (1 − u)i + (1 − u)j, u ∈ [0, 1] 1 7 y 2 dx + x2 dy = (u4 + 2u3 ) du = 10 0 C1 1 2 1 y 2 dx + x2 dy = −2(1 − u)2 du = − ; = + = 3 30 C2 C C1 C2 0
1 x ∂ 2 ∂ 2 1 y 2 dx + x2 dy = (b) (x ) − (y ) dx dy = 2(x − y) dy dx = ∂x ∂y 30 2 C 0 x Ω
5. 3y dx + 5x dy = (5 − 3) dxdy = 2A = 2π C
Ω
6. 5x dx + 3y dy = 0 dx dy = 0 C
Ω
a x2 dy = 7. (ab) = a2 b 2x dxdy = 2xA = 2 2 C Ω
8. y 2 dx = −2y dx dy = −ab2 C
Ω
[(2y + 10x) − (3x + 2y)] dxdy (3xy + y 2 ) dx + (2xy + 5x2 ) dy =
9.
C
Ω
=
7x dxdy = 7 xA = 7(1)(π) = 7π Ω
10. (xy + 3y 2 ) dx + (5xy + 2x2 ) dy = (3x − y) dx dy = (3x − y)A = (3 + 2)π = 5π. C
Ω
2 2 2 2 11. (2x + xy − y ) dx + (3x − xy + 2y ) dy = [(6x − y) − (x − 2y)] dxdy C
Ω
(5x + y) dxdy = (5x + y)A = (5a + 0)(πr2 ) = 5aπr2
= Ω
(x2 − 2xy + 3y 2 ) dx + (5x + 1) dy = 12. (5 + 2x − 6y) dx dy = (5 + 2x − 6y)A = (5 − 6b)πr2 C
Ω
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SECTION 18.5 ex sin y dx + ex cos y dy = 13. [ex cos y − ex cos y] dxdy = 0 C
Ω
14. ex cos y dx + ex sin y dy = 2ex sin y dx dy = C
1
0
Ω
π
2ex sin y dy dx = 4(e − 1)
0
2 15. 2xy dx + x dy = [2x − 2x] dxdy = 0 C
Ω
16. y 2 dx + 2xy dy = 0 dx dy = 0 C
Ω
17. C : r(u) = a cos u i + a sin u j; u ∈ [ 0, 2π ] 2π −y dx = A= (−a sin u)(−a sin u) du = a2 C
0
2π
sin2 u du = a2
0
18. C : r(u) = a cos3 u i + a sin3 u j, u ∈ [0, 2π] 2π (−a sin3 u)(−3a cos2 u sin u) du = 3a2 A= −y dx = C
0
2π
1 1 u − sin 2u 2 4
sin4 u cos2 u du =
0
2π = π a2 0
3 2 πa 8
19. A = x dy, where C = C1 ∪ C2 ; C
4 C1 : r(u) = u i + j, 1 ≤ u ≤ 4; C2 : r(u) = (4 − 3u) i + (1 + 3u) j, 0 ≤ u ≤ 1. u 4 4 −4 1 du = −4 ln 4; du = −4 u x dy = 2 u u C1 1 1 1 1 15 x dy = . (4 − 3u)3 du = (12 − 9u) du = 2 C2 0 0 Therefore, 20. A =
A=
15 2
− 4 ln 4.
1 x dy − y dx, where C = C1 ∪ C2 ; 2 C √ √ √ √ C1 : r(u) = 5 tan u i + 5 sec u j, tan−1 −2/ 5 ≤ u ≤ tan−1 2/ 5 C2 : (2 − 4u) i + 3 j, 0 ≤ u ≤ 1 1 5 1 x dy − y dx = ln 5, x dy − y dx = 6 2 C1 2 2 C2
Therefore,
21. (ay + b) dx + (cx + d) dy = (c − a) dxdy = (c − a)A C
Ω
15 22. F(r) · d r = (−5) dx dy = −5A = − πa2 (by Exercise 18) 8 C Ω
A=6+
5 2
ln 5.
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SECTION 18.5
23. We take the arch from x = 0 to x = 2πR. (Figure 9.11.1) Let C1 be the line segment from (0, 0) to (2πR, 0) and let C2 be the cycloidal arch from (2πR, 0) back to (0, 0). Letting C = C1 ∪ C2 , we have A = x dy = x dy + x dy = 0 + x dy C
C1
0
=
C2
C2
R(θ − sin θ)(R sin θ) dθ
2π
2π
= R2 = R2
(sin2 θ − θ sin θ) dθ
0
θ sin 2θ − + θ cos θ − sin θ 2 4
2π = 3πR2 . 0
3 3 2 2 24. y dx + (3x − x ) dy = (3 − 3x − 3y ) dx dy = 3 (1 − x2 − y 2 ) dx dy C
Ω
Ω
The double integral is maximized by Ω : 0 ≤ x2 + y 2 ≤ 1. (This is the maximal region on which the integral is nonnegative.) The line integral is maximized by the unit circle traversed counterclockwise. 25. Taking Ω to be of type II (see Figure 18.5.2), we have d ψ2 (y) ∂Q ∂Q (x, y) dxdy = (x, y) dx dy ∂x ∂x c ψ1 (y) Ω
d
=
{Q[ψ2 (y), y] − Q[ψ1 (y), y]} dy
c
(∗) =
d
d
Q[ψ2 (y), y] dy −
c
Q[ψ1 (y), y] dy. c
The graph of x = ψ2 (y) from x = c to x = d is the curve u ∈ [c, d ].
C4 : r4 (u) = ψ2 (u) i + u j, The graph of x = ψ1 (y) from x = c to x = d is the curve
u ∈ [c, d ].
C3 : r3 (u) = ψ1 (u) i + u j, Then
Q(x, y) dy = C
Q(x, y) dy −
C4
=
d
Q(x, y) dy C3
d
Q[ψ2 (u), u] du −
c
Q[ψ1 (u), u] du. c
Since u is a dummy variable, it can be replaced by y. Comparison with (∗) gives the result. 26. Let h(r) = f (r)∇g(r) + g(r)∇f (r). Then h = ∇(f g)
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SECTION 18.5
943
27. Suppose that f is harmonic. By Green’s theorem, 2 ∂f ∂ f ∂f ∂2f − 2 − 2 dx − dy = dxdy = 0 dxdy = 0. ∂x ∂ x ∂ y C ∂y Ω Ω
28. −
λ 3 λ y dx = − (−3y 2 ) dx dy = λy 2 dx dy = Ix 3 3 Ω
Ω
λ 3 λ 2 x dy = 3x dx dy = λx2 dx dy = Iy 3 3 Ω
Ω
= + 29. C1
C2
C3
30. Let Ω be the region enclosed by C. Then f (x) dx + g(y) dy f (x) dx + g(y) dy = ± C
C
=±
0
∂ ∂ [g(y)] − [f (x)] ∂x ∂y
dx dy = 0
Ω
31.
∂P −2xy ∂Q = 2 = 2 2 ∂y (x + y ) ∂x
except at
(0, 0)
(a) If C does not enclose the origin, and Ω is the region enclosed by C, then x y dx + dy = 0 dxdy = 0. 2 2 x2 + y 2 C x +y Ω
(b) If C does enclose the origin, then
= C
where Ca : r(u) = a cos u i + a sin u j, In this case
= C
2π
0
u ∈ [ 0, 2π ]
Ca
is a small circle in the inner region of C.
2π a cos u a sin u (−a sin u) + (a cos u) du = 0 du = 0. a2 a2 0
The integral is still 0. 32. (a) −
y3 xy 2 dx + 2 dy = 0 dy dx = 0 (x2 + y 2 )2 (x + y 2 )2 C Ω (b) By Green’s theorem, = , where C is a circle about the origin. = C
0
C 2π
C
(sin4 u + sin2 u cos2 u) du = 0
2π
sin2 u du = π
r(u) = a cos u i + a sin u j.
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SECTION 18.6
33. If Ω is the region enclosed by C, then ∂φ ∂ ∂φ ∂φ ∂ ∂φ dx + dy = − dxdy v · dr = ∂y ∂x ∂y ∂y ∂x C C ∂x Ω
=
0 dxdy = 0. Ω
equality of mixed partials 34. r(u) = [x1 + (x2 − x1 )u]i + [y1 + (y2 − y1 )u]j, u ∈ [0, 1] 1 −y dx + x dy = {[−y1 − (y2 − y1 )u] (x2 − x1 ) + [x1 + (x2 − x1 )u] (y2 − y1 )} du C
0
=
1
(x1 y2 − x2 y1 ) du = x1 y2 − x2 y1 .
0
1 (−y dx + x dy) 2 C
= + +···
35. A =
C1
C2
Cn
Now (−y dx + x dy) = Ci
1
{[yi + u(yi+1 − yi )] (xi+1 − xi ) + [xi + u(xi+1 − xi )] (yi+1 − yi )} du
0
= xi yi+1 − xi+1 yi , i = 1, 2, . . . , n; xn+1 = x1 , yn+1 = y1 1 Thus, A = [(x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + · · · + (xn y1 − x1 yn )] 2 7 1 [0 + (8 − 1) + 0] = 2 2 1 (b) A = [0 + (12 − 2) + (12 − 0) + (0 + 6) + 0] = 14 2
36. (a) A =
SECTION 18.6 1. 4[(u2 − v 2 )i − (u2 + v 2 )j
3. 2(j − i)
2. u k
+ 2uv k] 4. sin u sin v i + cos u cos v j + (sin2 u sin2 v − cos2 u cos2 v) k 5. r(u, v) = 3 cos u cos v i + 2 sin u cos v j + 6 sin v k, 6. r(θ, z) = 2 cos θ i + 2 sin θ j + z k,
u ∈ [ 0, 2π ], v ∈ [ 0, π/2 ]
θ ∈ [0, 2π], z ∈ [1, 4].
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SECTION 18.6 7. r(u, v) = 2 cos u cos v i + 2 sin u cos v j + 2 sin v k, 8. r(s, θ) = s cos θ i + s sin θ j + (s cos θ + 2) k,
945
u ∈ [ 0, 2π ], v ∈ ( π/4, π/2 ]
s ∈ [0, 1], θ ∈ [0, 2π].
9. The surface consists of all points of the form (x, g(x, z), z) with (x, z) ∈ Ω. This set of points is given by r(u, v) = u i + g(u, v) j + v k, (y, z) ∈ Γ
10. r(y, z) = h(y, z) i + y j + z k, 11. x2 /a2 + y 2 /b2 + z 2 /c2 = 1; 12. z =
y2 x2 + 2; 2 a b
(u, v) ∈ Ω.
ellipsoid
elliptic paraboloid.
13. x2 /a2 − y 2 /b2 = z;
hyperbolic paraboloid
14. (a) See Exercise 53, Section 14.2. (b)
(c)
15. For each v ∈ [a, b ], the points on the surface at level z = f (v) form a circle of radius v. That circle can be parametrized: R(u) = v cos u i + v sin u j + f (v)k,
u ∈ [ 0, 2π ].
Letting v range over [a, b ], we obtain the entire surface: r(u, v) = v cos u i + v sin u j + f (v)k;
0 ≤ u ≤ 2π,
a ≤ v ≤ b.
16. For the parametrization given in the answer to Exercise 15 N (u, v) = −vf (v) cos u i + v j − vf (v) sin u k, N(u, v) = v 2π b Therefore 2 A= v 1 + [f (v)] dv du 0
= 2π a
1 + [f (v)]2 .
a
b
2 v 1 + [f (v)] dv = a
b
2πx 1 + [f (v)]2 dx
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SECTION 18.6
17. Since γ is the angle between p and the xy-plane, γ is the angle between the upper normal to p and k. (Draw a figure.) Therefore, by 18.6.5, area of Γ = sec γ dxdy = (sec γ)AΩ = AΩ
sec γ.
Ω
γ is constant 18. n = i + j + k
is an upper normal.
n·k 1 cos γ = √ = √ , 3 3
√
sec γ =
3
A=
√ 3πb2
19. The surface is the graph of the function x y c f (x, y) = c 1 − − = (ab − bx − ay) a b ab defined over the triangle Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ b(1 − x/a). Note that Ω has area 12 ab. A=
[fx (x, y)]2 + [fy (x, y)]2 + 1 dxdy
Ω
=
c2 /a2 + c2 /b2 + 1 dxdy
Ω
1 2 2 = a b + a2 c2 + b2 c2 ab
dx dy =
1 2 2 a b + a2 c2 + b2 c2 . 2
Ω
20. f (x, y) = x2 + y 2 , Ω : 0 ≤ x2 + y 2 ≤ 1 √ √ A= [fx (x, y)]2 + [fy (x, y)]2 + 1 dx dy = 2 dx dy = 2π Ω
Ω
21. f (x, y) = x2 + y 2 ,
Ω : 0 ≤ x2 + y 2 ≤ 4
A=
4x2 + 4y 2 + 1 dx dy
Ω
2π
= 0
= 2π 22. f (x, y) =
2xy,
2
4r2 + 1 r dr dθ
0 1 2 12 (4r
[ change to polar coordinates ]
+ 1)3/2
2 0
√ = 16 π(17 17 − 1)
Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ b x+y 1 √ A= dx dy = √ ( x/y + y/x) dx dy 2xy 2 Ω
Ω
a b 1 = √ x/y + y/x dy dx 2 0 0 √ √ 2 = 2(a + b) ab 3
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SECTION 18.6 23. f (x, y) = a2 − (x2 + y 2 ), A=
947
Ω : 14 a2 ≤ x2 + y 2 ≤ a2
Ω
4x2 + 4y 2 + 1 dxdy
[ change to polar coordinates ]
a 1 4r2 + 1 dr dθ = 2π (4r2 + 1)3/2 12 0 a/2 a/2
π = (4a2 + 1)3/2 − (a2 + 1)3/2 6
=
2π
a
r
1 24. f (x, y) = √ (x + y)3/2 , Ω : 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 − x 3 2 2−x 1 464 1 √ 3x + 3y + 2 dx dy = √ 3x + 3y + 2 dy dx = A= 135 2 2 0 0 25. f (x, y) = 13 (x3/2 + y 3/2 ), A= Ω
Ω : 0 ≤ x ≤ 1,
0≤y≤x
1 x + y + 4 dxdy 2
x 1 (x + y + 4)3/2 dx 3 0 0 0 0
1 1
1 1 1 2 (2x + 4)3/2 − (x + 4)3/2 dx = (2x + 4)5/2 − (x + 4)5/2 = 3 5 5 0 3 0 √ √ 1 (36 6 − 50 5 + 32) = 15
=
1
x
1 x + y + 4 dy dx = 2
Ω : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 1 4y 2 + 1 dx dy = A=
1
26. f (x, y) = y 2 ,
0
Ω
1
4y 2 + 1 dy dx =
0
√ 1 √ 2 5 + ln(2 + 5) 4
27. The surface x2 + y 2 + z 2 − 4z = 0 is a sphere of radius 2 centered at (0, 0, 2): x2 + y 2 + z 2 − 4z = 0
⇐⇒
x2 + y 2 + (z − 2)2 = 4.
The quadric cone z 2 = 3(x2 + y 2 ) intersects the sphere at height z = 3: x2 + y 2 + z 2 − 4z = 0 z 2 = 3(x2 + y 2 )
3(x2 + y 2 ) + 3z 2 − 12z = 0 4z 2 − 12z = 0
=⇒ z = 3.
(since z ≥ 2)
The surface of which we are asked to find the area is a spherical segment of width 1 (from z = 3 to z = 4) in a sphere of radius 2. The area of the segment is 4π. (Exercise 27, Section 9.9.) A more conventional solution. The spherical segment is the graph of the function f (x, y) = 2 + 4 − (x2 + y 2 ), Ω : 0 ≤ x2 + y 2 ≤ 3.
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SECTION 18.6 Therefore
2 2 −y ! −x + + 1 dxdy 4 − x2 − y 2 4 − x2 − y 2
A= Ω
2 dxdy 4 − (x2 + y 2 )
= Ω
2π
√
3
2r dr dθ 4 − r2 0 0
√ 3 = 2π −2 4 − r2 = 4π =
√
[ changed to polar coordinates ]
0
28. The spherical segment is the graph of the function f (x, y) = a + a2 − (x2 + y 2 ), Therefore
A= Ω
2π
a a2
√
− (x2 + y 2 )
2ab−b2
√
= 0
0
Ω ≤ x2 + y 2 ≤ 2ab − b2 .
dx dy
[change to polar coordinate]
ar dr dθ − r2
a2
= 2πab.
"
29. (a)
2
∂g (y, z) ∂y
∂g (y, z) + ∂z
2
+ 1 dydz =
sec [α(y, z)] dydz
Ω
Ω
where α is the angle between the unit normal with positive i component and the positive x-axis
"
(b)
∂h (x, z) ∂x
2
2
∂h (x, z) + ∂z
+ 1 dxdz =
sec [ β(x, z)] dxdz
Ω
Ω
where β is the angle between the unit normal with positive j component and the positive y-axis 30. (a) (b)
N(u, v) = ru × rv = a cos u i + a sin u j ru = −a sin u i + a cos u j; rv = k 2π l A= N(u, v) du dv = a du dv = a dv du = 2πla Ω
31. (a)
0
Ω
N(u, v) = v cos u sin α cos α i + v sin u sin α cos α j − v sin2 α k
(b)
0
N(u, v) dudv =
A= Ω
v sin α dudv Ω
2π
= 0
0
s
v sin α dv du = πs2 sin α
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SECTION 18.6 32. (a) Set x = a cos u sin v, y = a sin u sin v, z = b cos v. Then x2 y2 z2 + 2 + 2 =1 2 a a b (b)
(c) N(u, v) = −ab cos u sin2 v i − ab sin u sin2 v j − a2 sin v cos v k,
2π
N(u, v) du dv =
A=
0
Ω
π
a2 b2 sin4 v + a4 sin2 v cos2 v dv du
0
π
= 2πa
sin v
b2 sin2 v + a2 cos2 v dv
0
33. (a) Set x = a cos u cosh v,
y = b sin u cosh v, 2
z = c sinh v. Then, 2
x y z2 + − = 1. a2 b2 c2 (b)
N(u, v) dv du
(c) A = Ω
= 0
2π
ln 2
− ln 2
64 cos2 u cosh2 v + 144 sin2 u cosh2 v + 36 cosh2 v sinh2 v dv du
34. (a) Set x = a cos u sinh v,
y = b sin u sinh v,
z = c cosh v. Then,
x2 y2 z2 + − = −1. a2 b2 c2
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SECTION 18.7 (b)
(c) Assuming c > 0, z = c cosh v > 0 for all v. 35. A =
A1 2 + A 2 2 + A 3 2 ;
the unit normal to the plane of Ω is a vector of the form cos γ1 i + cos γ2 j + cos γ3 k.
Note that A1 = A cos γ1 ,
A2 = A cos γ2 ,
A3 = A cos γ3 .
Therefore A1 2 + A2 2 + A3 2 = A2 [cos2 γ1 + cos2 γ2 + cos2 γ3 ] = A2 . 36. We can parametrize the surface by setting R(r θ) = r cos θ i + r sin θ j + f (r, θ) k,
(r, θ) ∈ Ω.
The integrand is N(r, θ) . 37. (a) (We use Exercise 36.) f (r, θ) = r + θ; Ω : 0 ≤ r ≤ 1, 0 ≤ θπ r2 [fr (r, θ)]2 + [fθ (r, θ)]2 + r2 drdθ = 2r2 + 1 drdθ A= Ω
π
Ω 1
2r2 + 1 dr dθ =
= 0
0
Ω : 0 ≤ r ≤ a,
θ
f (r, θ) = re ;
(b)
r
A=
√
1√ 2π 4
√ √ 6 + ln ( 2 + 3)
0 ≤ θ ≤ 2π
2π
2e2θ + 1 drdθ =
2e2θ + 1 dθ
0
Ω
a
r dr 0
√ √ √ √ = 12 a2 [ 2e4π + 1 − 3 + ln (1 + 3) − ln (1 + 2e4π + 1)] 38. r(u, v) = x(u, v) i + y(u, v) j, (u, v) ∈ Ω Straightforward calculation shows that N(u, v) = |J(u, v)|. SECTION 18.7 For Exercises 1–6 we have sec [ γ(x, y)] = 1.
1
y 2 + 1 dx dy =
dσ = S
1
0
0
y 2 + 1. 1
N(x, y) = −yj + k, so N (x, y) =
y 2 + 1 dy =
0
√ 1√ [ 2 + ln (1 + 2)] 2
y 2 + 1.
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SECTION 18.7
2.
1
2
1
x2
x dσ = 0
y 2 + 1 dy dx
0
S
1
=
2
x dx 0
3.
1
1
3y dσ =
3y 0
S
y2
+ 1 dy
√ 1 √ 2 + ln(1 + 2) 6
=
0
y2
1
1
+ 1 dy dx =
3y
0
1 √ y 2 + 1 dy = (y 2 + 1)3/2 = 2 2 − 1 0
0
1
(x − y) dσ =
4.
0
x y 2 + 1 dy dx −
0
S
1
1
=
x dx
0
1
y2
1
y
0 1
0
y 2 + 1 dy dx
1
+ 1 dy −
y
0
√ 1√ 1 √ [ 2 + ln(1 + 2)] − (2 2 − 1) 4 3 √ 1 1 5√ = − 2 + ln(1 + 2) 3 12 4
y 2 + 1 dy
0
=
5.
√
2z dσ =
y dσ =
S
6.
1 √ (2 2 − 1) 3
(Exercise 3)
S
1 + y 2 dσ = 0
S
1
1
(1 + y 2 ) dy dx =
0
1
(1 + y 2 ) dy =
0
4 3
7.
xy dσ;
S : r(u, v) = (6 − 2u − 3v) i + u j + v k,
0≤u≤3−
3 2
S
N(u, v) = (−2 i + j) × (−3 i + k) = xy dσ =
√
0≤v≤2
v, √ 14
14
x(u, v)y(u, v) du dv Ω
S
=
√
(6 − 2u − 3v)u du dv
14 Ω
=
√ 14
2
0
3−3v/2
0
√ = 14 3 3 −
8. S
(6u − 2u2 − 3uv) du dv
3 2
v
2
−
2 3
3−
3 2
v
3
−
3 2
v 3−
3 2
v
2
dv =
is given by z = f (x, y) = 1 − x − y, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x 1 1−x xyz dσ = xy(1 − x − y) (−1)2 + (−1)2 + 1 dy dx S
0
0
√ = 3 0
1
0
1−x
√ 3 xy(1 − x − y) dy dx = 120
9√ 14 2
951
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SECTION 18.7 x2 z dσ;
9.
0 ≤ u ≤ π,
S : r(u, v) = (cos u i + v j + sin u k,
0 ≤ v ≤ 2.
S
# # # i # N(u, v) = ##− sin u # # 0
# # k # # cos u## = − cos u i − sin u k # 0 #
j 0 1
2
x z dσ =
π
cos u sin u du dv = 0
Ω
S
10.
2
2
and N(u, v) = 1.
cos2 u sin u du dv =
0
4 3
√ [x2 + y 2 + (x + 2)2 ] 2 dx dy
(x2 + y 2 + z 2 ) dσ = x2 +y 2 ≤1
S
2π
1
= 0
√ [r2 + (r cos θ + 2)2 ] 2 r dr dθ
0 2π
= 0
0
1
√ 19 2 (r + r cos θ + 4r cos θ + 4r) dr dθ = π 4 3
3
2
2
(x2 + y 2 ) dσ;
11.
0 ≤ u ≤ π/2,
S : r(u, v) = cos u cos v i + cos u sin v j + sin u k,
0 ≤ v ≤ 2π.
S
# # i # # N(u, v) = ##− sin u cos v # #− cos u sin v
j − sin u sin v cos u cos v
# # k # # cos u## = − cos2 u cos v i + cos2 u sin v j − sin u cos u k; # 0 #
N(u, v) = cos u.
2
2
2
(x + y ) dσ =
cos u cos u du dv = 0
Ω
S
12.
2π
0
1
r3
2π
0
x2 +y 2 ≤1
= 2π
cos3 u du dv =
(x2 + y 2 ) 4x2 + 4y 2 + 1 dx dy =
(x2 + y 2 ) dσ = S
π/2
4r2 + 1 dr =
0
4 π 3
1
r2
4r2 + 1 r dr dθ
0
√ 25 5 + 1 π 60
For Exercises 13–16 the surface S is given by: f (x, y) = a − x − y; 13. M =
a
λ(x, y, x) dσ = S
0
0
a−x
0 ≤ x ≤ a, 0 ≤ y ≤ a − x √ k 3 dy dx = 0
a
and
sec [γ (x, y)] =
√ √ 1 k 3 (a − x) dx = a2 k 3 2
√ 3.
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SECTION 18.7
14.
0
S
a
a−x
λ(x, y, z) dσ = 0
S
0
a
A=
0
a
(a2 − x2 ) dx =
0 a
1√ 3 3a k 3
√ 1 4 √ a k 3 k 3x2 (a − x) dx = 12
a−x
xdσ = S
√ x 3 dy dx
0
√ a 1√ 3 = 3 (ax − x2 ) dx = 3a 6 0 a a−x √ dσ = 3 dy dx 0
S
= x = xA/A =
17.
0
xA =
√ k(x + y) 3 dy dx
0
√ kx2 3 dy dx =
16.
a−x
1 √ k 3 2
=
15. M =
a
k(x + y) dσ =
M=
0
√ a 1√ 2 3 (a − x) dx = 3a 2 0 1 a; 3
similarly y = z =
S : r(u, v) = a cos u cos v i + a sin u cos v j + a sin v k with 0 ≤ u ≤ 2π,
1 a 3 0 ≤ v ≤ 12 π.
By a previous
calculation N(u, v) = a2 cos v. x = 0,
y=0
(by symmetry) zA = z dσ = z(u, v) N(u, v) dudv = S
z = 12 a
π/2
a3 sin v cos v dv du = πa3
0
A = 2πa2
18. N(u, v) = 2 i + 2 j − 2 k A= dσ = N(u, v) du dv = S
2π 0
Ω
since
1
0
Ω
1
√ √ 2 3 du dv = 2 3
0
19. N(u, v) = (i + j + 2 k) · (i − j) = 2 i + 2 j − 2 k N v· dσ = [v(x(u), y(u), z(u)) · N(u, v)] dudv flux in the direction of N = N S
Ω
[(u + v)i − (u − v)j] · [2 i + 2 j − 2 k] dudv.
= Ω
=
1
4v dudv = 4 Ω
953
1
v dv du = 2 0
0
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SECTION 18.7
20. sec[γ(x, y)] =
x2 + y 2 + 1
1
= = n=
xy 0
S
For Exercises 21–23:
1
kxy dσ = k
M=
1 k 3
x2 + y 2 + 1 dy dx
0
1
x(x2 + 2)3/2 − x(x2 + 1)3/2 dx
0
√ 1 √ (9 3 − 8 2 + 1)k 15
1 (x i + y j + z k) a 0 ≤ u ≤ 2π,
S : r(u, v) = a cos u cos v i + a sin u cos v j + a sin v k with
− 12 π ≤ v ≤ 12 π
N(u, v) = a2 cos v 21. With
v = zk
1 (v · n) dσ = a
flux = S
2π
= a3 0
22. With
1 z dσ = a
2
S
π/2
−π/2
S
Ω
(sin2 v cos v) du dv =
v = xi + yj + zk (v · n) dσ = a flux =
(a2 sin2 v)(a2 cos v) dudv
4 3 πa 3
dσ = aA = 4πa3
S
v = yi − xj
23. With
(v · n) dσ =
flux =
1 a
S
√ (y + z ) dσ = 2 3 2
24. Ix =
2
√ 2 2 (x + z ) dσ = 2 3
S
√ 2 2 (x + z ) dσ = 4 3
Iz =
1
1
0
S Iy =
1
S
(yx − xy) dσ = 0 0
1
√ (5u2 − 2uv + v 2 ) dv du = 3 3
1
√ (5u2 + 2uv + v 2 ) dv du = 5 3
0
0
0
S
0 1
(u2 + v 2 ) dσ =
0
8√ 3 3
For Exercises 25–27 the triangle S is the graph of the function f (x, y) = a − x − y The triangle has area A =
1 2
√
3a2 .
on
Ω : 0 ≤ x ≤ a,
0 ≤ y ≤ a − x.
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SECTION 18.7 25. With
v = xi + yj + zk flux = (v · n) dσ = (−v1 fx − v2 fy + v3 ) dxdy S
Ω
[−x(−1) − y(−1) + (a − x − y)] dxdy = a
=
dxdy = aA =
Ω
26. With
Ω
v = (x + z) k
(v · n) dσ =
flux = S
Ω
a−x
(a − y) dy dx =
0
1 3 a 3
v = x2 i − y 2 j flux = (v · n) dσ = (−v1 fx − v2 fy + v3 ) dxdy S
Ω
a
[−x (−1) − (−y )(−1) + 0] dxdy = 2
=
2
0
Ω a
= 0
a−x
(x2 − y 2 ) dy dx
0
a 1 3 1 4 1 1 ax2 − x3 − (a − x)3 dx = ax − x + (a − x)4 = 0 3 3 4 12 0
v = −xy 2 i + z j
(−v1 fx − v2 fy + v3 ) dx dy
(v · n) dσ =
flux = S
1
= 0
29. With
0
Ω
28. With
a
(a − y) dx dy =
=
(−v1 fx − v2 fy + v3 ) dx dy
27. With
1√ 3 3a 2
Ω 2
(xy 3 − x2 y) dy dx =
0
4 3
v = xz j − xy k
(v · n) dσ =
flux = S
(−v1 fx − v2 fy + v3 ) dxdy
Ω
(−x3 y − xy) dxdy =
=
0
Ω
=
0
1
−2(x3 + x) dx = −
3 2
1
0
2
(−x3 y − xy) dy dx
955
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SECTION 18.7
30. With
v = x2 y i + z 2 k
S
1
0
1 (x i + y j) a
(−x2 y 2 + x2 y 2 ) dy dx = 0
0
(v · n) dσ =
flux =
1 a
[(x i + y j + z k) · (x i + y j)] dσ
S
S
1 = a
2
2
dσ = a (area of S) = a (2πal) = 2πa2 l
(x + y ) dσ = a S
S
32.
r r GmM 3 · dσ = GmM r r
flux = S
1 GmM dσ = a2 a2
= GmM
(v · n) dσ =
GmM (4πa2 ) = 4πGmM a2
0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x
(−v1 fx − v2 fy + v3 ) dx dy.
S
Ω
v = x i − y j + 32 z k flux = (v · n) dσ = (−v1 fx − v2 fy + v3 ) dxdy = 2y 3/2 dxdy S
1
Ω
1−x
= 0
Ω
1
2y 3/2 dy dx =
0
0
4 8 (1 − x)5/2 dx = 5 35
v = x2 i,
1
1−x
flux = 0
35. With
S
2 3/2 (x + y 3/2 ), 3
flux =
34. With
1 dσ r2
S
In Exercises 33–36, S is the graph of f (x, y) = We use
dσ =
S
33. With
(−v1 fx − v2 fy + v3 ) dx dy Ω
2
=
31. n =
(v · n) dσ =
flux =
v = y2 j
(v · n) dσ =
flux = S
1
= 0
0
0
4 . 63
(−v1 fx − v2 fy + v3 ) dxdy =
Ω 1−x
−x5/2 dy dx = −
−y 5/2 dy dx = 0
−y 5/2 dσ Ω
1
2 4 − (1 − x)7/2 dx = − 7 63
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SECTION 18.7 v = yi−
36. With
√ xy j,
1
1−x
flux = 0
(−yx1/2 +
√
957
xyy 1/2 ) dy dx = 0.
0
y=0 by symmetry. You can verify that N(u, v) = v sin α. 2π s zA = z dσ = (s cos α)(v sin α) dudv = sin α cos α v 2 dv du = 23 π sin α cos α s3
37. x = 0,
S
0
Ω
z = 23 s cos α
A = πs2 sin α
since
38.
k
M =
Ω
√ =k 2
2π
x2
+
1
0
39. f (x, y) = xM = 0,
x2 + y 2
yM = 0
0
y2
dσ =
k
√ x2 + y 2 2 dx dy
Ω
r2 dr dθ =
0
on Ω : 0 ≤ x2 + y 2 ≤ 1;
2√ 2πk 3
λ(x, y, z) = k
x2 + y 2
(by symmetry) zλ(x, y, z) dσ = k(x2 + y 2 ) sec [γ(x, y)] dxdy zM M = S
Ω
√ =k 2 (x2 + y 2 ) dxdy Ω
√ =k 2
2π
0
zM =
3 4
since
M=
40. (a)
Ix =
k
2 3
1
r3 dr dθ =
0
√
2πk
1√ 2πk 2
(Exercise 38)
x2 + y 2 (y 2 + z 2 ) dσ =
Ω
k
x2 + y 2 (y 2 + x2 + y 2 ) dσ
Ω
√ 2 2 =k 2 y (x + y 2 )1/2 + (x2 + y 2 )3/2 dx dy Ω
√ =k 2
2π
0
(b) (c)
Iy = Ix
1
(r4 sin2 θ + r4 ) dr dθ =
0
√ 3 2 πk 5
by symmetry Iz =
k
Ω
x2
+
y 2 (x2
2
k(x2 + y 2 )3/2 dσ
+ y ) dσ = S
√ √ =k 2 (x2 + y 2 )3/2 dx dy = k 2 Ω
0
2π
0
1
r4 dr dθ =
2√ 2πk 5
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SECTION 18.7
41. no answer required 42.
√ k(y 2 + z 2 ) dσ = 2 3k [(u − v)2 + 4u2 ] du dv
M = S
√ = 2 3k
1
0
43.
S
1
√ (5u2 − 2uv + v 2 ) dv du = 3 3k
0
xM M =
kx(y 2 + z 2 ) dσ
xλ (x, y, z) dσ =
S S √ = 2 3k (u + v) (u − v)2 + 4u2 dudv Ω
√ = 2 3k
1
0
√ = 2 3k 0
xM =
11 9
1
(5u3 − 2u2 v + uv 2 + 5u2 v − 2uv 2 + v 3 ) dv du
0
1
1 2 5 1 5u3 − u2 + u + u2 − u + 3 2 3 4 √ M = 3 3k
since
2
Iz =
2
k(y 2 + z 2 )(x2 + y 2 ) dσ
λ(x, y, z)(x + y ) dσ = S
√
S
(u − v)2 + 4u2 (u + v)2 + (u − v)2 du dv
= 2 3k Ω
√ = 4 3k
11 √ 3k 3
(Exercise 42)
44.
du =
0
1
0
1
√ 82 3 (5u − 2u v + 6u v − 2uv + v ) du dv = k. 15 4
3
2 2
3
4
45. Total flux out of the solid is 0. It is clear from a diagram that the outer unit normal to the cylindrical side of the solid is given by n = x i + y j in which case v · n = 0. The outer unit normals to the top and bottom of the solid are k and −k respectively. So, here as well, v · n = 0 and the total flux is 0. 46. The flux through the upper boundary is 0: (yi − xj) · k = 0. The flux through the lower boundary is 0: v·
∂f ∂f i+ j−k ∂x ∂y
= (y i − x j) · (2x i + 2y j − k) = (2xy − 2xy) = 0.
Thus the total flux out of the solid is 0.
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SECTION 18.7 47. The surface z =
959
2 − (x2 + y 2 ) is the upper half of the sphere x2 + y 2 + z 2 = 2. The surface intersects
the surface z = x2 + y 2 in a circle of radius 1 at height z = 1. Thus the upper boundary of the solid, √ √ call it S1 , is a segment of width 2 − 1 on a sphere of radius 2. The area of S1 is therefore 2π √ √ 2( 2 − 1). (Exercise 27, Section 9.9). The upper unit normal to S1 is the vector 1 n = √ (x i + y j + z k). 2 Therefore
1 (v · n) dσ = √ 2
flux through S1 = S1
√ = 2
2
(x2 + y 2 + z 2 ) dσ
S1
√ √ dσ = 2 (area of S1 ) = 4π( 2 − 1).
S1
The lower boundary of the solid, call it S2 , is the graph of the function f (x, y) = x2 + y 2
on Ω : 0 ≤ x2 + y 2 ≤ 1.
Taking n as the lower unit normal, we have
(v · n) dσ =
flux through S2 = S2
Ω
v1 fx + v2 fy − v3 dxdy
(x2 + y 2 ) dxdy =
=
2π
0
Ω
1
r3 dr dθ =
0
1 π. 2
√ √ 1 7 The total flux out of the solid is 4π( 2 − 1) + π = (4 2 − )π. 2 2
48.
face
n
v·n
x=0
−i
−xz = 0
0
x=1
i
xz = 1
1 2
y=0
−j
−4xyz 2 = 0
0
y=1
j
4xyz 2 = 4xz 2
2 3
z=0
−k
−2z = 0
0
z=1
k
2z = 2
2
flux
total flux =
1 2
+
2 3
+2=
19 6
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SECTION 18.8
SECTION 18.8
1.
∇ · v = 2,
∇×v = 0
2.
∇ · v = 0,
3.
∇ · v = 0,
∇×v = 0
4.
∇ · v=−
5.
∇ · v = 6,
∇×v = 0
6.
∇ · v = 0,
7. ∇ · v = yz + 1,
4xy , + y 2 )2
∇×v = 0
∇ × v = (2xy − y 2 ) i − y 2 j − x2 k
∇×v = 0
10. ∇ · v = ex (3 + x),
∇ × v = −ex z j + ex yk 2
11. ∇ · v = 2(x + y + z)er , 12. ∇ · v = 0,
(x2
∇ × v = −x i + xy j + (1 − x)z k
8. ∇ · v = 2y(x + z), 9. ∇ · v = 1/r2 ,
∇×v = 0
2
∇ × v = 2er [(y − z)i − (x − z) j + (x − y) k]
2
2 2 ∇ × v = −2 zez i + xex j + yey k
13. ∇ · v = f (x),
∇×v = 0
14. each partial derivative that appears in the curl is 0
15. use components. 16. ∇ · F = −GmM ∇ · (r−3 r) = −GmM (0) = 0 linearity (Exercise 15)
(17.8.8)
∇ × F = −GmM [∇ × (r−3 r)] = −GmM (0) = 0 linearity (Exercise 15)
(17.8.8)
17. ∇ · v =
∂P ∂Q ∂R + + =2+4−6=0 ∂x ∂y ∂z
18. ∇ · v =
∂ ∂ ∂ (3x2 ) + (−y 2 ) + (2yz − 6xz) = 6x − 2y + (2y − 6x) = 0 ∂x ∂y ∂z
∇×v =
2(y 2 − x2 ) k (x2 + y 2 )2
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SECTION 18.8 # #i # # # 19. ∇ × F = ## ∂ # ∂x # #x
j
# # # # ∂ ## = 0 # ∂z # # −2z #
k
∂ ∂y y
20. F(x, y, z) = (2x + y + 2z)i + (x + 4y − 3z)j + (2x − 3y − 6z)k # # # # i j k # # # # # # ∂ ∂ ∂ # ∇ × F = ## # ∂x ∂y ∂z # # # # # 2x + y + 2z x + 4y − 3z 2x − 3y − 6z # = (−3 + 3)i − (2 − 2)j + (1 − 1)k = 0
21. ∇2 f = 12(x2 + y 2 + z 2 )
22. ∇2 f = ∇ · ∇f = ∇ · (yzi + xzj + xyk) = 0
23. ∇2 f = 2y 3 z 4 + 6x2 yz 4 + 12x2 y 3 z 2
24. Note that for r = Then
x2 + y 2 + z 2 ,
∂2 ∂ (cos r) = ∂x2 ∂x
−x sin r r
∂r x = ∂x r =
With similar formulas for y and z.
−r2 sin r − x2 r cos r + x2 sin r , r3
Therefore
∇2 f = =
∂2 ∂2 ∂2 cos r + cos r + cos r ∂x2 ∂y 2 ∂z 2 −3r2 sin r − (x2 + y 2 + z 2 )r cos r + (x2 + y 2 + z 2 ) sin r r3
= − cos r − 2r−1 sin r 25. ∇2 f = er (1 + 2r−1 )
26.
∂2 ∂ x r2 − 2x2 = ln r = , ∂x2 ∂x r2 r4 Then
27. (a)
∇2 f =
2r2
with similar formula for y and z.
3r2 − 2(x2 + y 2 + z 2 ) 1 = 2 r4 r
(b) −1/r
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SECTION 18.8 (u · ∇)r = (u · ∇x)i + (u · ∇y)j + (u · ∇z)k
28. (a)
= (u · i)i + (u · j)j + (u · k)k = u (r · ∇)u = (r · ∇yz)i + (r · ∇xz)j + (r · ∇xy)k
(b)
= [r · (zj + yk)]i + [r · (zi + xk)]j + [r · (yi + xj)]k = (yz + zy)i + (xz + zx)j + (xy + yx)k = 2(yzi + xzj + xyk) = 2u 29. ∇2 f = ∇2 g(r) = ∇ · (∇g(r)) = ∇ · g (r)r−1 r = (∇g (r)) · r−1 r + g (r) ∇ · r−1 r =
g (r)r−1 r · r−1 r + g (r)(2r−1 )
= g (r) + 2r−1 g (r) ∂ ∂ ∂ (f v1 ) + (f v2 ) + (f v3 ) ∂x ∂y ∂z ∂v1 ∂v2 ∂v3 ∂f ∂f ∂f = f + v1 + f + v2 + f + v3 ∂x ∂x ∂y ∂y ∂z ∂z ∂f ∂v1 ∂f ∂f ∂v2 ∂v3 = i+ j+ k · v+f + + ∂x ∂y ∂z ∂x ∂y ∂z
30. (a) ∇ · (f v) =
= (∇f ) · v + f (∇ · v) (b) ∇ × (f v)
∂ ∂ ∂ ∂ ∂ ∂ = (f v3 ) − (f v2 ) i + (f v1 ) − (f v3 ) j + (f v2 ) − (f v1 ) k ∂y ∂z ∂z ∂x ∂x ∂y
∂v3 ∂v2 ∂f ∂f = f + v3 − f − v2 i + etc. ∂y ∂y ∂z ∂z # # # i j k ## # # ∂ ∂ ∂ ## ∂v1 ∂v2 ∂v3 ∂v2 ∂v3 ∂v1 # − i+ − j+ − k (c) ∇ × v = # #= # ∂x ∂y ∂z # ∂y ∂z ∂z ∂x ∂x ∂y # # # v1 v2 v3 # ∂v1 ∂ ∂v1 ∂v3 ∂ ∂v2 − − − i-component of ∇ × (∇ × v) = ∂y ∂x ∂y ∂z ∂z ∂x ∂ 2 v2 ∂ 2 v1 ∂ 2 v1 ∂ 2 v3 − − + 2 2 ∂y∂x ∂y ∂z ∂z∂x 2 ∂ v1 ∂ ∂v2 ∂v3 ∂ 2 v1 = + − + ∂x ∂y ∂z ∂y 2 ∂z 2 =
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SECTION 18.9
963
∂ 2 v1 , we get ∂x2 2 ∂ v1 ∂ ∂v1 ∂ ∂ 2 v1 ∂ 2 v1 ∂v2 ∂v3 = + + + + − (∇ · v) − ∇2 v1 = the i-component of ∇2 v. 2 2 2 ∂x ∂x ∂y ∂z ∂x ∂y ∂z ∂x Adding and subtracting
Equality of the other components can be obtained in a similar manner. 31.
∂f ∂2f = 2; = 2x + y + 2z, ∂x ∂x2
∂f ∂2f = 4; = 4y + x − 3z, ∂y ∂y 2
∂f ∂2f = −6; = −6z + 2x − 3y, ∂z ∂z 2 ∂2f ∂2f ∂2f + + =2+4−6=0 ∂x2 ∂y 2 ∂z 2 32. f (r) = Then
1 . r
1 −r2 + 3x2 ∂ −x = , = 3 r ∂x r r5
∂2 ∂x2
with similar formulas for y and z
−3r2 + 3(x2 + y 2 + z 2 ) = 0. r5
∇2 f =
33. n = −1 34. Since ∇ · (∇f ) = ∇2 f = 0,
the gradient field ∇f is solenoidal.
∇f is irrotational by Theorem 18.8.4
SECTION 18.9
(v · n) dσ =
1.
(∇ · v) dxdydz =
S
T
(v · n) dσ =
2.
3 dxdydz = 3V = 4π T
(∇ · v) dx dy dz =
(−3) dx dy dz = −3V = −4π
S
T
T
(v · n) dσ =
3. S
(∇ · v) dxdydz = T
2(x + y + z) dxdydz. T
The flux is zero since the function f (x, y, z) = 2(x + y + z) satisfies the relation f (−x, −y, −z) = −f (x, y, z) and T is symmetric about the origin.
(v · n) dσ =
4. S
(∇ · v) dx dy dz =
T
(−2x + 2y + 1) dx dy dz =
T
dx dy dz = V = T
by symmetry
4 π 3
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SECTION 18.9 n
v·n
x=0
−i
0
0
x=1
i
1
1
y=0
−j
0
0
y=1
j
1
1
z=0
−k
0
0
z=1
k
1
1
face
5.
flux
total flux = 3
(∇ · v) dxdydz =
T
3 dxdydz = 3V = 3 T
face
n
v·n
x=0
−i
−xy = 0
0
x=1
i
xy = y
1/2
y=0
−j
−yz = 0
0
y=1
j
yz = z
z=0
−k
−xz = 0
z=1
k
xz = x
6.
flux
total flux =
1/2 0 1/2
(∇ · v) dxdydz =
(y + z + x) dxdydz = (y + z + x)V =
T
3 2
1 1 1 + + 2 2 2
T
7.
face
n
v·n
flux
x=0
−i
0
0
x=1
i
1
1
y=0
−j
xz
y=1
j
−xz
z=0
−k
0
0
z=1
k
1
1
fluxes add up to 0
(∇ · v) dxdydz =
T
total flux = 2
2 (x + z) dxdydz = 2 (x + z)V = 2 ( 12 + 12 )1 = 2 T
(1) =
3 . 2
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SECTION 18.9 8.
face
n
v·n
x=0
−i
−x = 0
0
x=1
i
x=1
1
y=0
−j
−xy = 0
0
y=1
j
xy = x
z=0
−k
−xyz = 0
0
z=1
k
xyz = xy
1/4
flux
total flux =
1
1
1
(1 + x + xy) dxdydz =
T
7 4
1/2
(∇ · v) dxdydz =
965
(1 + x + xy) dx dy dz = 0
T
0
0
7 . 4
9. flux = (1 + 4y + 6z) dxdydz = (1 + 4y + 6z)V = (1 + 0 + 3) 9π = 36π T
10.
(∇ · v) dxdydz =
flux =
1−x
1−x−y
(y + z + x) dz dy dx 0
T
1
0
0
1 3 1 = (x + y + z)V = = 4 6 8 11.
(2x + x − 2x) dxdydz
flux =
x dxdydz
T
T
1
1−x
1−x−y
=
x dz dy dx 0
0 1
0 1−x
= 0
x − x2 − xy dy dx
1
0
1−x 1 2 dx xy 2 0 0 1 1 1 1 3 2 = x−x + x dx = 2 2 24 0
xy − x2 y −
=
(∇ · v) dxdydz =
12. flux = T
4y dx dy dz = 4yV = 4(1) T
32 3
=
128 3
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SECTION 18.9
13. flux =
4
2
2π
2(x + y + z) dxdydz =
2(r cos θ + r sin θ + z)r dθ dr dz 0
T
0 4
=
0 2
4π rz dr dz 0
=
0 4
8π z dz = 64π 0
14. flux =
(v · n) dσ =
1 2
(∇ · v) dxdydz =
1 2
S
T
(2 + 2x) dx dy dz = V =
32 π 3
T
15. flux = (2y + 2y + 3y) dxdydz = 7yV = 0 T
(∇ · v) dxdydz =
16. flux =
7y dx dy dz = 7 y V = 7
T
a
a3 =
2
7 4 a 2
T
17. flux = (A + B + C) dxdydz = (A + B + C)V T
(∇f · n) dσ =
18. S
[∇ · (∇f )] dxdydz T
(∇2 f ) dxdydz =
= T
0 dxdydz = 0 T
19. Let T be the solid enclosed by S and set n = n1 i + n2 j + n3 k.
S
(i · n) dσ =
n1 dσ = S
(∇ · i) dxdydz =
T
0 dxdydz = 0. T
Similarly
n2 dσ = 0
S
and
n3 dσ = 0. S
20. (a) The identity follows from setting v = ∇f in (17.8.6). (f fn ) dσ = (f ∇f · n) dσ = [∇ · (f ∇f )] dxdydz S
S
T
=
∇f 2 +f (∇2 f ) dxdydz
T ∇f 2 dxdydz
= T
since
∇2 f = 0
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SECTION 18.9
(b)
S
(g∇f · n) dσ =
(gfn ) dσ = S
[∇ · (g∇f )] dxdydz T
{(∇g · ∇f ) + g[∇ · (∇f )]} dxdydz
= T
[(∇g · ∇f ) + g(∇2 f )] dxdydz
= T
21. A routine computation shows that ∇ · (∇f × ∇g) = 0. Therefore [(∇f × ∇g) · n] dσ = [ ∇ · ( ∇f × ∇g)] dxdydz = 0. S
T
22. Since ∇ · r = 3, we can write 1 r dxdydz = r · n dσ, V = ∇· dxdydz = 3 3 T
T
by the divergence theorem.
S
23. Set F = F1 i + F2 j + F3 k.
[ρ(z − c)i · n] dσ =
F1 = S
[ ∇ · ρ(z − c)i] dxdydz T
= T
∂ [ρ(z − c)] dxdydz = 0. ∂x
Similarly F2 = 0.
[ ρ(z − c)k · n] dσ =
F3 = S
[ ∇ · ρ(z − c)k] dxdydz T
=
∂ [ρ(z − c)] dxdydz ∂z
T
=
ρ dxdydz = W. T
τT ot · i =
24.
{[r × ρ(c − z)n] · i} dσ S
(12.5.6) =−
[(i × r) · ρ(c − z)n] dσ S
(z − c)[(i × r) · n] dσ
=ρ S
divergence theorem
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SECTION 18.10 [∇ · (z − c)(i×r)] dxdydz
=ρ T
i × r = i×(xi + yj + zk) = −zj + yk (z − c)(i×r) = (z − c)(−zj + yk) = (cz − z 2 )j + (yz − cy)k ∇ · (z − c)(i×r) = y τT ot · i = ρ
y dxdydz = ρyV = y(ρV ) = yW T
(r × F) · i = [(xi + yj + zk) × W k] · i = (−xW j + yW i) · i = yW = τT ot · i Equality of the other components can be shown in a similar manner.
PROJECT 18.9 1. For r = 0,
∇ · E = ∇ · qr−3 r = q(−3 + 3)r−3 = 0 by (17.8.8)
2. By the divergence theorem,
(∇ · E) dx dy dz =
flux of E out of S = T
0 dx dy dz = 0 T
r q r r q 3. On Sa , n = , and thus E · n = q 3 · = 2 = 2 r r r r a q q q Thus flux of E out of Sa = (E · n) dσ = dσ = 2 (area of Sa ) = 2 (4πa2 ) = 4πq. 2 a a a Sa
Sa
SECTION 18.10
For Exercises 1–4:
n = xi + yj + zk
S
C : r(u) = cos u i + sin u j,
(0 · n) dσ = 0 S
(b) S is bounded by the unit circle C : r(u) = cos u i + sin u j, $ v(r) · dr = 0 since v is a gradient. C
u ∈ [ 0, 2π ].
[(∇ × v) · n] dσ =
1. (a)
and
u ∈ [ 0, 2π ].
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SECTION 18.10
[(∇ × v) · n] dσ =
2. (a)
(−2k · n) dσ = −2
S
z dσ = −2zA = −2
S
1 2
969
2π = −2π
S
Exercise 17, Section 17.7 $
$
v(r) · dr =
(b) C
C
(− sin2 u − cos2 u) du = −2π
0
[(∇ × v) · n] dσ =
3. (a)
2π
y dx − x dy =
S
(−3y 2 i + 2zj + 2k) · n dσ
S
(−3xy 2 + 2yz + 2z) dσ
= S
(−3xy 2 ) dσ +
= S
2yz dσ +2
S
0
1 z dσ = 2zV = 2( )2π = 2π 2
S
0
Exercise 17, Section 17.7 $
$
$
v(r) · dr =
(b) C
2
z dx + 2x dy = C
2π
2x dy = C
[(∇ × v) · n] dσ =
4. (a)
2 cos2 u du = 2π
0
S
[(−6yi + 6xj − 2xk) · n] dσ S
=
(−2xz) dσ = 0
by symmetry
S
$
v(r) · dr = C
6xz dx − x dy = − 2
x2 dy
C
C
=−
2π
cos3 u du = −
0
2π
For Exercises 5–7 take S : z = 2 − x − y with 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 − x and C as the triangle (2, 0, 0), (0, 2, 0), (0, 0, 2). Then C = C1 ∪ C2 ∪ C3 C1 : r1 (u) = 2(1 − u) i + 2uj,
n=
3
3(i + j + k)
[(∇×v) · n] dσ =
S
v(r) · dr =
(b) C
+
C1
1√ 1√ 3 dσ = 3A = 2 3 3
S
$
u ∈ [ 0, 1 ],
C3 : r3 (u) = 2(1 − u) k + 2ui, u ∈ [ 0, 1 ]. √ area of S : A = 2 3 centroid: 23 , 23 , 23
5. (a)
with
u ∈ [ 0, 1 ],
C2 : r2 (u) = 2(1 − u) j + 2uk, √ 1
(cos u − sin2 u cos u) du = 0
0
v(r) · dr = −2 + 2 + 2 = 2
+ C2
C3
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SECTION 18.10
[(∇ × v) · n] dσ =
6. (a) S
[(−2xj − 2yk) · n] dσ S
=−
2√ 3 3
(x + y) dσ = −
2√ 16 3(x + y)A = − 3 3
S
$ v(r) · dr =
(b) C
+
C1
+ C2
S
v(r) · dr = C
y dσ =
S
$ (b)
C3
8 8 16 v(r) · dr = − + 0 − = − 3 3 3
1√ (yk · n) dσ = 3 3
[(∇×v) · n] dσ =
7. (a)
S
+ C1
1√ 4 3yA = 3 3
v(r) · dr =
+ C2
C3
4 32 − 3 5
+
8. By (18.10.2) v is a gradient: v = ∇φ. Therefore v(r) · dr = [(∇φ) · dr] = 0 C
32 4 +0= 5 3
by (18.2.2).
C
9. The bounding curve is the set of all (x, y, z) with x2 + y 2 = 4
and
z = 4.
Traversed in the positive sense with respect to n, it is the curve −C where u ∈ [0, 2π].
C : r(u) = 2 cos u i + 2 sin u j + 4k,
By Stokes’s theorem the flux we want is − v(r) · dr = − y dx + z dy + x2 z 2 dz C
=−
C 2π
−4 sin2 u + 8 cos u du = 4π.
0
10. The bounding curve is the set of all (x, y, z) with x2 + z 2 = 9,
y = −8.
Traversed in the positive direction with respect to n, it is the curve −C where C : r(u) = 3 cos u i − 8 j + 3 sin u k,
u ∈ [0, 2π].
By Stokes’s theorem the flux we want is 1 y dx + 2xz dy − 3x dz v(r) · dr = − − C C 2 2π =− (12 sin u − 27 cos2 u) du = 27π. 0
11. The bounding curve C for S is the bounding curve of the elliptical region Ω :
1 2 4x
+ 19 y 2 = 1. Since
∇ × v = 2x2 yz 2 i − 2xy 2 z 2 j is zero on the xy-plane, the flux of ∇ × v through Ω is zero, the circulation of v about C is zero, and therefore the flux of ∇ × v through S is zero.
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SECTION 18.10
971
12. Let T be the solid enclosed by S. By our condition on v, ∇×v is continuously differentiable on T . Therefore by the divergence theorem
[(∇ × v) · n] dσ =
S
[∇ · (∇ × v)] dxdydz. T
This is zero since the divergence of a curl is zero. 13. C bounds the surface S: z =
1 − 12 (x2 + y 2 ),
(x, y) ∈ Ω
with Ω : x2 + (y − 12 )2 ≤ 14 . Routine calculation shows that ∇ × v = yk. The circulation of v with respect to the upper unit normal n is given by
(yk · n) dσ =
y dxdy = yA =
S
1 π 1 = π. 2 4 8
Ω
(18.7.9) If −n is used, the circulation is − 18 π.
Answer: ± 18 π.
14. ∇ × v = i − 2j − 2k. Since the plane x + 2y + z = 0 passes through the origin, it intersects the sphere in a circle of radius a. The surface S bounded by this circle is a disc of radius a with upper unit normal n=
1√ 6(i + 2j + k). 6
The circulation of v with respect to n is given by
5√ 5√ 5√ 6 dσ = − 6A = − 6πa2 . 6 6 6 S S √ √ 5 2 If −n is used, the circulation is 6 6πa . Answer: ± 56 6πa2 .
−
[(∇ × v) · n] dσ =
15. ∇ × v = i + 2j + k. The paraboloid intersects the plane in a curve C that bounds a flat surface S that projects onto the disc x2 + (y − 12 )2 = 14 in the xy-plane. The upper unit normal to S is the vector √ n = 12 2 (−j + k). The area of the base disc is 14 π. Letting γ be the angle between n and k, we √ √ √ have cos γ = n · k = 12 2 and sec γ = 2. Therefore the area of S is 14 2π. The circulation of v with respect to n is given by 1√ 1√ 1 [(∇ × v) · n] dσ = − 2 dσ = − 2 (area of S) = − π. 2 2 4 S
If −n is used, the circulation is 14 π.
S
Answer:
± 14 π.
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SECTION 18.10
16. ∇ × v = −yi − zj − xk. The curve C bounds a flat surface S that projects onto the disc x2 + y 2 = b2 √ in the xy-plane. The upper unit normal to S is the vector n = 12 2(j + k). The area of the base disc √ √ is πb2 . Letting γ be the angle between n and k, we have cos γ = n · k = 12 2 and sec γ = 2. √ Therefore the area of S is πb2 2. The circulation of v with respect to n is given by
1√ [(∇×v) · n] dσ = − 2 2
S
1√ (x + z) dσ = − 2 2
S
z dσ = −
1√ 2zA. 2
S
by symmetry √ It’s clear by symmetry that z = a2 , the height at which S intersects the xz-plane. Since A = πb2 2, the circulation is −πa2 b2 . If −n is used, the circulation becomes πa2 b2 ,
Answer: ±πa2 b2 .
17. Straightforward calculation shows that ∇ × (a × r) = ∇ × [(a2 z − a3 y) i + (a3 x − a1 z) j + (a1 y − a2 x)k] = 2a. 18. ∇ × (φ∇ψ) = (∇φ × ∇ψ) + φ[∇ × ∇ψ] = ∇φ × ∇ψ (18.8.7) since the curl of a gradient is zero. Therefore the result follows from Stokes’s theorem. 19. In the plane of C, the curve C bounds some Jordan region that we call Ω. The surface S ∪ Ω is a piecewise–smooth surface that bounds a solid T. Note that ∇ × v is continuously differentiable on T. Thus, by the divergence theorem, [ ∇ · (∇ × v)] dxdydz = [(∇ × v) · n] dσ T
S∪Ω
where n is the outer unit normal. Since the divergence of a curl is identically zero, we have [(∇ × v) · n] dσ = 0. S∪Ω
Now n is n1 on S and n2 on Ω. Thus [(∇ × v) · n1 ] dσ + [(∇ × v) · n2 ] dσ = 0. S
Ω
This gives
[(∇ × v) · n1 ] dσ =
S
$ [(∇ × v) · (−n2 )] dσ =
Ω
v(r) · dr C
where C is traversed in a positive sense with respect to −n2 and therefore in a positive sense with respect to n1 . (−n2 points toward S.)
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SECTION 18.10 20. By the chain rule
d ∂x dx ∂x = [x(u(t), v(t))] = u (t) + v (t). dt dt ∂u ∂v
Thus
b
C1
dx dt
v1
v1 dx = a
b
v1
dt = a
=
v1 Cr
∂x ∂x u (t) + v1 v (t) dt ∂u ∂v
∂x ∂x du + v1 dv ∂u ∂v
by Green’s theorem =
∂ ∂u
∂x v1 ∂v
∂ − ∂v
∂x v1 ∂u
du dv.
Γ
The integrand can be written ∂2x ∂2x ∂v1 ∂x ∂v1 ∂x ∂v1 ∂x ∂v1 ∂x + v1 − − v1 = − . ∂u ∂v ∂u∂v ∂v ∂v ∂v∂u ∂u ∂v ∂v ∂u equality of partials Thus we have
v1 dx = C
∂v1 ∂x ∂v1 ∂x − du dv. ∂u ∂v ∂v ∂u
Γ
By our previous choice of unit normal,
n = N/ N .
[(∇ × v1 i) · n] dσ =
S
Note that
Therefore
[(∇ × v1 i) · N] du dv. Γ
# # # # i j k # # j k # # # ∂x ∂y ∂z ## ∂ ∂ ·# # # # ∂u ∂u ∂u # ∂y ∂z # ∂x ∂y ∂z ## # 0 0 # # ∂v ∂v ∂v
∂v1 ∂x ∂z ∂x ∂y ∂x ∂y ∂v1 ∂x ∂z = j− k · − j+ − k ∂z ∂y ∂v ∂u ∂u ∂v ∂u ∂v ∂v ∂u
# # i # # # (∇ × v1 i) · N = ## ∂ # ∂x # # v1
∂v1 = ∂z =
# # # # # # # # # #
∂x ∂z ∂x ∂z − ∂v ∂u ∂u ∂v
∂v1 ∂z ∂v1 ∂y + ∂z ∂u ∂y ∂u
∂v1 + ∂y
∂x − ∂v
∂x ∂y ∂x ∂y − ∂v ∂u ∂u ∂v
∂v1 ∂z ∂v1 ∂y + ∂z ∂v ∂y ∂v
∂x · ∂u
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REVIEW EXERCISES Now, by the chain rule, ∂v1 ∂v1 ∂x ∂v1 ∂y ∂v1 ∂z = + + , ∂u ∂x ∂u ∂y ∂u ∂z ∂u
∂v1 ∂v1 ∂x ∂v1 ∂y ∂v1 ∂z = + + . ∂v ∂x ∂v ∂y ∂v ∂z ∂v
Therefore
∂v1 ∂v1 ∂v1 ∂x ∂x ∂v1 ∂x ∂x (∇ × v1 i) · N = − − − ∂u ∂x ∂u ∂v ∂v ∂x ∂v ∂u ∂v1 ∂x ∂v1 ∂x = − ∂u ∂v ∂v ∂u and, as asserted,
[(∇ × v1 i) · n] dσ = S
∂v1 ∂x ∂v1 ∂x − du dv. ∂u ∂v ∂v ∂u
Γ
REVIEW EXERCISES 1 1 1. (a) r(u) = ui + uj, 0 ≤ u ≤ 1; h · dr = (u3 − u2 ) du = − 12 0 C 1 11 8 7 h · dr = (2u − 3u ) du = − (b) 72 0 C
π/2
h · dr =
2. (a)
0
C
h · dr =
(b)
(− cos3 u sin u + sin3 u cos u) du = 0
π/2
(−3 cos11 u sin u + 3 sin11 u cos u) du = 0
0
C
3. Since h(x, y) = ∇f where f (x, y) = x2 y 2 + 12 x2 − y, 119 h(r) · dr = f (2, 4) − f (−1, 2) = 2 C for any curve C beginning at (−1, 2) and ending at (2, 4). ∂P ∂Q y 2 − x2 = = 2 ; ∂y (x + y 2 )2 ∂x
4. h(x, y) is a gradient:
h(x, y) = ∇ arctan (y/x).
Therefore the integrals in (a), (b) and (c) all have the same value. (a) r(u) = 2 cos u i + 2 sin u j,
h · dr =
C
0
3π/4
0 ≤ u ≤ 34 π;
3π/4 −2 sin u 3π 2 cos u 1 du = (−2 sin u) + (2 cos u) du = 4 4 4 0
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REVIEW EXERCISES 5. h(x, y, z) = sin y i + xexy j + sin z k; x(u) = u2
r(u) = u2 i + u j + u3 k, x (u) = 2u,
y(u) = u z(u) = u3 ,
y (u) = 1,
975
u ∈ [ 0, 3 ] z (u) = 3u2
h(r(u)) · r (u) = 2u sin u + u2 eu + 3u2 sin u3 3 3 2u sin u + u2 eu + 3u2 sin u3 du h(r) · dr = 3
C
0
= − 2u cos u + 2 sin u + =
− 6 cos 3 + 2 sin 3 +
2 3
1 3
3
eu − cos u3
1 3
3 0
e27 − cos 27
6. h(x, y, z) = x2 i + xy j + z 2 k; r(u) = cos u i + sin u j + u2 k, π/2 1 π 6 5 h(r(u)) · r (u) = 2u ; h(r) · dr = 2u5 du = 3 2 0 C
u ∈ [0, π/2]
r(u) = u i + u2 j + u3 k. 2 1 5 2 2685 W = (u3 + 5u6 )du = u4 + u7 = −1 4 7 28 −1
7. F(x, y, z) = xy i + yz j + xz k; F(r(u)) · r = u3 + 5u6 ;
8. F(x, y) = x i + (y − 2) j;
r(u) = (u − sin u) i + (1 − cos u) j, 0 ≤ u ≤ 2π.
F(r(u)) · r = u − u cos u − 2 sin u; 2π 1
2π W = (u − u cos u − 2 sin u)du = u2 + u sin u + 3 cos u = 2π 2 0 2 0 9. A vector equation for the line segment is: r(u) = (1 + 2u) i + 4u k, u ∈ [ 0, 1 ]. 1 (20u + 2) 2 + 20u √ F(r(u)) · r = C √ ; F · dr = C du = 4C 1 + 4u + 20u2 1 + 4u + 20u2 0 C 10. Suppose that the path C of the object is given by the vector function r = r(u), a ≤ u ≤ b. Then r = v is the velocity of the object and F · v = 0. The work done by F is
F(r) · dr =
C
11.
a
b
F(r(u)) · r (u) du =
b
F(r(u)) · v(u) du = 0.
a
∂(yexy + 2x) ∂(xexy − 2y) = exy + xyexy = =⇒ h is a gradient. ∂y ∂x 2 3 3 (a) h(r(u)) · r = 3u2 eu − 4u3 + 2u; 3u2 eu − 4u3 + 2u du = e8 − 13 h · dr = 0 C xy 2 2 h · dr = f (2, 4) − f (0, 0) = e8 − 13 (b) Let f (x, y) = e + x − y . Then ∇f = h and C
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January 4, 2007
REVIEW EXERCISES ∂Q ∂P = 4xy + 2 = =⇒ h is a gradient. ∂y ∂x 1
1 (a) h(r(u)) = 6 + 66u + 216u2 + 576u3 du = 6u + 33u2 + 72u3 + 144u4 = 255 h · dr = 0 C 0 (b) Let f (x, y) = (x2 y 2 + 2xy). Then ∇f = h and h · dr = f (3, 5) − f (0, 1) = 255 C
13. h(x, y, z) = ∇f where f (x, y, z) = x4 y 3 z 2 . (a) h(r(u)) = 4u15 i + 3u14 j + 2u13 k; r (u) = i + 2u j + 3u2 k 1 h(r) · dr = 16 u15 du = 1. C 0 (b) h(r) · dr = f (r(1)) − f (r(0)) = f (1, 1, 1) − f (0, 0, 0) = 1. C
14. (a) r(u) = ui + 4uj, 0 ≤ u ≤ 2 2 y 2 dx + (x2 − xy) dy = [16u2 + 4(u2 − 4u2 )] du = C
0
2
4u2 du =
0
32 3
0 ≤ u ≤ 2; C2 : r(u) = 2i + uj, 0 ≤ u ≤ 8 8 y 2 dx + (x2 − xy) dy = y 2 dx + (x2 − xy) dy + y 2 dx + (x2 − xy) dy = 0 + (4 − 2u)du = −32 (b) C1 : r(u) = ui,
C
C1
C2
0
(c) C : r(u) = ui + u3 j, 0 ≤ u ≤ 2 2 608 y 2 dx + (x2 − xy)dy = (3u4 − 2u6 ) du = − 35 C 0 15. (a) r(u) = (1 − u)i + u j, 0 ≤ u ≤ 1. 1/2 1/2 2xy dx + yx dy = C
2(1 − u)u1/2 (−1) + u(1 − u)1/2 du
1
0
1
= −2 =−
(1 − u)u
1/2
1
du +
0
0
1
4 15
(1 − u)u1/2 du = −
0
u(1 − u)1/2 du
(b) r1 = i + u j, 0 ≤ u ≤ 1; r2 = (1 − u) i + j 1 1 1 1/2 1/2 2xy dx + yx dy = u du + −2(1 − u) du = − 2 C 0 0 (c) r = cos u i + sin u j, 0 ≤ u ≤ π/2 π/2 2 1/2 1/2 −2 sin3/2 u cos u + cos3/2 u sin u du = − 2xy dx + yx dy = 5 C 0
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REVIEW EXERCISES 16.
2π
zdx + xdy + ydz =
977
(a2 cos2 u − au sin u + a sin u)du = πa2 + 2πa
0
17.
ye
xy dx + cos x dy + ( ) dz = z
xy
C
2
3 u2 eu + 2u cos u + 3u2 du
0 1 3
= =
3
eu + 2u sin u + 2 cos u + u3
2 0
1 8 17 e + + 4 sin 2 + 2 cos 2 3 3
18. r = cos u i + sin u j; λ(x, y) = k; s (u) = r = 1. π (a) M = λ(x, y) ds = k du = kπ C
0
By symmetry, xM = 0. yM M = yλ(x, y) ds = C
π
π k sin u du = − k cos u = 2k;
0
(b)
C
π
= 0
k k cos u du = 2
Then, C = C1 + C2 . $ 2 2 xy dx − x y dy =
π
(1 + sin 2u) du = 0
0 ≤ u ≤ 1;
1
xy dx − x y dy +
1
=
−u5 du −
0
1 2
0
−(1 − u)2 + 12 (1 − u)2 du
0 1
xy 2 dx − x2 y dy
2
1 (1 − u)2 du = − 16 u6 + 16 (1 − u)3 = − 13 0
Q = −x2 y
(b) P = xy 2 ; $
1
xy dx − x y dy = 2
2
C
√
x
x2
(x + y ) dx + (x − y ) dy =
20. (a)
2
2
$
(x2 + y 2 ) dx + (x2 − y 2 ) dy =
C
Ω
1 2x2 − 2x5 dx = − 3
0
(2x − 2y) dx dy =
2
C
(b)
1
(−4xy) dy dx = 0
$ 2
0 ≤ u ≤ 1.
C2
1
(u − 2u ) du + 5
0
√ 1 − u j,
2
5
1 kπ 2
C2 : r(u) = (1 − u) i +
C1
=
kx2 ds C
2
19. (a) Set C1 : r(u) = u i + u2 j,
2 π
λ(x, y)R2 (x, y) ds =
I=
C
yM =
0
0
2π
1
(2r cos θ − 2r sin θ)r dr dθ = 0
0
2π
(− sin u + cos 2u cos u) du = 0 0
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REVIEW EXERCISES
21. P = x − 2y 2 ; Q = 2xy $ (x − 2y 2 ) dx + 2xy dy =
$ xy dx + C
1
0
2
1
6y dy dx = 6
0
C
22.
2
2 x + xy dy =
y dx dy =
Ω
1
−1
√
1−x2 2
y dy dx =
1
−1
0
1 8 (1
− x2 )dx =
1 6
∂Q ∂P 2x − 2y − = 2 ∂x ∂y x + y2 $ 2x − 2y ln(x2 + y 2 ) dx + ln(x2 + y 2 ) dy = dx dy 2 2 C Ω x +y π 2 2r cos θ − 2r sin θ = r dr dθ r2 0 1 π 2 =2 (cos θ − sin θ) dr dθ = −4
23. P = ln(x2 + y 2 );
Q = ln(x2 + y 2 );
0
1
∂Q ∂P 1 1 − =− 2 + 2 ∂x ∂y x y $ −2 (1/y) dx + (1/x) dy = −x + y −2 dxdy =
24. P = 1/y, Q = 1/x,
C
4
4
1
Ω
25.
−2ydxdy =
y dx = Ω
2π
0
1+sin θ
2π
−2r sin θ dr dθ = 2
0
0
∂Q ∂P − = −2ey cos x ∂x ∂y $ y y y e cos x dx − e sin x dy = (−2e cos x) dxdy =
−x−2 + y −2 dy dx
3 −x−3/2 − x−1/2 + x−2 + 1 dx = 4
1
2
x
1
=
$
√
(− 23 )(1 + sin θ)3 sin θ dθ = −
26. P = ey cos x, Q = −ey sin x,
C
Ω
0
π/2
1
(−2ey cos x) dy dx = 2(1 − e)
0
27. C1 : r(u) = −u i + (4 − u2 ) j, −2 ≤ u ≤ 2; C2 : r(u) = u i, −2 ≤ u ≤ 2; C = C1 ∪ C2 1 1 1 A= (−y dx + x dy) = (−y dx + x dy) + (−y dx + x dy) 2 C 2 C1 2 C2 1 2 1 2 = −(4 − u2 )(−1) du − u(−2u) du + 0 du 2 −2 2 −2 2 32 = (4 + u2 ) du = 3 −2
5π 2
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REVIEW EXERCISES 28. C1 : r(u) = (3 − 2u) i + (1 + 2u) j, 0 ≤ u ≤ 1;
979
C2 : r(u) = u i + (3/u) j, 1 ≤ u ≤ 3;
C = C1 ∪ C2 1 (−y dx + x dy) + (−y dx + x dy) 2 C2 C1 1 1 1 3 = [−(1 + 2u)(−2) + (3 − 2u)2] du + [−(3/u) + u(−3/u2 )] du 2 0 2 1 1 1 1 3 = 8 du + (−6/u) du = 4 − 3 ln 3 2 0 2 1
1 A= 2
1 (−y dx + x dy) = 2 C
29. By symmetry, it is sufficient to consider the upper part of the sphere: ∂z −x , = ∂x 4 − x2 − y 2
z=
4 − x2 − y 2
∂z −y = ∂y 4 − x2 − y 2
Let Ω be the projection of the sphere onto the xy plane, then 2 2 2 S=2 (zx ) + (zy ) + 1 dx dy = 2 dx dy 4 − x2 − y 2 Ω Ω π/2 2 cos θ 1 √ =4 r dr dθ 4 − r2 −π/2 0 π/2 =4 2 − 2 1 − cos2 θ dθ = 8(π − 2) −π/2
4−x−y and zx = − 12 , zy = − 12 . 2 2π 2 √ 3 2 2 area of S = zx + zy + 1 dx dy = r dr dθ = 2 6π. 2 0 0 Ω
30. From x + y + 2z = 4, we get z =
31.
∂z x = , 2 ∂x x + y2
y ∂z = . 2 ∂y x + y2
The projection Ω of the surface onto the xy plane is the disk x2 + y 2 ≤ 9. (zx )2 + (zy )2 + 1 dx dy =
S= Ω
A=2 Ω
2π
zx2 + zy2 + 1 dxdy = 2
=2 0
=
3
0
1+
0
π (373/2 − 1) 3
3
3
0
2 r dr dθ = 9 π
4x2 + 4y 2 + 1 dxdy
1 + 4r2 r dr
r dr dθ = 4π
√
0
Ω
4r2
2π
2 dx dy =
Ω
32.
√
√
2
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REVIEW EXERCISES
33.
√ yz dσ = 2
S
1
√ r sin θ(r sin θ + 4) dr dθ =
0
√ x(1 − x − y) 3 dx dy = 3 √
x(1 − x − y) dσ =
S
2π 4
2
xz dσ =
√
0
34.
2π
S
Ω
0
1
1−x
x(1 − x − y)dy dx =
0
3 24
35. The cylindrical surface S1 is parametrized by: x = u, y = 2 cos v, z = 2 sin v, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π. N(u, v) = −2 cos v i − 2 sin v j,
x2 + y 2 + z 2 dσ =
S1
2
0
2π
0
|N(u, v)| = 2 128π u2 + 4 2 dv du = 3
The disc S2 : x = 0, y 2 + z 2 is parametrized by: x = 0, y = u cos v, z = u sin v, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π. |N(u, v)| = u;
N = u i,
x2 + y 2 + z 2 dσ =
S2
2
0
2π
0 + u2 u dv du = 8π
0
The disc S3 : x = 2, y 2 + z 2 is parametrized by: x = 2, y = u cos v, z = u sin v, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π. |N(u, v)| = u;
N = u i, Thus, S
2
2
x +y +z
2
2
2π
dσ =
S2
0
4 + u2 u dv du = 24π
0
128π 224π x2 + y 2 + z 2 dσ = + 8π + 24π = 3 3
36. The cylindrical surface S is parametrized by: x = 2 cos u, y = 2 sin u, z = v, 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1. 1 2π N(u, v) = 2 cos u i − 2 sin u j, |N(u, v)| = 2; xz dσ = 2v cos u(2) dv du = 0. S
37. ∇ · v = 4x,
∇ × v = 2yk
39. ∇ · v = 1 + xy,
38 ∇ · v = 0,
0
0
∇×v=0
∇ × v = (xz − x)i − yzj + zk
40. ∇ · v = yz + x sin xy,
∇ × v = x cos xyi + (xy − y cos xy)j + (y sin xy − xz)k
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REVIEW EXERCISES 41. (a) ∇ · v = z − x + y 1 1 1 1 (z − x + y)dzdydx = 2 0 0 0 (b) at x = 0,
1
n = −i, v · n = 0,
0dydz = 0 0
at x = 1,
1
n = i, v · n = z,
0 1
zdydz = 1/2 0
at y = 0,
1
0
1
n = −j, v · n = xy = 0,
0dxdz = 0 0
at y = 1,
n = j, v · n = −xy = −x, 0
at z = 0,
1
n = −k, v · n = 0,
0 1
1
−xdxdz = −1/2
0
1
0dydx = 0 0
at z = 1,
1
n = k, v · n = yz,
1
0
1
ydydx = 1/2 0
0
The sum is 1/2 42. (a) ∇ · v = 3 3 dxdydz =
4
(b) at x = 0,
1
3r dr dθ dx = 4(2π)( 32 ) = 12π n = −i, v · n = −z, −z dydz = 0 (by symmetry) 0
T
2π
0
0
S
at x = 4,
n = i, v · n = 4 + z,
(4 + z) dydz =
S
for z =
4 dydz = 4π S
1 − y 2 , 0 ≤ x ≤ 4, n = −y j + 1 − y 2 k and v · n = 1 − 2y 2 − y
1
−1
4
(1 − 2y 2 − y
0
1 − y2 + x 1 − y2
16 + 4π 1 − y 2 + x 1 − y 2 )dxdy = 8 − 3
for z = − 1 − y 2 , 0 ≤ x ≤ 4, n = y j + 1 − y 2 k and v · n = −1 + 2y 2 − y
1
−1
The sum is 12π
0
4
(−1 + 2y 2 − y
1 − y2 + x 1 − y2
16 + 4π 1 − y 2 + x 1 − y 2 )dxdy = −8 + 3
981
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REVIEW EXERCISES
43. The projection of S onto the xy-plane is: Ω : x2 + y 2 ≤ 9. 2 v · n dσ = 4x + 2xyz + z 2 dx dy S
Ω
2 4x2 + 2xy 9 − x2 − y 2 + 9 − x2 − y 2 dx dy
= Ω
2π
3
= 0
44. On x = 0, n = −i, on x = a, n = i,
4r2 cos2 θ + r2 (9 − r2 ) sin 2θ + (9 − r2 )2 rdr dθ = 324π
0
v · n = −x2 = 0,
v · n = a2 ,
the flux is 0;
the flux is a4 ;
on y = 0, n = −j,
v · n = xz,
a
xz dx dz = 0
on y = a, n = j,
v · n = −xz,
0
a
on z = a, n = k,
a
v · n = 0,
the flux is 0;
v · n = a2 ,
the flux is a4 .
1 2 a ; 4
1 −xz dx dz = − a2 ; 4
the flux is 0
on z = 0, n = −k,
a
the flux is
0
Hence the total flux is 2a4 . 45. (a) (∇ × v) · n = (i + j + k) · S
1 1 − x− y+ 2 2
1 1 − xi − yj + 2 2
4 − x2 − y 2 2
4 − x2 − y 2 4 − x2 − y 2 1 1 k =− x− y+ 2 2 2 2
1 1 − x− y+ 2 2
dσ = S
=
2π
2
= 0
(b) r(θ) = 2 cos θ i + 2 sin θ j,
0
4 − x2 − y 2
4 − x2 − y 2 2
−y
−
4 − x2 − y 2
v(r) · dr =
C
4 − x2 − y 2
+ 1 dx dy
r cos θ r sin θ −√ −√ + 1 r dr dθ = 4π 4 − r2 4 − r2
$ [(∇ × v) · n]dσ =
2
0 ≤ θ ≤ 2π
S
−x
0
2π
4 cos2 θdθ = 4π
dx dy
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REVIEW EXERCISES
983
2x i + 2y j + k n= ; 1 + 4x2 + 4y 2 1 [(∇ × v) · n] dσ = (4xy + 6yz 2 + 1) dσ 1 + 4x2 + 4y 2 S S = (4xy + 6yz 2 + 1) dx dy
46. (a) v = z 3 i + x j + y 2 k;
Ω
[4xy + 6y(9 − x2 − y 2 )2 + 1] dx dy
=
Ω 2π
= 0
=
3
4r2 cos θ sin θ + 6r sin θ(9 − r2 )2 + 1 r dr dθ
0 3
2πr dr = 9π 0
(b) The boundary of the surface is the curve x2 + y 2 = 9, z = 0; v(r(u)) = 3 cos u j + 9 sin2 u k; r (u) = −3 sin u i + 3 cos u j $ 2π v · dr = 9 cos2 u du = 9π. C
0
r(u) = 3 cos u i + 3 sin u j + 0 k;
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SECTION 19.1
CHAPTER 19 SECTION 19.1 1. y + xy = xy 3
=⇒
y −3 y + xy −2 = x.
Let v = y −2 ,
v = −2y −3 y .
1 − v + xv = x 2 v − 2xv = −2x e−x v − 2xe−x v = −2xe−x 2
2
2
e−x v = e−x + C 2
2
2
v = 1 + Cex 1 y2 = . 1 + Cex2 2. y − y = −(x2 + x + 1)y 2
=⇒
y −2 y − y −1 = −(x2 + x + 1).
Let v = y −1 ,
−v − v = −(x2 + x + 1) 2 v + v = x +x+1 ex v = ex (x2 + x + 1) dx = x2 ex − xex + 2ex + C
v = x2 − x + 2 + Ce−x 1 y= 2 . x − x + 2 + Ce−x 3. y − 4y = 2ex y 2 1
=⇒
y − 2 y − 4y 2 = 2ex . 1
1
1
Let v = y 2 ,
v =
1 −1 y 2y . 2
2v − 4v = 2ex v − 2v = ex e−2x v − 2e−2x v = e−x e−2x v = −e−x + C v = −ex + Ce2x y = (Ce2x − ex )2 . 4. y =
1 y + 2xy 2x
=⇒
yy −
1 2 1 y = . 2x 2x
Let v = y 2 ,
1 1 v − v 2 2x 1 v − v x 1 1 v − 2v x x 1 v x v
v = 2yy .
1 2x 1 = x 1 = 2 x 1 = − +C x = Cx − 1
=
y 2 = Cx − 1.
v = −y −2 y .
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SECTION 19.1 5. (x − 2)y + y = 5(x − 2)2 y 2 1
y− 2 y + 1
=⇒
1 1 y 2 = 5(x − 2). x−2
1 v x−2 1 v + v 2(x − 2) √ 1 x − 2v + √ v 2 x−2 √ x − 2v 2v +
1
Let v = y 2 ,
= 5(x − 2) 5 (x − 2) 2 3 5 = (x − 2) 2 2 =
5
= (x − 2) 2 + C
v = (x − 2)2 + √
C x−2
C y = (x − 2) + √ x−2 2
6. yy − xy 2 + x = 0.
2 .
v = 2yy .
Let v = y 2 ,
1 v − xv = −x 2 v − 2xv = −2x e−x v − 2xe−x v = −2xe−x 2
2
2
e−x v = e−x + C 2
2
2
v = 1 + Cex √ y = 1 + Cex2 . 7. y + xy = y 3 ex
2
=⇒
y −3 y + xy −2 = ex . 2
Let v = y −2 ,
v = −2y −3 y .
1 2 − v + xv = ex 2 2 v − 2xv = −2ex e−x v − 2xe−x v = −2 2
2
e−x v = −2x + C 2
2
2
v = −2xex + Cex y −2 = Cex − 2xex . 2
C=4 8. y +
=⇒
2
y −2 = 4ex − 2xex .
ln x 2 1 y= y x x
2
=⇒
2
y −2 y +
1 −1 ln x y = . x x
Let v = y −1 ,
ln x 1 v= x x ln x 1 v − v = − x x 1 1 ln x v − 2v = − 2 x x x −v +
v = −y −2 y .
v =
1 −1 y 2y . 2
985
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SECTION 19.1 1 v=− x
ln x 1 dx = (ln x + 1) + C x2 x
v = ln x + 1 + Cx 1 . ln x + 1 + Cx 1 =⇒ y = . ln x + 1
y= 1=
1 ln 1 + 1 + C
=⇒
9. 2x3 y − 3x2 y = y 3
C=0 y −3 y −
=⇒
3 −2 1 y = 3 . Let v = y −2 , 2x 2x 1 3 1 v= 3 − v − 2 2x 2x 1 3 v + v = − 3 x x
v = −2y −3 y .
x3 v + 3x2 v = −1 x3 v = −x + C
1=
1 C −x
=⇒
C=2
10. y + tan x y = y 2 sec3 x
y2 =
=⇒
=⇒
v=
C −x x3
y2 =
x3 C −x
x3 . 2−x
y −2 y + tan xy −1 = sec3 x.
Let v = y −1 ,
−v + tan xv = sec3 x v − tan x v = − sec3 x cos xv − sin x v = − sec2 x cos x v = − tan x + C cos x = − tan x + C y cos 0 = − tan 0 + C 3 11. y −
y ln y = xy x
=⇒
=⇒
C=
1 3
=⇒
y 1 − ln y = x. y x
cos x 1 = − tan x. y 3 Let u = ln y,
u =
1 u=x x 1 1 u − 2u = 1 x x 1 u = x+C x u −
u = x2 + Cx ln y = x2 + Cx.
y . y
v = −y −2 y .
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SECTION 19.1 y + yf (x) ln y = g(x)y y + f (x) ln y = g(x) y u + f (x)u = g(x).
12. (a)
(b) cos y y + g(x) sin y = f (x).
u = cos y y .
Let u = sin y,
Thus we have u + g(x)u = f (x). (tx)2 + (ty)2 t2 (x2 + y 2 ) x2 + y 2 x2 + y 2 ; f (tx, ty) = = = = f (x, y) 2xy 2(tx)(ty) t2 (2xy) 2xy Set vx = y. Then, v + xv = y and
13. f (x, y) =
v + xv = v−
x2 + v 2 x2 1 + v2 = 2vx2 2v
1 + v2 + xv = 0 2v
v 2 − 1 + 2xvv = 0 1 2v dx + 2 dv = 0 x v −1 1 2v dx + dv = C x v2 − 1 ln |x | + ln |v 2 − 1 | = K Replacing v by y/x, we get x
14. f (tx, ty) =
x(v 2 − 1) = C
or
y2 −1 x2
=C
or
y 2 − x2 = Cx
y2 (ty)2 = = f (x, y). 2 (tx)(ty) + (tx) xy + x2
Set vx = y. Then, v + xv = y and v + xv =
dx + x
v2 1+v
v+1 dv = C v
ln |x| + v + ln |v| = C v + ln |xv| = C y + ln |y| = C x x−y (tx) − (ty) t(x − y) x−y ; f (tx, ty) = = = = f (x, y) x+y tx + ty t(x + y) x+y Set vx = y. Then, v + xv = y and
15. f (x, y) =
v + xv =
1−v x − vx = x + vx 1+v
v 2 + 2v − 1 + x(1 + v)v = 0
987
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SECTION 19.1 1 1+v dx + 2 dv = 0 x v + 2v − 1 1 1+v dx + dv = C x v 2 + 2v − 1 ln | x | +
1 2
ln |v 2 + 2v − 1 | = K
√ x v 2 + 2v − 1 = C
or
Replacing v by y/x, we get x 16. f (tx, ty) =
y2 y +2 −1=C x2 x
y 2 + 2xy − x2 = C
or
x+y tx + ty = = f (x, y). tx − ty x−y
Set vx = y. Then, v + xv = y and v + xv =
ln |x| +
dx + x
1+v x + vx = x − vx 1−v
v−1 dv = C1 v2 + 1
1 ln |v 2 + 1| − arctan v = C1 2
ln x2 + ln(v 2 + 1) − 2 arctan v = C
(= 2C1 )
ln[x2 (v 2 + 1)] − 2 arctan v = C ln[x2 + y 2 ] − 2 arctan x2 ey/x + y 2 ; 17. f (x, y) = xy
y x
=C
t2 x2 ey/x + y 2 (tx)2 − e(ty)/(tx) + (ty)2 f (tx, ty) = = = f (x, y) (tx)(ty) t2 (xy)
Set vx = y. Then, v + xv = y and v + xv =
x2 ev + v 2 x2 ev + v 2 = 2 vx v v 2 + xvv = ev + v 2 −ev + xvv = 0 1 dx = ve−v dv x 1 dx = ve−v dv x ln |x | = −ve−v − e−v + C
Replacing v by y/x, and simplifying, we get y + x = xey/x (C − ln |x |)
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SECTION 19.1 18. f (tx, ty) =
(tx)2 + 3(ty)2 x2 + 3y 2 = = f (x, y). 4(tx)(ty) 4xy
Set vx = y. Then, v + xv = y and v + xv =
dx + x
x2 + 3x2 v 2 4x2 v
4v dv = C1 −1
v2
ln |x| + 2 ln |v 2 − 1| = C1 x(v 2 − 1)2 = C (= eC1 ) (y 2 − x2 )2 = Cx3 y (ty) y + sin(y/x); f (tx, ty) = + sin[(ty/tx)] = + sin(y/x) = f (x, y) x tx x Set vx = y. Then, v + xv = y and vx + sin[(vx)/x] = v + sin v v + xv = x xv = sin v 1 csc v dv = dx x 1 csc v dv = dx x
19. f (x, y) =
ln | csc v − cot v | = ln | x | + K
or
csc v − cot v = Cx
Replacing v by y/x, and simplifying, we get 1 − cos(y/x) = Cx sin(y/x) y y ty y ty y 1 + ln ; f (tx, ty) = 1 + ln = 1 + ln = f (x, y) x x tx tx x x Set vx = y. Then, v + xv = y and vx vx 1 + ln = v(1 + ln v) v + xv = x x xv = v ln v
20. f (x, y) =
1 1 dv = dx v ln v x 1 1 dv = dx v ln v x ln |ln v| = ln |x| + K y ln = Cx x y = eCx x
or
y = xeCx
21. The differential equation is homogeneous since f (x, y) =
y 3 − x3 ; xy 2
f (tx, ty) =
y 3 − x3 (ty)3 − (tx)3 t3 (y 3 − x3 ) = = = f (x, y) (tx)(ty)2 t3 (xy 2 ) xy 2
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SECTION 19.2 Set vx = y. Then, v + xv = y and v + xv =
(vx)3 − x3 v3 − 1 = v 2 x3 v2 1 + xv 2 v = 0
1 dx + v 2 dv = 0 x 1 dx + v 2 dv = 0 x 1 ln | x | + v 3 = C 3 Replacing v by y/x, we get y 3 + 3x3 ln | x | = Cx3 Applying the side condition y(1) = 2, we have 8 + 3 ln 1 = C 22.
1 y dy = + . dx sin(y/x) x
=⇒
C=8
and y 3 + 3x3 ln | x | = 8x3
Set y = vx. Then y = v + xv and
dx − x
v + xv =
1 +v sin v
sin v dv = C
ln |x| + cos v = C
ln |x| + cos xy = C y =1 y(1) = 0 =⇒ 0 + cos 0 = C =⇒ C = 1 =⇒ ln |x| + cos x
SECTION 19.2 1.
∂P ∂Q = 2xy − 1 = ; ∂y ∂x ∂f = xy 2 − y ∂x
=⇒
the equation is exact on the whole plane. f (x, y) =
1 2
∂f = x2 y − x + ϕ (y) = x2 y − x ∂y Therefore
f (x, y) =
1 2
x2 y 2 − xy + ϕ(y) =⇒
x2 y 2 − xy,
ϕ (y) = 0
=⇒
ϕ(y) = 0 (omit the constant)∗
and a one-parameter family of solutions is: 1 2
x2 y 2 − xy = C
∗ We will omit the constant at this step throughout this section. 2.
∂ x ∂ x (e sin y) = ex cos y = (e cos y); ∂y ∂x f (x, y) = ex sin y,
the equation is exact on the whole plane.
and ex sin y = C is a one-parameter family of solutions.
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SECTION 19.2 3.
∂P ∂Q = ey − ex = ; the equation is exact on the whole plane. ∂y ∂x ∂f = ey − yex =⇒ f (x, y) = xey − yex + ϕ(y) ∂x ∂f = xey − ex + ϕ (y) = xey − ex =⇒ ϕ (y) = 0 =⇒ ϕ(y) = 0 ∂y Therefore
f (x, y) = xey − yex ,
and a one-parameter family of solutions is: xey − yex = C
4.
∂ ∂ (sin y) = cos y = (x cos y + 1); ∂y ∂x f (x, y) = x sin y + y,
5.
and x sin y + y = C is a one-parameter family of solutions.
∂P 1 ∂Q = + 2x = ; ∂y y ∂x ∂f = ln y + 2xy =⇒ ∂x ∂f x = + x2 + ϕ (y) = ∂y y Therefore
the equation is exact on the whole plane.
the equation is exact on the upper half plane. f (x, y) = x ln y + x2 y + ϕ(y) x + x2 y
ϕ (y) = 0
=⇒
f (x, y) = x ln y + x2 y,
=⇒
ϕ(y) = 0
and a one-parameter family of solutions is: x ln y + x2 y = C
6.
2x ∂ ∂ (2x arctan y) = = ∂y 1 + y2 ∂x f (x, y) = x2 arctan y,
7.
x2 1 + y2
;
the equation is exact on the whole plane.
and x2 arctan y = C is a one-parameter family of solutions.
1 ∂Q ∂P = = ; the equation is exact on the right half plane. ∂y x ∂x ∂f y = + 6x =⇒ f (x, y) = y ln x + 3x2 + ϕ(y) ∂x x ∂f = ln x + ϕ (y) = ln x − 2 ∂y Therefore
=⇒
ϕ (y) = −2
f (x, y) = y ln x + 3x2 − 2y,
=⇒
ϕ(y) = −2y
and a one-parameter family of solutions is:
y ln x + 3x2 − 2y = C 8.
1 1 ∂ x ∂ x y (e + ln y + ) = + = ( + ln x + sin y); ∂y x y x ∂x y the equation is exact in the first quadrant, not including the axes. f (x, y) = ex + x ln y + y ln x − cos y
and ex + x ln y + y ln x − cos y = C is a
one-parameter family of solutions. 9.
∂P ∂Q = 3y 2 − 2y sin x = ; ∂y ∂x
the equation is exact on the whole plane.
∂f = y 3 − y 2 sin x − x ∂x
f (x, y) = xy 3 + y 2 cos x −
=⇒
1 2
x2 + ϕ(y)
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SECTION 19.2 ∂f = 3xy 2 + 2y cos x + ϕ (y) = 3xy 2 + 2y cos x + e2y ∂y Therefore
f (x, y) = xy 3 + y 2 cos x −
1 2
x2 +
1 2
ϕ (y) = e2y
=⇒
ϕ(y) =
1 2
e2y
e2y , and a one-parameter family of solutions is:
xy 3 + y 2 cos x − 10.
=⇒
1 2
x2 +
1 2
e2y = C
∂ ∂ 2y (e − y cos xy) = 2e2y − cos xy + xy sin xy = (2xe2y − x cos xy + 2y); ∂y ∂x the equation is exact on the whole plane. f (x, y) = xe2y − sin xy + y 2
and xe2y − sin xy + y 2 = C is a
one-parameter family of solutions. 11. (a) Yes:
∂ ∂ [p(x)] = 0 = [q(y)]. ∂y ∂x
1 is an integrating factor. p(y)q(x)
(b) For all x, y such that p(y)q(x) = 0, Multiplying the differential equation by
1 , we get p(y)q(x)
1 1 + y =0 q(x) p(y) which has the form of the differential equation in part (a). 12. Mimic the proof of the first part. 13.
∂P ∂Q = ey−x − 1 and = ey−x − xey−x ; the equation is not exact. ∂y ∂x
y−x 1 ∂P ∂Q 1 Since − = y−x xe − 1 = 1, μ(x) = e dx = ex Q ∂y ∂x xe −1 an integrating factor. Multiplying the given equation by ex ,
is
we get
(ey − yex ) + (xey − ex ) y = 0 This is the equation given in Exercise 3. A one-parameter family of solutions is: xey − yex = C 1 14. w = P factor.
∂P ∂Q − ∂y ∂x
=
1 (ey + x) = 1 x + ey
1 (xe−y + 1) − x2 e−y y = 0 is exact; 2 1 2 −y is 2 x e + x = C. 15.
∂Q ∂P = 6x2 y + ey = ; ∂y ∂x ∂f = 3x2 y 2 + x + ey ∂x
doesn’t depend on x, so e
= e−y is an integrating
f (x, y) = 12 x2 e−y + x and a one-parameter family of solutions
the equation is exact. =⇒
− dy
f (x, y) = x3 y 2 +
1 2
x2 + xey + ϕ(y)
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SECTION 19.2 ∂f = 2x3 y + xey + ϕ (y) = 2x3 y + y + xey ∂y f (x, y) = x3 y 2 +
Therefore
16.
x2 + xey +
1 2
y2 ,
x3 y 2 +
1 2
x2 + xey +
=⇒
ϕ(y) =
∂P = 3y 2 ∂y Since
1 2
y2
and a one-parameter family of solutions is: 1 2
y2 = C
∂ ∂ (sin 2x cos y) = − sin 2x sin y = −2 sin x cos x sin y = (− sin2 x sin y); ∂y ∂x f (x, y) = sin2 x cos y
17.
1 2
ϕ (y) = y
=⇒
993
exact.
sin2 x cos y = C is a one-parameter family of solutions.
and
∂Q = 0; the equation is not exact. ∂x 1 ∂P ∂Q 1 − = 2 (3y 2 ) = 1, μ(x) = e dx = ex Q ∂y ∂x 3y and
is an
an integrating factor. Multiplying the given equation by ex , we get
3 x
y e + xex + ex + 3y 2 ex y = 0 ∂f = y 3 ex + xex + ex ∂x
∂f = 3y 2 ex + ϕ (y) = 3y 2 ex ∂y Therefore
f (x, y) = y 3 ex + xex + ϕ(y)
=⇒
ϕ (y) = 0
=⇒
f (x, y) = y 3 ex + xex ,
=⇒
ϕ(y) = 0
and a one-parameter family of solutions is: y 3 ex + xex = C
1 18. v = Q
∂P ∂Q − ∂y ∂x
=
xe2x+y + 1
is an integrating factor.
∂Q ∂P =1= ; ∂y ∂x ∂f = x2 + y ∂x
− 2 = −2,
and xey + ye−2x = C
independent of y,
so
=⇒
f (x, y) =
f (x, y) =
1 3
=⇒
1 3
is a one-parameter family of solutions.
x3 + xy + ϕ(y) ϕ (y) = ey
x3 + xy + ey ,
Setting x = 1, y = 0, we get C = 1 3
4 3
=⇒
ϕ(y) = ey
and a one-parameter family of solutions is: 1 3
x3 + xy + ey = C
and
x3 + xy + ey =
4 3
e
is exact.
the equation is exact.
∂f = x + ϕ (y) = x + ey ∂y Therefore
−2xe
2x+y
(ey − 2ye−2x ) + (xey + e−2x )y = 0
Thus
f (x, y) = xey + ye−2x , 19.
1
or
x3 + 3xy + 3ey = 4
−2 dx
= e−2x
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SECTION 19.2 ∂ ∂
(3x2 − 2xy + y 3 ) = −2x + 3y 2 = 3xy 2 − x2 ; ∂y ∂x
exact
f (x, y) = x3 − x2 y + xy 3 =⇒ x3 − x2 y + xy 3 = C. Substituting x = 1, y = −1 we get 1 + 1 − 1 = C =⇒ x3 − x2 y + xy 3 = 1 21.
∂P = 4y ∂y Since
∂Q = 2y; the equation is not exact. ∂x 1 ∂P ∂Q 1 1 − = (2y) = , μ(x) = e (1/x) dx = eln x = x Q ∂y ∂x 2xy x and
is an
integrating factor. Multiplying the given equation by x, we get
2xy 2 + x3 + 2x + 2x2 y y = 0 ∂f = 2x2 y =⇒ f (x, y) = x2 y 2 + ϕ(x) ∂y ∂f = 2xy 2 + ϕ (x) = 2xy 2 + x3 + 2x =⇒ ϕ (x) = x3 + 2x =⇒ ϕ = 14 x4 + x2 ∂x Therefore f (x, y) = x2 y 2 + 14 x4 + x2 , and a one-parameter family of solutions is: x2 y 2 + Setting x = 1, y = 0, we get C = x2 y 2 + 1 22. v = Q
∂P ∂Q − ∂y ∂x
=
1 4
5 4
1 4
x4 + x2 = C
and
x4 + x2 =
2 − 6xy 2 2 =− 2 2 3x y − x x
5 4
or
4x2 y 2 + x4 + 4x2 = 5
doesn’t depend on y,
so e
2 −x dx
= x−2
is an integrating factor. Thus (1 + yx−2 ) + (3y 2 − x−1 )y = 0 is exact. y y f (x, y) = x − + y 3 =⇒ x − + y 3 = C x x y Substituting x = 1, y = 1 we get 1 − 1 + 1 = C =⇒ x − + y = 1. x 23.
∂P = 3y 2 ∂y Since
∂Q the equation is not exact. = y2 ; ∂x 1 ∂P ∂Q 1 2 − = 3 (2y 2 ) = , w(y) = e− (2/y) dy = e−2 ln y = y −2 P ∂y ∂x y y and
is an
integrating factor. Multiplying the given equation by y −2 , we get
y + y −2 + x y = 0 ∂f = y =⇒ f (x, y) = xy + ϕ(y) ∂x ∂f 1 = x + ϕ (y) = y −2 + x =⇒ ϕ (y) = y −2 =⇒ ϕ(y) = − ∂y y 1 1 Therefore f (x, y) = xy − , and a one-parameter family of solutions is: xy − = C y y 1 Setting x = −2, y = −1, we get C = 3 and the solution xy − = 3. y
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SECTION 19.2 24.
25.
995
∂ ∂ (x + y)2 = 2(x + y) = (2xy + x2 − 1); exact. ∂y ∂x x3 x3 + x2 y + xy 2 − y =⇒ + x2 y + xy 2 − y = C f (x, y) = 3 3 x3 4 Setting x = 1, y = 1, we get C = 43 and the solution + x2 y + xy 2 − y = . 3 3 ∂Q ∂P = −2y sinh(x − y 2 ) = ; ∂y ∂x
the equation is exact.
∂f = cosh(x − 2y 2 ) + e2x ∂x
f (x, y) = sinh(x − y 2 ) +
=⇒
∂f = −2y cosh(x − y 2 ) + ϕ (y) = y − 2y cosh(x − y 2 ) ∂y Therefore
f (x, y) = sinh(x − y 2 ) +
1 2
e2x +
1 2
y2 ,
sinh(x − y 2 ) + Setting x = 2, y =
1 2
e2x + ϕ(y) ϕ (y) = y
=⇒
=⇒
ϕ(y) =
1 2
y2
and a one-parameter family of solutions is: 1 2
e2x +
1 2
y2 = C
√ 2, we get C = 12 e4 + 1 and the solution sinh(x − y 2 ) +
1 2
e2x +
1 2
y 2 = 12 e4 + 1
26. Write the linear equation as p(x)y − q(x) + y = 0. Then P (x) = p(x)y − q(x), Q(x) = 1 1 ∂P ∂Q and v = − = p(x) depends only on x. Therefore v = e p(x) dx is Q ∂y ∂x an integrating factor.
27. (a)
∂P = 2xy + kx2 ∂y
and
∂P = e2xy + 2xye2xy ∂y
(b)
∂Q = 2xy + 3x2 ∂x and
=⇒
k = 3.
∂Q = ke2xy + 2kxye2xy ∂x
28. (a) We need g (y) sin x = y 2 f (x). (b) We need g (y)ey + g(y)ey = y,
=⇒
k = 1.
Take g(y) = 13 y 3 and f (x) = − cos x d [g(y)ey ] = y. dy
that is
It follows that g(y)ey = 12 y 2 + C, =⇒ g(y) = e−y ( 12 y 2 + C). 29. y = y 2 x3 ;
the equation is separable. y −2 dy = x3 dx
=⇒
−
1 = y
1 4
x4 + C
=⇒
y=
−4 x4 + C
30. y y = 4xe2x+y = 4xe2x ey =⇒ the equation is separable. ye−y dy = 4xe2x dx =⇒ −ye−y − e−y = 2xe2x − e2x + C 4 y = x4 ; the equation is linear. x 4 H(x) = (4/x) dx = 4 ln x = ln x4 , integrating factor: eln x = x4
31. y +
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SECTION 19.3 x4 y + 4x3 y = x8 d dx
x4 y = x8
x4 y =
1 9
x9 + C
y=
1 9
x5 + Cx−4
2 32. y + 2xy = 2x3 ; the equation is linear with integrating factor e 2x dx = ex d x2 2 2 2 2 (e y) = 2x3 ex =⇒ ex y = ex (x2 − 1) + C =⇒ y = x2 − 1 + Ce−x . =⇒ dx 33.
∂Q ∂P = exy + xyexy = ; the equation is exact. ∂y ∂x ∂f = yexy − 2x =⇒ f (x, y) = exy − x2 + ϕ(y) ∂x ∂f 2 2 = xexy + ϕ (y) = + xexy =⇒ ϕ (y) = =⇒ ∂y y y f (x, y) = exy − x2 + 2 ln | y |,
Therefore
ϕ(y) = 2 ln | y |
and a one-parameter family of solutions is:
exy − x2 + 2 ln | y | = C 34. w =
1 P
Then
∂P ∂Q − ∂y ∂x
1 (1 − 2y) depends only on y, so an integrating factor is y 1 e −w(y) dy = e [2−(1/y)] dy = e2y−ln y = e2y . y 1 e2y dx + 2xe2y − dy = 0 is exact. y =
f (x, y) = xe2y − ln y,
and a one-parameter family of solutions is xe2y − ln y = C .
SECTION 19.3 1. y = y
y = Cex .
=⇒
Also, y(0) = 1
=⇒
C=1
Thus y = ex and y(1) = 2.71828 (a) 2.48832, relative error= 8.46%. (b) 2.71825, relative error= 0.001%. 2. y = x + y
=⇒
y = Cex − x − 1, y(0) = 2
=⇒
C=3
Thus y = 3e − x − 1 and y(1) 6.15485 x
(a) 5.46496, relative error = 11.2%. (b) 6.15474, relative error = 0%. 3.
(a) 2.59374, relative error= 4.58%. (b) 2.71828, relative error= 0%.
4.
(a) 5.78124, relative error= 6.07%. (b) 6.15482, relative error= 0%.
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SECTION 19.3 5. y = 2x
y = x2 + C.
=⇒
Also, y(2) = 5
=⇒
C=1
2
Thus y = x + 1 and y(1) = 2. (a) 1.9, relative error= 5.0%. (b) 2.0, relative error= 0%. 6. y = 3x2
=⇒
y = x3 + C.
Also, y(1) = 2
=⇒
C=1
3
Thus y = x + 1 and y(0) = 1. (a) 0.84500, relative error= 15.5%. (b) 1.0, relative error= 0%. 7.
9.
1 2y √ √ Thus y = x and y(2) = 2 1.41421. y =
1 3y 2 1 Thus y = x 3 and y(2) 1.25992. y =
8.
(a) 1.42052, relative error= −0.45%.
(a) 1.26494, relative error= −0.4%.
(b) 1.41421, relative error= 0%.
(b) 1.25992, relative error= 0%.
(a) 2.65330, relative error= 2.39%.
10.
(a) 5.95989, relative error= 3.17%.
(b) 2.71828, relative error= 0%.
(b) 6.15487, relative error= 0%.
PROJECT 19.3 1. (a) and (b) y' = y 3
2.5
2
1.5
1
0.5
0
−0.5
−1 −3
(c) y − y = 0
−2.5
H(x) =
−2
−1.5
−1
−0.5
−dx = −x;
0
d −x (e y) = 0 dx e−x y = C y = Cex =⇒
C = 1.
1
1.5
integrating factor: e−x
e−x y − e−x y = 0
y(0) = 1
0.5
Thus y = ex .
997
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SECTION 19.3
2. (a) and (b) y' = x+2y z
8
7
6
5
4
3
2
1
0 −1 −1
(c) y − 2y = x
−0.5
0
H(x) =
0.5
1
1.5
−2 dx = −2x;
2
integrating factor: e−2x
e−2x y − 2e−2x y = xe−2x d −2x (e y) = xe−2x dx 1 1 e−2x y = − xe−2x − e−2x + C 2 4 1 1 y = Ce2x − x − 2 4 y(0) = 1
=⇒
C=
5 . 4
Thus y =
5 2x 1 1 e − x− . 4 2 4
3. (a) and (b) y' = 2xy 8
7
6
5
4
3
2
1
0
−1 −1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
3
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SECTION 19.4 (c) y − 2xy = 0
H(x) =
−2x dx = −x2 ;
999
integrating factor: e−x
2
e−x y − 2xex y = 0 d −x2 (e y) = 0 dx 2
2
e−x y = C 2
2
y = Cex y(0) = 1
=⇒
2
Thus y = ex .
C = 1.
4. (a) and (b) y' = 4x/y 3
2
1
0
−1
−2
−3 −2
(c) y = −
−1.5
−1
−0.5
0
0.5
1
1.5
2
1 2 1 y = −2x2 + C or x2 + y 2 = C 2 4 5 =⇒ C = . Thus 4x2 + y 2 = 5. 4
4x ; y
y(1) = 1
SECTION 19.4 1. First consider the reduced equation. The characteristic equation is: r2 + 5r + 6 = (r + 2)(r + 3) = 0 and
u1 (x) = e−2x ,
u2 (x) = e−3x
are fundamental solutions. A particular solution of the given
equation has the form y = Ax + B. The derivatives of y are:
y = A,
y = 0.
Substituting y and its derivatives into the given equation gives 0 + 5A + 6(Ax + B) = 3x + 4.
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SECTION 19.4
Thus, 6A = 3 5A + 6B = 4 A = 12 , B = 14 ,
The solution of this pair of equations is: 2. The constant
yp = − 12
and
y=
1 2
x + 14 .
is a solution.
3. First consider the reduced equation. The characteristic equation is: r2 + 2r + 5 = 0 and
u1 (x) = e−x cos 2x,
u2 (x) = e−x sin 2x
are fundamental solutions. A particular solution of
the given equation has the form y = Ax2 + Bx + C. y = 2Ax + B,
The derivatives of y are:
y = 2A.
Substituting y and its derivatives into the given equation gives 2A + 2(2Ax + B) + 5(Ax2 + Bx + C) = x2 − 1. Thus, 5A = 1 4A + 5B = 0 2A + 2B + 5C = −1 4 27 A = 15 , B = − 25 , C = − 125 ,
The solution of this system of equations is: y= 4. We try
1 5
x2 −
4 25
x−
and
27 125 .
y = Ax3 + Bx2 + Cx + D :
y + y − 2y = (6Ax + 2B) + (3Ax2 + 2Bx + C) − 2(Ax3 + Bx2 + Cx + D) = x3 + x. =⇒ −2A = 1,
3A − 2B = 0,
6A + 2B − 2C = 1,
17 =⇒ A = − 12 , B = − 34 , C = − 11 4 , D =− 8 ;
2B + C − 2D = 0
yp = − 12 x3 − 34 x2 −
11 4 x
−
17 8 .
5. First consider the reduced equation. The characteristic equation is: r2 + 6r + 9 = (r + 3)2 = 0 and
u1 (x) = e−3x ,
u2 (x) = xe−3x
are fundamental solutions. A particular solution of the given
equation has the form y = Ae3x . The derivatives of y are:
y = 3Ae3x ,
y = 9Ae3x .
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1001
Substituting y and its derivatives into the given equation gives 9Ae3x + 18Ae3x + 9Ae3x = e3x . Thus,
36A = 1
=⇒
1 , 36
A=
and
y=
1 36
e3x .
6. Since −3 is a double root of the characteristic equation r2 + 6r + 9 = 0, we try y = Ax2 e−3x .
Then y = A(−3x2 + 2x)e−3x , y = A(9x2 − 12x + 2)e−3x , and
[A(9x2 − 12x + 2) + 6A(−3x2 + 2x) + 9Ax2 ]e−3x = e−3x , Thus
A=
or 2Ae−3x = e−3x
and yp = 12 x2 e−3x .
1 2
7. First consider the reduced equation. The characteristic equation is: r2 + 2r + 2 = 0 and
u1 (x) = e−x cos x,
u2 (x) = e−x sin x
are fundamental solutions. A particular solution of the
given equation has the form y = Aex . y = Aex ,
The derivatives of y are:
y = Aex .
Substituting y and its derivatives into the given equation gives Aex + 2Aex + 2Aex = ex . Thus,
5A = 1
=⇒
A=
8. Try y = (A + Bx)e−x . A = −2, B = 1;
1 5
and y =
1 5
ex .
Substituting into y + 4y + 4y = xe−x gives
yp = (x − 2)e−x
9. First consider the reduced equation. The characteristic equation is: r2 − r − 12 = (r − 4)(r + 3) = 0 and
u1 (x) = e4x ,
u2 (x) = e−3x
are fundamental solutions. A particular solution of the given
equation has the form y = A cos x + B sin x. The derivatives of y are:
y = −A sin x + B cos x,
y = −A cos x − B sin x.
Substituting y and its derivatives into the given equation gives −A cos x − B sin x − (−A sin x + B cos x) − 12(A cos x + B sin x) = cos x. Thus, −13A − B = 1 A − 13B = 0
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SECTION 19.4 13 1 A = − 170 , B = − 170 ,
The solution of this system of equations is:
13 cos x − y = − 170
1 170
and
sin x.
is a particular solution of the complete equation. Substituting into y − y − 12y = sin x gives
10. Try y = A cos x + B sin x. A=
1 170 ,
B=
−13 170 ;
yp =
1 170
cos x −
13 170
sin x.
11. First consider the reduced equation. The characteristic equation is: r2 + 7r + 6 = (r + 6)(r + 1) = 0 u1 (x) = e−6x ,
and
u2 (x) = e−x
are fundamental solutions. A particular solution of the given
equation has the form y = A cos 2x + B sin 2x. y = −2A sin 2x + 2B cos 2x,
The derivatives of y are:
y = −4A cos 2x − 4B sin 2x.
Substituting y and its derivatives into the given equation gives −4A cos 2x − 4B sin 2x + 7(−2A sin 2x + 2B cos 2x) + 6(A cos 2x + B sin 2x) = 3 cos 2x. Thus, 2A + 14B = 3 −14A + 2B = 0 The solution of this system of equations is: y= 12. Try y = A cos 3x + B sin 3x. 1 2 A = − 15 , B = − 15 ;
A=
3 10
3 100 ,
cos 2x +
21 100
B=
21 100
and
sin 2x.
Substituting into y + y + 3y = sin 3x gives
1 yp = − 15 cos 3x −
2 15
sin 3x.
13. First consider the reduced equation. The characteristic equation is: r2 − 2r + 5 = 0 and
u1 (x) = ex cos 2x,
u2 (x) = ex sin 2x
are fundamental solutions. A particular solution of the
given equation has the form y = Ae−x cos 2x + Be−x sin 2x The derivatives of y are:
y = −Ae−x cos 2x − 2Ae−x sin 2x − Be−x sin 2x + 2Be−x cos 2x,
y = 4Ae−x sin 2x − 3Ae−x cos 2x − 4Be−x cos 2x − 3Be−x sin 2x. Substituting y and its derivatives into the given equation gives 4Ae−x sin 2x − 3Ae−x cos 2x − 4Be−x cos 2x − 3Be−x sin 2x− 2 (−Ae−x cos 2x − 2Ae−x sin 2x − Be−x sin 2x + 2Be−x cos 2x) + 5 (Ae−x cos 2x + Be−x sin 2x) = e−x sin 2x.
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1003
Equating the coefficients of e−x cos 2x and e−x sin 2x we get, 8A + 4B = 1 4A − 8B = 0 The solution of this system of equations is: y=
1 10
A=
1 10 ,
e−x cos 2x +
B= 1 20
1 20
and
e−x sin 2x.
14. Try y = e2x (A cos x + B sin x). Substituting into y + 4y + 5y = e2x cos x gives
1 1 1 1 A = 20 , B = 40 ; yp = e2x 20 cos x + 40 sin x . 15. First consider the reduced equation. The characteristic equation is: r2 + 6r + 8 = (r + 4)(r + 2) = 0 and
u1 (x) = e−4x ,
u2 (x) = e−2x
are fundamental solutions. A particular solution of the given
equation has the form y = Axe−2x . y = Ae−2x − 2Axe−2x ,
The derivatives of y are:
y = −4Ae−2x + 4Axe−2x .
Substituting y and its derivatives into the given equation gives
−4Ae−2x + 4Axe−2x + 6 Ae−2x − 2Axe−2x + 8Axe−2x = 3e−2x Thus,
2A = 3
=⇒
A=
3 2
3 2
y=
xe−2x .
Substituting into y − 2y + 5y = ex sin x gives
16. Try y = ex (A cos x + B sin x) . A = 0, B = 13 ;
and
yp = 13 ex sin x.
17. First consider the reduced equation: y + y = 0.
The characteristic equation is:
r2 + 1 = 0 and
u1 (x) = cos x,
u2 (x) = sin x
are fundamental solutions. A particular solution of the given
equation has the form y = Aex . The derivatives of y are:
y = y = Aex .
Substitute y and its derivatives into the given equation: Aex + Aex = ex
=⇒
The general solution of the given equation is:
A=
1 2
and y =
1 2
ex .
y = C1 cos x + C2 sin x +
1 2
ex .
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SECTION 19.4
18. r2 − 2r + 1 = 0 =⇒ r = 1 =⇒ y = C1 ex + C2 xex is the general solution of the reduced equation. To find a particular solution, we try y = A cos 2x + B sin 2x.
Substituting into
y − 2y + y = −25 sin 2x gives A = −4, B = 3, so yp = 3 sin 2x − 4 cos 2x. y = C1 ex + C2 xex + 3 sin 2x − 4 cos 2x.
Therefore the general solution is:
19. First consider the reduced equation: y − 3y − 10y = 0.
The characteristic equation is:
r2 − 3r − 10 = (r − 5)(r + 2) = 0 and
u1 (x) = e5x ,
u2 (x) = e−2x
are fundamental solutions. A particular solution of the given
equation has the form y = Ax + B. y = A,
The derivatives of y are:
y = 0.
Substitute y and its derivatives into the given equation: −3A − 10(Ax + B) = −x − 1
=⇒
A=
1 10 ,
B=
7 100
and y =
1 10
x+
7 100
The general solution of the given equation is: y = C1 e5x + C2 e−2x +
1 10
x+
7 100
20. r2 + 4 = 0 =⇒ r = ±2i =⇒ y = C1 cos 2x + C2 sin 2x, general solution of reduced equation. Particular solution: try y = x(A + Bx) cos 2x + x(C + Dx) sin 2x. 1 Substituting into y + 4y = x cos 2x gives A = 16 , B = 0, C = 0, D = 18 ; 1 1 x cos 2x + x2 sin 2x. General solution: y = C1 cos 2x + C2 cos 2x + yp = 16 8 + 18 x2 sin 2x.
21. First consider the reduced equation: y + 3y − 4y = 0.
1 16 x cos 2x
The characteristic equation is:
r2 + 3r − 4 = (r + 4)(r − 1) = 0 and u1 (x) = ex ,
u2 (x) = e−4x
are fundamental solutions. A particular solution of the given equa-
tion has the form y = Axe−4x . y = Ae−4x − 4Axe−4x ,
The derivatives of y are:
y = −8Ae−4x + 16Axe−4x .
Substitute y and its derivatives into the given equation:
−8Ae−4x + 16Axe−4x + 3 Ae−4x − 4Axe−4x − 4Axe−4x = e−4x . This implies
−5A = 1,
so
A = − 15
and y = − 15 xe−4x .
The general solution of the given equation is:
y = C1 ex + C2 e−4x −
1 5
xe−4x .
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1005
22. r2 + 2r = 0 =⇒ r = 0, −2 =⇒ y = C1 + C2 e−2x , general solution of reduced equation. Particular solution: try y = A cos 2x + B sin 2x. Substituting into y + 2y = 4 sin 2x gives A = B = − 12 ; General solution:
y = C1 + C2 e−2x −
1 2
cos 2x −
1 2
yp = − 12 cos 2x −
1 2
sin 2x.
sin 2x.
23. First consider the reduced equation: y + y − 2y = 0.
The characteristic equation is:
r2 + r − 2 = (r + 2)(r − 1) = 0 and
u1 (x) = e−2x ,
u2 (x) = ex
are fundamental solutions. A particular solution of the given
equation has the form y = x(A + Bx)ex . The derivatives of y are: y = (A + (2B + A)x + Bx2 )ex ,
y = (2A + 2B + (4B + A)x + Bx2 )ex .
Substitute y and its derivatives into the given equation: (2A + 2B + (4B + A)x + Bx2 + A + (2B + A)x + Bx2 − 2Ax − 2Bx2 )ex = 3xex . This implies
A = − 13
,B =
1 2
so
y = x(− 13 + 12 x)ex .
The general solution of the given equation is:
y = C1 e−2x + C2 ex − 13 xex + 12 x2 ex .
24. r2 + 4r + 4 = 0 =⇒ r = −2 =⇒ y = C1 e−2x + C2 xe−2x , general solution of reduced equation. Particular solution: try y = x2 (A + Bx)e−2x . Substituting into y + 4y + 4y = xe−2x gives A = 0, B = 16 ; General solution:
yp = 16 x3 e−2x .
y = C1 e−2x + C2 xe−2x + 16 x3 e−2x .
25. Let y1 (x) be a solution of y + ay + by = φ1 (x), and let z = y1 + y2 .
let y2 (x) be a solution of y + ay + by = φ2 (x),
Then
z + az + bz = (y1 + y2 ) + a(y1 + y2 ) + b(y1 + y2 ) = (y1 + ay1 + by1 ) + (y2 + ay2 + y2 ) = φ1 + φ2 . 26. (a)
1 x− y = − 15
2 225
is a particular solution of y + 2y − 15y = x
y = − 17 e2x is a particular solution of y + 2y − 15y = e2x Therefore (b)
1 y = − 15 x−
2 225
− 17 e2x is a particular solution of y + 2y − 15y = x + e2x .
y = − 14 e−x is a particular solution of y − 7y − 12y = e−x . y=
7 226
cos 2x −
4 113
sin 2x is a particular solution of y − 7y − 12y = sin 2x.
Therefore y = − 14 e−x +
7 226
cos 2x −
y − 7y − 12y = e−x + sin 2x.
4 113
sin 2x is a particular solution of
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SECTION 19.4
27. First consider the reduced equation: y + 4y + 3y = 0.
The characteristic equation is:
r2 + 4r + 3 = (r + 3)(r + 1) = 0 and
u1 (x) = e−3x ,
u2 (x) = e−x
are fundamental solutions. Since
cosh x =
1 2
(ex + e−x ),
particular solution of the given equation has the form y = Aex + Bxe−x y = Aex + Be−x − Bxe−x
The derivatives of y are:
y = Aex − 2Be−x + Bxe−x .
Substitute y and its derivatives into the given equation: Aex − 2Be−x + Bxe−x + 4 (Aex + Be−x − Bxe−x ) + 3 (Aex + Bxe−x ) = Equating coefficients, we get
A=
1 16 ,
B = 14 ,
The general solution of the given equation is: 28. r2 + 1 = 0 =⇒ r = ±i.
and so y =
ex +
y = C1 e−3x + C2 e−x +
Fundamental solutions: u1 = cos x,
1 16
1 2 1 4
1 16
(ex + e−x ) . xe−x . ex +
1 4
xe−x .
u2 = sin x.
Wronskian: W = u1 u2 − u1 u2 = 1; φ(x) = 3 sin x sin 2x u2 φ 3 dx = − 3 sin2 x sin 2x dx = −6 sin3 x cos x dx = − sin4 x, z1 = − W 2 u1 φ 3 3 z2 = dx = 3 cos x sin x sin 2x dx = sin2 2x dx = (4x − sin 4x). W 2 16 Therefore
yp = z1 u1 + z2 u2 = − 32 sin4 x cos x +
3 16 (4x
29. First consider the reduced equation y − 2y + y = 0.
− sin 4x) sin x. The characteristic equation is:
r2 − 2r + 1 = (r − 1)2 = 0 and u1 (x) = ex ,
u2 (x) = xex
are fundamental solutions. Their Wronskian is given by
W = u1 u2 − u2 u1 = ex (ex + xex ) − xex (ex ) = e2x Using variation of parameters, a particular solution of the given equation will have the form y = u 1 z 1 + u 2 z2 , where
xex (xex cos x) dx = − x2 cos x dx = −x2 sin x − 2x cos x + 2 sin x, e2x x x e (xe cos x) dx = x cos x dx = x sin x + cos x z2 = e2x
z1 = −
Therefore,
y = ex −x2 sin x − 2x cos x + 2 sin x + xex (x sin x + cos x) = 2ex sin x − xex cos x.
a
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SECTION 19.4 30. r2 + 1 = 0.
Fundamental solutions: u1 = cos x, u2 = sin x.
W = u1 u2 − u1 u2 = 1; φ(x) = csc x. u2 φ dx = − sin x csc x dx = −x, z1 = − W u1 φ z2 = dx = cos x csc x dx = cot x dx = ln(sin x) W
Wronskian:
Therefore
1007
[sin x > 0 since 0 < x < π].
yp = z1 u1 + z2 u2 = −x cos x + ln(sin x) sin x.
31. First consider the reduced equation y − 4y + 4y = 0.
The characteristic equation is:
r2 − 4r + 4 = (r − 2)2 = 0 and u1 (x) = e2x ,
u2 (x) = xe2x
are fundamental solutions. Their Wronskian is given by
W = u1 u2 − u2 u1 = e2x e2x + 2xe2x − xe2x (2e2x ) = e4x .
Using variation of parameters, a particular solution of the given equation will have the form y = u1 z1 + u2 z2 , where
Therefore,
1
1 dx = − 13 x, z1 = − dx = − e4x 3 2x 1 −1 2x e 1 1 3 x e z2 = dx = 13 ln |x|. dx = e4x 3 x
xe2x
3
x−1 e2x
y = e2x − 13 x + xe2x 13 ln |x| = − 13 xe2x + Note: Since
u = − 13 xe2x
1 3
x ln |x| e2x .
is a solution of the reduced equation, y=
1 3
x ln |x| e2x
is also a particular solution of the given equation. 32. r2 + 4 = 0 =⇒ r = ±2i.
Fundamental solutions: u1 = cos 2x, u2 = sin 2x.
φ(x) = sec2 2x Wronskian: W = u1 u2 − u1 u2 = 2 cos2 2x + 2 sin2 2x = 2; u2 φ sin 2x 1 dx = − dx = − sec 2x, z1 = − W 2 cos2 2x 4 cos 2x 1 1 u1 φ z2 = dx = dx = sec 2x dx = ln | sec 2x + tan 2x|. W 2 cos2 2x 2 4 Therefore yp = −
1 1 1 sec 2x cos 2x + ln | sec 2x + tan 2x| sin 2x = − (1 − ln | sec 2x + tan 2x| sin 2x) . 4 4 4
33. First consider the reduced equation y + 4y + 4y = 0.
The characteristic equation is:
r2 + 4r + 4 = (r + 2)2 = 0
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SECTION 19.4
and u1 (x) = e−2x ,
u2 (x) = xe−2x
are fundamental solutions. Their Wronskian is given by
W = u1 u2 − u2 u1 = e−2x e−2x − 2xe2x − xe−2x (−2e−2x ) = e−4x .
Using variation of parameters, a particular solution of the given equation will have the form y = u1 z1 + u2 z2 , where
xe−2x x−2 e−2x 1 dx = − dx = − ln |x| e−4x x −2x −2 −2x e x e 1 1 z2 = dx = dx = − −4x 2 e x x
z1 = −
Therefore,
Note: Since
1 = − e−2x ln |x| − e−2x . y = e−2x (− ln |x|) + xe−2x − x u = − e−2x
is a solution of the reduced equation, we can take y = − ln | x |e2x .
34. r2 + 2r + 1 = 0 =⇒ r = −1.
u1 = e−x , u2 = xe−x .
Fundamental solutions:
Wronskian: W = u1 u2 − u1 u2 = (1 − x)e−2x + xe−2x = e−2x ; φ(x) = e−x ln x. u2 φ 1 2 x2 xe−x e−x ln x dx = − x ln x dx = − ln x + dx = − x , z1 = − W e−2x 2 4 −x −x e e ln x u1 φ dx = dx = ln x dx = x ln x − x. z2 = W e−2x Therefore −x
yp = z1 u1 + z2 u2 = e =
x2 1 2 − x ln x + xe−x (x ln x − x) 4 2
1 2 −x x e (2 ln x − 3) 4
35. First consider the reduced equation y − 2y + 2y = 0.
The characteristic equation is:
r2 − 2r + 2 = 0 and u1 (x) = ex cos x,
u2 (x) = ex sin x
are fundamental solutions. Their Wronskian is given by
W = ex cos x [ex sin x + ex cos x] − ex sin x [ex cos x − ex sin x] = e2x Using variation of parameters, a particular solution of the given equation will have the form y = u 1 z 1 + u 2 z2 , where
ex sin x · ex sec x dx = − tan x dx = − ln | sec x| = ln | cos x| e2x x e cos x · ex sec x z2 = dx = dx = x e2x
z1 = −
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SECTION 19.4
1009
Therefore, y = ex cos x (ln | cos x|) + ex sin x(x) = ex cos x ln | cos x| + xex sin x. 36. v + (2k + a)v + (k 2 + ak + b)v = cn xn + cn−1 xn−1 + · · · + c1 x + c0 . 37. Assume that the forcing function F (t) = F0 (constant). Then the differential equation has a particular solution of the form i = A. The derivatives of i are:
i = i = 0. Substituting i and its derivatives
into the equation, we get 1 A = F0 C
=⇒
A = CF0
=⇒
i = CF0 .
The characteristic equation for the reduced equation is: √ √ −R ± R2 − 4L/C −R C ± CR2 − 4L 1 √ = 0 =⇒ r1 , r2 = = Lr2 + Rr + C 2L 2L C (a) If CR2 = 4L, then the characteristic equation has only one root: and u1 = e−(R/2L)t ,
u2 = t e−(R/2L)t
r = −R/2L,
are fundamental solutions.
The general solution of the given equation is: i(t) = C1 e−(R/2L)t + C2 t e−(R/2L)t + CF0 and its derivative is: i (t) = −C1 (R/2L)e−(R/2L)t + C2 e−(R/2L)t − C2 (R/2L)t e−(R/2L)t . Applying the side conditions i(0) = 0,
i (0) = F0 /L,
we get
C1 + CF0 = 0 (−R/2L)C1 + C2 = F0 /L The solution is
C1 = −CF0 ,
The current in this case is:
C2 =
F0 (2 − RC). 2L
i(t) = −CF0 e−(R/2L)t +
F0 (2 − RC) t e−(R/2L)t + CF0 . 2L
(b) If CR2 − 4L < 0 then the characteristic equation has complex roots: 4L − CR2 r1 = −R/2L ± iβ, where β = (here i2 = −1) 4CL2 and fundamental solutions are: u1 = e−(R/2L)t cos βt,
u2 = e−(R/2L)t sin βt.
The general solution of the given differential equation is: i(t) = e−(R/2L)t (C1 cos βt + C2 sin βt) + CF0 and its derivative is: i (t) = (−R/2L)e−(R/2L)t (C1 cos βt + C2 sin βt) + βe−(R/2L)t (− C1 sin βt + C2 cos βt) .
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SECTION 19.4 i (0) = F0 /L,
Applying the side conditions i(0) = 0,
we get
C1 + CF0 = 0 (−R/2L)C1 + βC2 = F0 /L The solution is
C1 = −CF0 ,
C2 =
The current in this case is: i(t) = e−(R/2L)t 38. (a)
F0 (2 − RC). 2Lβ
F0 (2 − RC) sin βt − CF0 cos βt + CF0 . 2Lβ
x2 y1 − xy1 + y1 = x2 · 0 − x · 1 + x = 0 : y1 is a solution.
x2 y2 − xy2 + y2 = x2 x1 − x(ln x + 1) + x ln x = 0 : y2 is a solution. W = y1 y2 − y1 y2 = x(ln x + 1) − 1(x ln x) = x
is nonzero on (0, ∞).
(b) To use the method of variation of parameters as described in the text, we first re-write the equation in the form y −
1 1 4 y + 2 y = ln x. x x x
Then, a particular solution of the equation will have the form yp = y1 z1 + y2 z2 , 1 x ln x · [(4/x) ln x] 4 dx = −4 (ln x)2 dx = − (ln x)3 z1 = − x x 3 and
z2 =
x · [(4/x) ln x] dx = 4 x
Thus, yp = − 43 x (ln x)3 + x ln x · 2(ln x)2
39. (a) Let y1 (x) = sin ln x2 . Then
2 y1 = cos ln x2 x
and
ln x dx = 2(ln x)2 x yp = 23 x (ln x)3 .
which simplifies to:
y1
=−
4 x2
where
2
sin ln x
−
2 x2
cos ln x2
Substituting y1 and its derivatives into the differential equation, we have
2 2 4 2 2 2 − + x + 4 sin ln x2 = 0 sin ln x cos ln x cos ln x x2 − 2 2 x x x The verification that y2 is a solution is done in exactly the same way. The Wronskian of y1 and y2 is: W (x) = y1 y2 − y2 y1
2 2 sin ln x2 − cos ln x2 cos ln x2 = sin ln x2 − x x = − x2 = 0 on (0, ∞) (b) To use the method of variation of parameters as described in the text, we first re-write the equation in the form y + x−1 y + 4x−2 y = x−2 sin(ln x).
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SECTION 19.5 Then, a particular solution of the equation will have the form y = y1 z1 + y2 z2 , cos(ln x2 )x−2 sin(ln x) dx z1 = − −2/x = 12 cos(2 ln x)x−1 sin(ln x) dx = 12 cos 2u sin u du (u = ln x) = 12 (2 cos2 u − 1) sin u du = − 13 cos3 u + and
1 2
sin u
sin(ln x2 )x−2 sin(ln x) dx −2/x = − 12 sin(2 ln x)x−1 sin(ln x) dx (u = ln x) = − 12 sin 2u sin u du =− sin2 u cos u du
z2 =
Thus, y = sin 2u − 13
= − 13 sin3 u
cos3 u + 12 sin u − cos 2u 13 sin3 u y=
1 3
sin u =
1 3
which simplifies to:
sin(ln x).
SECTION 19.5 1. The equation of motion is of the form x (t) = A sin (ωt + φ0 ) . The period is T = 2π/ω = π/4. Therefore ω = 8. Thus x (t) = A sin (8t + φ0 ) and v (t) = 8A cos (8t + φ0 ) . Since x (0) = 1 and v (0) = 0, we have 1 = A sin φ0
and
0 = 8A cos φ0 .
These equations are satisfied by taking A = 1 and φ0 = π/2. Therefore the equation of motion reads
x (t) = sin 8t + 12 π . The amplitude is 1 and the frequency is 8/2π = 4/π. 1 =2 2. x(t) = A sin(ωt + φ0 ). ω = 2πf = 2π π 0 = x(0) = A sin φ0 , −2 = x (0) = ωA cos φ0 =⇒ A = 1, Amplitude 1,
period T =
1 f
= π.
φ0 = π.
where
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SECTION 19.5
3. We can write the equation of motion as
x (t) = A sin
Differentiation gives v (t) =
2πA cos T
2π t . T
2π t . T
The object passes through the origin whenever sin [(2π/T )] = 0. Then cos [(2π/T ) t] = ±1
and v = ±2πA/T .
2π
2π t + φ0 , v = x (t) = 2π T A cos T t + φ0 .
T 2 Note that x2 + 2π v = A2 .
T 2 At x = x0 , v = ±v0 , so A = x0 2 + 2π v0 = (1/2π) 4π 2 x20 + T 2 v02 .
4. x(t) = A sin
T
5. In this case φ0 = 0 and, measuring t in seconds, T = 6. Therefore ω = 2π/6 = π/3 and we have x (t) = A sin
π t , 3
v (t) =
π πA cos t . 3 3
Since v (0) = 5, we have πA/3 = 5 and therefore A = 15/π. The equation of motion can be written x (t) = (15/π) sin 6. (a) (b)
A sin(ωt + φ0 ) = A cos(ωt + φ0 − π2 );
3 πt
φ1 = φ0 − 12 π.
A sin(ωt + φ0 ) = A cos φ0 sin ωt + A sin φ0 cos ωt = B sin ωt + C cos ωt.
7. x (t) = x0 sin 8. (a)
take
1
k/m t + 12 π
maximum speed at
x = 0.
x = ±x0 .
(b)
zero speed at
(c)
maximum acceleration (in absolute value) at x = ±x0 .
(d)
zero acceleration at x = 0
(when total force is zero).
9. The equation of motion for the bob reads
x (t) = x0 sin t k/m + 12 π .
(Exercise 7)
k/m t + 12 π , the maximum speed is k/m x0 . The bob takes on half of that speed where cos k/m t + 12 π = 12 . Therefore √ √ sin k/m t + 12 π = 1 − 14 = 12 3 and x (t) = ± 12 3 x0 .
Since v (t) =
k/m x0 cos
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SECTION 19.5 10. v = −
11.
k x0 sin m
1013
k k t has maximum value x0 , so the maximum kinetic energy is m m
1 1 k 1 mv 2 = m x0 2 = kx0 2 . 2 2 m 2 k/m t + 12 π KE = 12 m[v(t)]2 = 12 m(k/m)x0 2 cos2 = 14 kx0 2 1 + cos 2 k/m t + π . 1 Average KE = 2π m/k
2π
√
m/k
0
1 kx0 2 1 + cos 2 k/m t + π dt 4
= 14 kx0 2 . 12. v(t) = −
k x0 sin m
k k 2 t =± x0 − [x(t)]2 . m m
y (t) = x (t) − 2,
13. Setting
we can write
x (t) = 8 − 4x (t)
as
y (t) + 4y (t) = 0.
This is simple harmonic motion about the point y = 0; that is, about the point x = 2. The equation of motion is of the form y (t) = A sin (2t + φ0 ) . The condition x(0) = 0 implies y(0) = −2 and thus A sin φ0 = −2
(∗)
Since y (t) = x (t) and y (t) = 2A cos(2t + φ0 ), the condition x (0) = 0 gives y (0) = 0, and thus (∗∗)
2A cos φ0 = 0.
Equations (*) and (**) are satisfied by A = 2, φ0 = 32 π. The equation of motion can therefore be written
3 y(t) = 2 sin 2t + π . 2
The amplitude is 2 and the period is π. sin θ = 1, θ
sin θ ∼ = θ for small θ. g/L t + φ0 (b) The general solution is θ(t) = A sin (i) Here A = θ0 and φ0 = π2 , so θ(t) = θ0 sin g/L t + π2 = θ0 cos g/L t . g g (ii) 0 = θ(0) = A sin φ0 , − θ0 = θ (0) = A cos φ0 =⇒ A = θ0 , φ0 = π. L L g/L t Therefore, the equation of motion becomes θ(t) = −θ0 sin g L 2π g (c) ω = , T = = 2π = 2 =⇒ L = 2 ∼ = 3.24 feet or 0.993 meters. L ω g π
14. (a) Since lim
θ→0
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SECTION 19.5
15. (a) Take the downward direction as positive. We begin by analyzing the forces on the buoy at a general position x cm beyond equilibrium. First there is the weight of the buoy: F1 = mg. This is a downward force. Next there is the buoyancy force equal to the weight of the fluid displaced; this force is in the opposite direction: F2 = −πr2 (L + x) ρ. We are neglecting friction so the total force is
F = F1 + F2 = mg − πr2 (L + x) ρ = mg − πr2 Lρ − πr2 xρ. We are assuming at the equilibrium point that the forces (weight of buoy and buoyant force of fluid) are in balance: mg − πr2 Lρ = 0. Thus, F = −πr2 xρ. By Newton’s (force = mass × acceleration)
F = ma we have ma = −πr2 xρ
and thus
a+
πr2 ρ x = 0. m
Thus, at each time t, x (t) + (b) The usual procedure shows that
πr2 ρ x (t) = 0. m
x(t) = x0 sin r πρ/m t + 12 π .
The amplitude A is x0 and the period T is (2/r) mπ/ρ. 16. Uniform circular motion consists of simple harmonic motion in both x and y, the two being out of phase by
π 2.
17. From (19.5.4), we have x(t) = Ae(−c/2m)t sin(ωt + φ0 ) =
A e(c/2m)t
√ sin(ωt + φ0 )
where
ω=
4km − ω 2 2m
A ω decrease. If c increases, then both the amplitude, (c/2m)t and the frequency 2π e
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SECTION 19.5 18. Assume that r1 > r2 . If C1 = 0
or C2 = 0,
If both C1 and C2 are nonzero, then
x = C1 er1 t + C2 er2 t
then
r1 t
C1 e
r2 t
+ C2 e
1015
can never be zero. C2 = − . Since C1
(r1 −r2 )t
= 0 implies
e
C2 at most once. C1 can be zero at most once. Therefore the
e(r1 −r2 )t is an increasing function (r1 > r2 ), it can take the value − By the same reasoning,
x (t) = C1 r1 er1 t + C2 r2 er2 t
motion can change direction at most once. 19. Set x(t) = 0 in (19.5.6). The result is: C1 e(−c/2m)t + C2 te(−c/2m)t = 0
=⇒
C1 + C2 t = 0
=⇒
t = −C1 /C2
Thus, there is at most one value of t at which x(t) = 0. The motion changes directions when x (t) = 0: x (t) = −C1 (c/2m)e(−c/2m)t + C2 e(−c/2m)t − C2 (c/2m)te(−c/2m)t . Now, x (t) = 0
=⇒
−C1 (c/2m) + C2 − C2 t(c/2m) = 0
=⇒
t=
C2 − C1 (c/2m) C2 (c/2m)
and again we conclude that there is at most one value of t at which x (t) = 0. 20. If γ = ω, we try xp = A cos γt + B sin γt as a particular solution of x + ω 2 x = −γ 2 xp + ω 2 xp =
Substituting xp into the equation, we get F0 /m giving xp = 2 cos γt. ω − γ2
F0 m
F0 cos γt. m
cos γt,
F0 /m cos(γt) ω2 − γ 2 If ω/γ = m/n is rational, then 2πm/ω = 2πn/γ is a period.
21. x(t) = A sin(ωt + φ0 ) +
22. If γ = ω, we try xp = At cos ωt + Bt sin ωt as a particular solution of x + ω 2 x =
F0 cos ωt. m
Substituting xp into the equation, we have F0 cos ωt, m
(2Bω − Aω 2 t) cos ωt − (2Aω + Bω 2 t) sin ωt + ω 2 (At cos ωt + Bt sin ωt) = which gives
A = 0, B =
F0 , as required. 2ωm
23. The characteristic equation is r2 + 2αr + ω 2 = 0;
the roots are
r1 , r2 = − α ±
α2 − ω 2
Since 0 < α < ω, α2 < ω 2 and the roots are complex. Thus, u1 (t) = e− αt cos βt, √ where β = ω 2 − α2 are fundamental solutions, and the general solution is: √ 2 x(t) = e− αt (C1 cos βt + C2 sin βt); β = α − ω 2 24. Straightforward computation.
u2 (t) = e− αt sin βt,
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REVIEW EXERCISES
25. Set ω = γ in the particular solution xp given in Exercise 24. Then we have xp = c = 2αm → 0+ ,
As
26.
F0 2αγm → ∞
the amplitude
2 F0 /m (ω − γ 2 ) cos γt + 2αγ sin γt 2 2 2 2 − γ ) + 4α γ ω2 − γ 2 2αγ F0 /m cos γt + sin γt = (ω 2 − γ 2 )2 + 4α2 γ 2 (ω 2 − γ 2 )2 + 4α2 γ 2 (ω 2 − γ 2 )2 + 4α2 γ 2 2 ω − γ2 , this expression becomes. Setting φ = tan−1 2αγ (ω 2
27.
F0 sin γt 2αγm
F0 /m (ω 2 − γ 2 )2 + 4α2 γ 2
F0 /m
(sin φ cos γt + cos φ sin γt) =
(ω 2 − γ 2 )2 + 4α2 γ 2
sin(γt + φ)
2
2 ω − γ 2 + 4α2 γ 2 = ω 4 + γ 4 + 2γ 2 (2α2 − ω 2 ) increases as γ increases.
28. (a) To maximize the amplitude, we need to minimize (ω 2 − γ 2 )2 + 4α2 γ 2 = ω 4 − 2(ω 2 − 2α2 )γ 2 + γ 4 . This is a parabola in γ 2 , and the minimum occurs when γ 2 = ω 2 − 2α2 . √ Therefore the maximum amplitude occurs when γ = ω 2 − 2α2 2π 2π (b) f = =√ γ ω 2 − 2α2 F0 /m F /m √0 (c) When γ 2 = ω 2 − 2α2 , the amplitude is: = 2 2 2 2 2 2α ω 2 − α2 (ω − γ ) + 4α γ (d) Since 2α = c/m,
the resonant amplitude in (c) can be rewritten
F0 . 2 c ω − c2 /4m2
This gets large as c gets small.
REVIEW EXERCISES 1. The equation is linear:
H(x) =
1dx = x =⇒ eH(x) = ex
d x (e y) = 2e−x =⇒ ex y = −2e−x + C; dx 2. Since
∂(2x3 y + 4y 3 ) ∂(3x2 y 2 ) = , ∂x ∂y
f (x, y) =
the solution is: y = −2e−2x + Ce−x
the equation is exact.
3x2 y 2 dx = x3 y 2 + φ(y).
∂f = 2x3 y + φ (y) = 2x3 y + 4y 3 =⇒ φ (y) = 4y 3 ∂y The solution is:
x3 y 2 + y 4 = C
and φ(y) = y 4 .
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REVIEW EXERCISES 3. The equation is separable: 1 y dy = dx = sec2 x dx y2 + 1 cos2 x 1 ln(y 2 + 1) = tan x + C 2 The solution is:
ln(1 + y 2 ) = 2 tan x + C
4. The equation is separable: y ln y dy = xex dx y ln y dy = xex dx 1 2 1 y ln y − y 2 = xex − ex + C 2 4 The solution is:
1 1 2 y ln y − y 2 = xex − ex + C 2 4 y −
5. The equation can be written
2 1 y = 2 y2 x x
a Bernoulli equation.
2 −1 1 y = 2 x x Let v = y −1 . Then v = −y −2 y , and we get the linear equation y −2 y −
v + Integrating factor:
H(x) =
1 2 v=− 2 x x .
(2/x) dx = ln x2 and eH(x) = x2 . x2 v + 2xv = −1 x2 v = −x + C v=−
The solution for the original equation is y =
C −x C 1 + 2 = x x x2
x2 C −x
cos x 2 6. Rewrite the equation as y + y = 2 , a linear equation. x x 2 H(x) H(x) = (2/x) dx = ln x ; e = x2 . x2 y + 2xy = cos x x2 y = sin x + C y= 7. Since
C sin x + 2 2 x x
∂(y sin x + xy cos x) ∂(x sin x + y 2 ) = sin x + x cos x = , the equation is exact. ∂y ∂x f (x, y) = (y sin x + xy cos x) dx = xy sin x + φ(y).
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1018 ∂f ∂y
January 4, 2007
REVIEW EXERCISES = x sin x + φ (y) = y 2 + x sin x =⇒ φ (y) = y 2
φ(y) = 13 y 3
=⇒
The solution is x sin x + 13 y 3 = C 8.
1 1 1 (Py − Qx ) = (2y − y) = . Q x xy (1/x) dx
Therefore, e
= x is an integrating factor; (x3 + xy 2 + x2 ) dx + (x2 y dy) = 0
is exact. 1 f (x, y) = x2 y dy = x2 y 2 + ψ(x). 2 ∂f = xy 2 + ψ (x) = x3 + xy 2 + x2 =⇒ ψ (x) = x3 + x2 ∂x The solution is:
1 4 4x
and ψ(x) =
+ 13 x3 + 12 x2 y 2 = C or 3x4 + 4x3 + 6x2 y 2 = C
9. The equation is separable: 1+y dy = (x2 − 1) dx y 1 ln |y| + y = x3 − x + C 3 10. The equation is a Bernoulli equation. We write it as y −1/3 y −
3 2/3 y = x4 x
Let v = y 2/3 . Then v = 23 y −1/3 y , and get the linear equation v − H(x) = −
(2/x) dx = ln x−2 ;
2 2 v = x4 x 3
eH(x) = x−2 . 2 2 x 3 2 x−2 v = x3 + C 9 2 v = x5 + Cx2 9
2 5 2 3/2 y = 9 x + Cx .
x−2 v − 2x−3 v =
Therefore,
y 2/3 = 29 x5 + Cx2
or
11. The equation can be written as y + H(x) = ln x2 ;
2 y = x2 , a linear equation. x
eH(x) = x2 . x2 y + 2xy = x4 1 5 x +C 5 1 y = x3 + Cx−2 5
x2 y =
1 4 1 3 x + x . 4 3
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REVIEW EXERCISES
1019
dy 3y 2 + 2xy ; a homogeneous equation. = dx 2xy + x2 dy dv Set v = y/x. Then y = vx and =v+x . dx dx
12. Rewrite the equation as
v+x x
3v 2 x2 + 2x2 v 3v 2 + 2v dv = = 2 2 dx 2x v + x 2v + 1 3v 2 + 2v v2 + v dv = −v = dx 2v + 1 2v + 1
2v + 1 1 dv = dx 2 v +v x ln |v 2 + v| = ln |x| + C v 2 + v = Cx Replacing v by y/x, we get y 2 + xy = Cx3 . 13. The differential equation is homogeneous. dv dy =v+x . Set v = y/x. Then y = vx and dx dx v+x
dv x2 + x2 v 2 1 + v2 = = dx 2x2 v 2v
x
1 + v2 1 − v2 dv = −v = dx 2v 2v
2v 1 dv = dx 1 − v2 x − ln |1 − v 2 | = ln |x| + C 1 − v2 =
C x
Replacing v by y/x, we get x2 − y 2 = Cx. Applying the initial condition y(1) = 2 gives C = −3. The solution of the initial-value problem is: x2 + 3x − y 2 = 0. 2y = x; the equation is linear. 14. Write the equation as y + x H(x) = (2/x) dx = ln x2 and eH(x) = x2 . x2 y + 2xy = x3 1 4 x +C 4 1 y = x2 + Cx−2 4
x2 y =
Applying the initial condition y(1) = 0 gives C = −1/4
and
y=
1 x2 − 2 4 4x
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REVIEW EXERCISES
∂(x + y)2 ∂(2xy + x2 − 1) = 2x + 2y = , the equation is exact. ∂y ∂x 1 f (x, y) = (x + y)2 dx = (x2 + 2xy + y 2 ) dx = x3 + x2 y + xy 2 + φ(y). 3
15. Since
∂f = x2 + 2xy + φ (y) = 2xy + x2 − 1 =⇒ φ (y) = −1 ∂y The general solution is:
1 3 3x
φ(y) = −y
=⇒
+ x2 y + xy 2 − y = C
Applying the initial condition y(1) = 1 gives C = 4/3. The solution of the initial-value problem is: 1 3 3x
+ x2 y + xy 2 − y = 4/3
16. The equation is separable: y
dy = 4x dx +1 y 2 + 1 = 2x2 + C
y2
y 2 = (2x2 + C)2 − 1 √ Applying the initial condition y(0) = 1 gives C = 2. The solution of the initial-value problem is: √ y 2 = (2x2 + 2)2 − 1. 17. The equation is a Bernoulli equation; rewrite it as: y −2 y + xy −1 = x. Set v = y −1 . Then v = −y −2 y , and we have v − xv = −x, 2 a linear equation. H(x) = (−x) dx = − 12 x2 and eH(x) = e−x /2 e−x
2
/2
v − xe−x
/2
v = −xe−x
e−x
/2
v = e−x
2
2
2
2
/2
/2
+C 2
v = 1 + Cex y=
/2
1 1 + Cex2 /2
Applying the initial condition y(0) = 2 gives C = −1/2. The solution of the initial-value problem 2 is: y = . 2 − ex2 /2 18. The equation is separable: dy = dx y − y2 1 1 + dy = dx y 1−y y = x+C ln |y| − ln |1 − y| = ln 1 − y
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REVIEW EXERCISES
1021
Applying the initial condition y(0) = 2 gives C = ln 2. The solution of the initial-value problem is: y 2 = x + ln 2 or y = ln . 1−y 2 + e−x 19. The characteristic equation is: r2 − 2r + 2 = 0. The roots are: r1 , r2 = 1 ± i. The general solution is: ex (C1 cos x + C2 sin x). 20. The characteristic equation is: r2 + r +
1 4
= 0. The roots are: r1 = r2 = −1/2.
The general solution is: y = C1 e−x/2 + C2 xe−x/2 . 21. The characteristic equation for the reduced equation is: r2 − r − 2 = 0. The roots are: r = 2, −1. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation: z = A cos 2x + B sin 2x z = −2A sin 2x + 2B cos 2x z = −4A cos 2x − 4B sin 2x Substituting z, z , z into the differential equation yields the pair of equations: −6A − 2B = 0, 2A − 6B = 1 The general solution is: y = C1 e2x + C2 e−x +
1 20
=⇒
cos 2x −
A= 3 20
3 1 , B=− . 20 20
sin 2x
22. The characteristic equation is: r2 − 4r = 0. The roots are: r1 = 4, r2 = 0. The general solution is: y = C1 e4x + C2 . 23. The characteristic equation for the reduced equation is: r2 − 6r + 9 = 0. The roots are: r1 = r2 = 3. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation. Since z = e3x and z = xe3x are solutions of the reduced equation, set z = Ax2 e3x . z = Ax2 e3x z = 2Axe3x + 3Ax2 e3x z = 2Ae3x + 12Axe3x + 9Ax2 e3x Substituting z, z , z into the differential equation gives: 2A = 3
=⇒
The general solution is: y = C1 e3x + C2 xe3x + 32 x2 e3x .
A=
3 . 2
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REVIEW EXERCISES
24. The characteristic equation for the reduced equation is: r2 + 1 = 0. The roots are: r1 = i, r2 = −i. Use variation of parameters to find a particular solution of the nonhomogeneous equation. Set u1 = cos x and u2 = sin x. Then their Wronskian is W (x) = 1. 1 z1 = − sin x sec3 x dx = − sec2 x, z2 = cos x sec3 x dx = tan x 2 yp = − 12 sec x + tan x sin x. The general solution of the equation is:
y = C1 cos x + C2 sin x −
1 2
sec x + tan x sin x
25. The characteristic equation for the reduced equation is: r2 − 2r + 1 = 0. The roots are: r1 = r2 = 1. Use variation of parameters to find a particular solution of the nonhomogeneous equation. Set u1 = ex and u2 = xex . Then their Wronskian is W (x) = e2x . x xex (1/x)ex e (1/x)ex dx = − dx = −x z = dx = (1/x) dx = ln x z1 = − 2 e2x e2x yp = −xex + xex ln x The general solution of the equation is:
y = C1 ex + C2 xex + xex ln x, x > 0
26. The characteristic equation for the reduced equation is: r2 − 5r + 6 = 0. The roots are: r1 = 2, r2 = 3. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation: z = A cos x + B sin x + C z = −A sin x + B cos x z = −A cos x − B sin x Substituting z, z , z into the differential equation yields the equations: 5A − 5B = 0, 5A + 5B = 2, 6C = 4 The general solution is: y = C1 e2x + C2 e3x +
1 5
=⇒
cos x +
1 5
A=
1 1 2 , B= , C= . 5 5 3
sin x + 23 .
27. The characteristic equation for the reduced equation is: r2 + 4r + 4 = 0. The roots are: r1 = r2 = −2. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation: z = Ax2 e−2x + Be2x z = 2Axe−2x − 2Ax2 e−2x + 2Be2x z = 2Ae−2x − 8Axe−2x + 4Ax2 e−2x + 4Be2x Substituting z, z , z into the differential equation yields the equations: 2A = 4, 16B = 2
=⇒
A = 2, B =
The general solution is: y = C1 e−2x + C2 xe−2x + 18 e2x + 2x2 e−2x
1 . 8
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REVIEW EXERCISES
1023
28. The characteristic equation for the reduced equation is: r2 + 4 = 0. The roots are: r1 = 2i, r2 = −2i. Use variation of parameters to find a particular solution of the nonhomogeneous equation. Set u1 = cos 2x and u2 = sin 2x. Then their Wronskian is W (x) = 2. sin 2x tan 2x 1 1 − cos2 2x 1 1 dx = − dx = − ln | sec 2x + tan 2x| + sin 2x z1 = − 2 2 cos 2x 4 4 1 1 cos 2x tan 2x z2 = dx = sin 2x dx = − cos 2x 2 2 4 yp = − 14 cos 2x ln | sec 2x + tan 2x| The general solution of the equation is:
y = C1 cos 2x + C2 sin 2x −
1 4
cos 2x ln | sec 2x + tan 2x|
29. First find the general solution of the differential equation. The characteristic equation for the reduced equation is: r2 + r = 0. The roots are: r1 = −1, r2 = 0. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation: Set z = Ax2 + Bx z = Ax2 + Bx z = 2Ax + B z = 2A Substituting z, z , z into the differential equation yields the equations: 2A = 1, A + B = 0
=⇒
A = 1/2, B = −1.
The general solution of the differential equation is: y = C1 e−x + C2 + 12 x2 − x Applying the initial conditions y(0) = 1, y (0) = 0, we get the pair of equations C1 + C2 = 1, −C1 − 1 = 0, The solution of the initial-value problem is:
=⇒
C1 = −1, C2 = 2.
y = 2 − e−x + 12 x2 − x
30. First find the general solution of the differential equation. The characteristic equation for the reduced equation is: r2 + 1 = 0. The roots are: r1 = i, r2 = −i. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation: Set z = A cos 2x + B sin 2x + Cx cos x + Dx sin x z = A cos 2x + B sin 2x + Cx cos x + Dx sin x z = −2A sin 2x + 2B cos 2x + C cos x − Cx sin x + D sin x + Dx cos x z = −4A cos 2x − 4B sin 2x − 2C sin x − Cx cos x + 2D cos x − Dx sin x Substituting z, z , z into the differential equation yields the equations: −3A = 4, −3B = 0, −2C = −4, 2D = 0 ⇒ A = −4/3, B = 0, C = 2, D = 0.
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REVIEW EXERCISES
The general solution of the differential equation is: 4 cos 2x + 2x cos x 3
y = C1 cos x + C2 sin x −
Applying the initial conditions y(π/2) = −1, y (π/2) = 0, we get C1 = −π − 16/13, C2 = −25/13. The solution of the initial-value problem is: y = −π cos x −
7 4 sin x − cos 2x + 2x cos x 3 3
31. First find the general solution of the differential equation. The characteristic equation for the reduced equation is: r2 − 5r + 6 = 0. The roots are: r1 = 2, r2 = 3. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation: Set z = Axe2x z = Axe2x z = Ae2x + 2Axe2x z = 4Ae2x + 4Axe2x Substituting z, z , z into the differential equation gives: −A = 10, A = −10. The general solution of the differential equation is: y = C1 e2x + C2 e3x − 10xe2x Applying the initial conditions y(0) = 1, y (0) = 1, we get C1 = −8, C2 = 9. The solution of the initial-value problem is:
y = 9e3x − 8e2x − 10xe2x
32. First find the general solution of the differential equation. The characteristic equation is: r2 + 4r + 4 = 0. The roots are: r1 = r2 = −2. The general solution of the differential equation is: y = C1 e−2x + C2 xe−2x Applying the initial conditions y(−1) = 2, y (−1) = 1, yields the equations C1 e2 − C2 e2 = 2, −2C1 e2 + 3C2 e2 = 1, The solution of the initial-value problem is:
=⇒
C1 = 7e−2 , C2 = 5e−2 .
y = 7e−2(x+1) + 5xe−2(x+1)
33. Assume x(t) = A sin(wt + φ0 ). From T = 2π/ω = π/2, ω = 4 and x(t) = A sin(4t + φ0 ) x(0) = 2
=⇒
A sin(φ)0) = 2;
x (0) = 0
=⇒
4A cos(φ0 ) = 0
=⇒
φ0 =
π and A = 2. 2
Therefore, x(t) = 2 sin(4t + π/2);
amplitude A = 2;
frequency 2/π.
34. Assume x(t) = A sin(wt + φ0 ). The period T = 8. Therefore 2π/ω = 8 which implies ω = π/4 and x(t) = A sin
The condition x(0) = 0 implies that φ0 = 0. Therefore, x(t) = A sin 14 π t .
1
4π t
+ φ0 .
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REVIEW EXERCISES The condition x (0) = 8 =
π 4 A cos 0
1025
implies A = 32/π. Hence 1 32 sin πt . x(t) = π 4
35. Assume that the downward direction is positive. Then 1 4x (t) = −64x(t) + 8 sin 4t, x(0) = − , x (0) = 0 2 This equation can be written as x + 16x = 2 sin 4t The characteristic equation for the reduced equation is: r2 + 16 = 0 and the roots are r = ±4i. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation: Set z = At cos 4t + Bt sin 4t z = At cos 4t + Bt sin 4t z = A cos 4t − 4At sin 4t + B sin 4t + 4Bt cos 4t z = −8A sin 4t − 16At cos 4t + 8B cos 4t − 16Bt sin 4t Substituting z, z , z into the differential equation yields the equations, we get A = − 14 , B = 0. The general solution of the differential equation is: 1 x(t) = C1 cos 4t + C2 sin 4t − t cos 4t. 4 Applying the initial conditions x(0) = −1/2, x (0) = 0, we get C1 = −1/2, C2 = 1/16. The equation of motion is: 1 1 1 sin 4t x(t) = − cos 4t − t cos 4t + 2 4 16 36. Assume that the downward direction is positive. Then 10x (t) = −60x(t) − 50x (t) + 4 sin t, x(0) = 0, x (0) = −1. The differential equation can be written as x + 5x + 6x =
2 sin t 5
The characteristic equation for the reduced equation is: r2 + 5r + 6 = 0 and the roots are: r1 = −2, r2 = −3. Use undetermined coefficients to find a particular solution of the nonhomogeneous equation: Set z = A cos t + B sin t z = A cos t + B sin t z = −A sin t + B cos t z = −A cos t − B sin t
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REVIEW EXERCISES
Substituting z, z , z into the differential equation gives A = −1/25, B = 1/25. The general solution of the differential equation is: x(t) = C1 e−2t + C2 e−3t −
1 25
cos t +
1 25
sin t.
Applying the initial conditions x(0) = 0, x (0) = −1, we get C1 = −23/25, C2 = 24/25. The equation of motion is: x(t) = −
1 23 −2t 24 −3t 1 e cos t + sin t. + e − 25 25 25 25
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