Calculus II 9781461250265, 1461250269

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Geometry Formulas A

Area of rectangle

lw

=

circle A = wr 2 A = ! bh

triangle

= 4wr 2

A

Surface Area of sphere

cylinder A

2wrh

=

V = lwh

Volume of box sphere

V

= 1wr3

cylinder

V

= wr 2h

t

V=

cone

x (height)

(area of base)

Trigonometrie Identities Pythagorean

cos 2(} + sin 2(} Parity sin(-(})

=

csc( - (})

= -

Co-relations

cos ()

=

sin(

=

1, 1 + tan 2(}

=

sec 2(}, cot 2(} + 1 = csc 2(}

-sin(}, cos(-(})

=

cos(}, tan(-(})

=

csc (), sec( - (})

=

sec (), cot( - (})

= -

i -() ), csc ()

=

sec(

i -() ), cot ()

=

-tan(} cot ()

tan(

i -() )

Addition formulas sin( () + 0)

~O)

o fore\.' (F~O)

Gravity tron~cr

than prll1g

x

Gravlty ba lances prll1g

t

Upward forc< (/-'< 01

x

GrJvity we.kcr than pring.

a minus sign in the formula for F because the force, being directed toward the equilibrium, has the opposite sign to x. Substituting formula (2) into Newton's law (1) gives

_kx=m d2x or d 2x = -(k)x. dt 2 dt 2 m It is convenient to write the ratio k/m as (./, where w =Jk/m constant. This substitution gives us the spring equation:

IS

a new

2 An example of a physical problem in which F depends on both x and t is the motion of a charged partic1e in a time-varying electric or magnetic field-see Exercise n, Scction 14.7.

8.1 Oscillations

371

d 2x 2 -=-wx. (3) dt 2 Since x is an unknown function of t, we cannot find dx I dt by integrating the right-hand side. (In particular, dxldt is not - !W 2X 2 + C, since it is trather than x which is the independent variable.) Instead, we shall begin by using trial and error. A good first guess, guided by the observation that weights on springs bob up and down, is x = sinto Differentiating twice with respect to t, we get 2 . -d x = -slnt = -x.

dt 2 The factor w2 is missing, so we may be tempted to try x we get d 2x dt 2

= -w 2sint

= w2sint. In this case,

'

which is again - X. To bring out a new faetor when we differentiate, we must take advantage of the chain rule. If we set x = sinwt, then dx dt

d(wt) dt

d 2x dt 2

= -w 2 sinwt = -w 2x

- = coswt-- = wcoswt and '

which is just what we wanted. Looking back at our wrong guesses suggests that it would not hurt to put a constant factor in front, so that x = Bsinwt is also a solution for any B. Finally, we note that coswt is another solution. In fact, if A and Bare any two constants, then x = A coswt + B sinwt

(4)

is a solution of the spring equation (3), as you may verify by differentiating (4) twice. We say that the solution (4) is a superposition of the two solutions A sinwt and Bcoswt. Example 1

Solution

Let x = f(t) = A coswt + Bsinwt. Show that xis periodic with period 277lw; that is, f(t + 2771 w) = f(t). Substitute t + 2771 w for t:

f( t +

2:) = = =

A cos[

w( t +

2: )]

+ B sin[ w( t +

2: )]

+ 277 ] + B sin[ wt + 277 ] A coswt + B sinwt = f(1). A cos[ wt

Here we used the fact that the sine and eosine functions are themselves periodic with period 277 . .& The constants A and Bare similar to the constants which arise when antiderivatives are taken. For any value of A and B, we have a solution. If we assign particular values to A and B, we get a partieular solution. The choice of particular values of A and B is often determined by specifying initial eonditions.

372

Chapter 8 Differential Equations

Example 2

Solution

Find a solution of the spring equation d 2x I dt 2 = dx I dt = 1 when t = O.

- w2x

for which x

= 1 and

In the solution x = A coswt + B sinwt, we have to find A and B. Now x = A when t = 0, so A = 1. Also dx I dt = wB cos wt - wA sin wt = wB when t = O. To make dxldt = 1, we choose B = l/w, and so the required solution is x = coswt + (1/w) sinwt . ... In general, if we are given the initial conditions that x = Xo and dx I dt = Vo when t = 0, then Vo . x = xocoswt + - smwt (5) w is the unique function of the form (4) which satisfies these conditions.

Example 3

Solution

Solve for x: d 2xl dt 2 =

-

= 0 and dxl dt = 1 when t = O. Here Xo = 0, Vo = 1, and w = I, so x = xocoswt + (volw)sinwt = sinto x, x

...

Physicists expect that the motion of a particle in a force field is completely determined once the initial values of position and velocity are specified. Our solution of the spring equation will meet the physicists' requirements if we can show that every solution of the spring equation (3) is of the form (4). We turn to this task next. In deriving formula (5) we saw that there are enough solutions of the form (4) so that x and dx I dt can be specified arbitrarily at t = O. Thus if x = f(t) is any solution of equation (3), then the function g(t) = f(O)cos wt + [f'(O)/w]sinwt is a solution of the special form (4) with the same initial conditions as f: g(O) = f(O) and g'(O) = 1'(0). We will now show that f(t) = g(t) for all t by using the following fact: if h(t) is any solution of equation (3), then the quantity E = t {[h'(t)]2 + [wh(t)f} is constant over time. This expression is called the energy of the solution h. To see that E is constant over time, we differentiate using the chain rule:

~~

= h'(t)h"(t)

+ w2h(t)h'(t) = h'(t){ h"(t) + w2h(t)}.

(6)

This equals zero since h" + w 2h = 0; thus E is constant over time. Now if f and gare solutions of equation (3) with j(O) = g(O) and 1'(0) = g'(O), then h(t) = f(t) - g(t) is also a solution with h(O) = 0 and h'(O) = O. Thus the energy E = t {[h,(t)]2 + [wh(t)f} is constant; but it vanishes at t = 0, so it is identically zero. Thus, since two non-negative numbers which add to zero must both be zero: h'(t) = 0 and wh(t) = O. In particular, h(t) = 0, and so f(t) = g(t) as required. The solution (4) of the spring equation can also be expressed in the form

x

= acos(wt - 8),

where a and 8 are constants. In fact, the addition formula for cosine gives

= a coswt cos8 + a sinwtsin8. This will be equal to A cos wt + B sin wt if acos8 = A and asin8 = B. a cos(wt - 8)

(7)

Thus a and () must be the polar coordinates of the point whose cartesian coordinates are (A,B), and so we can always find such an a and 8 with a ;;;. O. The form (7) is convenient for plotting, as shown in Fig. 8.1.2. In Fig. 8.1.2 notice that the solution is a cosine curve with amplitude a which is shifted by the phase shift 81 W. The number w is called the angular

8.1 Oscillations

Figure 8.1.2. The graph of x = a cos(wt - 0).

!L w

373

= phase shift

frequeney, since it is the time rate of change of the "angle" wt - (} at which the eosine is evaluated. The number of oscillations per unit time is the frequeney w/2rr (= l/period). The motion described by the solutions of the spring equation is called simple harmonie motion. It arises whenever a system is subject to a restoring force proportional to its displacement from equilibrium. Such oscillatory systems occur in physics, biology, electronics, and chemistry.

Simple Harmonie Motion Every solution of the spring equation d 2x = -w 2x has the form dt 2 where A and Bare constants. The solution can also be written

x

x = A coswt + B sinwt,

= O'cos(wt - 0),

where (0',0) are the polar coordinates of (A,B). [This function is graphed in Fig. 8.1.2.] If the values of x and dx / dt are specified to be X o and Vo at t = 0, then the unique solution is

x = xocoswt

Example 4 Solution

+ (vo/w)sinwt.

Sketch the graph of the solution of d 2x/ dt 2 + 9x = 0 satisfying x = 1 and dx/dt = 6 when t = O. Using (5) with w = 3,

Xo =

1, and

Vo =

6, we have

x = cos(3t) + 2sin(3t) = O'cos(3t - 0). Since (A,B) 0'

= (1,2), and (0',0) are its polar coordinates,

= ~12 + 22 = ß ~ 2.2

and

(} = tan -12 ~ 1.1 radians (or 63°), so 0/ w ~ 0.37. The period is shown in Fig. 8.1.3. ~

7'

= 2rr / w ~ 2.1. Thus we can plot the graph as

374

ehapter 8 Differential Equations x 1---2.1~

2

3

I

Figure 8.1.3. The graph of x = 2.2 cos(3 t - 1.1).

As usual, the independent variable need not always be called t, nor does the dependent variable need to be called x. Example 5

Solution

(a) Solve for y: dy / dx 2 + 9y = 0, Y = 1 and dy / dx = -I when x = O. (b) Sketch the graph of y as a function of x. (a) Here Yo = I, Vo = -I, and w = J9 = 3 (using x in place of t and y in place of x), so y = yocoswx

. + -Vo smwx = cos 3x w

I' 3 - 3sm x.

Jl+l79

(b) The polar coordinates of (I, -t) are given by oe = = fo/3 ~ 1.05 and O=tan-I(-n~-0.32 (or -18°). Hencey=acos(wx-O) becomes y = 1.05 cos(3x + 0.32), which is sketched in Fig. 8.1.4. Here 0/ w = -0.1 and 2.,,/w = 2.1. • y

x

Figure 8.1.4. The graph of

y = (1.05)cos(3x + 0.32).

ARemark on Notation. Up until now we have distinguished variables, which

are mathematical objects that represent "quantities," and funetions, which represent relations between quantities. ThUS, when y = f(x}, we have written j'(x) and dy / dx but not y', df/ dx, or y(x). It is common in mathematical writing to use the same symbol to denote a function and its dependent variable; thus one sometimes writes y = y(x) to indicate that y is a function of x and then writes "y' = dy / dx," "y(3) is the value of y when x = 3," and so on. Beginning with the next example, we will occasionally drop our scruples in distinguishing functions from variables and will use this abbreviated notation. Example 6

Let M be a weight with mass 1 gram on aspring with spring constant ~. Let the weight be initially extended by a distance of I centimeter moving at a velocity of 2 centimeters per second. (a) How fast is M moving at t = 3? (b) What is M's acceleration at t = 4? (c) What is M's maximum displacement from the rest position? When does it occur? (d) Sketch a graph of the solution.

8.1 Oscillations

Solution

375

Let x = x(t) denote the position of M at time t. We use the spring equation (3) with w = ~k Im, where k is the spring constant and m is the mass of M. Since k = ~ and m = 1, w is b/2. At t = 0, M is extended by a distance of 1 centimeter and moving at a velocity of 2 centimeters per second, so X o = 1 and Vo= 2. Now we have all the information we need to solve the spring equation. Applying formula (5) gives

x(t)

=

cosb/2 t

+ _2_ sinb/2 t. b/2

(a) x'(t)

= - ~3/2 sin~3/2 t + 2cosb/2 t. Substituting t = 3 gives

x' (3)

=-

b 12 sin 3b 12 + 2 cos 3b 12 ~ - 1.1 centimeters per second.

(Negative velocity represents upward motion.) (b) x"(t)

= ~( - { f sin{f t+ 2COS{f t) = -

icos{f t -{6 sin{f t

and thus the acceleration at t = 4 is

x"(4)

= - ~cos2{6

- {6 sin2{6 ~ 2.13 centimeters per second2

(c) The simp1est way to find the maximum displacement is to use the "phase-amplitude" form (7). The maximum displacement is the amplitude

= ~A 2 + B 2 , where A and Bare the coefficients of sin and cos in the solution. Here, A = 1 and B = 2/b/2, so 0: 2 = 1 + 4/(3/2) = 1 + 8/3 = 11/3, so 0: = 113 ~ 1.91 centimeters, which is a little 1ess than twice the initial displacement. (d) To sketch a graph we also need the phase shift. Now O=tan-'(BIA), which is in the first quadrant since A and B are positive. Thus 0 = tan-'(2/b/2)~ 1.02, and so the maximum point on the graph (see Fig. 8.1.2) occurs at 01 w = 1.021 b 12 ~ 0.83. The period is 2'IT I w ~ 5.13. The graph is shown in Fig. 8.1.5 . • 0:

b

x(t)

Figure 8.1.5. The graph of x(t) = cos{i t R::

+ [i sin{i t

1.91 cos( {i t - 1.02).

Supplement to Sectlon 8.1: Llnearlzed Oscillations The spring equation can be applied to determine the approximate motion of a1!Y system subject to a restoring force, even if the force is not linear in the

376

Chapter 8 Differential Equations

displacement. Such forces occur in more realistic models for springs and in equations for electric circuits. Suppose that we wish to solve the equation of motion m d 2x dt 2

= j(x),

(8)

where the force function j(x) satisfies the conditions: (i) j(x o) = 0; and (ii) 1'(xo) < 0, for some position x o. The point X o is an equilibrium position since the constant function x(t) = Xo satisfies the equation of motion, by condition (i). By condition (ii), the force is positive when x is near X o and x < xo, while the force is negative when x is near X o and x > x o. Thus the particle is being pushed back toward X o whenever it is near that point, just as with the spring in Figure 8.1.1. Rather than trying to solve equation (8) direcdy, we shall replace j(x) by its linear approximation j(x o) + 1'(xo)(x - xo) at x o. Since j(xo) = 0, the equation (8) becomes

(8') which is called the linearization of equation (8) at xo. If we write k for the positive number - 1'(xo) and y = x - X o for the displacement horn equilibrium, then we get

md~ dt

=- -ky,

which is precisely the spring equation. We thus conclude that, to the degree that the linear approximation of the force is valid, the particle oscillates around the equilibrium point X o with period 2'lT / -/ - 1'( xo) / m. It can be shown that the particle subject to the exact force law (8) also oscillates around x o, but with aperiod which depends upon the amplitude of the oscillations. As the amplitude approach es zero, the period approaches 27T/-/- 1'(xo)/m, which is the period for the linearized equation. Here is an application of these ideas: By decomposing the gravitational force on a pendulum of mass m and length I into components parallel and perpendicular to the pendulum's axis, it can be shown that the displacement angle 0 of the pendulum from its equilibrium (vertical) position satisfies the differential equation m(d~ / dt 2) = - m(g/ l)sinO, where g = 9.8 meters per second2 is the gravitational constant. (See Fig. 8.1.6). The force function is j(O) = -(mg/l)sinO. Since j(O) = 0, 0=0 is an equilibrium point. Since 1'(0) = - (mg/ l)sin'(O) = - mg/ I, the linearized equation is md~/dt2= -(mg//)O. The period of oscillations for Figure 8.1.6. The forces acting on a pendulum.

the linearized equation is thus 2'lT /-/(mg/ I)m = 2'lTffg. (See Review Exerci se 83 for information on the solution of the nonlinear equation.) A point X o satisfying the conditiöns above is called a stable equilibrium point. The word stable refers to the fact that motions which start near Xo with small initial velocity stay near XO•3 If j(x o) = 0 but 1'(xo) > 0, we have an unstable equilibrium point (see Section 8.3). 3 We have only proved stability for the linearized equations. Using conservation of energy, one can show that the motion for the exact equations stays near Xo as weil. (See Exercise 33.)

8.1 Oscillations

377

Exercises for Section 8.1 1. Show that f(t) = cos(3t) is periodic with period 2'!T13. 2. Show that f(l) = 8 sin( '!Tt) is periodic with period 2.

3. Show that f(t) = cos(6t) + sin(3t) is periodic with period 2'!T 13. 4. Show that f(t) = 3 sin( '!Tt 12) + 8 cos( '!Tt) is pe· riodie wi th period 4. In Exercises 5-8, find the solution of the given equation with the prescribed values of x and x' = dx I dt at t = O. 5. x" + 9x = 0, x(O) = 1, x'(O) = -2. 6. x" + 16x = 0, x(O) = - I, x'(O) = -1. 7. x" + 12x = 0, x(O) = 0, x'(O) = -1. 8. x" + 25x = 0, x(O) = I, x'(O) = O. In Exercises 9-12, sketch the graph of the given func· tion and find the period, amplitude, and phase shift. 9. x = 3 cos(3t - I). 10. x = 2cos(5t - 2). 11. x = 4cos(t + I). 12. x = 6 cos(3t + 4). In Exercises 13-16, solve the given equation for x and sketch the graph. 13. d 2x dt 2 2 14. -d x dt 2

d~

15. dt 2 2 d 16. - x

+ 4x = 0; x = - land dx = 0 when t = o. dt

+

16x = 0, x = I an d -dx = 0 when t = O. dt

+ 25x = 0 x = 5 and ~ dt = 5 when t = O.

4 d 2x dt 2

= _

when t = '!T14. 17. Find the solution of d); dt 2

t=

3 when t

r

= - 4y for which y = 1

= O.

18. Find y = fex) if + 4]= 0 and ](0) = 0,1'(0) = -1. 19. Suppose that fex) satisfies + 16f = 0 and f(O) = 2,1'(0) = o. Sketch the graph y = fex). 20. Suppose that z = ger) satisfies 9z" + z = 0 and z(O) = - 1, z'(O) = o. Sketch the graph z = ger). 21. A mass of 1 kilogram is hanging from aspring. If x = 0 is the equilibrium position, it is given that x = land dxl dt = 1 when t = o. The weight is observed to oscillate with a frequency of twice a second. (a) What is the spring constant? (b) Sketch the graph of x as a function of t, indicating the amplitude of the motion on yourdrawing. 22. An observer sees a weight of 5 grams on aspring undergoing the motion x(t) = 6.1 cos(2t - '!T 16). (a) What is the spring constant?

r

x 3 + x 2 - X + 1.

(b) What is the frequency of Iinearized oscillations? *27. An atom of mass m in a linear molecule is subjected to forces of attraction by its neighbors given by fex)

= k(x -

x()3+ k 2(x - X2)3, k( ,k2

'

__ 5 + 25x = 0 x = 5 at t = 0, and dx ~2' ~

and

(b) What is the force acting on the weight at t = O? At t = 2? 23. What happens to the frequency of oscillations if three equal masses are hung from aspring where there was one mass before? 24. Find a differential equation of the "spring" type satisfied by the function y( t) = 3 cos( t I 4) sin(t 14). *25. A "flabby" spring exerts a force fex) = -3x + 2x 3 when it is displaced a distance x from its equilibrium state, x = o. (a) Write the equation of motion for an object of mass 27, vibrating on this spring. (b) Write the linearized equation of motion at xo= O. (c) Find the period of Iinearized oscillations. *26. (a) Find the equilibrium position of an object which satisfies the equation of motion

> 0, 0< x( < X < X2.

(a) Compute the equilibrium position. (b) Show that motion near this equilibrium is unstable. *28. The equation for aspring with friction is m d 2x dt 2

= _

kx _

I)

dx dt

(spring equation with damping). (a) If 1)2 < 4km, check that a solution is x(t)

=

e-&/2m(A coswt + Bsinwt),

where w2 = k/m - 1) 2/4m 2 > O. (b) Sketch the general appearance of the graph of the solution in (a) and define the "period" of oscillation. (c) If the force - kx is replaced by a function fex) satisfying f(O) = 0,1'(0) < 0, find a formula for the frequency of damped linearized oscillations. *29. Suppose that x = f(l) satisfies the spring equation. Let g(t) = at + b, where a and bare constants. Show that if the composite function fog satisfies the spring equation (with the same w), then a = ± 1. What about b? *30. (a) Suppose that f(t) is given and that y = g(t) satisfies d 2y I dt 2 + w); = f(t). Show that

378

Chapter 8 Differential Equations

+ A sin wt + B cos wt represents the general solution of d 2x I dt 2 + w2x = f; that is, x is a solution and any solution has this form. One calls y a particular solution and x the general solution. (h) Solve d 2xl dt 2 + w2x = k if x = land dxl dt = - I when t = 0; k is a nonzero constant. (c) Solve d 2x I dl 2 + w2x = w2t if x = -1 and dxldl = 3 when 1=0. Exercises 31 and 32 outline the complete proof of the following theorem using the "method of variation of constants": Let x = f{l) be a twice-differentiable function of I such thai (d 2x I dt 2) + w 2x = O. Then x = A cos wt + B sin wl for conslanls A and B. *31. Some preliminary calculations are done first. Write x = y

x = A (t)COSWI

+ B(t)sinwt.

(10)

This equation is obtained by differentiating (9) pretending that A (t) and B(t) are constants. Since this is what we are trying to prove, we should be very suspicious here of circular reasoning. But push on and see what happens. Show that . B(/) = x smwl

dxldt w

+ - - coswt.

dxldl . A(t)=xcoswt---smwt. w

(12)

*32. Use the calculations in Exercise 31 to give the proof of the theorem, making sure to avoid circular reasoning. We are given x = f(l) and w such that (d 2xldt 2)+w 2X=0. Define A(t) and B(t) by equations (11) and (12). Show that A (t) and B(t) are in fact constants by differentiating (11) and (12) to show that A'(/) and B'(t) are identically zero. Then rewrite formulas (11) and (12) as . B = x(t)smwt

+ --coswt,

dxldt w

(13)

dxldt . --smwt. w

(14)

and

(9)

It is possible to choose A (I) and B(/) in many ways, since for each t either sin wl or cos wl is nonzero. To determine A (t) and B(t) we add a second equation:

~~ = -wA(t)sinwt + wB(t)coswt.

Similarly, show that

A

= x(t)coswt -

Use these formulas to show A coswt + Bsinwt = x, which proves the theorem. *33. Suppose that m(d 2xldt 2) = fex), where f(xo) = 0 and j'(xo) < O. Let V(x) be an antiderivative of -' f. (a) Show that Xo is a local minimum of V. (h) Show that dEI dl = 0, where the energy E is given by E = !m(dxldti + V(x). (c) Use conservation of energy from (b) to show that if dx I dl and x - Xo are sufficiently small at t = 0, then they both remain smalI.

(11)

8.2 Growth and Decay The solution oJ the equation Jor population growth may be expressed in terms oJ exponential functions.

Many quantities, such as bank balances, populations, the radioactivity of ores, and the temperatures of hot objects change at a rate which is proportional to the current value of the quantity. In other words, if J( t) is the quantity at time t, then J satisfies the differential equation

f'( t) = yJ( t),

(1)

where y is a constant. For example, in the specific case of temperature, it is an experimental fact that the temperature of a hot object decreases at arate proportional to the difference between the temperature of the object and that of its surroundings. This is called Newton's law oJ cooling. Example 1

The temperature of a hot bowl of porridge decreases at a rate 0.0837 times the difference between its present temperature and room temperature (fixed at 20°C). Write down a differential equation for the temperature of the porridge. (Time is measured in minutes and temperature in oe.)

8.2 Growth and Decay

Solution

Let T be the temperature eq of the porridge and let f{l) temperature above 20 D C. Then 1'(t) = dT j dt and so 1'(1) =

=T-

379

20 be its

- (0.0837)f(t)

i.e.,

dT dt = - (0.0837)( T -

20).

The minus sign is used because the temperature is decreasing when T is greater than 20; y = - 0.0837. Ä We solve equation (1) by guesswork, just as we did the spring equation. The answer must be a function which produces itself times a constant when differentiated once. It is reasonable that such a function should be related to the exponential since e' has the reproductive property (dj dt)e ' = e ' . To get a factor y, we replace t by yt. Then (dj dt)e Y1 = ye Y1 , by the chain rule. We can also insert a constant factor A to get

~ (Ae Y1 ) = y(Ae Y1 ). Thus f(/) = Ae yl solves equation

(1). If we pick 1 = 0, we see that A = f(O). This gives us a solution of equation (1); we shall show below that it is the only solution.

The Solution of l' =

"11

Given j(O), there is one and only one solution to the differential equation 1'(/) = yf(t), namely

f( t)

= f(O)e yl

(2)

To show that formula (2) gives the only solution, let us suppose that g(t) also satisfies g'(t) = yg(t) and g(O) = f(O). We will show that g(t) = f(0)e Y1 . To do this, consider the quotient

h(t) = g(/) = e-Y1g(t) e yl

and differentiate:

h'(t) = -ye-Y1g(t) + e-Y1g'(t) = -ye-Y1g(t) + ye-Y1g(t) Since h'(t) = 0, we may conclude that his constant; but h(O) so e-Y1g(t) = h(t) = f(O), and thus

= O.

= e-Og(O) = f(O),

g( t) = f(O)e yl = f( t), as required.

Example 2 Solution Example 3

= 3x, and x = 2 at t = 0, find x for all t. If x = f{l), then f(O) = 2 and l' = 3j, so y = 3 in formula (2), x = f(t) = 2e 31 . Ä

If dx j dt

the box above. Hence, by

Find a formula for the temperature of the bowl of porridge in Example 1 if it starts at 80 D C. Jane Cool refuses to eat the porridge when it is too coldnamely, if it falls below SODe. How long does she have to come to the table?

380

Chapter 8 Differential Equations

Solution

Let j(t) = T - 20 as in Example l. Then 1'(1) = -0.0837j(1) and the initial condition is j(O) = 80 - 20 = 60. Therefore

j(1)

=

60e-O.0837/.

Hence T = j(1)

+ 20 = 60e-O.0837/ + 20. When T = 50,

we have

50 = 60e-O.0837/ + 20

1 = 2e -0.0837/ eO.0837/ = 2

0.0837t = In2 = 0.693. Thus t = 8.28 minutes. Jane has a little more than 8 minutes before the temperature drops to 50°C. • Note how the behavior of solution (2) depends on the sign of y. If y > 0, then e y/ ~ 00 as t ~ 00 (growth); if y < 0, then e y/ ~ 0 as t ~ 00 (decay). See Fig. 8.2.1.

1'>0 ~--==~----~---1'=O

Figure 8.2.1. Growth occurs if y > 0, decay if

Large y < 0

y 0.

The general solution is y = A cosh( ~1'( x o) t) + Bsinh( ~l' (x o) t), with A and B depending on the initial values of y and dy / dt, as in formula (10). We can also write this solution as y = (A + B)eJJ'(Xo)1 + (A - B)e-JJ'(Xo)l. No matter how small the initial values, unless they are chosen so carefully that A + B = 0, the solution will approach + 00 or - 00 as t~ 00. We say that the point Xo is an unstable equilibrium position. In contrast to the stable case treated in Seetion 8.1, the linearization is not useful for all t, since most solutions eventually leave the region where the linear approximation is valid. Still, we can correct1y conelude that solutions starting arbitrarily elose to Xo do not usually stay elose, and that there are special solutions which approach X o as t~

00. 4

This analysis is useful in more advanced studies of differential equations. See, for instance, Elementary Differential Equations and Boundary Value Problems, by W. Boyce and R. DiPrima, Chapter 9, Third Edition, Wiley (1977) and Differential Equations and Their Applications, by M. Braun, Chapter 4, Third Edition, Springer-Verlag (1983).

4

8.3 The Hyperbolic Functions

391

For instance, let us find the unstable equilibrium point for the pendulum equation

= - m..K sinD = /(0). dt 2 I First note that j(O) = 0 at 0 = 0 and '11, corresponding to the bottom and top of pendulum swing (see Fig. 8.1.6). At 0='11, f'(O)=-(mgjl)cosO= - (mg j I) . ( - 1) = mg j I > 0, so the top position is an unstable equilibrium point. At that point, the linearized equation is given by (d 2 j dt 2)(0 - 00) = m d'O

(mg/I)(O - 00 ), with solutions 0 - 00 = ce-Jmiiir + De--Jmiiir• For most initial conditions, C will not equal zero, and so the pendulum will move away

from the equilibrium point. If the initial conditions are chosen just right, we will have C = 0, and the pendulum will gradually approach the top position as t ~ 00, but it will never arrive. A very different application of hyperbolic functions, to the shape of a hanging cable, is given in Example 6, Section 8.5.

Exerclses for Sectlon 8.3 Prove the identities in Exercises 1-8. I. 2. 3. 4. 5. 6.

tanh2x + sech2x = I. cothl,x = I + csch2x. cosh(x + y) = cosh x cosh Y + sinh x sinh y. sinh2x = (cosh2x - 1)/2.

fx coshx fx cothx

= sinhx. == -csch2x.

d 7. dx sechx = -sechx tanhx. d 8. dx cschx = -cschx cothx.

Differentiate the functions in Exercises 9-24. 9. sinh(x 3 + x 2 + 2) 10. tan- 1(coshx) 11. sinh x sinh 5x 12. sinhx 1 + coshx 13. sinh(cos(8x» 14. cos(sinh(x 2» 15. sinh2x + cosh2x 16. sinh4x + cosh4x 17. coth3x 18. (tanhx)(sechx) 19. exp(tanh2x) 20. sin- 1(tanhx) 21. coshx 1+ tanhx 22. (csch2x)(1 + tanx) 23. (COshX)(f

dx ) 1+ tanh2x

24. (SinhX)(f

dx ) 1+ sech 2x

Solve the differential equations in Exercises 25-28. 25. y" - 9y = 0, y(O) = 0,1'(0) = I. 26.1"- 8Ij= 0, j(O) = 1,1'(0) = -I. 27. g" - 3g = 0, g(O) = 2, g'(O) = O. 28. h" - 9h = 0, h(O) = 2, h'(O) = 4. 29. Find the solution of the equation d 1x/dt 2 - 9x = 0, for which x = 1 and dx / dt = 1 at t = O. 30. Solve dy / dt 1 - 25y = 0, where y == 1 and dy / dt = - 1 when t = O. 31. Find j(l) if 1" ... 36j, and j(O) = 2,1'(0) = O. 32. Find g(t) if g"(t) = 25g(t), and g(O) = 0, g'(O) = I. Sketch the graphs of the functions in Exercises 33-36. 33. Y = 3 + sinh x 34. y = (coshx) - I 35. Y = tanh 3x 36. y = 3 cosh 2x Compute the integrals in Exercises 37-46. 37. fcosh3xdx 38. flcschl(2x) + (3/x)Jdx

f coth x dx 40. f x tanh(x1)dx 41. f sinh x dx 42. f cosh 9y dy 43. f e"'sinhxdx 44. f e cosh 2t dt 39.

2

2

21

45. f cosh 2x sinh x dx 46.

f cosh sinh x dx x 3

392

Chapter 8 Differential Equations

where k 2 > k l > 0, 0
0, but since our initial condition is x = 0 where cosx = I, this is justified.) Thus

+ (tanx)y ] =

cosx,

A. [~]= dx cosx

cosx,

_1_ [y' cosx

~ =sinx+ C.

cosx Since y = 1 when x = 0, C = l. Thus Y = cosx sinx + cosx. It may be verified that this solution is valid for all x . •

Example 3

(Electric circuits) In Example 4, Seetion 8.5, replace E by the sinusoidal voltage E

= Eosinwt

with L, R, and E o constants, and solve the resulting equation.

Solution

The equation is dI RI dt = - L

Eo . sm wt.

+L

We follow the procedure in the preceding box with x replaced by t and y by I:

l. P(t) = -

f, a constant, so JP(t)dt = - tRI L.

2. dI dt

=

RI

+L

Eo . sm wt.

L

tR )] ( dI RI ) = L E o exp ( L tR ) smwt. . 3. [ exp ( L dt + L

8.6 Linear First-Order Equations

4.

411

~ {exp ( t: )1 } = ~o exp ( t: )sinwt.

5. exp ( t:)1 =

~o

J

exp ( t: )sinwtdt.

This integral may be evaluated by the method of Example 4, Section 7.4, namely integration by parts twice. One gets exp ( t:)1 =

~{

(R/:;::

w2

(

f sinwt - wcoswt) } + C.

6. Solving for I,

1= Eo

1 ( R sinwt - wcoswt) L (R/L)2+ w2 L

+ Ce- tR / L •

The constant C is determined by the value of I at t = O. This expression for I contains an oscillatory part, oscillating with the same frequency w as the driving voltage (but with a phase shift; see Exercise 10) and a transient part Ce- tR / L which decays to zero as t~ 00. (See Fig. 8.6.1.) •

Figure 8.6.1. The graph of the solution of a sinusoidally forced electric circuit containing a resistor and an inductor.

Example 4

(pollution) A small lake contains 4 X 107 liters of pure water at t = O. A polluted stream carries 0.67 liter of pollutant and 10 liters of water into the lake per second. (Assume that this mixes instantly with the lake water.) Meanwhile, 10.67 liters per second pf the lake flow out in a drainage stream. Find the amount of pollutant in the lake as a function of time. What is the limiting value?

Solution

Let y(t) denote the amount of pollutant in liters in the lake at time t. The amount of pollutant in one liter of lake water is thus y(t) 4 X 107

•

The rate of change of y(t) is the rate at which pollutant flows out, which is - 1O.67y(t)/4 X 107 = -2.67 x 1O-;,(t) liters per second, plus the rate at which it flows in, which is 0.67 liter per second. Thus

y'

= -2.67 X

10-;' + 0.67.

The solution of dy / dt e- at :

e-at(y' - ay)

= ay + b, y(O) = 0, is found using the integrating factor

= be-at,

~ (e-ay) = be-at, e-ay

= - !! e- at + C. a

412

Chapter 8 Differential Equations

Since y

= 0 at t = 0,

e-ay =

C = b/ a. Thus

!!.. (I - e- at ), a

y=!!..(eat_l).

a Here a = -2.67 X 10- 7 and b = 0.67, so Y = (1 - e-2.67XIO-7t)(2.51 X 106) liters. For t smalI, y is relatively smalI; but for larger t, y approaches the (steady-state) catastrophic value of 2.51 X 106 liters, that is, the lake is well over half pollutant. (See Exercise 13 to find out how long this takes.) •

Terminal speed

Example 5

(Falling object in a resisting medium) The downward force acting on a body of mass m falling in air is mg, where g is the gravitational constant. The force of air resistance is yv, where y is constant of proportionality and v is the downward speed. If a body is released from rest, find its speed as a function of time t. (Assurne it is released from a great enough height so that it has not hit the ground by time t.)

Solution

Since mass times acceleration is force, and acceleration is the time derivative of velocity, we have the equation m( dv / dt) = mg - yv or, equivalently, dv / dt = -(y/m)v + gwhich is a linear first-order equation. Its general solution is

Slope = g

~--I t = 0 = moment of release

Figure 8.6.2. The speed of an object moving in a resisting medium.

Example 6

v

= e- ytlm

f ge ytlm dt= e- ytlm ( ~g e ytlm + C ).

If v = 0 when t = 0, C must be - mgjy, so v = (mgjy)(1 - e- ytlm ). Note that as t -+ 00, e - yt I m -+ and so v -+ mg / y the terminal speed. (See Fig. 8.6.2.) For small t the velocity is approximately gt, which is what it would be if there were no air resistance. As t increases, the air resistance slows the velocity and a terminal velocity is approached. •

°

°

(Rocket propulsion) A rocket with an initial mass Mo (kilograms) blasts off at time t = (Fig. 8.6.3). The mass decreases with time because the fuel is being Velocity = v

Figure 8.6.3. A rocket blasting off at t = O.

spent at a constant burn rate r (kilograms per second). Thus, the mass at time is M = Mo - rt. If the thrust is a constant force F, and the velocity is v, Newton's second law gives t

d dt(Mv)=F-Mg,

(7)

where g = 9.8 is the gravitational constant. (We neglect air resistance and assume the motion to be vertical.) (a) Solve equation (7). (b) If the mass of the rocket at burnout burnout.

IS

MI' compute the velocity at

8.6 Linear First-Order Equations

Solution

(a) Substituting M

413

= Mo - rt into equation (7) gives

~ [(Mo - rt)v]

F- (Mo - rt)g.

=

Although this is a linear equation in v, it is already in a form which we can directly integrate:

(Mo - rt)v= f(F- g(Mo - rt)dt+ C = Ft - g ( Mot -

rt2 ) T + C.

Solving for v,

v=

Ft _ g (M t _ rt 2 Mo-rt Mo-rt 0 2

Since v =

v=

°

+

C

(Mo-rt)'

at t = 0, C = 0, so the solution is

Ft g (M t _ rt 2 ) Mo - rt Mo - rt 0 2'

(b) At burnout, Mo - rt V

)

= F(Mo -

rM I

MI)

= MI' so

-..K. [Mo( Mo - MI) _ (Mo - M I)2] r

MI

= F(Mo- MI) _ g(MJill l

2ill l

M~) =

2r

Mo- MI [Fill l

g( Mo + MI)] 2

is the velocity at burnout. •

Exercises for Section 8.6 In Exercises 1-4, solve the given differential equation by the method of this section. dy y 2 1. - = - - + - - + 3. dx I-x I-x dy . 2 . 2. dx = ysmx smx. 3. Yx =

xy - x3.

11. The equation R(dl/ dt) + (I/ C) = E describes the current 1 in a circuit containing a resistor with resistance R (a constant) and a capacitor with capacitance C (a constant) as shown in Fig. 8.6.4. If E = E o, a constant, and 1 = 0 at t = 0, find 1 as a function of time. Discuss your solution.

4. x(Yx)=y+xlnx.

In Exercises 5-8, solve the given equation with the stated conditions. 5. y' = ycosx + 2cosx,y(0) = o. 6. y' =

l

x

+x,y(l) = 1.

7. xl' = eX - y,y(1) = o. 8. y' = Y + cos5x,y(0) = o. 9. Rework Example 3 assuming that the voltage is E = Eosinw! + E lo i.e., a sinusoidal plus a constant voltage. 10. In Example 3, use the method on p. 373 of Section 8.1 to determine the amplitude and phase of the oscillatory part:

Eo T'

I2 (R/L) +w 2

{R-L smwt . -

}

wcoswt .

Figure 8.6.4. A resistor and a capacitor in an electric circuit. 12. Repeat Exercise 11 for the case 1 = 10 at t = O. 13. In Example 4, show that the lake will reach 90% of its limiting pollution value within 3.33 months. 14. Mixing problem. A lake contains 5 X 108 liters of water, into which is dissolved 10" kilograms of salt at t = O. Water flows into the lake at a rate of 100 liters per second and contains 1% salt; water flows out of the lake at the same rate. Find the amount of salt in the lake as a function of time. When is 90% of the limiting amount of salt reached?

414

Chapter 8 Differential Equations

15. Two-stage mixing. Suppose the pollutants from

16. 17.

18.

19.

20.

21.

22.

the lake in Example 4 empty into a second smaller lake rather than the stream. At t = 0, the smaller lake contains 107 liters of pure water. The second lake has an exit stream carrying the same .volume of fluid as entered. Find the amount of pollution in the second lake as a function of time. Repeat Exercise 15 with the size and flow rates of the lake in Example 4 replaced by those in Exercise 14. The terminal speed of a person in free fall in air is about 64 meters per second (see Example 5). How long does it take to reach 90% of terminal speed? How far has the person fallen in this time? (g = 9.8 meters per second 2). The terminal speed of a person falling with a parachute is about 6.3 meters per second (see Example 5). How long does it take to reach 90% of terminal speed? How far has the person fallen in this time? (g = 9.8 meters per second2). Falling object with drag resistance. Redo Example 5 assuming that the resistance is proportional to the square of the velocity. An object of mass 10 kilograms is dropped from a balloon. The force of air resistance is 0.07v, where v is the ve1ocity. What is the object's velocity as a function of time? How far has the object traveled before it is within 10% of its terminal ve1ocity? What is the acce1eration of the rocket in Exampie 6 just before burnout? How high is the rocket in Example 6 at burnout?

23. Sketch the direction field for the equation dy 1dx = Y + 2x and solve the equation. 24. Sketch the direction field for the equation dy 1dx = - 3Y + x and solve the equation. *25. Assuming that P(x) and Q(x) are continuous functions of x, prove that the problem y' = P(x)y + Q(x), y(O) = Yo has exactly one solution. 26. Express the solution of the equation y' = xy + I, y(O) = 1 in terms of an integral. *27. Bernoulli's equation. This equation has the form dyldx = P(x)y + Q(x)yn, n = 2,3,4, .... (a) Show that the equation satisfied by w = i -n is linear in w. (b) Use (a) to solve the equation x(dyldx) =xj3- y. *28. Riccati equation. This equation is dy 1dx = P(x) + Q(x)y + R(X)y2. (a) Let Yl(X) be a known solution (found by

inspection). Show that the general solution is y(x) = Yl(X) + w(x), where w satisfies the Bernoulli equation (see Exercise 27) dw 1dx = [Q(x)

+ 2R(x)Yl(X)]W + R(x)w 2.

(b) Use (a) to solve the Riccati equation y' = y 1x + xY - x 5, takingYl(x) = x. *29. Redo the rocket propulsion Example 6 adding air resistance proportional to velocity. *30. Solve the equation dYldt = -AY + r, where I. and rare constants. Write a two page report on how this equation was used to study the Van Meegeren art forgeries which were done during World War 11. (See M. Braun, Difjerential Equations and their Applications, Third Edition, Springer-Verlag, New York (1983), Section 1.3).

Review Exercises for Chapter 8 Solve the differential equations with the given conditions in Exercises 1-22. 1. 2.

t t=

= 3y,y(0) =

y, y(O)

1.

+ 3y = 0, y(O) = 0, y'(O) =

4. dy dt 2

+ 9y = 0, y(O) = 2; y'(0) = O.

6.

7. 8.

9.

t t t= t+

=3y+ l,y(O)

= (cost)y

= 1.

+ cost,y(O) = O.

tY,y(O)

= 1.

lOy = O,y(O) = 1.

l' = 4j; j(O) = 1.

l' = - 4j, j(O) = 1.

r = 4j; j(O) = 1,1'(0) = 1. 12. r = - 4j; j(O) = 1,1'(0) = 1. d2~ + x = 0;

13.

= 1.

3. dy dt 2

5.

10.

11.

1.

dt

t=

7T

14. + 6x = 0;

2

14. d x dt 2

t

15.

=

dt d 2x

17.

18. ':;;

9x = 0, x(O) = 0, x'(O) = 1.

+ 16x = 0,

ix = dt 2

x = I when t = 0, x = 6 when

1.

d 2x -2 -

16. -

x = 1 when t = 0, x = 0 when

eX+Y,y(O)

x(O) = 1, x'(O) = -1.

= 1.

= 8e 3y , y(O) =

1.

19. ':;; = -4x; x = 1 when t = O.

Review Exercises for Chapter 8

20.

!

22. dx = _x_ dt 8-t

1

+ t-I ,y(O)=O. + _3_, x(O) = 1. 8-t

23. Solve for g(t): 3(d 21dt 2 )g(t) = -7g(t), g(O) = I, g'(O) = - 2. Find the amplitude and phase of g(t). Sketch. 24. Solve for z = J(t): d 2z 1dP + 5z = 0, J(O) = - 3, 1'(0) = 4. Find the amplitude and phase of J( t). Sketch. 25. Sketch the graph of y as a function of x if -dYldx 2 = 2y; Y = Land dyldx = t when x=O. 26. Sketch the solution of d 2x 1dt 2 + 9x = 0, where x(O) = I, x'(O) = O. 27. Sketch the graph of the solution of y' = - 8y, Y = I when x = O. 28. Sketch the graph of y = J(x) if l' = 2J + 3, and J(O) = O. 29. Sketch thegraph of the solution to dxldt = - x + 3, x(O) = 0 and compute limHoox(t). 30. Sketch the graph of the solution to dx 1dt = -2x + 2, x(O) = 0 and compute limHOOx(t). 31. Solve d 2xl dt 2 = dxl dt, x(O) = 1, x'(O) = I by letting y = dx 1dt. 32. Solve d 2xldt 2 =[I/{l + t)]dxldt, x(O) = I, x'(O) = I by letting y = dx 1dt. 33. Solve dy 1dx 2 + dy 1dx = x, y(O) = 0, y'(O) = I. [Hint: Let w = dyldx]. 34. Solve y" + 3yy' = 0 for y(x) if y(O) = I,y'(O) = 2. [Hint: yy' = (y2/2)'.] 35. Solve dYldx 2 = 25y,y(0) = O,y(I) = I. 36. Solve dy 1dx 2 = 36y, y(O) = I, y(l) = O. Differentiate the functions in Exercises 37-44. 37. sinh(3x 2) 38. tanh(x 3 + x) 39. cosh-l(x 2 + I) 40. tanh -1(x 4 - 1) 41. (sinh-lx)(cosh3x) 42. (cosh-13x)(tanhx 2) 43. exp(l - cosh -1(3x» 44. 3(cosh -1(5x 2 ) + I) Calculate the integrals in Exercise 45-50. 45. coshx dx 1 + sinh2x 46. sech2x tanh xJ2 + tanh 2x dx

J J

47.J~2 9- x dx

J rxr+4 49. Jx sinhx dx 50. Jx coshx dx

48.

52. A weight hanging on aspring oscillates with a frequency of two cycles per second. Find the displacement x(t) of the mass if x(O) = land x'(O) = o. 53. An observer sees a weight of 10 grams on a spring undergoing the motion x(t) = lOsin(8t). (a) What is the spring constant? (b) What force acts on the weight at t = ." 116? 54. A 3-foot metal rod is suspended horizontally from aspring, as shown in Fig. 8.R.I. The rod bobs up and down around the equilibrium point, 5 feet from the ground, and an amplitude of I foot and a frequency of two bobs per second. What is the maximum length of its shadow? How fast is its shadow changing length when the rod passes the middIe of its bob?

= Y/lny,y(l) = e.

dy y 21. dl = 1-1

51. A weight of 5 grams hangs on aspring with spring constant k = 2.1. Find the displacement x(t) of the mass if x(O) = land x'(O) = o.

415

1-4 ft-J

Shadow

Figure 8.R.l. Study the movement of the shadow of the bobbing rod. 55. Simple Harmonie Motion with Damping. Consider the equation x" + 2ßx' + w2x = 0, where 0 < ß < w. (a) Show that y = eß1x satisfies a harmonic oscillator equation. (b) Show that the solution is of the form x = e-ß'(A cos wlt + B sin Wlt) where WI = fw 2 - ß2, and A and Bare constants. (c) SoIve x" + 2x' + 4x = 0; x(O) = 1, x'(O) = 0, and sketch. 56. Foreed oscil/ations. (a) Show that a solution of the differential equation x" + 2ßx' + w2x = Jocoswot is

XI(t) =

~o

(w 2 - W5) + 4W5ß2

[ 2woß sinw ot

+ (w 2 -

w~)coswotl

(b) Show that the general solution is x(t) = Xl(t) + xo(t) where Xo is the solution found in Exercise 55. (c) Resonanee. Show that the "amplitude" Jo/[(w 2 - w5)2 + 4w5ß 2] of the solution is largest when Wo is near w (the natural frequency) for ß (the friction constant) smalI, by maximizing the amplitude for fixed Jo, w,

416

57. 58. 59. 60.

61. 62.

63.

64.

Chapler 8 Differential Equations

ß and variable Wo. (This is the phenomenon responsible for the Tacoma bridge disaster ... somewhat simplified of course; see Section 12.7 for further information.) If a population doubles every 10 years and is now 100,000, how long will it take to reach 10 million? The half-life of a certain radioactive substance is 15,500 years. What percentage will have decayed after 50,000 years? A certain radioactive substance decreases at a rate of 0.00128% per year. What is its half-life? (a) The population of the United States in 1900 was about 76 million; in 1910, about 92 million. Assurne that the population growth is uniform, so f(t) = e 1'f(0), t in years after 1900. (i) Show that y l':::: 0.0191. (ii) What would have been a reasonable prediction for the population in 1960? In 1970? (iii) At this rate, how long does it take for the population to double? (b) The actual U.S. population in 1960 was about 179 million and in 1970 about 203 million. (i) By what fraction did the "growth rate factor" ("y") change between 1900-1910 and 1960-1970? (ii) Compare the percentage increase in population from 1900 to 1910 with the percentage increase from 1960 to 1970. If an object cools from 100°C to 80°C in an environment of 18°C in 8 minutes, how long will it take to cool from 100° to 50°C? "Suppose the pharaohs had buHt nuclear energy plants. They might have elected to store the resulting radioactive wastes inside the pyramids they buHt. Not a bad solution, considering how weil the pyramids have lasted. But plutonium-239 stored in the oldest of them-some 4600 years ago--would today still exhibit 88 percent of its initial radioactivity." (see G. Hardin, ''The Fallibility Factor," Skeptic 14 (1976): 12.) (a) What is the half-life of plutonium? (b) How 10ng will it take for plutonium stored today to have only 1% of its present radioactivity? How long for l~? (a) The oil consumption rate satisfies the equation C(t) = Coe rt , where Co is the consumption rate at t = 0 (number of barrels per year) and r is a constant. If the consumption rate is Co = 2.5 X 1010 barrels per year in 1976 and r = 0.06, how long will it take before 2 X 1012 barrels (the total world's supply) are used up? (b) As the fuel is alm ost used up, the prices will probably skyrocket and other sources of energy will be turned to. Let S(t) be the supply left at time t. Assurne that dS / dt = - aS, where a is a constant (the panic factor). Find S(t). (a) A bank advertises "5% interest on savingsbut you earn more because it is compounded

continuously." The formula for computing the amount M(t) of money in an account t days after M(O) dollars is deposited (and left untouched) is M(t) = M(0)eO.OSt/36S. What is the percentage increase on an amount M(O) left untouched for 1 year? (b) A bank wants to compute its interest by the method in part (a), but it wants to give only $5 interest on each $100 that is left untouched for I year. How must it change the formula for that to occur? 65. A certain electric circuit is governed by the equation Ldl/ dt + RI = E, where E, R, and L are constants. Graph the solution if 1o < EI R. 66. If a savings account containing P dollars grows at a rate dP I dt = rP + W (interest with continuous deposits), find P in terms of its value Po at t = O. 67. (a) Sketch the direction field for the equation y' = -9xly. Solve the equation exactly. (b) Find the orthogonal trajectories for the solutions in (a). 68. Consider the predator-prey model in Example 5, Section 8.5. Solve this explicitly if r = 0 (Le., ignore deaths of the prey). 69. Suppose your car radiator holds 4 gallons of fluid two thirds of which is water and one third is old antifreeze. The mixture begins flowing out at a rate of t gallon per minute whHe fresh water is added at the same rate. How long does it take for the mixture to be 95% fresh water? Is it faster to wait until the radiator has drained before adding fresh water? 70. An object falling freely with air resistance has a terminal speed of 20 meters per sec. Find a formula for its velocity as a function of time. 71. The current 1 in a certain electric circuit is governed by dlldt= -3/+2sin('/Tt), and 1= I at t = O. Find the solution. 72. In Example 6, Section 8.6, suppose that the burnout mass is 10% of the initial mass Mo, the burnout time is 3 minutes, and the rocket thrust i8 3Mog. Calculate the acceleration of the rocket just before burnout in terms of Mo and g. 73. Sketch the direction field for the equation y' = 3y + 4 and solve it. 74. Sketch the direction field for the equation y' = - 4y + land solve it. 75. Let x(t) be the solution of dxl dt = x 2 - 5x + 4, x(O) = 3. Find limHoox(t). 76. Let x satisfy dx / dt = x 3 - 4x 2 + 3x, x(O) = 2. Find limHoox(t). 1i177. Test the accuracy of the Euler method by using a ten-step Euler method on the problem of finding y(l) if y' = y and y(O) = 1. Compare your answer with the exact solution and with a twenty-step Euler method.

Review Exercises for Chapter 8

1i178. Solve y' = ese y approximately for 0 0;;; x 0;;; I with y(O) = 1 using a ten-step Euler method. 1i179. Numerieally solve for y(2) if y' = y2 and y(O) = I, using a twenty-step Euler method. 00 you deteet some numerieal trouble? What do you think is going wrong? 180. Solve for y(l) if y' = eos(x + y) and y(O) = 0, using a ten-step Euler method. 81. Solve y' = ay + b, given eonstants a and b, by (a) introdueing w = y + bl a and a differential equation for w; (b) treating it as a separable equation; and (e) treating it as alinear equation. Are your answers the same? *82. Let x(t) be the solution of dxldt = -x + 3 sin '/TI, x(O) = 1. Find a, wand 0 such that limHoo[x(t) - aeos(wt + 0)] = O. *83. Simple pendulum. The equation of motion for a simple pendulum (see Fig. 8.R.2) is d2fJ = _ 1 sinO. dt 2 L

/

,,/

Figure 8.R.2. A simple pendulum. (a) Let w(t) = dO 1dt. Show that

!L ( w2 ) dt 2

=

1 !L eos 0 Ldt

'

and so 2

w T

=

rg (eosO -

eosOo),

---

417

where w = 0 when 0 = 00 (the maximum value of 0). (b) Conc1ude that 0 is implieitly determined by

I joo,.~"".,

-

fit

(e) Show that the period of oscillation is T=2

f2Li1JO

Vg

0

ILI'IT/2

=

4V g

0

dO

~eosO -

~I

eosOo

drp - k 2sin2rp ,

where k = sin(00/2). [Hint: Write eos 0 = I - 2 sin2(0 12).] The last integral is ealled an elliptic integral of the first kind, and eannot be evaluated explicitly. (d) How does the answer in (e) eompare with the predietion from linearized oseillations for 00 smalI? *84. A photographer dips athermometer into a developing solution to determine its temperature in degrees Centigrade. The temperature O(t) registered by the thermometer satisfies a differential equation dO 1dt = - k( 0 - 1/), where 1/ is the true temperature of the solution and k is a eonstant. How ean the photographer determine when 0 is eorreet to within O.IOC? *85. (a) Find y = fex) so that

f

f(x)dx=

f~l + [j'(x)f

dx

for all a and b. (b) What geometrie problem leads to the problem (a).

Chapter9

Applications of Integration Many physieal and geometrie quantities ean be expressed as integrals.

Our applications of integration in Chapter 4 were limited to area, distancevelocity, and rate problems. In this chapter, we will see how to use integrals to set up problems involving volumes, averages, centers of mass, work, energy, and power. The techniques developed in Chapter 7 make it possible to solve many of these problems completely.

9.1 Volumes by the Slice Method The volume of a solid region is an integral of its cross-seetional areas.

By thinking of a region in space as being composed of "infinitesimally thin slices," we shall obtain a formula for volumes in terms of the areas of slices. In this section, we apply the formula in a variety of special cases. Further methods for calculating volumes will appear when we study multiple integration in Chapter 17. We will develop the slice method for volumes by analogy with the computation of areas by integration. If fand gare functions with fex) ..;; g(x) on [a, b], then the area between the graphs of fand g is f~[ g(x) - f(x)]dx (see Section 4.6). We recall the infinitesimal argument for this formula. Think of the region as being composed of infinitesimally thin strips obtained by cutting with lines perpendicular to the x axis. Denote the vertical line through x by b x Lx; the intersection of Lx with the region between the graphs has length lex) = g(x) - fex), and the corresponding "infinitesimal rectangle" with thickness dx has area l(x)dx (= height X width) (see Fig. 9.1.1). The area of the Figure 9.1.1. The area of entire region, obtained by "summing" the infinitesimal areas, is the shaded region is f~l(x) dx.

Given a region surrounded by a closed curve, we can often use the same formula f~l(x)dx to find its area. To implement this, we position it conveniently with respect to the axes and determine a and b by noting where the ends of the region are. We determine lex) by using the geometry of the situation at hand. This is done for a disk of radius r in

420

Chapter 9 Applications of Integration

Fig. 9.1.2. We may evaluate the integral r_r2~r2 - x 2 dx by using integral tables to obtain the answer 'TTr 2 , in agreement with elementary geometry. (One can also readily evaluate integrals of this type by using the substitution x = rcosO.) y

x

Figure 9.1.2. Area of the disk = r-r2~ dx.

To find the volume of a solid region, we imagine it sliced by a family of parallel planes: Tbe plane Px is perpendicular to a fixed x axis in space at a distance x from a reference point (Fig. 9.1.3). Tbe plane Px cuts the solid in a plane region Rx ; the corresponding "infinitesimal piece" of the solid is a slab whose base is a region R x and whose thickness is dx (Fig. 9.1.4). The volume of such a cylinder is equal to the area Referenc. point

x x

Figure 9.1.3. The plane Px is at distance x horn Po.

Figure 9.1.4. An infinitesirnally thin slice of asolid.

of the base R x times the thickness dx. If we denote the area of R x by A (x), then this volume is A (x) dx. Tbus the volume of the entire solid, obtained by summing, is the integral f~A (x)dx, where tue limits a and bare determined by the ends of the solid.

The Slice Method Let S be asolid and Px be a family of parallel planes such that: 1. S lies between Pa and Pb; 2. the area of the slice of S cut by P x is A (x). Tben the volume of S is equal to .rA(x)dX.

9.1 Volumes by the Slice Method

421

The slice method can also be justified using step functions. We shall see how to do this below. In simple cases, the areas A (x) can be computed by elementary geometry. For more complicated problems, it may be necessary to do a preliminary integration to find the A (x)'s themselves. Find the volume of abalI! of radius r.

Example 1

Draw the ball above the x axis as in Fig. 9.1.5.

Solution

region R.. of are. A (x)

- I - - - - + l f - - ! - + t - -- x

Figure 9.1.5. The area of the slice at x of a ball of radius r is A (x) = 77(r 2 - x 2).

Let the plane Po pass through the center of the ball. The ball lies between P -r and Pr' and the slice Rx is a disk of radius ~r2 - x 2 . The area of the slice is 'Tl X (radiusf; i.e., A (x) = 'Tl(~r2 - x 2 )2 = 'Tl(r 2 - x 2). Thus the volume is

frA (x)dx= f r'Tl(r Example 2

~3 )[r = 1'Tlr 3• A

x 2)dx='Tl(r2x -

Find the volume of the conical solid in Fig. 9.1.6. (The base is a circle.)

1

Figure9.1.7. IDEI/IABI

h

=

Figure 9.1.6. Find the volume of this oblique circular cone.

Solution

2-

IGEI/IGBI = IGEl/IGCI

h

by similar triangles. But IABI = r, IGe! = h, and IGFI = h - x, and so IDEI = [(h - x)/ h]r.

We let the x axis be vertical and choose the family Px of plimes such that Po contains the base of the coneand Px is at distance x above Po . Then the cone lies between Po and Ph' and the plane section by Px is a disk with radius [(h - x)/h]r and area 'Tl[(h - x)/hfr 2 (see Fig. 9.1.7). By the slice method, IA sphere is the set of points in space at a fixed distance from a point. A ball is the solid region enclosed by a sphere, just as a disk is the plane region enclosed by a circle.

422

ehapter 9 Applications of Integration

the volume is

Example 3

Find the volume of the solid W shown in Fig. 9.1.8. It can be thought of as a wedge-shaped piece of a cylindrical tree of radius , obtained by making two saw cuts to the tree's center, one horizontally and one at an angle o. x

Figure 9.1.8. Find the volume of the wedge W.

Solution

With the setup in Fig. 9.1.8, we slice W by planes to produce triangles Rx of area A (x) as shown. The base b of the triangle is b = ~,2 - x 2 , and its height is h = btanO =~ tanO. Thus, A(x) = !bh = !(,2 - x 2)tanO. Hence, the volume is

frA (x)dx=

f, t (,2 - x 2)tanOdx= t (tanO)(,2x -

~3 )[,

= t(tanO)(2,3 - 2{) = 2;3 tanO. Notice that even though we started with a region with a circular boundary, 71 does not occur in the answer! A Example 4

A ball of radius, is cut into three pieces by parallel planes at a distance of ,/3 on each side of the center. Find the volume of each piece.

Solution

The middle piece lies between the planes P -,/3 and Pr / 3 of Example I, and the area function is A (x) = 71(,2 - x 2 ) as before, so the volume of the middle piece is

f rl

3

-r13

71(,2 _ x 2 ) dx = 71(,2X

_

.

,3 = 71 ( "3

X

3

3 ,3

)

I

r 3 /

-r/3 ,3

,3)

52

3

- TI + "3 - TI = TI 71' .

*

This leaves a volume of (t - )71,3 = H71,3 to be divided between the two outside pieces. Since they are congruent, each of them has volume H71,3. (you may check this by computing J;/371(,2 - x 2)dx.) A One way to construct asolid is to take a plane region R,. as shown in Fig. 9.1.9 and revolve it around the x axis so that it sweeps out a solid region S. Such solids are common in woodworking shops (lathe-tooled table legs), in pottery

9.1 Volumes by the Slice Method

423

)'

Revolve

Figure 9.1.9. S is the solid of revolution obtained by revolving the plane region R about the x axis.

x

x

studios (wheel-thrown pots), and in nature (unicellular organisms).2 They are called solids oj revolution and are said to have axial symmetry. Suppose that region R is bounded by the lines x = a, x = b, and y = 0, and by the graph of the function y = j(x). To compute the volume of S by the slice method, we use the family of planes perpendicular to the x axis, with Po passing through the origin. The plane section of S by Px is a circular disk of radius j(x) (see Fig. 9.1.10), so its area A (x) is '7I"[f(x)f By the basic formula of the slice method, the volume of S is

ibA(X)dX= i b'7l"[j(X)Ydx= '71"i b[j(X)]2 dx . [(x)

x

Figure 9.1.10. The volume of asolid of revolution obtained by the disk method.

We use the term "disk method" for this special case of the slice method since the slices are disks.

Volume

0'Disk a Solid 0' Revolution: Method

The volume of the solid of revolution obtained by revolving the region under the graph of a (non-negative) functionj(x) on [a,b] about the x aXls IS

Example 5

The region under the graph of x 2 on [0, I] is revolved about the x axis. Sketch the resulting solid and find its volume.

Solution

The solid, which is shaped something like a trumpet, is sketched in Fig. 9.1.11. )'

Figure 9.1.11. The volume of this solid of revolution is

x

'lTfA YI), (X2' Y2), and (X3' Y3)' Show that their center of mass is at the intersection point of the medians of the triangle at whose vertices the masses are located. Find the center of mass of the regions in Exercises 17-22. 17. The region under the graph of 4/ x 2 on [1,3]. 18. The region under the graph of 1 + x 2 + x 4 on [-I, I]. 19. The region under the graph of ~ on [0, I]. 20. The region under the graph of ~I - x 2 / a2 on [ -a,a]. 21. The triangle with vertices at (0,0), (0,2), and (4,0). 22. The triangle with vertices at (1,0), (4,0), and (2,3). 23. If, in formula (3), we have a .;; Xi .;; b for all Xi' show that a .;; x .;; b as weil. Interpret this statement geometrically. 24. Let a mass mi be placed at position Xi on a line (i = I, ... ,n). Show that the function fex) = 2:7=lmi (x - xi is minimized when X is the center of mass of the n partic1es.

9.5 Energy, Power and Work

25. Suppose that masses mj are located at points Xj on the line and are moving with velocity Vj = dxJ dt (i = I, ... , n). The total momentum of the particles is defined to be P = m,v, + m2v2 + ... + mnvn • Show that P = Mv, where M is the total mass and v is the velocity of the center of mass (i.e., the rate of change of the position of the center of mass with respect to time). 26. A mass mj is at position Xj = }';(1) at time t. Show that if the force on mj is Fj(t), and F,(t) + Flt) = 0, then the center of mass of m, and m2 moves with constant velocity. 27. From a disk of radius 5, a circular hole with radius 2 and center 1 unit from the center of the disk is cut out. Sketch and find the center of mass of the resulting figure. 28. Suppose thatj(x) ..;; g(x) for all x in [a,b). Show that the center of mass of the region between the graphs of j and gon [a,b) is located at (x,y),

445

where _

x=

y=

f~x[g(x) - j(x)]dx f~[ g(x) - j(x)] dx H~[g(x)

, and

+ j(x)][g(x) -

j(x)]dx

---~-----::----

J~[g(x) - j(x)]dx

29. Find the center of mass of the region between the graphs of sin x and cos x on [0, 'IT /4). [Hint: Find the center of mass of each infinitesimal strip making up the region, or use Exercise 28.) 30. Find the center of mass of the region between the graphs of - x 4 and x 2 on [-I, I). (See the hint in Exercise 29.) *31. Find the center of mass of the triangular region with vertices (x" y,),.(X2, Y2), and (X3, 13). (For X2 .;; X3, convenience, you may assurne that y, .;; 13, and Y2 .;; Y3·) Compare with Exercise

x, . ;

16.

9.5 Energy, Power, and Work Energy is the integral of power over time, and work is the integral of force over distance. Energy appears in various forms and can often be converted from one form into another. For instance, a solar cell converts the energy in light into electrical energy; a fusion reactor, in changing atomic structures, transforms nuc1ear energy into heat energy. Despite the variety of forms in which energy may appear, there is a common unit of measure for all these forms. In the MKS (meter-kilogram-second) system, it is the joule, which equals 1 kilogram meter per second2• Energy is an "extensive" quantity. This means the following: the longer a generator runs, the more electrical energy it produces; the longer a light bulb burns, the more energy it consumes. The rate (with respect to time) at which some form of energy is produced or consumed is called the power output or input of the energy conversion device. Power is an instantaneous or "intensive" quantity. By the fundamental theorem of calculus, we can compute the total energy transformed between times a and b by integrating the power from a to b.

Power and Energy Power is the rate of change of energy with respect to time:

P

= dE dt

.

The total energy over a time period is the integral of power with respect to time:

E

= Lbpdt.

446

ehapler 9 Applications of Integration

A common unit of measurement for power is the watt, which equals 1 joule per second. One horsepower is equal to 746 watts. Tbe kilowatt-hour is a unit of energy equal to the energy obtained by using 1000 watts for 1 hour (3600 seconds}--that is, 3,600,000 joules.

Example 1

Solution

Tbe power output (in watts) of a 6O-cyc1e generator varies with time (measured in seconds) according to the formula P = Pasin2(120wt), where Po is the maximum power output. (a) What is the total energy output during an hour? (b) What is the average power output during an hour? (a) Tbe energy output, in joules, is

E

= fo36OOPasin2( 120wl) dt.

Using the formula sinlg = (1 - cos29)/2, we find

1

E = "2 PoSo

3600

(1 - cos240wt)dt=

1 [1

"2 Po t - 240n sin240ntJI3600 0

(b) Tbe average power output is the energy output divided by the time (see Section 9.3), or 1800 Po/3600 = t Po; in this case, half the maximum power output. ... A common form of energy is mechanical energy-the energy stored in the movement of a massive object (kinetic energy) or the energy stored in an object by virtue of its position (potential energy). Tbe latter is illustrated by the energy we can extract from water stored above a hydroelectric power plant. We accept the following principles from physics:

t

1. Tbe kinetic energy of a mass m moving with velocity v is mv 2• 2. Tbe (gravitational) potential energy of a mass m at a height h is mgh (here g is the gravitational acceleration; g = 9.8 meters/{second)2 = 32 feet/(second)2. Tbe total force on a moving object is equal to the product of the mass m and the acce1eration dv / dt = d 2x / dt 2• Tbe unit of force is the newton which is 1 kilogram meter per second2• If the force depends upon the position of the object, we may calculate the variation of the kinetic energy K = mv 2 with position. We have

t

dK dx

= dK/dt = (d/dt)(tmv 2) = dx/ dt

v

mvdv/dt v

= m dv = F

dt·

Applying the fundamental theorem of calculus, we find that the change t::.K of kinetic energy as the partic1e moves from a to b is J:F dx. Often we can divide the total force on an object into parts arising from identifiable sources (gravity, friction, fluid pressure). We are led to define the work W done by a particular force Fon a moving object (even if there are other forces present) as W= J:F dx. Note that if the force F is constant, then the work done is simply the product of F with the displacement t::.x = b - a. Accordingly, 1 joule equals 1 newton-meter.

9.5 Energy, Power and Work

447

Force and Work Tbe work done by a force on a moving object is the integralof the force with respect to position:

W= LbFdx. If the force is constant, Work = Force X Displacement.

If the total force F is a sum F\

tlK= Lb(F\

+ . . . + Fn, then we have

+ ... + Fn)dx= LbF\dx+

...

+ LbFndx.

Tbus the total change in kinetic energy is equal to the sum of the works done by the individual forces.

Example 2

Tbe acc~leration of gravity near the earth is g = 9.8 meters/(second)2. How much work does a weight-lifter do in raising a 50-kilogram barbell to a height of 2 meters? (See Figure 9.5.1.) 25 kg

25 kg

1 2m

Flgure 9.5.1. How much

work did the weight-lifter do?

Solution

We let x denote the height of the barbell above the ground. Before and after the lifts, the barbell is stationary, so the net change in kinetic energy is zero. Tbe work done by the weight-lifter must be the negative of the work done by gravity. Since the pull of gravity is downward, its force is -9.8 meters per second2 X 50 kilograms = - 490 kilograms . meters per second2 = - 490 newtons; tlx = 2 meters, so the work done by gravity is - 980 kilograms . meters2 per second2 = - 980 joules. Thus the work done by the weight-lifter is 980 joules. (If the lift takes s seconds, the average power output is (980/ s) watts.)

Example 3

Show that the power exerted by a force F on a moving object is Fv, where v is the velocity of the object.

Solution

Let E be the energy content. By our formula for work, we have tlE = f~F dx, so dE / dx = F. To compute the power, which is the time derivative of E, we use the chain rule:

•

dE P = dt

=

dE dx dx . dt

= Fv.

(In pushing a child on a swing, this suggests it is most effective to exert your force at the bottom of the swing, when the velocity is greatest. Are there any complicating factors?) •

448

Chapler 9 Applications ot Integration

Example 4

A pump is to empty the conical tank of water shown in Fig. 9.5.2. How much energy (in joules) is required for the job? (A cubic meter of water has mass 103 kilograms. )

Figure 9.5.2. To calculate the energy needed to empty the tank, we add up the energy needed to remove slabs of thickness dx.

Solution

Consider a layer of thickness dx at depth x, as shown in Fig. 9.5.2. By similar triangles, the radius is r = -?o(1O - x), so the volume of the layer is given by 'TT' 160(10 - x?dx and its mass is 103 • 'TT' 160(10 - x?dx = 90'TT(10 - xfdx. To lift this layer x meters to the top of the tank takes 90'TT(10 - x? dx . g' x joules of work, where g = 9.8 meters per second2 is the acceleration due to gravity (see Example 2). Thus, the total work done in emptying the tank is

90~fo'°(10 -

~3 + ~4 Je

90~[ 100 ~2 -

20

=

90~(104)[ ~ -

t+~]

=

(90)(9.8)('TT)(104)(

x)2x dx=

I~ )

~ 2.3 X 106 joules. •

!lil Example 5 The pump which is emptying the conical tank in Example 4 has apower output of 105 joules per hour (i.e., 27.77 watts). What is the water level at the end of 6 minutes of pumping? How fast is the water level dropping at this time? Solution

The total energy required to pump out the top h meters of water is

90~ fo\1O -

x)2x dx=

90~( 100 ~2

_

20 ~3 +

~4)

~2770h2(50 - 230 h + ~2). At the end of 6 minutes (1ö hour), the pump has produced IQ4 joules of energy, so the water level is h meters from the top, where h is the solution of 2770h 2( 50 - 230 h +

~2 ) = 104 •

Solving this numerically by the method of bisection (see Section 3.1) gives h ~ 0.27 meter.

At the end of t hours, the total energy output is lOSt joules, so

90~ fo\1O - x)2 x dx=

lOSt,

where h is the amount pumped out at time t. Differentiating both sides with

9.5 Energy, Power and Work

449

respect to t gives

90g'1T(1O - h)2h ~~

=

105,

dh = _----=1;...:;..04_--::dt 9g'1T(1O - h)2h .

SO

when h = 0.27, this is 1.41 meters per hour....

Supplement to Sectlon 9.5: Integratlng Sunshlne

We will now apply the theory and practice of integration to compute the total amount of sunshine received du ring a day, as a function of latitude and time of year. If we have a horizontal square meter of surface, then the rate at which solar energy is received by this surface-that is, the intensity of the solar radiation-is proportional to the sine of the angle A of elevation of the sun above the horizon. 6 Thus the intensity is highest when the sun is direct1y overhead (A = 'lT /2) and reduces to zero at sunrise and sunset. The total energy received on day T must therefore be the product of a constant (which can be determined only by experiment, and which we will ignore) and the integral E = J:~msinA dt, where t is the time of day (measured in hours from noon) and to(T) and tl(T) are the times of sunrise and sunset on day T. (When the sun is below the horizon, although sinA IS negative, the solar intensity is simply zero.) We presented a formula for sinA (formula (1) in the Supplement to Chapter 5, to be derived in the Supplement to Chapter 14), and used it to determine the time of sunset (formula (3) in the Supplement to Chapter 5). The time of sunrise is the negative of the time of sunset, so we have 7

E=

JS-ssinA dt,

(1)

where

. smA = cos I

1 - sm . 2 2'lTt ) + sm . I sma . cos ( 2'lTT a cos 2( 2'lTT) 365 cos ( 24 365 )

and

s = 224 cos- I [ -tan! 'lT

sin a cos(2'lTT/365)

VI - sin a cos (2'lTT /365) 2

1.

(2)

2

Here a:::::; 23.5 0 is the inclination of the earth's axis from the perpendicular to the plane of the earth's orbit; I is the latitude of the point where the sunshine is being measured. The integration will be simpler than you may expect. First of all, we simplify notation by writing k for the expression sin a cos(2 'lTT /365), which appears so often. Then we have

E= =

LSA cosl~ cos( ;7) + (Sin!)k] dt coslb - k LSscos( ;7 )dt+ (sin/)k LSs dt. 2

6 We will justify this assertion in the Supplement to Chapter 14. We also note that, strictly speaking, it applies only if we neglect absorption by the atmosphere or assume that our surface is at the top of the atmosphere.

7 All these calculations assume that there is a sunrise and sunse!. In the polar regions during the summer, the calculations must be altered (see Exercise 5 below).

450

Chapter 9 Applications of Integration

Integration gives

E=

COS/~(

24 sin 2'Tlt /s ) + 2Sksinl 2'TT 24- s

2 S - Sin2'Tl(-S») + 2Sksinl = cos/VI~(24)( - k2 sin~ 2'Tl

24

24

= :: cos/~1 - k 2 sin 2~S + 2Sk sin I. The expression sin(2'TlS /24) can be simplified. Using the formula cos(2'TlS/24) = -(tanl)(k/~), we get sin 2'TlS 24

=

1 - cos2 2'TlS 24

=

1 - k 2 - (tan 2/)k 2

1- k 2 1 - (1

+ tan 2/)k 2

1- k2 and so, finally, we get,

E=

24cOs/~I-(sec2/)k2 + 24ksin/cos-l['Tl

'Tl

(tan/)k k2

~1 _

l.

Since both k and ~ appear, we can do even better by writing k = sin D (the number Dis important in astronomy-it is called the declination), and we get

E=

: : [ cos I~l

- sec21sin2D + sin I sinD cos -1( - tanltan D)

l

Since we have already ignored a constant factor in E, we will also ignore the factor 24/'Tl. Incorporating cosl into the square root, we obtain as our final result

E = ~cos2l- sin2D + sinlsinD cos- 1( - tanltanD),

(3)

where sin D = sin a cos(2'TlT/365). Plotting E as a function of I for various values of T leads to graphs like those in Figs. 3.5.4 and 9.5.3. Example

When does the equator receive the most solar energy? The least?

Solution

At the equator, 1 = 0, so we have

E = ~l - sin2D

=

1 - sin2a cos2 ( 2'TlT) 365

We see by inspection that E is largest when cos2(2'TTT/365) = O-that is, when T /365 = ~ or i; that is, on the first days of fall and spring: on these days E = 1. We note that E is smallest on the first days of summer and winter, when cos2(2'Tl /365) = 1 and we have E = sin2a = cosa = cos23.5° = 0.917, or about 92% of the maximum value . .A.

b-

9.5 Energy, Power and Work

451

Using this example we can standardize units in which E can be measured. One unit of E is the total energy received on a square meter at the equator on the first day of spring. All other energies may be expressed in terms of this unit.

Figure 9.5.3. Computergenerated graph of the daily sunshine intensity on the earth as a funetion of day of the year and latitude.

Exercises for the Supplement to Section 9.5 I. Compare the solar energy reeeived on lune 21 at the Aretie Circle (l = 90° - 0:) with that reeeived at the equator. 2. What would the inelination of the earth need to be in order for E on lune 21 to have the same value at the equator as at the latitude 90° - o:? 3. (a) Express the total solar energy reeeived over a whole year at latitude I by using summation notation. (b) Write down an integral which is approximately equal to this sumo Can you evaluate it? 4. Simplify the integral in the solution of Exereise 3(b) for the eases 1=0 (equator) and 1= 90° - 0: (Aretie Circle). In eaeh ease, oneof the two terms in the integrand ean be integrated explieitly: find the integral of this term.

5. Find the total solar energy reeeived at a latitude in the polar region on a day on whieh the sun never sets. 6. How do you think the climate of the earth would be affeeted if the inelination 0: were to beeome: (a) 10°? (b) 40°? (In eaeh ease, diseuss whether the North Pole reeeives more or less energy during the year than the equator-see Exercise 5.) 7. Consider equation (3) for E. For D = 7r /8, eompute dE / dl at I = 7r /4. Is your answer eonsistent with the graph in Fig. 9.5.3 (look in the plane of eonstant T)? 8. Determine whether a square meter at the equator or at the North Pole reeeives more solar energy: (a) during the month of February, (b) during the month of April, (e) during the entire year.

452

Chapter 9 Applications of Integration

Exerclses for Sectlon 9.5 1. The power output (in watts) of a 6O-cycle generator is P = 1050 sin2(120'/Tt), where t is measured in seconds. What is the total energy output in an hour? 2. A worker, gradually becoming tired, has apower output of 30e- 2t watts for 0 ... t ... 360, where t is the time in seconds from the start of a job. How much energy is expended during the job? 3. An electric motor is operating with power 15 + 2 sin(t'/T /24) watts, where t is the time in hours measured from midnight. How much energy is consumed in one day's operation? 4. The power output ora solar cell is 25sin('/Tt/12) watts, where t is the time in hours after 6 A.M. How many joules of energy are produced between 6 A.M. and 6 P.M? In Exercises 5-8, compute the work done by the given force acting over the given interval. 5. F.= 3x; 0 ... x ... 1. 6. F = k/ x 2 ; I ... x ... 6 (k a constant). 7. F= 1/(4+ x 2); 0 ... x ... 1. 8. F = sin3x cos2x; 0 ... X ... 2. [Hint: Write sin 3x = sinx(l- cos2x).) 9. How much power must be applied to raise an object of mass 1000 grams at a rate of 10 meters per second (at the Earth's surface)? 10. The gravitational force on an object at a distance r from the center of the earth is k/r 2, where k is a constant. How much work is required to move the object: (a) From r = I to r = 1O? (b) From r = I to r = looo? (c) From r = I to r = 1O,ooo? (d) From r = I to "r = oo"? 11. A particle with mass 1000 grams has position x = 3t 2 + 4 meters at time t seconds. (a) What is the kinetic energy at time t? (b) What is the rate at which power is being supplied to the object at time 1 = 1O? 12. A particle of mass 20 grams is at rest at 1 = 0, and power is applied at the rate of 10 joules per second. (a) What is the energy at time I? (b) If all the energy is kinetic energy, what is the velocity at time I? (c) How far has the particle moved at the end of t seconds? (d) What is the force on the particle at time t? 13. A force F(x) = -3x newtons acts on a particle between positions x = I and x = O. What is the change in kinetic energy of the particle between these positions? 14. A force F(x) = 3xsin('/Tx/2) newtons acts on a particle between positions x = 0 and x = 2. What is the increase in kinetic energy of the particle between these positions? 15. (a) The power output of an electric generator is 25 cos2(l20'/Tt) joules per second. How much en-

ergy is produced in I hour? (b) The output of the generator in part (a) is converted, with 80% efficiency, into the horizontal motion of a 250-gram object. How fast is the object moving at the end of I minute? 16. The generator in Exercise 15 is used to lift a 500-kilogram weight and the energy is converted via pulleys with 75% efficiency. How high can it lift the weight in an hour? Exercises 17-20 refer to Figure 9.5.4. 17. How much energy is required to pump all the water out of the swimming pool? 18. Suppose that a mass equal to that of the water in the pool were moving with kinetic energy equal to the result of Exercise 17. What would its velocity be? 19. Repeat Exercise 17 assuming that the pool is fi1led with a liquid three times as dense as water. 20. Repeat Exercise 18 assuming that the pool is filled with a liquid three times as dense as water.

2m

~

T

15 m

"'l"

lsm~

Figure 9.5.4. How much energy is required to empty this pool of water? 21. Suppose that aspring has a naturallength of 10 centimeters, and .that a force of 3 newtons is required to stretch it to 15 centimeters. How much work is needed to compress the spring to 5 centimeters? 22. If all the energy in the compressed spring in Exercise 21 is used to fire a ball weighing 20 grams, how fast will the ball travel? 23. How much work is required to fill the tank in Figure 9.5.5 with water from ground level? 24. Asolid concrete monument is built in the pyramid shape of Fig. 9.5.6. Assurne that the concrete weighs 260 pounds per cubic foot. How much work is done in erecting the monument? (g = 32 feet per second2 ; express your answer in units of pound-feetl per second 2 :)

Review Exercises for Chapter 9

453

Figure 9.5.6. How much energy is needed to erect this monument?

Figure 9.5.5. How much energy is needed to fill this tank?

Review Exerclses for Chapter 9 In Exercises 1-4, find the volume of the solid obtained by rotating the region under the given graph about (a) the x axis and (b) the y axis. l. y = sinx, x.;;; 'Ir 2. Y = 3 sin2x, x .;;; 'Ir /4 3. Y = e X , x .;;; In2 4. y = 5e 2x , x';;; In4

°.;° ; .; ; °° .; ; .; ;

5. A cylindrical hole of radius ~ is drilled through the center of a ball of radius l. What is the volume of the resulting solid? 6. A wedge is cut in a tree of radius I meter by making two cuts 10 the center, one horizontally, and one at an angle of 20° to the first. Find the volume of the wedge. 7. Find the volume of the "football" whose dimensions are shown in Fig. 9.R.l. The two ares in the figure are segments of parabolas.

:;;;.r.-t I

I~

f

p=". :=:q~ 3"

Figure 9.R.l. Find the volume of the football.

8. Imagine the "football" in Fig. 9.R.I, formed by revolving a parabola, to be solid. A hole with radius 1 inch is drilled along the axis of symmetry. How much material is removed? In Exercises 9-12, find the average value of each function on the stated interval.

°.; ; °.; ;

9. 1 + t 3,

10. t sin(t 2), 'Ir .;;; t .;;; 3'Ir /2 1 12. - - 2 ' 1 .;;; x .;;; 3 11. xeX, x .;;; 1 l+x 13. If fä/(x)dx = 4, what is the average value of g(x) = 3j(x) on [O,2]? 14. Ifj(x) = kg(ex) on [a,b], how is the average ofj on [a,b] related to that of gon [ae,be]?

t .;;; I

15. Show that for some x in [O,'Ir], 2 + :osx is equal to

L +dU w

o 2

cos

IJ·

16. (a) Prove that

1/12 .;;;f[ dt/{t!+T] .;;; (b) Prove that

f[ dt / {t!+T]

I.

= sin IJ for some IJ,

'Ir/4.;;; IJ.;;; 'Ir/2. In Exercises 17-22, let p. be the average value of j on [a, b]. Then the average value of [f(x) - p.]2 on [a, b] is called the lIarianee of j on [a, b], and the square root of the variance is called the standard deviation of j on [a, b] and is denoted CI. Find the average value, variance, and standard deviation of each of the following functions on the interval specified. 17. x 2 on [0,1] 18. 3 + x 2 on [0, 1] 19. xe x on [0, I] 20.sin 2x on [0, 4'1r] 21. j(x) = I on [0, 1] and 2 on (1,2]. 2 on [0, 1] 22. j(x) = { 3 I

on (1,2] on (2,3]

5

on (3,4]

23. Let the region under the graph of a positive function j(x), a .;;; x .;;; b, be revolved about the x axis to form asolid S. Suppose this solid has a mass density of p(x) grams per cubic centimeter at a distance x along the x axis. (a) Find a formula for the mass of S. (b) If j(x) = x 2 , a = and b = I, and p(x) = (I + x 4 ), find the mass of S. 24. A rod has linear mass density p.(x) grams per centimeter at the point x along its length. If the rod extends from x = a to x = b, find a formula for the location of the rod's center of mass. In Exercises 25-28, find the center of mass of the region under the given graph on the given interval. 25. y = x 4 on [0, 2] 26. Y = x 3 + 2 on [0, 1] 27. Y = In(l + x) on [0,1] 28. y = e X on [1,2]

°

454

Chapter 9 Applications of Integration

29. Find the center of mass of the region between the graphs of y = x 3 and y = - Xl between X = 0 and X = 1 (see Exercise 28, Section 9.4). 30. Find the center of mass of the region composed of the region under the graph y = sinx, 0.;;; x .;;; 'TT, and the circle with center at (5,0) and radius 1. 31. Over a time period 0.;;; t .;;; 6 (t measured in minutes), an engine is consuming power at a rate ~f 20 + 5te- t watts. What is (a) the total energy consumed? (b) The average power used? 32. Water is being pumped from a deep, irregularly shaped well at a constant rate of 3! cubic meters per hour. At a certain instant, it is observed that the water level is dropping at a rate of 1.2 meters per hour. What is the cross-sectional area of the well at that depth? 33. A force F(x) = 30sin('TTx/4) newtons acts on a particle between positions x = 2 and x = 4. What is the increase in kinetic energy (in joules) between these positions? 34. The engine in Fig. 9.R.2 is using energy at a rate of 300 joules per second to lift the weight of 600 kilograms. If the engine operates at 60% efficiency, at what speed (meters per second) can it raise the weight?

Figure 9.R.2. The engine for Problem 34. 35. Find a formula for the work required to empty a tank of water which is asolid of revolution about a vertical axis of symmetry. 36. How much work is required to empty the tank shown in Fig. 9.R.3? [Hint: Use the result of Exercise 35.]

Parabol3

Figure 9.R.3. How much energy is needed to empty the tank?

37. The pressure (force per unit area) at a depth h below the surface of a body of water is given by p = pgh = 9800h, measured in newtons per square meter. (This formula derives from the fact that the force needed to support a column of water of cross-sectional area A is (volume) X (density) X (g) = Ahpg, so the force per unit area is pgh, where p = 103 kilograms per cubic meter, and g = 9.8 meters per second per second). (a) For the dam shown in Fig. 9.R.4(a), show that the total force exerted on it by the water is F = H~pg[f(x)f dx. [Hint: First calculate the force exerted on a vertical rectangular slab.] (b) Make up a geometric theorem relating F to the volume of a certain solid. (c) Find the total force exerted on the dam whose face is shown in Fig. 9.R.4(b).

~ SOm

(b) Dam face

Figure 9.R.4. Calculate the force on the dam.

38. (a) Pappus' theorem Jor volumes. Use the shell method to show that if a region R in the xy plane is revolved around the y axis, the volume of the resulting solid equals the area of R times the circumference of the circle obtained by revolving the center of mass of R around the y axis. (b) Use Pappus' theorem to do Exercise 21(a) in Section 9.2. (c) Assuming the formula v= ~'TTr3 for the volurne of a ball, use Pappus's theorem to find the center of mass of the semicircular region xl + y2 .;;; r 2, x ;;;. O. *39. See the instructions for Exercises 17-22. (a) Suppose that J(t) is a step function on [a, b], with value k i on the interval (t; - I ' tJ belong-

Review Exercises for Chapter 9

(b) (c)

(d) (e) *40. (a)

ing to a partition (to, t l , ••• ,tn ). Find a fonnula for the standard deviation of j on [a,b]. Simplify your fonnula in part (a) for the case when all the t:.t;'s are equal. Show that if the standard deviation of a step function is zero, then the function has the same value on all the intervals of the partition; Le., the function is constant. Give a definition for the standard deviation of a list ab ... , an of numbers. What can you say about a list of numbers if its standard deviation is zero? Prove, by analogy with the mean value theorem for integrals, the second mean value theorem: If j and gare continuous on [a, b] and g(x);;. 0, for x in [a,b], then there is a point to in [a,b] such that

t

j(t)g(t)dt= j(to)

f

g(t)dt.

455

(b) Show that the mean value theorem for integrals is a special case of the result in part (a). (c) Show by example that the conclusion of part (a) is false without the assumption that g(t) ;;. O. *41. Show that if f is an increasing continuous function on [a,b], the mean value theorem for integrals implies the conclusion of the intennediate value theorem. *42. Show that in the context of Exercise 35, the work needed to empty the tank equals Mgh, where M is the total mass of water in the tank and h is the distance of the center of mass of the tank below the top of the tank. *43. Let j(x) > 0 for x in [a,b]. Find a relation between the average value of the logarithmic derivative j'(x)jj(x) of fon [a, b] and the values of j at the endpoints.

Chapter10

Further Techniques and Applications of Integration Some simple geometrie problems require advaneed methods of integration. Besides the basie methods of integration associated with reversing the differentiation rules, there are speeial methods for integrands of partieular forms. Using these methods, we ean solve some interesting length and area problems.

10.1 Trigonometrie Integrals The key to evaluating many integrals is a trigonometrie identity or substitution. The integrals treated in this seetion fall into two groups. First, there are some purely trigonometrie integrals that ean be evaluated using trigonometrie identities. Seeond, there are integrals involving quadratie funetions and their square roots whieh ean be evaluated using trigonometrie substitutions. We begin by eonsidering integrals of the form J sinmx cosnx dx,

where m and n are integers. The ease n = 1 is easy, for if we let u = sinx, we find m+ 1 sinm+ 1( x) __ +C= +C J sinmxeosxdx=Jumdu= _u m+1 m+l (or lnlsinxl + C, if m = -I). The ease m = 1 is similar:

Jsinxcosnxdx=-

eosn+ 1(x) n+1 +C

(or -lnleosxl + C, if n = - I). If either m or n is odd, we ean use the identity sin2x + eos2x = 1 to reduee the integral to one of the types just treated.

Example 1 Solution

Evaluate

f sin x eos x dx. 2

3

fsin 2xeos 3xdx = fsin 2xcos 2xeosxdx = f(sin 2x)(1 - sin 2x)eosxdx, which ean be integrated by the substitution u = sinx. We get

l. 3 I· 5 C .... f 2( I -u2) u=T-s+ C=3smx-Ssmx+ U

d

u3

u5

458

Chapler 10 Further Techniques and Applications of Integration

If m = 2k and n = 2/ are both even, we can use the half-angle formulas sin2x = (I - cos2x)j2 and cos2x = (I + cos2x)j2 to write

=

! J(

1-

~s Y ) k ( 1 + ~os Y ) 1dy,

where y = 2x. Multiplying this out, we are faced with a sum of integrals of the form fcos'jldy, with m ranging from zero to k + I. The integrals for odd m can be handled by the previous method; to those with even m we apply the half-angle formula once again. The whole process is repeated as orten as necessary until everything is integrated.

Example 2 Solution

Evaluate J sin2x cos2x dx.

J sin2x cos2x dx = J ( 1 - ~s 2x )( 1 + ~os 2x ) dx

= .!. - ! 4

4

J 1 + cos4x dx= .!. - .!. - ! Jcos4xdx 2 4 8 8

Trigonometrie Integrals To evaluate fsinmxcosRxdx: 1. If m is odd, write m

= 2k + I, and

f sinmx cosRx dx = f sin2kx cosRx sin x dx = J(I- cos2x)kCoslixsinxdx.

Now integrate by substituting u = cosx. 2. If n is odd, write n = 2/ + I, and

J sinmx cosRx dx = J sinmx COS21X cos x dx

= Jsinmx(1

- sin2x)/cosxdx.

Now integrate by substituting u = sinx. 3. (a) If m and n are even, write m = 2k and n = 2/ and

Jsin2kxcos2t.xdx= J( 1- ~os2x

f(

1 + ~S2x)1 dx.

Substitute y = 2x. Expand and apply step 2 to the odd powers of cosy. (b) Apply step 3(a) to the even powers of cos y and continue until the integration is completed.

10.1 Trigonometrie Integrals

Example 3

Evaluate: (a) fo21Tsin4x cos2x dx (b) f (sin 2x (c)

Solution

f tan

459

+ sin3x cos2x)dx.

3(J sec3(J dO.

(a) Substitute sin2x

= (l

- cos2x)/2 and cos 2x

= (l + cos2x)/2 to get

. 4 2 f (1 - COS2X)2 (1 + cos2x) f smxcosxdx= 4 2 dx

t = t f (l -

= f (1- 2cos2x + cos22x)(1 + cos2x)dx

= 2x.

+ cos32x) dx

/6 f (1- cos y - cosy + cosy)dy,

=

where y

cos 2x - cos22x

Integrating the last two terms gives

y sin2y f cosydy= f (1 + cos2y) 2 dy= '2 + -4- +c 2..

and • 3

f cosydy= f(1- siny)cosydy= siny - Sl~Y

+ C.

Thus

f sin4x COS2X dx= ...l. 16

(Y - sin y -

= ...l.(1. _ sin2y 2

16

4

= ...l. (x _ sin4x 16

and so fo21T sin4x eos2x dx (b)

=

i.

4

1. - sin2y + sin y _ 2

siny ) 3

4

_ Sin y ) 3

+C

_ sin 32x ) 3

+C

+C

'

f(sin 2x + sin3x eos2x) dx= fsin2xdx+ fsin3xeos2xdx =

f ( ~os 2x ) dx + f ( cos2x )cos x sin x dx 1-

1-

= =

2

I - sini x - f(l- u )u du 2

2

+ cos5x + C

:!. _ sin 2x _ cos3x 2 4 3

5

(u

=

eosx)

.

(e) Method l. Rewrite in terms of sec 0 and its derivative tan 0 sec 0:

f tan30see30 dO = f (tan Osee 0 )(tanZO secZO) dO

= f(tanOseeO)(seeZO - l)seeZOdO

(l

+ tan 2 e = see2 e)

=

(u

= secO)

f (u

=

2-

1) u2 du

u u sec 0 '5 - T +c= -55

3

5

see30 -3-

+c.

460

Chapter 10 Further Techniques and Applications of Integration

Method 2. Convert to sines and cosines:

(u = cosO) =

-5

~

5

-3

-

~

3

5(}

3(}

+ C = sec _ sec + C .. 5

3

.

Certain other integration problems yield to the use of the addition formulas:

+ y) = sinx cos y + cosx sin y, cos( x + y) = cos x cos y - sin x sin y sin( x

(la) (lb)

and the product formulas:

= Hsin( x sinxsiny = H cos(x cosxcos Y = Hcos(x sin x cos y

Example 4 Solution

+ sin( x + y)], y) - cos(x + y)J,

(2b)

y)

(2c)

y)

(2a)

+ cos(x + y) J.

Evaluate (a) fsinxcoil2xdx and (b) fcos3xcos5xdx. (a)

Jsinxcos2xdx=

t t

J(sin3x - sinx)dx= - co~3x + co;x +C.

(see product formula (2a». (b)

J cos3xcos5xdx=

J(cos8x + cos2x)dx= si~:x + sin42x + C

(see product formula (2c» ... Example 5 Solution

Evaluate J sin ax sin bx dx, where a and bare constants. If we use identity (2b), we get

f sinax sinbxdx= t f[ cos(a -

b)x - cos(a + b)xJ dx

1 sin(a - b)x a-b

2

1 sin(a + b)x a+b

-2

+c

if a'i= ± b,

x l · 2 ax+ C ---sm 2 4a '

if a = b,

-.L sin2ax - ~ + C

if a

4a

2'

= -b.

[The difference between the case a 'i= ± band the other two should be noted. The first case is "pure oscillation" in that it consists of two sine terms. The others contain the nonoscillating linear term x /2, called a secular term. This example is related to the phenomena of resonance: when an oscillating system is subjected to a sinusoidally varying force, the oscillation will build up indefinitely if the force has the same frequency as the oscillator. See Review Exercise 56, Chapter 8 and the discussion in the last part of Section 12.7, following equation (14).] ..

10.1 Trigonometrie Integrals

461

Many integrals eontaining faetors of the form ~a2 ± x 2 , ~X2 - a2 , or a2 + x 2 ean be evaluated or simplified by means of trigonometrie substitutions. In order to remember what to substitute, it is useful to draw the appropriate right-angle triangle, as in the following box.

Trigonometrie Substitutions l. If ~a2 - x 2 oeeurs, try x = asinlJ; then dx = aeosOdlJ and ~a2 - x 2

= aeoslJ; (a > 0 and IJ is an aeute angle).

2. If ~x2 - a 2 oeeurs, try x ~X2 - a 2 = atanlJ.

= aseeO;

then dx

= atanOseelJdO

and

x=ase~

~ Jx1-a 1 =atanO a

3. If ~a2 + x 2 or a2 + x 2 oeeurs, try x = atanlJ; then dx = asee1JdO and ~a2 + x 2 = a seelJ (one ean also use x = a sinhlJ; then ~a2 + x 2

= a eoshlJ).

P:?j

atanO = x

9

a

Example 6

Evaluate: (a) J

F?

dx, (b)

X

Solution

(a) Let x

J

f

dx

~4x2 - 1

.

= 3sinlJ, so h - x 2 = 3eosO. Thus dx = 3eosIJdO and

~ 2 x

dx=J 3eosO 3eosIJdIJ 9sin1J = J eos1J dlJ= J 1 - sin 21J dlJ sin1J sin1J = Jl

=

The terms with a quadratic denominator may be integrated by a manipulation which makes the derivative of the denominator appear in the numerator, together with completing the square (see ExampIes 3 and 6).

470

Chapler 10 Further Techniques and Applications of Integration

Example4 Solution

First we divide out the fraction to get 1 )dx=!x3+f dx . f xS-X4+1dx=f(x2+ 3 x 3- x 2 x3_ x2 x3_ x2

The denominator x 3 - x 2 is of degree 3, and the numerator is of degree zero. Thus we proceed to step 2 and factor: x 3 - x 2 = x 2( x-I). Here x

=x

- 0 occurs to the power 2, so by step 3, we write down

a)

a2

x

x

-+-. 2 We also add the term bd(x - 1) for the second factor. a)

a2

b)

-+-+--. x x 2 x-I Since there are no quadratic factors, we omit step 4. By step 5, we equate the preceding expression to 1/(x 3 - x 2): a)

a2

b)

1

- + - 2 + - - = ---,---=-X x x-I x 2( x-I) . Then we muItipIy by x 2(x - 1): a)x(x - 1) + a 2(x - 1) + b)x 2 = 1. Setting x = 0, we get a2 = -1. Setting x = 1, we get b) = 1. Comparing the coefficients of x 2 on both sides of the equation gives a) + b) = 0, so a) = - b) = -1. Thus a 2 = -1, a) = -1, and b) = 1. (We can check by substitution into the preceding equation: the left side is (-I)x(x - 1) - (x - 1) + x 2, which is just 1.) Thus

and so f x 3 dx _ x2

= 1-lnlxl + Inlx - 11 + c = !X + X

Inl x-I 1+ C. X

Finally, 4 f XSx-3 _x x+2 1 dx= !3 x 3 + !x Example 5 Solution

Integrate f

+ Inl

x-I x

1+ c.•

x 2 2 dx. (x 2 - 2)

The denominator factors as (x -fii(x +fii, so we write x2

a)

a2

(x 2 - 2)2

X -fi

(x -fi)

_..:.:.....---=-=--+

b)

b2

x +fi

(x +fi)

2+--+

2'

10.2 Partial Fractions

471

Thus

2

+ b2( x - ,;2) . We substitute values for x:

=,;2 : 2 = 8a 2

x

*; b = *.

so a2 =

x = -,;2 : 2 = 8b2

so

2

Therefore

x 2 = a l(x 2 - 2)(x

+,;2) + !(x 2 + 2,;2x + 2)

+b t (x 2 - 2)(x -,;2) + !(x 2 (al

=

2,;2 x + 2)

+ b l )X 3 + (,;2 a l +! -,;2b l )X 2

+ (-2a l - 2b l )x - 2,;2 al + 1 + 2,;2b l , and so

a l + b l = 0 and

,;2 al + ! -,;2 b l = 1.

Thus al

= _1_,

1 bl = - - ·

4,;2

4,;2

Hence

+

1

4,;2 (x -,;2)

1 4(x _ ,;2)2

1 ----':....-+ 4,;2 (x +,;2)

1

4(x

+,;2)

2 '

and so

f (x

x2 2 -

2)2

dX=_I_ ln 4.;2

= Example 6 Solution

Integrate

f (x - 1)(x

lx -,;2I_ x +.;2

I _ 1 4(x -.;2) 4(x +.;2)

x

2(x 2 -

_I_lnl x -.;21_ +,;2 4,;2

x3 2

+ 2x + 2)

2

x

2)

+C

+c..

dx.

For the factor x-I we write

al x-I ' and for (x 2 + 2x + 2)2 (which does not factor further since x 2 + 2x + 2 does not have real roots since b 2 - 4ac = 4 - 4· 1 ·2= -4< 0) we write Alx+B I X

2

+ 2x + 2

+

A 2x+B2 (x 2 + 2x

+ 2)

2 •

472

Chapter10 Further Techniques and Applications of Integration

We then set

a) A)x+B) A 2x+B 2 x3 --1+2 + 2= 2 X X + 2x + 2 (x 2 + 2x + 2) (x - 1)(x 2 + 2x + 2) and multiply by (x - l)(x 2 + 2x + 2)2:

a)(x 2 + 2x + 2)2 + (A)x + B))(x - 1)(x 2 + 2x + 2) +(A 2x + B 2)(x - 1) = x 3• Setting x = 1 gives a)(25)

= 1 or a) = E' Expanding the left-hand side, we get:

E(X 4 + 4x 3 + 8x 2 + 8x + 4) + A)x 4 + (AI + B I )X 3 + B)x 2 -2A)x - 2B) + A 2x 2 + (B 2 - A 2)x - B2 = x 3• Comparing coefficients:

+A ) = O',

x 4 ••

...l.. 25

x 3:

fs+ (A) + B) = 1;

(3) (4)

x 2 : fs+B) +A 2 =0; x:

fs -

(5)

2A 1+ (B 2 - A 2 )

xo(= 1):

fs -

= 0;

(6)

2B) - B 2 = O.

(7)

Thus

A)=

-E

(from equation (3»); (from equation (4»);

B)=* A2 = -

fs

(from equation (5»);

B2 =

~

(from equation (6»).

T"

At this stage you may check the algebra by substitution into equation (7). Algebraic errors are easy to make in integration by partial fractions. We have thus far established

x3 2 (x - I)(x + 2x + 2)2

= ~ [ _1_ 25

+

x-I

- x + 22 + - 30x - 40 ] x 2 + 2x + 2 (x 2 + 2x + 2/ .

We compute the integrals of the first two terms as follows:

J_1_ x-I

J

dx

= Inlx - 11 + c

-x + 22 dx =J -x - 1 + 23 dx x 2 + 2x + 2 x 2 + 2x + 2 = -

1J

= -

Ilnlx 2 + 2x + 21 + 23J

= -

!lnlx 2 + 2x + 21 + 23tan-)(x + 1) + C.

2

2

2x + 2 dx + 23J dx x 2 + 2x + 2 x 2 + 2x + 2 (x

dx

+ 1)2 + 1

10.2 Partial Fractions

473

Finally, for the last term, we rearrange the numerator to make the derivative of the quadratic polynomial in the denominator appear: J

- 30x - 40

---"-':':':"'---'~2

(x 2 + 2x + 2)

dx =

J - 15(2x + 2) - 10

=

2

(x 2 + 2x + 2)

15 .

2 1 - 10J 1 (x + 2x + 2) [ (x + 1)2 +

Let x + 1 = tan9, so dx = sec'l(}d9 and (x + J

[(x +

1 1)2 +

t]

dx

t]

2

dx.

li + 1 = sec'l(). Then

2 dx=J sec'l(}d9 = Jcos'l(}d9 sec4fJ =J 1 + cos29 d9= !l.. + sin29 + C 224

!tan- 1(x + I) + !sin9cos9 + c

=

v'(X+I)~+1 x+1

=ltan- 1(x+I)+1. x+1 +C 2 2 (x + 1)2 + 1

o

I

Flgure 10.2.1. Geometry of the substitution

x+l

(see Fig. 10.2.1). Adding the results obtained above, we find

=tan(J.

J

x3 (x - 1)( x 2 + 2x

+ 2)

dx

= 215 [lnlx - 11- !ln(x2 + 2x + 2) + 23tan- 1(x + I) +15

1 -5tan- 1(x+I)-5 x+1 ]+c x 2 + 2x + 2 x 2 + 2x + 2

Ix-li )+18tan- 1(x+I)+ 1O-5x ]+c. • =...!..[ln( 25 ~X2 + 2x + 2 x 2 + 2x + 2 Integrands with a single power (x - a)' in the denominator may appear to require partial fractions but are actually easiest to evaluate using a simple substitution. Example 7 Solution

Integrate J x 3 + 2x + I dx. (x - 1)5 Let u = x-I so du = dx and x = u + l. Then

+ 2x + 1 f (u + 1)3 + 2( u + I) + I du dx = J""---'--""-'-'::-"(x - 1)5 u X3

5

=J u3 +3u 2 +5u+4 du=J(...!.. + 1.- + 2- + .1.)dU u5 u2 u 3 u4 u5

=-l--L-~-....i..+C U

2u 2

3u 3

4u 4

=_[_1_+ 3 + 5 + 1 ]+c.. x-I 2(x - 1)2 3(x - 1)3 (x _ 1)4

474

Chapter 10 Further Techniques and Applications of Integration

To conclude this seetion, we present a couple of techniques in which an integrand is converted by a substitution into a rational function which can then be integrated by partial fractions. The fi.rst such technique, called the method of rationalizing substitutions, applies when an integrand involves a fractional power. The idea is to express the fractional power as an integer power of a new variable.

Example 8 Solution

Eliminate the fractional power from J

(1 + x)2 / 3 1 + 2x

To get rid of the fractional power, substitute u and 3u 2 du = dx, so the integral becomes J

U2

1 + 2( u 3 - 1)

•

dx.

= (1 + x)1/3.

Then u 3 = 1 + x

3u 2 du =J 3u 4 du ... 2u 3 - 1

After the rationalizing substitution as made, the method of partial functions can be used to evaluate the integral. (Evaluating the integral above is left as Exercise 24).

Example 9

u= o/x + 4

Try the substitution

2

(b) J

and

Solution

in the integrals: 2x 7 dx

Vx 2 +4

.

We have u 3 = x 2 + 4 and 3u 2 du = 2xdx, so integral (a) becomes J X4 u

•

3u 2 du 2x

=J l

2

ux 3 du,

from which we eannot eliminate x without introducing a new fraetional power. However, (b) is J

2:

7

•

~~

du

= J 3ux 6 du= J

3u(u 3 - 4)3 du

(which can be evaluated as the integral of a polynomial). The reason the method works in case (b) lies in the special relation between the exponents of x inside and outside the radical (see Exereise 27) ... The second general technique applies when the integrand is built up by rational operations from sinx and cosx (and hence from the other trigonometrie functions as weil). The substitution u = tan(x/2) turns such an integrand into a rational function of u by virtue of the following trigonometrie identities:

sinx=~

and 11

Figure 10.2.2. With the substitution tan(x/2) = u, .sin(x 12) = u 1..[l+U2.

1 + u2

'

cosx = 1 - u 2 1 + u2

'

(8) (9)

dx=~.

(10)

1+ u 2 To prove equation (8), use the addition formula sinx

= sin(

I + I) = 2sin( I )cos( I)

=2( ~ u )( I ) =~ ~ l+u 2

(see Fig. 10.2.2) .

10.2 Partial Fractions

475

Similarly we derive equation (9). Equation (10) holds since

du 1 2X dx = 2sec 2 = Example 10 Solution

(x) ]= 2(1 1 +u )

21[ tan 2 2 + 1

2

Evaluatef 2 dx . + cosx Using equations (8), (9), and (10), we convert the integral to

2du 1 + u2

f

•

1 f 2du f 2du 2+[(1- u2 )/(1 + u2 )] = 2+2u 2 + 1- u2 = 3+ u2

'

which is rational in u. No partial fraction decomposition is necessary; the· substitution u = /3 tan IJ converts the integral to

f

2·/3 sec2IJdIJ = ~ fdIJ= 2IJ + C 3 + 3 tan2IJ /3 /3

(using the identity I + tan 2IJ = sec 2IJ). Writing the answer in terms of x, we get

for our final answer ....

Rational Expressions In sinx and cosx If j(x) is a rational expression in sinx and cosx, then substitute u = tan(x/2). Using equations (8), (9), and (10), transform jj(x)dx into

the integral of a rational function of u to which the method of partial fractions can be applied.

Example 11 Solution

Find !o"/\ecIJdIJ. First we find jsecIJdIJ= j[dIJ/coslJ]. We use equations (9) and (10) (with x replaced by IJ) to get

dIJ - J 1 + u2 2 du - 2J du J cos IJ 1 - u2 I + u2 1 - u2 =

J( l~u +

l~u)du=lnll+ul-lnll-ul+C

1+ u 1 1 1 + tan( IJ /2) 1 =ln 1 1- u +C=ln 1-tan(IJ/2) +c. (Compare this procedure with the method we used to find j sec IJ dIJ m Example 6(b), Section 10.1.) Finally, {"/4

Jo

_( II+tan(IJ/2)1)1'lT 14_ (l+tan('1T/8»)_ secIJdIJ- In l-tan(IJ/2) -ln l-tan('1T/8) lnl o

1+ tan('1T/8») = In ( 1 _ tan('1T/8) ~0.881. ...

476

Chapter 10 Further Techniques and Applications of Integration

Exerclses for Sectlon 10.2 Evaluate the integrals in Exercises 1-12 using the rnethod of partial fractions. 1 1. dx (x - 2)2(X2 + 1)2 . 2.

3.

J J

X4+ 2x 3

Lo

+6 3 dx.

(x - 4)

I

x4

(x 2

+ 1)

2

dx

4. (I 2x 3 - 1 dx Jo x 2 + 1 x2

J(x - 2)(x + 2x + 2) 6 J dx (x - 2)(x + 3x + . 5.

2

2

dx.

1)

+ 1 dx J2 x 3 - 1

7. (4 8.

foo

l

x3

dx

8x 3 + 1

x dx. x 4 + 2x 2 - 3 10. J 2x 2 - X + 2 dx x 5 + 2x 3 + x

J /:. x dx 15. JxV7+! dx

14. J _ x - dx

v'x+T

16.

Jx

3

V+I dx

Evaluate the integrals in Exercises 17-20.

dx

18. J

·1+sinx 19. (,,/4 dfJ Jo 1+ tanfJ

dx

1+2cosx 20. (,,/2 dfJ J"/4 1- cosfJ

21. Find the volume of the solid obtained by revolving the region under the graph of the function y = 1/[(1 - x)(1 - 2x)] on [5,6] about the y axis. 22. Find the center of mass of the region under the graph of 1/(x 2 + 4) on [1,3]. (I + x)3/2 23. Evaluate J x dx. *24. Evaluate the integral in Example 8. 25. A chemical reaction problem leads to the following equation: x)

+I

=~ +

13.

J(80 - X~(60 -

(a) Integrate to get a logarithmic formula involving x and t (x = 0 when t = 0). (b) Convert the answer to an exponential formula for x (assurne x < 60). (c) How much reaction product is present after 15 minutes, assuming x = 20 when t = 107 26. Partial fractions appear in electrical engineering as a convenient means of analyzing and describing circuit responses to applied voItages. By means of the Laplace transform, circuit responses are associated with rational funcitons. Partial fraction methods are used to decompose these rational functions to elementary quotients, which are recognizable to engineers as arising from standard kinds of circuit responses. For example, from

(s + 2)(S2 + 1)(s2 + 4)

11. ("11/2 cosxdx J"I1/6 sinx + sin 3x "l1/4 (sec2x + l)sec 2xdx 12. fo o 1 + tan3x Evaluate the integrals in Exercises 13-16 using a rationalizing substitution.

J

action.

s

9. J

17

of reaction product present after t minutes, starting with 80 kilograms and 60 kilograms of two reacting substances which obey the law of mass

J

=k

dt,

k = constant.

In this formula, x(t) is the number of kilograms

s+2

Bs + C + Ds + 2E S2 + 1 s2 + 4 '

an engineer can easily see that this rational function represents the response

Ae- 2t + Bcost + Csint + Dcos2t + Esin2t. Find the constants A, B, C, D, E. *27. (a) Try evaluating f(x m + by/qx r dx, where m, p, q, and rare integers and b is a constant by the substitution u = (x m + b)l/q. (b) Show that the integral in (a) becomes the integral of a rational function of u when the number r - m + I is evenly divisible by m. *28. Any rational function which has the form p(x)/(x - a)m(x - b)", where deg p < m + n, can be integrated in the following way: (i) Write

p(x)

q(x)

rex)

(x_a)m

(x-b)"

------=+

,

where deg q < m and deg r < n. (ii) Integrate each term, using the substitutions u = x - a and v = x - b. (a) Use this procedure to find

J

dx 2

(x - 2) (x - 3)

3 •

(b) Find the same integral by the ordinary partial fraction method. (c) Compare answers and the efficiency of the two methods.

10.3 Are Length and Surfaee Area

477

10.3 Are Length and Surfaee Area Integration ean be used to find the length of graphs in the plane and the area of surfaees of revolution.

In Sections 4.6, 9.l, and 9.2, we developed formulas for areas under and between graphs and for volumes of solids of revolution. In this section we continue applying integration to geometry and obtain formulas for lengths and areas. The length of a piece of curve in the plane is sometimes called the are length of the curve. As we did with areas and volumes, we assurne that the length exists and will try to express it as an integral. For now, we confine our attention to curves which are graphs of functions; general curves are considered in the next section. We shall begin with an argument involving infinitesimals to derive the formula for are length. Following this, a different derivation will be given using step functions. The second method is the "honest" one, but it is also more technical. We consider a curve that is a graph y = fex) from x = a to x = b, as in Fig. 10.3.1. The curve may be thought of as being composed of infinite1y

IC::t::==~ "f~) Figure 10.3.1. An

I I

I I

a

b

I

"infinitesimal segment" of the graph of f.

I

x

many infinitesimally short segments. By the theorem of Pythagoras, the length

~dX2 + dl. But dy / dx = j'(x), so dy = and dS=~dx2+[JI(x)rdx2 =~J + [f'(x)r dx. To get the total

ds of each segment is equal to 1'(x)dx

length, we add up all the infinitesimallengths:

i

b ds

=

i

b

~l + [1'( x)]2

dx.

Length 01 Curves Suppose that the function f is continuous on [a, b], and that the derivative l' exists and is continuous (except possibly at finitely many points) on [a,b). Then the length of the graph offon [a,b] is: L

i JI + [I'(X) t

= b

dx.

(1)

Let us check that formula (I) gives the right result for the length of an are of a circle.

478

Chapter 10 Further Techniques and Applications of Integration

Example 1

Solution

b-

Use integration to find the length of the graph of fex) = x 2 on [0, b], where 0 < b < 1. Then find the length geometrieally and eompare the results. By formula (1), the length is

lob ~l + [j'(X)]2

dx, wheref(x)

=b - x 2. We

have j'(x)

=

-x

~l - x 2

1 + [j'(x)Y= _1_2 I-x

,

•

Henee

Examining Fig. 10.3.2, we see that sin-I(b) is equal to (J, the angle intereepted y

p

Figure 10.3.2. The length of the are PQ is () = sin -Ib.

-I

x

b

by the are whose length we are computing. By the definition of radian measure, the length of the arc is equal to the angle (J = sin - I( b), which agrees with our calculation by means of the integral. ... Example 2 Solution

Find the length of the graph of fex)

= (x -

1)3/2 + 2 on [1,2].

We are given fex) = (x - 1)3/2 + 2 on [1,2]. Since j'(x) = i(x - 1)1/2, the length of the graph is

LbJI + [J'(X)]2 i2~I + ~(X -1) dx=

_- - I

18

=

i

4

l3

u 1/2du

dx=

~ i 2-19 x -

5 dx

(where u = 9x - 5)

f,-(13 3/ 2 - 8) ~ 1.44 ....

Due to the square root, the integral in formula (1) is often difficult or even impossible to evaluate by elementary means. Of course, we can always approximate the result numerically (see Section 11.5 for speeific examples). The following example shows how a simple-Iooking funetion can lead to a eompIieated integral for are length. Example 3 Solution

Find the length of the parabola y We substitute fex) L

=

=

=

x 2 from x = 0 to x = 1.

x 2 and rex) = 2x into formula (1):

fo' ~1 + (2X)2 dx= 2fo' ~U)2 + x 2 dx.

10.3 Are Length and Surfaee Area

Now substitute x = !tanO and

f

~O)2+X2

~(1/2)2 + x 2

479

= !secO:

dx= f(tsecO)(tseclgdO)=ifsec30dO.

We evaluate the integral of sec30 using the following trickery3:

f sec30dO= f secOseclgdO= fsecO(tanlg + l)dO = f secOtanlgdO+ f secOdO

f

= (secOtanO)tanOdO+ InlsecO + tanOI· (see Example 10, Section 10.2.) Now integrate by parts:

f(secOtanO)tanOdO= f ~ (secO)tanOdO = sec 0 tan 0 -

f sec Oseclg dO

= secOtanO - fsec30dO. Substituting this formula into the last expression for Jsec30 dO gives

fsec30dO= secOtanO - fsec30dO+ InlsecO + tanOI; so

f sec30dO=

t (secOtanO + InlsecO + tanOI) +

Since 2x = tan 0 and sec 0 = 2 . ~( I /2)2 + x 2 =

C.

b + 4x 2 , we can express the

integral Jsec30 dO in terms of x as

xJt + 4x 2 + ! Inl2x +

b + 4x 2 I + c.

[One mayaiso evaluate the integral J~( ! )2 + x 2 dx using integral formula (43) from the endpapers.] Substitution into the formula for L gives L =!(

xb + 4x

2

+ !lnl2x +

b + 4X21)1~

=! [ß + !ln(2 +ß)] ~ 1.479.....

Example 4 Solution

Express the length of the graph of j(x) =

b- kx

2 2

on [0, bJ as an integral.

We get 1 + (k 4

-

k 2 )X 2

1- k 2x 2 3

We ean also write eosD dD=J cosD dD=J ~ Jsee DdD=J cos~ {l - sin'D)2 (I _ 3

U 2)2

The last integral may now be evaluated by partial fractions.

(u = sinD).

480

Chapter 10 Further Techniques and Applications of Integration

Thus,

It turns out that the antiderivative

J

eannot be expressed (unless k 2 = 0 or 1) in terms of algebraie, trigonometrie, or exponential funetions. It is a new kind of funetion ealled an elliptic funetion. (See Review Exereises 85 and 92 for more examples of sueh funetions.) ... We now turn to the derivation of formula (I) using step funetions. Our first prineiple for are length is that the length of a straight line segment is equal to the distanee between its endpoints. Thus, if fex) = mx + q on [a,b], the endpoints (see Fig. 10.3.3) of the graph are (a,ma + q) and y

y=mx+q

Figure 10.3.3. The length of the dark segment is (b - a)~, where m is the slope.

a

(b, mb

b

x

+ q), and the distanee between them is

~(a -

bf

+ [(ma + q) - (mb + q)]2 = ~(a - b)2 + m 2(a - b)2

= (b - a)~l + m 2

•

(Sinee a < b, the square root of (a - bf is b - a.) Our strategy, as in Chapter 9, will be to interpret the are length for a simple eurve as an integral and then use the same formula for general eurves. In the ease of the straight line segment, fex) = mx + q, whose length between x = a and x == bis (b - a)V! + m 2 , we ean interpret m as the derivative 1'(x), so that Length = (b - a)V! + m 2 =

L~1 + b

[1'(X)]2 dx.

Sinee the formula for the length is an integral of 1', rather than of f, it is natural to look next at the funetions for whieh l' is a step funetion. If l' is eonstant on an interval, f is linear on that interval; thus the funetions with whieh we will be dealing are the piecewise linear (also ealled ramp, or polygonal) funetions. To obtain a pieeewise linear funetion, we ehoose a partition of the interval [a,b], say, (XO,XI"'" x n) and speeify the values (YO'YI"" ,Yn) of

10.3 Are Length and Surfaee Area

481

the funetion j at these points. For eaeh i = 1,2, ... , n, we then conneet the point (xi-)o Yi-\) to the point (Xi' y;) by a straight line segment (see Fig. 10.3.4). y

Figure 10.3.4. The graph of a piecewise linear function.

a

b

x

The funetion j(x) is differentiable on eaeh of the intervals (Xi_I' x;), where its derivative is eonstant and equal to the slope (Yi - Yi-I)/(Xi - Xi-I)' Thus the funetion I

~I + [J'(X)]2

+ ( Yi -

is a step funetion [a,b), with value

I)2

Yi Xi-X i _ 1

n

=~

i=1

Note that the ith term in this sum, ~(Xi - xi-li + (Yi - Yi_I)2, is just the length of the segment of the graph of j between (Xi-I' Yi-I) and (Xi' y;). Now we invoke a seeond principle of are length: if n curves are placed end to end, the length of the total curve is the sum of the lengths of the pieees. Using this principle, we see that the preceding sum is just the length of the graph of j on [a, b). So we have now shown, for piecewise linear functions, that the length of the graph of j on [a,b) equals the integral Example 5

i VI + b

[J'(x)Y dx.

Let the graph of j consist of straight line segments joining (1,0) to (2, I) to (3,3) to (4,1). Verify that the length of the graph, as computed direcdy, is given by the formula

i ~I + b

[J'( X)]2 dx.

Solution The graph is sketehed in Fig. 10.3.5. The length is dl

+ d2 + d2

=.fl+I2 + b + 22 + VI + ( - 2)2 =fi + 2{5 .

Actually, ~l + [!'(x)f is not defined at the points Xo. XI' .•• ,x.' but this does not matter when we take its integral, since the integral is not affected by changing the value of the integrand at isolated points.

4

482

Chapter 10 Further Techniques and Applications of Integration y (3,3)

(4,1) (1,0)

Figure 10.3.5. The length of this graph is d 1 + d2 + d3 •

x

On the other hand,

f'(X)~{ ~

-2

on (1,2) on (2,3) on (3,4)

(and is not defined at x = 1,2,3,4). Thus, by the definition of the integral of a step function (see Section 4.3),

~4~1 + [f'(x)]2

dx=

([I+l2)'1 + (b + 22 ).1 + [-/1 + (-2/

].

1

=ji +2{5 which agrees with the preceding answer. • Justifying the passage from step functions to general functions is more complicated than in the case of area, since we cannot, in any straightforward way, squeeze a general curve between polygons as far as length is concerned. Neverthe1ess, it is plausible that any reasonable graph can be approximated by a piecewise linear function, so formula (1) should carry over. These considerations lead to the technical conditions stated in conjunction with formula (1).

If we revolve the region Runder the graph of fex) (assumed nonnegative) on [a,b], about the x axis, we obtain asolid of revolution S. In Section 9.1 we saw how to express the volume of such asolid as an integral. Suppose now that instead of revolving the region, we revolve the graph y = fex) itself. We obtain a curved surface ~, called a surface of revolution, which forms part of the boundary of S. (The remainder of the boundary consists of the disks at the ends of the solid, which have radii f(a) and f(b); see Fig. 10.3.6.) Our next goal is to obtain a formula for the area of the surface ~. Again we give the argument using infinitesimals first. y y

Figure 10.3.6. The boundary of the solid of revolution S consists of the surface of revolution 2: obtained by revolving the graph, together with two disks.

=[(xl

,0 Revolve

--+--+Q- - - + . -

Referring to Figure 10.3.7, we may think of a smooth surface of revolution as being composed of infinite1y many infinitesimal bands, as in Fig.

10.3 Are Length and Surfaee Area

y

Figure 10.3.7. The surface of revolution may be considered as composed of infinitely many infinitesimal frustums.

Area

=2f{!(x) Vdx l

483

+ dyl

Revolve

o

dx

x

b

10.3.7. The area of each band is equal to its circumference 2'1Tj(x) times its width ds = Jdx 2 + dy 2 , so the total area is

L b 2'1Tj(x)Jdx 2 + dy 2 since dy

=

= Lb2'1Tj(X)~1 + [j'(x)]2 dx

j'(x)dx.

Area of a Surface of Revolution about the x Axls The area of the surface obtained by revolving the graph of j( x) (> 0) on [a, b] about the x axis is dy 1+ ( dx

)2

dx.

(2)

We now check that formula (2) gives the correct area for a sphere. Example 6 Solution

Find the area of the spherical surface of radius r obtained by revolving the graph of y = ~r2 - x 2 on [ - r, r] about the x axis. As in Example I, we have

2 J -rr2'1T~r2 - x

r

~I + [j'(X)]2

~ vr - x-

=

r/~r2 -

x 2 , so the area is

dx= 2'1TrJr dx= 2'1Tr' 2r = 4'1Tr 2 -r

which is the usual value for the area of a sphere .... H, instead of the entire sphere, we take the band obtained by restricting x to [a ,b] (-r';;;; a ';;;; b.;;;; r), the area is 2'1Tf~rdx = 2'1Tr(b - a). Thus the area obtained by slicing a sphere by two parallel planes and taking the middle piece is equal to 2'1Tr times the distance between the planes, regardless of where the two planes are located (see Fig. 10.3.8). Why--doesn't the "longer" band around the middle have more area? Figure 10.3.8. Bands of equal width have equal area.

As with are length, the factor

~rl-+-[j-'-(X-)-]-2

in the integrand sometimes

makes it impossible to evaluate the surface area integrals by any means other than numerical methods (see Section 11.5). To get a problem which can be solved, we must choose j carefully.

484

Chapter 10 Further Techniques and Applications of Integration

Example 7 Solution

Find the area of the surface obtained by revolving the graph of x 3 on [0, I] about the x axis. We find that 1'(x) = 3x 2 and JI

= .!!....

(9(1 + V)1/2dv 18 Jo

=

+ [1'(x)]2 = b + 9x 4 ,

SO

(u = Iv du = I dV) 9' 9

~ [ j (l + V)3/2K = ;

(lW/ 2 - I) ~ 3.56....

By a method similar to that for deriving equation (2), we can derive a formula for the area obtained by revolving the graph y = j(x) about the y axis for a .:;; x .:;; b. Referring to Figure 10.3.9, the area of the shaded band is Width x Circumference = ds . 2'llX = 2'llx...jdx 2 + dy 2 Thus the surface area is ib2'llxJl

+ [1'( x)]2 dx.

Figure 10.3.9. Rotating

y = f(x) about they axis.

Area 01 a Surface 0' Revolution about the y Axls The area of the surface obtained by revolving the graph of j(x) (;;' 0) on [a,b] about they axis is (3)

Example 8 Solution

Find the area of the surface obtained by revolving the graph y = x 2 about the

y axis for 1 .:;; x .:;; 2.

If j(x) = x 2, 1'(x) = 2x and JI

+ [1'(x)

r = b + 4x2 . Then (u

=

1 + 4x 2, du = 8xdx)

10.3 Are Length and Surfaee Area

485

Finally we sketch how one derives formula (2) using step functions. The derivation of formula (3) is similar (see Exercise 41). If j(x) = k, a constant, the surface is a cylinder of radius k and height b - a. Unrolling the cylinder, we obtain a rectangle with dimensions 27Tk and b - a (see Fig. 10.3.10), whose area is 27Tk(b - a), so we can say that the area of the cylinder is 27Tk(b - a).

y Circumference

y=k

21rk

Rcvolve (

~Q- - - +, - -, ij

Cut here

.nd unroll

'------v------'

Figure 10.3.10. The area of the shaded cylinder is

!- b- o-j

2wk(b - a).

y

x

Figure 10.3.11. A frustum of a cone.

Figure 10.3.12. The area of the frustum, found by cutting and unrolling it, is

wS(" + '2)'

Heigh' b - u

Next we look at the case where j(x) = mx + q, a linear function. The surface of revolution, as shown in Fig. lO.3.11,is a frustum of a cone-that is, the surface obtained from a right circular cone by cutting it with two planes perpendicular to the axis. To find the area of this surface, we may slit the frustum along a line and unroll it into the plane, as in Fig. 10.3.12, obtaining a circular sector of radius rand angle 0 with a concentric sector of radius, - s removed. By the definition of radian measure, we have 0, = 27T'2 and O(r - s) = 2wr., so Os = 27T(r2 - 'I)' or 0 = 27T[(r2 - '1)/ s]; from this we find that r = '2s/(r2 - r l ). The area of the figure is

!7T [7Tr2 -

7T(r - sn

=

!

[r 2 -

(,2 - 2,s + S2)] = ! (2rs -

S2)

486

Chapler 10 Further Techniques and Applications of Integration

(Notiee that the proof breaks down in the ease " = '2 = k, a eylinder, sinee , is then "infinite." Nevertheless, the resulting forrnula 271ks for the area is still eorreet!) We now wish to express 71S(', + '2) as an integral involving the funetion fex) = mx + q. We have " = ma + q, '2 = mb + q, and S = JI + m 2 (b - a) (see Fig. 10.3.3), so the area is 7TJI

+ m 2 (b - a)[ m(b + a) + 2q] = 71b + m 2 [m(b 2 - a2 ) + 2q(b - a) ] = 27Tb + m2 [ m b2 2. a2 + q(b - a)] =

271JI +

Sinee m(x 2 /2)

2

+ qx is the antiderivative of mx + q, we have

271b + m2 ( m =

m(m ~ + qx)[ ~2 + qx)[

27Tlb ~ (mx + q)dx

= 271lb[ ~l + f'(x)2 ]f(x)dx, so we have sueeeeded in expressing the surfaee area as an integral. Now we are ready to work with general surfaees. If fex) is piecewise linear on [a, b], the surfaee obtained is a "eonoid," produeed by pasting together a . finite sequenee of frustums of cones, as in Fig. 10.3.13. y

Figure 10.3.13. The surface obtained by revolving the graph of a piecewise linear function is a "conoid" consisting of several frustums pasted together.

Revolve

a

b

x

x

The area of the eonoid is the surn of the areas of the eomponent frusturns. Sinee the area of eaeh frustum is given by the integral of the fune-

tion 271~1 + [f'(X)]2 fex) over the appropriate interval, the additivity of the integral implies that the area of the eonoid is given by the same forrnula:

We now assert, as we did for are length, that this formula is true for general funetions f. [To do this rigorously, we would need apreeise definition of surfaee area, whieh is rather eomplieated to give (mueh more eornplieated, even, than for are length).]

10.3 Are Length and Surfaee Area

Example 9

Solution

487

The polygonalline joining the points (2,0), (4,4), (7, 5), and (8, 3) is revolved about the x axis. Find the area of the resulting surface of revolution. The function f whose graph is the given polygon is

j

2m .

sin 4> = sin 4>msin ß (which defines ß),

T = 'Ir,ff

2 sIn 2 '4< m. *86. An engineer is studying the impact of an infinite bar by a short round-headed bar, making a maximum indentation al' Applying Hertz' theory of impact, she obtains the equation ~ pcoOa' = k(aV 2 - a 3 / 2 ) for the indentation a at time t. The symbols p, Co, 0, k are constants. The equation is solved by an initial integration to get

1= peoO (a/a, du . 2k{cX; Jo I - u3 / 2 (a) Evaluate the integral by making the substitution v = /Ü, followed by the method of partial fractions. (b) Substitute s = (4tk{cX;)/(3pcoO) to obtain

9s = 2'1r + 2lnl l + y + / .[3 (I -

yi

-4.[3tan- l ( 2Y;

I

I ),

wherey =~a/al' *87. Find a general formula for Jdx / ~ax2 + bx + c ; a =1= O. There will be two cases, depending upon the sign of a. *88. Let f(x) = x·, 0< a ..; x ..; b. For which rational values of n can you evaluate the integral occurring in the formula for: (a) The area under the graph of f! (b) The length of the graph of f! (c) The volume of the surface obtained by revolving the region under the graph of f about the x axis? The y axis?

Review Exercises for Chapter 10

*89. *90. *91.

*92.

(d) The area of the surfaee of revolution obtained by revolving the graph of f about the x axis? The y axis? Evaluate these integrals. Same as Exereise 88, but with f(x) = I + x n • Same as Exereise 88, but withf(x) = (l + x 2t. (a) Find the formula for the area of the surfa~e obtained by revolving the graph of r = f(8) about the x axis, a .;;; 8 .;;; ß. (b) Find the area of the surfaee obtained by revolving r=eos28, -'1T/4.;;; 8.;;; '1T/4 about the x axis (express as an integral if neeessary). Consider the integral

J ~(I_ x2~ _ k2x2)

.

(a) Show that, for k = 0 and k = I, this integral ean be evaluated in terms of trigonometrie and exponential funetions and their inverses. (b) Show that, for any k, the integral may be transformed to one of the form

J..jJ - d8esin2fJ .

(This integrand oeeurs in the sunshine formula-see the supplement to Seetion 9.5.) (e) Show that the integral

Jb - k sin2fJ d8 2

(which also oeeurs in the sunshine formula) arises when one tries to find the are length of an ellipse x 2 / a 2 + / / b 2 = I. Express k in terms of a and b. Due to the result of part (e), the integrals in parts (a), (b), and (e) are ealled elliptie integrals. *93. Consider the parametrie eurve given by x = cosmt,y = sinnt, when m and n are integrals. Such a eurve is ealled a Lissajous figure (see Example 6, Seetion 10.4). (a) Plot the eurve for m = land n = 1,2,3,4. (b) Describe the general behavior of the eurve if m = I, for any value of n. Does it matter whether n is even or odd? (e) Plot the eurve for m = 2 and n = 1,2,3,4,5. (d) Plot the curve for m = 3 and n = 4, 5. *94. (Lissajous figures eontinued). The path x = x(t), y = y(t) of movement of the tri-suspension pendulum of Fig. IO.R.I produces a Lissajous figure of the general form x = Aleos(wlt + 81), Y = A 2sin(w2t + 80 ),

507

y

x

Figure lO.R.1. The bob on this pendulum traees out a Lissajous figure. (a) Draw the Lissajous figures for WI = W2 = I, AI =A 2, for some sampie values of 81,82 , The figures should eome out to be straight lines, eircJes, ellipses.· (b) When w\ = I, W2 = 3, AI = A 2 = I, the bob retraees its path, but has two se1f-interseetions. Verify this using the results of Exercise 93. Conjecture what happens when W2 / W 1 is the ratio of integers. (c) When W2/WI = '1T, AI = A 2 = I, the bob does not retraee its path, and has infinitely many self-interseetions. Verify this, graphically. Conjecture what happens when wdwi is irrational (not the quotient of integers). *95. Consider the eurve r = f(8) for 0 .;;; 8 .;;; 2'1T as a parametrie eurve: x = f(t)eost, y = f(t)sint. Assuming that f(8) > 0 for all 8 in [0,2'1T] and that f(2'1T) = f(O), show that the area encJosed by the eurve is given by

-

Lo ydxdt- dt 2 ..

(A)

as weil as by J5"x( dy / dt) dt and by the more symmetrie furmula

.!

(2"[x dy _ dX] dt. 2 Jo dt y dt

[Hint: Substitute the definitions of x and y into (A), integrate by parts, and use the formula for area in polar coordinates.] These formulas are in fact valid for any cJosed parametrie curve. (See Section 18.4.) *96. (a) If r is a non-repeated root of Q(x), show that the portion of the partial fraetion expansion of P(x)/ Q(x) corresponding to the faetor x - r is A /(x - r) where A = P(r)/ Q'(r). (b) Use (a) to ealculate J[(x 2 + 2)/(x 3 - 6x + lIx - 6)]dx.

Chapter11

Limits, L'Höpital's Rule, and Numerical Methods Limits are used in both the theory and applications of calculus.

Our treatment of limits up to this point has been rather casual. Now, having learned some differential and integral calculus, you should be prepared to appreciate a more detailed study of limits. Tbe chapter begins with formal definitions for limits and a review of computational techniques for limits of functions, ineluding infinite and onesided limits. The next topic is I'Hopital's rule, which employs differentiation to compute limits. Infinite limits are used to study improper integrals. The chapter ends with some numerical methods involving limits of sequences.

11.1 Limits of Functions There are many kinrls of limits, but they all obey similar laws.

In Section 1.2, we discussed on an intuitive basis what limx~xJ(x) means and why the limit notion is important in understanding the derivative. Now we are ready to take a more carefullook at limits. Recall that the statement limx~xJ(x) = 1 means, roughly speaking, that fex) comes elose to and remains arbitrarily elose to I as x comes elose to x o. Tbus we start with a positive "tolerance" e and try to make !fex) - Illess than e by requiring x to be elose to xo. The eloseness of x to X o is to be measured by another positive number-mathematical tradition dictates the use of the Greek letter {} for this number. Here; then, i8 the famous e-{} definition of a limit-it was first stated in this form by Karl Weierstrass around 1850.

The e-8 Definition of lim j(x) x~xo

Let f be a function defined at all points near x o, except perhaps at X o itself, and let 1 be areal number. We say that 1 is the limit of fex) as x approaches X o if, for every positive number e, there is a positive number {} such that If(x) - II < e whenever Ix - xol < {} and x 1= xo' We write limHxJ(x) = I. Tb€: purpose of giving the e-{} definition is to enable us to be more precise in dealing with limits. Proofs of some of the basic theorems in this chapter and

510

Chapler 11 Limits, L'Hopital's Rule, and Numerical Methods

the next require this definition; however, practical computations can often be done without a full mastery of the theory. Your instructor should tell you how much theory you are expected to know. The e-8 definition of limit is illustrated in Figure 11.1.1. We shade the region consisting of those (x, y) for which: 1. Ix - xol > 8 (region I in Fig. 11.1.1(b»; 2. x = X o (the vertical.line 11 in Fig. 11.1.1(b»; 3. x =1= x o, Ix - xol < 8, and Iy -11< e (region III in Fig. 11.1.1(b». y

y

~--------~

Figure 11.1.1. When limx-Ho!(x) = I, we can, for any f > 0, catch the graph of! in the shaded region by making 8 small enough. The value of! at Xo is irrelevant, since the line x = Xo is always "shaded."

11

~--~

x

(b)

(a) {j

not small enough

(j

small enough

If limx->xJ(x) = I, then we can catch the graph of j in the shaded region by making 8 small enough-that is, by making the unshaded strips sufficiently narrow. Notice the statement x =1= X o in the definition. This means that the limit depends only upon the values of j(x) for x near x o, and not on j(x o} itself. (In fact, j(x o) might not even be defined.) Here are two examples of how the e-8 condition is verified. Example 1

(a) Prove that lim x ->2(x 2 + 3x) = 10 using the E-8 definition. (b) Prove that limx->arx =

Solution

ra, where a > 0, using the e-8 definition.

°

°

(a) Here j(x) = x 2 + 3x, X o = 2, and 1= 10. Given e > we must find 8 > such that Ij(x) - II < e if Ix - xol < 8. A useful general rule is to write down j(x) = land then to express it in terms of x - X o as much as possible, by writing x = (x - x o) + x o. In our case we replace x by (x - 2) + 2: j(x) - 1= x 2 + 3x - 10 = (x -

= (x

2 + 2)2+ 3(x - 2 + 2) - 10

- 2)2+ 4(x - 2) + 4 + 3(x - 2) + 6 - 10

= (x -

2)2+ 7(x - 2).

Now we use the properties value to note that Ij(x) -

la + bl .;; lai + Ibl

and

la 21= lal 2 of

the absolute

11.;; Ix - 21 2 + 71x - 21·

If this is to be less than e, we should choose 8 so that 8 2 + 78.;; E. We may require at the outset that 8 .;; 1. Then 8 2 .;; 8, so 8 2 + 78 .;; 88. Hence we pick 8 so that 8 .;; 1 and 8 .;; e18. With this choice of 8, we shall now verify that If(x) - II < e whenever

11.1 Limits of Functions

511

Ix - xol < 8. In our case Ix - xol < 8 means Ix - 21 < 8, so for such an x, If(x) - I1 " Ix - 21 2 + 71x - 21 < 8 2 + 78 " 8 + 78 = 8ß "

Il,

so If(x) - I1 < Il. (b) Here fex) = IX ,Xo = a, and 1= Iä, Given Il > 0 we must find a. ß > 0 such that 1v'X -läl < e when Ix - al < ß. To do this we write v'X -Iä = (x - a)/(v'X +Iä). Since f is only defined for x ~ 0, we confine OUf

and

attention to these x's. Then

1v'X -läl = Ix - al "Ix - al ~+Iä Iä

(decreasing the denominator increases the fraction).

Thus, given Il > 0 we can choose ß =Iä Il; then < Il, as required. .&

Ix - al < ß implies 1v'X -läl

In practice, it is usually more efficient to use the laws of limits ,than the Il-ß definition, to evaluate limits. These laws were presented in Section 1.3 and are recalled here for reference.

Basic Properties 01 Limits Assume that limx-->x.!(x) and limx-->xog(x) exist:

Sum rule: lim [f(x)

x4xo

+ g(x)] = x4xo lim f(x) +

lim g(x).

x~xo

Produet rufe: lim [J(x)g(x)]

X4Xo

= lim

X4Xo

f(x) xlim g(x). 4 xo

Reeiproeal rule: lim [l/f(x)] = 1/ llm f(x) x-->xo x-->xo Constant junetioll rule:

if

tim e= e.

x--..»xo

Identity junetion rufe: lim x= X o .

x4xo

Repfaeement rufe: If the fuoctioos f aod g agree for all x oear Xo (oot oecessarily including x = x o), theo lim f(x)

X4Xo

= X-+Xo lim g(x).

Rational junetionaf rufe: Ir P aod Q are polynomials and Q(xo) '" 0, then P / Q is continuous at Xo; i.e.,

}~Jp(x)/Q(X)]

= P(xo)/Q(xo)·

Composite junetion rufe: If h is continuous at limx-->x.!(x), then tim h(f(x»

x-+xo

= h( X4xo lim f(x»).

512

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

The properties of limits can all be proved using the E-8 definition. The theoretically inclined student is urged to do so by studying Exercises 75-77 at the end of this section. Let us recall how to use the properties of limits in specific computations. Example2

Solution

Using the fact that lim ( 1 - cosO ) f).--?O 0

=

0, find lim cos( 1 - cosO ). f).--?O 0

Tbe composite function rule says that limx-?xoh(J(x» = h(limx->xJ(x» if h is continuous at limx->xJ(x). We let f(O) = (l - cosO)/O, and h(O) = cosO so that h(J(O» = cos[(1 - cosO)/O]. Hence the required limit is

limh(f(O)) = h( lim 1 - ~osO) = cosO = 1,

f).--?O

f)-?O

since cos is continuous at 0 = Example3

Solution

o.....

Find (a) lim ( x 2 - 5x + 6 ) and (b) lim ( x-?2 X - 2 x-?\

~ ).

Vx -

1

(a) Since the denominator vanishes at x = 2, we cannot plug in this value. Tbe numerator may be factored, however, and for any x =t= 2 our function is

x2-5x+6 (x-3)(x-2) x-2 x-2 Thus, by the replacement ruIe, "---'-'-'-::---'- =

= X -

3

.

Iim x 2 - 5x + 6 = lim (x - 3) = 2 - 3 = - 1. X- 2 x.--?2 (b) Again we cannot plug in x = 1. However, we can rationalize the denominator by multipIying numerator and denominator by Vx + 1. Tbus (if x =t= 1): x.--?2

x-I

(x - 1)(Vx + I)

Vx - 1

(Vx - 1)( Vx + 1)

--=

=

(x - 1)(Vx + 1) x-I

=Vx+1.

As x approaches 1, this approaches 2, so limx->,[(x - 1)/(Vx - 1)] = 2.... Limitsof the form limx-?±oof(x), called limits at infinity, are dealt with by a modified version of the ideas above. Let us motivate the ideas by a physical example. Let y = f{t) be the length, at time t, of aspring with a bobbing mass on the end. If no frictional forces act, the motion is sinusoidal, given by an equation of the form f{t) = Yo + a cos wt.' In reality, aspring does not go on bobbing forever; frictional forces cause damping, and the actual motion has the form y = f(t) = Yo

+ ae-btcoswt,

(1)

where bis positive. A graph of this function is sketched in Fig. 11.1.2. As time passes, we observe that the length becomes and remains arbitrarily near to the equilibrium length Yo. (Even though y = Yo already for t = 'TT /2w, this is not the same thing because the length does not yet remain near Yo.) We express this mathematical property of the function f by writing limHOOf(t) = Yo. The limiting behavior appears graphically as the fact that the I This is derived in Seetion 8.\., but if you have not studied that section, you should simply take Cor granted the formulas given here.

11.1 Limits of Functions

513

y

Figure 11.1.2. The motion of a damped spring has the form

y

= J(t) = Yo + ae-b,cos wt.

h w

4~

6~

8~

10~

w

w

w

w

graph of f remains eloser and eloser to the line y = Yo as we look farther to the right. The precise definition is analogous to that for limx-->xJ(x). As is usual in our general definitions, we denote the independent variable by x rather thl'.n t.

The f-A Definition of limX-HOO!(X) Let f be a function whose domain contains an interval of the form (a, (0). We say that areal number I is the limit of fex) as x approaches 00 if, for every positive number e, there is a number A > a such that If(x) - I1 < e whenever x > A. We write limx-->,x,j(x) = I. A similar definition is used for limx-->-oof(x) = I. When limx->oof(x) = I or limx--> _ oof(x) = I, the line y = I is called a horizontal asymptote of the graph y = fex).

We iIIustrate this definition in Figs. 11.1.3 and 11.l.4 by shading the region consisting of those points (x, y) for which x ,; ; ; A or for which x> A and Iy - I1 < e. If limx-->oof(x) = I, we should be able to "catch" the graph of fin this region by choosing A large enough-that is, by sliding the point A sufficiently far to the right. )'

Figure 11.1.3. When limx-->ooJ(x) = I, we can catch the graph in the shaded region by sliding the region sufficiently far to the righ t. This is true no matter how small E may be.

Figure 11.1.4. When it is not true that limx-->ooJ(x) = I, then for some E, we can never catch the graph of J in the shaded region, no matter how far to the right we slide the region.

y

~

/1.,)

.I

~

I

1I

(al

(bI

A

514

Chapler 11 Limits, L'Hopital's Rule, and Numerical Methods

There is an analogous definition for limX~_()(J(x) in which we require a number A (usually large and negative) such that If(x) - /1 < e if x< A. Example4

Solution

2

Prove that lim _x_ = I by using the e-A definition. x~oo 1+ x 2 Given e > 0, we must choose A such that Ix 2/(1 + x 2) have

I

2 11 :2X2 - 11 = x ; :

:2

x21 = 11 : x 21


A. We

:2 .

To make this less than e, we observe that 1/ x 2 < e whenever x may choose A = I/Fe. (See Fig. 11.1.5.) •

> 1/';;, so we

Figure 11.1.5. Illustrating

the fact that limx-->oo[x2/(1 + x 2)] = 1.

x2

.
ooekx = 0.

First of all, we note that fex) = e kx is a decreasing positive function. Given e > 0, we wish to find A such that x> A implies e kx < e. Taking logarithms of the last inequality gives kx < Ine, or x> (Ine)/k. So we may letA = (Ine)/k. (If e is smalI, In e is a large negative number.) • The examples ahove illustrate the e-A method, hut limit computations are usually done using laws analogous to those for limits as x ~ xo, which are stated in the hox on the facing page.

Example 6

Solution

Find (a) lim

x->oo

(1x + l2x + 5)and (h)

(a) We have lim

x~oo

(1X + lx 2 + 5) =

lim

+ 21 .

lim 8 x x->oo 3 x-

°

1 + 3( x->oo lim 1)2 + lim 5 = + 3 . 02 + 5 = X x->oo

x->oo X

5.

(h) We cannot simply apply the quotient rule, since the limits of the numerator and denominator do not exist. Instead we use a trick: if x =1= 0, we can multiply the numerator and denominator by I/x to obtain 8x + 2 3x-1

= 8 + (2/ x) 3-(I/x)

for x =1= 0.

By the replacement rule (with A = 0), we have . 8x + 2 . 8 + (2/ x) __ 8 + hm - - - = hm

x->oo

3x-1

x->oo

(The ~alues of (8x

3-(I/x)

+ 2)/(3x - I) for

°___8

3-0

3'

x = 102,104 ,106,108 are 2.682 ... ,

2.66682 ... ,2.6666682 ... ,2.666666682 .... ) •

11.1 Limits of Functions

limits 01 Functlons as x Approaches

515

00

Constant function rufe: lim c= c.

1/ x

x->oo

rufe: lim

1 = o.

x->oo X

Assuming that limHOC'!(x) and limx-> 00 g(x) exist, we have these additional rules:

Sum rufe:

lim [fex)

x->oo

+ g(x)] ... x->oo lim fex) + x->oo lim g(x).

Product rufe: lim [f(x)g(x)]

x~oo

=

lim fex) lim g(x).

x~oo

x~oo

Quotient rufe: If limx->oo g(x) =1= 0, then

r

x~~

[f(X)] g(x)

=

}~f(x)

..:.:.l-im--'---g-(x-) . x->oo

Repfacement rufe: If for some real number A, the funetions f( x) and g( x)

> A, then fex) = lim g(x).

agree for all x lim

x~oo

x~oo

Composite function rufe: If h is eontinuous at limHoof(x), then lim h(J(x))

x~oo

= h( x--iooo lim fex)).

All these rules remain true if we replace " < A" in the replaeement rule).

00

by -

00

(and

"> A"

by

The method used in Example 6 also shows that

r

x~~

anx n + an_IX n - 1+ ... + a1x + ao

an

bnx n

bn

+ bn_1x n - 1 + ... + b1x + bo =

as long as bn =1= o. Example 7

Solution

"x

Find lim x ->oo(Jx 2 + 1 - x). Interpret the result geometrieally in terms of right triangles. Multiplying the numerator and denominator by Jx 2 + 1

2 + I-x

~}I

x Figure 11.1.6. As the length x goes to 00, the difference ~ - x between the lengths of the hypotenuse and the long leg goes to zero.

Jx 2 + 1 - x

= ( Jx 2

+ 1 - x)

= x2 + 1-

+ x gives

R+l + X Jx 2 + 1 +x

x2

= _-----'=---_

Jx 2 + 1 +x

Jx 2 + 1 +x

As x ~ 00, the denominator beeomes arbitrarily large, so we find that lim x->oo(Jx 2 + 1 - x) = O. For a geometrie interpretation, see Fig. 11.1.6....

516

Chapler 11 Limits, L'Hopital's Rule, and Numerical Methods

Example 8 Solution

Find the horizontal asymptotes of j(x)

= _....:.x:........-_. Sketch.

R+T

We find _.:..:..x_ = lim -;:::=1===- = 1 + 1 x~+oo ~I + l/x2

lim

X~+OO ~X2

and

_-'-x'--_ x~-oo fx2 + 1 lim

=

lim _---!R_x_2 _ = lim -;::=-=I==x~-oo ~x2 + 1 x~-oo ~I + l/x2

= -1

R). Hence the horizontal

(in the second limit we may take x < 0, so x = asymptotes are the lines y = ± 1. See Fig. 11.1.7 .... y

------Figure 11.1.7. The curve y = x/R+T has the !inesy = -I andy = I as horizontal asymptotes.

y=1

-----x

y= -I

Consider the limits limx-->osin(l / x) and limx-->o(l / x 2). Neither limit exists, but the functions sin(l/x) and I/x 2 behave quite differently as x~o. (See Fig. 11.1.8.) In the first case, for x in the interval ( - ß, ß), the quantity I/x ranges y .

I

.J'=:;smx

Figure 11.1.8. limx-->oj(x) does not exist for either of these functions.

x

x

over all numbers with absolute value greater than 1/ ß, and sin(l / x) oscillates back and forth infinitely often. The function sin(l / x) takes each value between -1 and 1 infinitely often but remains elose to no particular number. In the case of 1/ x 2 , the value of the function is again near no particular number, but there is a definite "trend" to be seen; as x comes nearer to zero, 1/ x 2 becomes a larger positive number; we may say that limx-->o(l / x 2) = 00. Here is a precise definition.

The B-8 Definition of limx~xof(x) =

00

Let j be a function defined in an interval about xo, except possibly at X o itself. We say thatj(x) approaches 00 as x approaches Xo if, given any real number B, there is a positive number ß such that for all x satisfying Ix - xol < ß and x =/= x o, we have j(x) > B. We write Iimx-->x.!(x) = 00. The definition of limx-->x.!(x) = - 00 is similar: replace j(x) > B in the B-ß definition by j(x) < B.

11.1 Limits of Functions

Remarks

517

1. In the preceding definition, we usually think of S as being small, while B is Iarge positive if the limit is 00 and Iarge negative if the limit is - 00. 2. If Iimx->x.!(x) is equal to ± 00, we still may say that "limx->x.!(x) does not exist," since it does not approach any particular number. 3. One can define the statements limx->oof(x) = ± 00 in an analogous way. The following test provides a useful technique for detecting "infinite limits."

Reciprocal Test for limx-Ho!(x) =

00

Let f be defined in an open interval about x o, except possibly at X o itself. Then Iimx-->x.!(x) = 00 if:

1. For all x =1= X o in some interval about x o, fex) is positive; and 2. limx-->xJI/f(x)]=O. Similarly, if fex) is negative and limx-->xJI/f(x)] = 0, then limx-->x.!(x) = - 00. The complete proof of the reciprocal test is left to the reader in Exercise 79. However, the basic idea is very simple: fex) is very large if and only if 1/fex) is very small. A similar result is true for limits of the form Iimx->oof(x); namely, if fex) is positive for large x and limx->oo[l/f(x)] = 0, then limx->oof(x) = 00.

Example 9 Solution

1

.

Find the following limits: (a) tim ; (b) hm x->] (x - 1)2 x->oo

1- x2 -3/2 . x

(a) We note that I/(x - 1)2 is positive for all x =1= 1. We look at the reciprocaI: limx->I(x - If = 0; thus, by the reciprocal test, limx->I[I/(x - Ii] = 00. (b) For x > 1, (1 - x 2 )/ X 3 / 2 is negative. Now we have

· X / 11m = l'1m x->oo 1 - x 2 x->oo 3 2

X- 3/ 2 _ X I/ 2

· - 1 = 11m

x->oo

X 1/2

=

l'1m -1- ---:----1 x->oo X I/ 2 I/x 2 - 1

l'1m 1 x->oo 1/ x 2 - 1

= 0 . (- 1) = 0,

so Iimx->oo[(l - x 2) / X 3 / 2 ] = - 00, by the reciprocal test. .6. If we look at the function fex) = I/(x - 1) near X o = 1 we find that lim x -->I[I/f(x)] = 0, hut fex) has different signs on opposite sides of 1, so Iimx->I[I/(x - I)] is neither 00 nor - 00. This example suggests the introduction of the notion of a "one-sided limit." Here is the definition.

One-Sided Limits Let f be defined for all x in an interval of the form (x o' b). We say that

fex) approaches 1 as x approaches X o from the right if, for any positive number 10, there is a positive number S such that for all x such that x o < x < X o + S, we have If(x) - II < 10. We write Iimx-->xo+ fex) = I. A similar definition holds for the limit of fex) as x approaches Xo from the Ieft; this limit is written as limx-->xo- fex) = I. In the definition of a one-sided limit, only the values of fex) for x on one side of X o are taken into account. Precise definitions of statements Iike limx-->xo+ fex) = 00 are Ieft to you. We remark that the reciprocal test extends to one-sided limits.

518

Chapter 11 Limits. L'Hopital's Rule. and Numerical Methods

Example 10

Find (a) tim (I 1 )' (b) lim (I 1 )' x-->I+

(c) lim x-->o+

Solution

(x 2

-

x-->I-

X

+ 2)lxl .

, and (d) lim

-

x-->o-

X

X

(x 2

+ 2)l x l X

.

(a) For x > ,I, we find that 1/(1 - x) is negative, and we have limx-+I(I - x) = 0, so limx-+I+ [1/(1 - x)] = - 00. Similarly, limx-+I_II/(I - x)] = + 00, so we get + 00 for (b). (c) Fot x positive, lxii x = I, so (x 2 + 2)lxll x = x 2 + 2 for x > O. Thus the limit is Q2 + 2 = 2. (d) Für x< 0, Ixllx = -I, so

(x 2 + 2)l x l lim = - x-->olim [x 2 + 2] = - 2. • x-->ox If a one-sided limit of j( x) at Xo is equal to 00 or - 00, thenthe graph of j lies closer lind closer to the line x = x o; we call this line a vertical asymptote of the graph .. Example 11

Finp the vertical asymptotes and sketch the graph of j(x)

Solution

= (x _

1 I)(x

~ 2)2

Vertical asymptotes occur where limx-+xo±lllj(x)] = 0; in this case, they occ~r at X o = land X o = 2. We observe that j(x) is negative on (- 00" I), positive on (1,2), and positive on (2,00). Thus we have limx-+ I _ j(x) == - 00, limx-+ I+ j(x) = ::lO, limx-->2_ f(x) = 00, and limx-+2+ j(x) = 00. The graph of j is sketched in Fig. 11.1.9. • y

!~ I I I

,I

Figure 11.1.9. The graph = I/(x - I)(x - 2f has the lines x = I and x = 2 as vertical asymptotes.

I

y

I I I I

: I ,

,2

I I

x

I

I I

We conclude this section with an additionallaw of limits. In the next sections we shall consider various additional techniques and principles for evaluating limits.

1.

Comparlson Test If limx-+xJ(x) = 0 and Ig(x)1 ~ If(x)1 for all x

then limx-->xog(x) = O. 2. If limx-->oof(x) = 0 and limx-+oog(x) = O.

Ig(x)1

near Xo with x

* x o,

~ If(x)1 for aIl large x, then

Some like to call this the "sandwich principle" since g( x) is sandwiched between -lf(x)1 and If(x)1 which are squeezing down on zero as x ~ X o (or x ~ 00 in case 2).

11.1 Limits of Functions

Example 12

(a) Establish comparison test I using the e-~ definition of limit. (b) Show that limx->o[ x sin(

Solution

519

~ ) ] = 0.

°

(a) Given e > 0, there is a ~ > such that Ij(x)1 < e if Ix - xol < ~, by the assumption that limx->xJ(x) = 0. Given that e > 0, this same ~ also gives I g(x)1 < e if Ix - xol < ~ since I g(x)1 < Ij(x)l. Hence g has limit zero as x ~ X o as weIl. (b) Letg(x) = xsin(ljx) andj(x) = x. Then Ig(x)1 < lxi for all x *0, since Isin(l j x)1 < 1, so the comparison test applies. Since x approaches as x ~ 0, so does g(x), ...

°

Exerclses tor Section 11.1 Verify the limit statements in Exercises 1-4 using the e-ß definition. I. lim.x-->ax2 = a 2 2. lim x->3(x 2 - 2x + 4) = 7 3. lim x->3(x 3 + 2x 2 + 2) = 47 4. limx-->3(x3 + 2x) = 33 5. Using the fact that lim8-->o[(tan8)/8) = I, find lim8-->oexp[(3 tan 8)/ 8]. 6. Using the fact that lim8-->o[(sin8)/8) = I, find lim8-->ocos[('fT sin 8)/(48»). Find the limits in Exercises 7-12. 7. lim (x 2 - 2x + 2) x->3 . (x 2 - 4) 9. hm 2 x->2 (x - 5x + 6)

9 (3+X)2_ 11. lim --x->O X

10. lim

3-'x 3 _v_.II.-_

x~27 X - 27

12.II·m x-2 x->2 x 2 - 3x + 2

Verify the limit statements in Exercises 13-16 using the e-Adefinition. 13. lim I + x 3 = I x~oo

x3

16. lim _1_ = 0 x->oo Inx Find the limits in Exercises 17-24. 17. lim + ~ -2) x->oo x x2 19. lim IOx 2 - 2 x->oo 15x 2 - 3

18. lim (~- ~ +5) x->oo x 2 x3 20. lim -4x + 3 x->oo X + 2

21. lim 3x 2 + 2x + 4 x->oo 5x 2 + X + 7 . x + 2 + I/x 23. IIm x->oo 2x + 3 + 2/x

22. lim x 2 + xe-X x->oo 6x 2 + 2 x - 3 - 1/ x 2 24. lim x->oo 2x + 5 + l/x 2

rxr+7 -

25. Find limx-->co[ x) and interpret your answer geometrically. 26. Find limx-->oo[ {C 2X2 + I-ex) and interpret your answer .geometrically. 27. Find the horizontal asymptotes of the graph of (x + I). Sketch. 28. Find the horizontal asymptotes of the graph y =(x + 1)/R+2. Sketch.

/T+T -

· X2 + 2 31 · II m--

· m X4 + 8 32. II -x->oo X5/2

IX

X-1'OO

Find the one-sided limits in Exercises 33-40. 33.

x2 - 4 tim 2 x->2+ (x - 2)

34·

x2 - 4 · IIm 2 x->2- (x - 2)

35.

. (x-I)(x-2) hm x->O- x(x + l)(x + 2)

36.

x(x

lim

+ 3)

x->l+ (x - I)(x - 2)

37.

. (x 3 - 1)lxl hm x->O+ x

39.

2x - I lim x->~- ~(2x _ 1)2

40.

li~ x-+'!+

14. lim ~ =0 x->oo x 2 + 2

15. lim (1 + e- 3x ) = I x->oo

(1

Find the limits in Exercises 29-32 using the reciprocal test. x2 · I 29 · IIm 30. lim 4 2 x->2 (x - 2) x->2 (x - 2)

....j(2x - 1)2 x _ 1/2

Find the vertical and horiwntal asymptotes of the functions in Exercises 41-44 and sketch their graphs. 1 1 41. j(x) = x 2 _ 5x + 6 42. j(x) = 2x + 3

x2 44. j(x) = - x - I x2 - 1 45. (a) Establish the comparison test 2 using the' e-A definition of limit. (b) Use (a) to find

43. j(x)

= -2 I-

lim x->oo

[1. sin( 1. )]. X

X

46. (a) Use the B-ß definition of limit to show that If limx-->x.f(x) = 00 and g(x) ;;. j(x) for x elose to xo. x xo. then limx->xog(x) = 00. (b) Use (a) to show that 1imx-->I[(l + cos2x)/(1 - xi) = 00.

"*

520

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

Find the limits in Exercises 47-60. 4x 48 . I'1m x 3- 1 · 3 +47 . I1m I x~1 4 + 5x x~1 x· m x3 -- 1 50 . I'1m 2 x - 2 49 . I1 x-> 1 x 2 - 1 x->2 X + 3x + 2 · x2 + 2x 3 52 xn - 1 51. I1m . I'1m -I x 2 +x-6

x->-3

53.

hm

x--> -I

55. lim x 2: 3x x-->2

7 5. 59.

x->I x-

+1 x+1

X 2n + I

.

X -

t

H2{X-12

56. Iim sin( 1. )

6

+

X

x----+oo

2 4) 58 . I'1m sm . ('lTX - 2+ -x-->oo 6x + 9

· '--1 eX - I 11m

x-->1

x - 2

54. Iim

X -

x~~- ~n~~

60.

x.!}IE

00

In(x 2)

Find the horizontal and vertical asymptotes of the functions in E:ercises 61-64. (x + 1)( x - I) 61.

63.

Y=

62. Y = (x - 2)x(x

x2_ I

= e + 2x

64

X

Y

eX

-

2x

.Y

= Inx Inx

+ 2)

I

+1

65. Let j(x) and g(x) be polynomials such that limHoo[f(x)/ g(x)] = I. Prove that the limit Iimx~_ooj(x)/g(x) is equal to 1 as weil. What happens if 1 = 00 or - oo? 66. How elose to 3 does x have to be to ensure that Ix 3 - 2x - 21)1 < ,doo? 67. Let j(x) = Ix\. (a) Find j (x) and sketch its graph. (b) Find Iim H o- j'(x) and limx~o+ j'(x). (c) Does limHoj'(x) exist? 68. (a) Give a precise definition of this statement: Iim Hoo j(x) = - 00. (b) Draw figures like Figs. Il.l.l, 11.1.3, and 1l.l.4 to ilIustrate your definition. 69. Draw figures like Figs. 11.1.1, 11.1.3, and 11.1.4 to i1lustrate the definition of these statements: (a) limx~xo+ j(x) = I; (b) limHXo + j(x) = 00. [Hinl: The shaded region should inelude all points with x .;;; xo.] 70. (a) Graph y = j(x), where j(x) = {lxi/x,

0,

X'''i= 0, x =0.

Does Iim H oj(x) exist? (b) Graph y = g(x), where g(x) = {x.+ I, 2x - I,

O.

x

Does limHog(x) exist? (c) Let j(x) be as in part (a) and g(x) as in part (b). Graph y = j(x) + g(x). Does limx-->o[f(x) + g(x)] exist? Conelude that the limit of a sum can exist even though the limits of the summands do not. 71. Tbe number N(t) of individuals in a population at time 1 is given by

N(/)

= No

e 31 . (3/2) + e 31

Find the value of Iim,-->ooN(/) and discuss its biological meaning. 72. Tbe current in a certain RLC circuit is given by 1(/) = {[(l/3)sin 1 + cos 1]e-- 1/ 2 + 4} amperes. Tbe value of limHooI(t) is called the sleady~slale current; it respresents the current present after a long period of time. Find it. 73. Tbe temperature T(x, t) at time t at position x of a rod located along 0 .;;; x .;;; I on the x axis is given by the rule T(x,t)=Ble-p,tsin).lx + B2e-P2Isin).2x + B3e-P3Isin).3x, where f.ll, f.l2, f.l3, ).1' ).2' ).3 are all positive. Show that lim,~", T(x, t) = 0 for each fixed location x along the rod. Tbe model applies to a rod without heat sources, with the heat allowed to radiate from the right end of the rod; zero limit means all heat eventually radiates out the right end. 74. A psychologist doing some manipulations with testing theory wishes to replace the reliability factor

R=

nr

1 + (n - I)r

(Spearman- Brown jormula)

by unity, because someone told her that she could do this for large extension factors n. She formally replaces n by I/x, simplifies, and then sets x = 0, to obtain 1. What has she done, in the language of limits? * 75. Study this e-8 proof of the sum rule: Let limx-->x.!(x) = L and limx~xog(x) = M. Given e > 0, choose 8] > 0 such that Ix - xol < 8] and x "i= Xo implies Ij(x) - LI < e/2; choose 82 > 0 such that Ix - xol < 82 , X"i= Xo, implies that Ig(x) - MI < e/2. Let 8 be the smaller of 8 1 and 82 , Then Ix - xol < 8, and x "i= Xo implies l(j(x) + g(x» - (L + M)I .;;; Ij(x) - LI + Ig(x) - MI (by the triangle inequality Ix + yl .;;;

lxi + Iyl). This is less than e/2 + e/2 = e, and therefore IimHX Jj(x) + g(x)] = L + M. Now prove that limHxo[aj(x) + bg(x)] = a lim.Hxoj(x)

+ b limx-->xog(x).

*76. Study this e-8 proof of the product rule: If limx-->x.!(x) = Land limx-->xog(x) = M, then limHxoj(x)g(x) = LM.

Prooj: Let e > 0 be given. We must find a number 8 > 0 such that If(x)g(x) - LMI < e whenever Ix - xol < 8, x "i= Xo. Adding and subtractingj(x)M, we have If(x)g(x) - LMI

= Ij(x)g(x) -

j(x)M

.;;; Ij(x)I'lg(x) - MI

+ j(x)M + Ij(x) -

LMI

LIIMI·

Tbe eloseness of g(x) to M and j(x) to L must depend upon the size of j(x) and IMI. Choose 81 such that Ij(x) - LI < (e/2)M whenever Ix - xol < 81> x "i= xo· Also, choose 82 such that Ix - xol < 8~, x"i= Xo, implies that Ij(x) - LI

°

< 1, which in turn implies that If(x)1 < ILI + 1 (since If(x)1 = If(x) - L + LI ..; If(x) - LI + ILI < 1 + ILI). Finally, choose 03> 0 such that Ig(x) - MI < (eI2)(ILI + 1) whenever Ix - xol < 03' x *- xo. Let be the smallest of 0\,02, and 03' If Ix - xol < 0, x*- xo, then Ix - xol < 0\, Ix - xol < 02' and Ix - xol < 03' so by the choice of 0\, 02' 03' we have

°

< (lLI + 1) 2(ILj + 1) + 21~1 ·IMI

*78. *79.

= E,

*80.

and so If(x)g(x) - LMI < e. Now prove the quotient rule for limits. 77. Study the following proof of the one-sided composite function rule: If limx->xo+ fex) = Land g is continuous at L, then g(j(x» is defined for all x in some interval of the form (xo, b), and limx->xo+ g(j(x» = geL). Proof: Let e > O. We must find a positive

521

number such that whenever Xo < x < Xo + 0, g(j(x» is defined and Ig(j(x» - g(L)1 < e. Since g is continuous at L, there is a positive number p such that whenever Iy - LI < p, g(y) is defined and Ig(y) - g(L)1 < f. Now since limx->xo+ fex) = L, we can find a positive number such that whenever Xo < x < Xo + 0, If(x) - LI < p. For such x, we apply the previously obtained property of p, with Y = fex), to conclude that g(j(x» is defined and that Ig(j(x» - g(L)1 < e. Now prove the composite function rule . Use the e-A definition to prove the sum rule for limits at infinity. Use the B-o definition to prove the reciprocal test for infinite limits. Suppose that a function f is defined on an open interval I containing xo, and that there are numbers m and K such that we have the inequality If(x) - f(xo) - m(x - xo)1 ..; Klx - xol 2 for all x in I. Prove that f is differentiable at Xo with derivative j'(xo) = m. Show that Iimx->oof(x) = 1 if and only if limy->o+ f(I/y) = I. (This reduces the computation of limits at infinity to one-sided limits at zero.)

°

If(x)g(x) - LMI ..; If(x)llg(x) - MI + If(x) - LIIMI

11.2 L'Hopital's Rule

*

*81.

11.2 L'Hopital's Rule Differentiation can be used to evaluate limits, L'Hopital's rule 2 is a very efficient way of using differential calculus to evaluate limits. It is not necessary to have mastered the theoretical portions of the previous sections to use l'Hopital's rule, but you should review some of the computational aspects of limits from either Section 11.1 or Section 1.3. L'Hopital's rule deals with limits of the form limx->x.[j(x)/ g(x)], where limx-->x.!(x) and limx-->xog(x) are both equal to zero or infinity, so that the quotient rule cannot be applied. Such limits are called indeterminate forms. (One can also replace X o by 00, X o +, or X o -.) Our first objective is to calculate limx->x.[f(x}/ g(x}] if f(x o} = 0 and g(x o) = O. Substituting x = X o gives us g, so we say that we are dealing with an indeterminate form of type S. Such forms occurred when we considered the derivative as a limit of difference quotients; in Section 1.3 we used the limit rules to evaluate some simple derivatives. Now we can work the other way around, using our ability to calculate derivatives in order to evaluate quite complicated limits: l'Hopital's rule provides the means for doing this. The following box gives the simplest version of I'Hopital's rule. In 1696, Guillaume F. A. I'Hopital published in Paris the first calculus textbook: Analyse des Infiniment Petits (Analysis 0/ the infinitely smalI). Included was a proof of what is now referred to as I'Hopital's rule; the idea, however, probably came from J. Bernoulli. This rule was the subject of some work by A. Cauchy, who clarified its proof in his· Cours d'Analyse (Course in analysis) in 1823. The foundations were in debate until almost 1900. See, for instance, the very readable article, "The Law of the Mean and the Limits ~, ~," by W. F. Osgood, Annals 0/ Mathematics, Volume 12 (1898-1899), pp. 65-78. 2

522

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

L'H6pital's Rule: Preliminary Version Let fand g be differentiable in an open interval containing x o; assume that f(x o) = g(xo) = O. If g'(xo) =1= 0, then

. f( x) 1'( x o) hm--=--.

X4xo

g(x)

g'(xo)

To prove this, we use the fact that f(x o) = 0 and g(x o) = 0 to write

fex) g(x)

fex) - f(xo) g(x)-g(x o)

- - = ---:--:----:--':-

[fex) - f(xo)J/(x - X o) [g(x)-g(xo)]!(x-xo)'

As x tends to x o, the numerator tends to j'(x o), and the denominator tends to g'(x o) =1= 0, so the result follows from the quotient rule for limits. Let us verify this rule on a simple example.

Example 1 Solution

Find lim [ x 3 -li ]. x~1 xHere we take Xo = I, f(x) = x 3 - I, and g(x) = x-I. Since g'(I) = I, the preliminary version of I'Hopital's rule applies to give . x 3 -1 1'(1) 3 hm - - I = -'-(1) = -I = 3. x~1 X g

We know two other ways (from Chapter I) to calculate this limit. First, we can factor the numerator:

x 3 _ I (x - 1)( x 2 + X + I) 2 =x +x+I (x =1= 1). x-I x-I Letting x ~ I, we again recover the limit 3. Second, we can recognize the function (x 3 - I)/(x - I) as the different quotient [h(x) - h(l)]/(x - I) for h(x) = x 3• As x ~ 1, this different quotient approaches the derivative of h at x = I, namely 3. • --=

The next example begins to show the power of I'Hopital's rule in a more difficult limit. Example 2 Solution

Find Iim cos.x - 1 . x~O

smx

We apply l'Hopital's rule with f(x) = cosx - 1 and g(x) = sinx. We have f(O) = 0, g(O) = 0, and g'(O) = I =1= 0, so 1'(0) - sin(O) . cos x-I hm . = -- = . = O. • smx g'(O) cos(O)

x~o

This method does not solve all find

g problems. For example, suppose we wish to

lim sinx - x . x3

x~O

If we differentiate the numerator and denominator, we get (cosx - I)/3x 2 , which becomes g when we set x = O. ThIS suggests that we use l'Hopital's rule

11,2 L'Hopital's Rule

523

again, but to do so, we need to know that limx->xo [f(x)/g(x») is equal to limHxJ1'(x)/ g'(x»), even when 1'(xo)/ g'(xo) is again indeterminate. The following strengthened version of I'Hopital's rule is the result we need. Its proof is given later in the section.

L'H6pltal's Rule Let fand g be differentiable on an open interval containing x o, except perhaps at X o itself. Assume: (i) g(x) i= 0, (ii) g'(x) i= 0 for x in an interval about x o, x i= x o, (iii) fand gare continuous at X o with f(x o) = g(xo) = 0, and

1'(x) (iv) !im - , X~XO g (x)

= I.

. f(x) Then hm - - = I. x~xo g(x)

Example 3

Solution

Calculate lim cos x-I . x~O x2 This is in

Hform,

so by I'Hopital's rule,

lim cosx - 1 x2

x~O

=

lim - sinx 2x

x~o

if the laUer limit can be shown to exist. However, we can use I'Hopital's rule again to write · - -sinx 11m - = I'1m - cosx . 2x x~O 2

x~O

Now we may use the continuity of cosx to substitute x = 0 and find the last limit to be - t; thus lim cosx - 1 x2

x~O

=_1. 2

To keep track of what is going on, some students like to make a table: form

1 g

l'

t

f"

g"

type

limit

cosx - 1 x2

~ indeterminate

?

-smx

~ indeterminate

?

- cosx 2

determinate

~

[]

Each time the numerator and denominator are differentiated, we must check the type of limit; if it is H, we proceed and are sure to stop when the limit becomes determinate, that is, when it can be evaluated by substitution of the limiting value.

524

ChapIer 11 Limits, L'Hopital's Rule, and Numerical Methods

Warnlng

Example4 Solution

If I'Hopital's rule is used when the limit is determinate, incorrect answers can result. For example, lim x -->o[(x 2 + 1)/ x] = 00 but I'Hopital's rule would lead to limx -->o(2x/I) which is zero (and is incorrect). Find lim sinx - x . x~O

This is in

tanx - x

8 form, so we use I'Hopital's rule: form

type

limit

1

sinx - x tanx - x

0

0

?

l'

cosx - I sec2x - I

0

0

?

0

?

determinate

t]

g

g'

1"

- sinx

g"

0

2 secx(secx tanx)

f'" g"'

4 sec2x

-cosx tan2x + 2 sec4x

Thus lim sinx - x x-->o

tanx - x

= _ ! .• 2

L'Hopital's rule also holds for one-sided limits, limits as x ~ 00, or if we have indeterminates of the form ~. To prove the rule for the form g in case x~ 00, we use a trick: set t = I/x, so that x = I/t and t~O + as x~ +00. Then

.

1'(x) g'(x)

1'(I/t) g'(I/t)

.

hm - - = hm

x~+oo

1-->0+

= = =

- t~'(l/t) - t 2g' (1/ t)

.

hm

1-->0+

. (d/ dt)f(l/t) hm

(by the chain rule)

f(I/t) g( 1/ t)

(by I'Hopital's rule)

I~O+

(d/ dt)g(I/ t)

tim 1-->0+

=

f(x) g(x)

lim - - .

x~+oo

It is tempting to use a similar trick for the not work. If we write

f(x) g(x)

--=

~

form as x ~ xo, but it does

I/g(x) I/f(x) '

which is in the

g form, we get

.

f(x) . - g'(x)/[ g(X)]2 = tim ---.,;:......:..-=--~-=--~ x~xo g(x) x~xo - 1'(x)/[J(x)Y lun - -

which is no easier to handle. For the correct proof, see Exercise 42. The use of I'Hopital's rule is summarized in the following display.

11.2 L'Hopital's Rule

525

L'H6pltal's Rule To find limHxJf(x)/ g(x)] where limHxJ(x) and limx-hog(x) are both zero or both infinite, differentiate the numerator and denominator and take the limit of the new fraction; repeat the process as many times as necessary, checking each time that l'Hopital's rule applies. If limHxJ(x) = limx_hog(x) = 0 (or each is ± 00), then lim fex) = lim f'(x) X-Ho g(x) X-Ho g'(x)

(x o may be replaced by ±

00

or X o ± ).

The result of the next example was stated at the beginning of Section 6.4. The solution by l'Hopital's rule is much easier than the one given in Review Exercise 90 of Chapter 6. Example 5

Solution

Find lim lnpx, where p

x-->oo

X

This is in the form find

> O.

*. Differentiating the numerator and denominator, we

lim lnx = lim ~ = lim _1_ = 0, x-->oo x P x-->oo px p- I x-->oo px P since p

> O....

Certain expressions which do not appear to be in the form f(x)/ g(x) can be put in that form with some manipulation. For example, the indeterminate form 00·0 appears when we wish to evaluate limx-->xJ(x)g(x) where limx-->xJ(x) = 00 and limx->Xog(x) = O. This can be converted to g or form by writing

*

f(x)g(x) Example6

Solution

g(x)

= I/f(x)

or f(x)g(x)

fex)

= I/g(x) .

FindlimHo+xlnx. We write x In x as (In x) / (1/ x), which is now in lim xlnx= lim

x-->o+

x-->o+

lln/x = lim X

X-->O+ -

1/ x

1/ x 2

* form. Thus

= lim (- x) = o.... X-->O+

.

Indeterminate forms of the type 00 and 100 can be handled by using logarithms: Example7

Solution

Find (a) limx-->o+

XX

and (b) limx-->Ixl/(I-x).

(a) This is of the form 0°, which is indeterminate because zero to any power is zero, while any number to the zeroth power is 1. To obtain a form to wh ich I'Hopital's rule is applicable, we write XX as exp(x Inx). By Example 6, we have limx -+ o+ xlnx = O. Since g(x) = exp(x) is continuous, the composite function rule applies, giving limx-+ o+ exp(xlnx) = exp(1imx -+o+ xlnx) = eO = 1, so limx -+ o+ XX = 1. (Numerically, 0.1°.1 = 0.79, 0.001°.001 = 0.993, and

0.00001°.00001 = 0.99988.)

526

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

(b) This has the indeterminate form 100 • We have applying l'Hopital's rule gives lim Inx x-I

x~ 1

= lim

= e(lnx)/(x-I);

=1

Ilx

1

x~ 1

Xl/(x-I)

'

so lim Xl/(x-I) = lim e(Jn x)/(x-l) x~l

= e1imx_,[(Jn x)/(x-I») = e l = e.

x~l

= 1 + (1In), then x-; 1 when n-; 00; we have I/(x - I) = n, so the limit we just calculated is limn-->oo(l + 1I nt. Thus l'Hopital's rule gives another proof of the limit formula limn-->oo(l + 1I nt = e. ...

If we set x

The next example is a limit of the form Example 8 Solution

00 -

00.

Find lim ( -~- - -1-). x~o xsmx x2 We can convert this limit to denominator:

~

form by bringing the expression to a common

form _1__ -1xsinx x2 x - sinx x 2sinx

1 g

l' g'

1- cosx 2x sinx

+ x 2cosx

type

limit

00-00

?

'0

0

?

'0

0

?

f" g"

sinx 2sinx + 4xcosx - x 2sinx

'0

0

?

f'"

cosx 6cosx - 6xsinx - x 2cosx

determinate

[TI

7'

Thus lim ( -~- - ...L ) x~O xsmx x2

= ! .... 6

Finally, we shall prove I'Hopital's rule. The proof relies on a generalization of the mean value theorem. Cauchy's mean value theorem

Suppose that fand gare continuous on [a,b] and differentiable on (a,b) and that g(a) =1= g(b). Then there is a number c in (a,b) such that

'( )f(b) - f(a) g c g(b) _ g(a) Proof

= f'(

) c.

First note that if g(x) = x, we recover the mean value theorem in its usual form. The proof of the mean value theorem in Seetion 3.6 used the function l(x)

= f(a) + (x -

a) f(bl

=~(a)

.

For the Cauchy mean value theorem, we replace x - a by g(x) - g(a) and look at h(x)

= f(a) + [g(x) -

f(b) - f(a) g(a)] g(b) _ g(a) .

11.2 L'Hopital's Rule

527

Notice that f(a) = h(a) and f(b) = h(b). By the horserace theorem (see Section 3.6), there is a point c such that j'(c) = h'(c); that is,

, , [f( b) - f( a) ] f(c)=g(c) g(b)-g(a) , which is what we wanted to prove. • We now prove the final version of l'Hopital's rule. = 0, we have

f(x) g(x)

=

f(x) - f(x o) _ j'(cx) g(x) - g(x o) - g'(cx) '

(1)

where Cx (which depends on x) lies between x and x o. Note that Cx ~ X o as x ~ x o. Since, by hypothesis, limx .... xJj'(x)/ g'(x)] = I, it follows that we also have limx->x.[j'(cx)/ g'(cx)] = I, and so by equation (1), limx .... x.[j(x)/ g(x)]

= I. •

Exercises tor Section 11.2 Use the pre!iminary version of I'HÖpital's rule to evaluate the !imits in Exercises 1-4. I. lim x 4 - 81 2. lim 3x 2 - 2 x ....3 x - 3 x ....2 x2 3 r x + 2x 4 r x 3 + 3x - 4 • x~ x~ sin( x-I)

1

----smx

25. !im

x .... oo

.

27.

Use the final version of I'HÖpital's rule to evaluate the limits in Exereises 5-8.

29.

5. lim eos3x - I x ....o 5x 2 7. !im sin 2x - 2x x ....o

x3

Evaluate the 9. lim

x .... oo

~

x ....o

~ X375

10. !im

3x 4 + 2x 2 + I ei/x 12. lim - x .... o+ I/x x .... oo

x- 2

Evaluate the O·

x3

forms in Exereises 9-12.

11. lim lnx X~O

6. !im eos IOx - I x ....o 8x 2 8. !im sin 3x - 3x

00

forms in Exereises 13-16.

13. !im [x 4lnx]

14. !im [tan 'lT2x Inx]

x ....o

x"" I

15. !im [x"e-"X] x ....o

Evaluate the limits in Exereises 17-36. 17. lim [(tanxYJ x ....o

19. lim(esex-eotx)20. lim [lnx-ln(x-I)] x-+oo

x-+o

21. !im

x ....o

~sm x

1--x-2 23. I I· m x .... 1 1+ x 2

I 22. !im

~

X"" I eos('lTx/2)

24

•

r x+sin2x x~ 2x + sin3x

x2

lim

x .... s+

x + I

26. x!im .... 00

Jx 2 - 25 x-5

lim x 2 +2x+ I x 2 -1 x-I + x 2/2 . eos 31. IIm ----:---'-x ....o x4

(2x + 1)3

x3 + 2 Jx 2 - 25 28. lim x+5 x .... s+ 30.

x ..... -I

!im x l /(I-x 2) x .... 1-

32. lim ~ x"" I e X - I 33. lim 1+ eosx x-+w

34.

x - ",

lim (x - :::'2 )tanx

x"'" ,,/2

. sinx - x + (1/6)x 3 35. hm ---~--'-X-)oO

x5

x 3 + lnx + 5 36. lim x .... oo 5x 3 + e- x + sinx

37. Find limx .....o+ xPlnx, where pis positive. 38. Use I'Höpital's rule to show that as x ~ 00, xn/ex~O for any integer n; that is, e X goes to infinity faster than any power of x. (This was proved by another method in Seetion 6.4.) *39. Give a geometrie interpretation of the Cauchy mean value theorem. [Hint: Consider the curve given in parametrie form by y = J(t), x = g(t).] *40. Suppose thatJ is continuous at x = xo, that j'(x) exists for x in an interval about xo, x =1= xo, and that limx_"Xoj'(x) = m. Prove that j'(xo) exists and equals m. [Hint. Use the mean value theorem.] *41. Graph the function J( x) = x X, x > O.

528

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

to conclude that

*42. Prove I'Höpital's rule for Xo = 00 as folIows: (i) Let fand g be differentiable on (a, 00) with g(x) #- 0 and g'(x) #- 0 for all x> a. Use Cauchy's mean value theorem to prove that for every e > 0, there is an M > a such that for y>x>M,

f(X)

I g(x)

1

-I < e.

(iii) Complete the proof using (ii).

- f(x) - 11 < e If(Y) g(y) - g(x)

*43. (a) Find lim

Q~O

(b) Find \im

(ii) Write

a-4tX;l

.! In( e

Q

a

.!a In( e

1)

-

a Q

a -

I )

(c) Are YOUT Tesults consistent with the computations of Exercise 30, Section 9.47

f(x)= \im f(x) - f(y) g(x) Y-CD g(x) - g(y) and choose y sufficiently large,

11.3 Improper Integrals The area of an unbounded region is defined by a limiting process.

Tbe definite integral f~(x)dx of a function f which is non-negative on the interval [a, b] equals the area of the region under the graph of f between a and b. If we let b go to infinity, the region becomes unbounded, as in Fig. 11.3.1. One's first inclination upon seeing such unbounded regions may be to assert that their areas are infinite; however, examples suggest otherwise. y

Figure 11.3.1. Tbe region under the graph of fon [a, (0) is unbounded.

x

Q

Example 1 Solution

Find (b ~ dx. What happens as b goes to infinity? J1 x We have (bdx = (b x - 4dx = x- 3 l b= l/b 3 -1 = 1-I/b3

J1 x 4

- 3 1

- 3

3

As b becomes larger and larger, this integral always remains less than furthermore, we have

y

lim

b-+et:J x

Figure 11.3.2. Tbe region under the graph of I j x 4 on [l, (0) has finite area. It is

fi(dxj x 4 ) = t·

JI

i

b

1

t;

d 1- l/b 3 I ...;!=lim 3 =-3· ... 4 x b-+et:J

Example I suggests that t is the area of the unbounded region consisting of those points (x, y) such that 1 < x and 0< Y < 1/ x 4• (See Fig. 11.3.2.) In accordance with our notation for finite intervals, we denote this area by ff( dx / x 4). Guided by this example, we define integrals over unbounded intervals as limits of integrals over finite intervals. Tbe general definition folIows.

11.3 Improper Integrals

529

Integrals over Unbounded Intervals Suppose that for a fixed,fis integrable on [a,b] for all b > a.1f the limit limb-Hof::t(x)dx exists, we say that the improper integral r:f(x)dx is convergent, and we define its value by

i

i oo

f(x)dx= lim b f(x)dx. b--'>oo a Similarly, if, for fixed b, f is integrable on [a, b] for all a define a

J~oof(x)dx= aj}~oo

i

b

< b,

we

f(x)dx,

if the limit exists. Finally, if fis integrable on [a, b] for all a < b, we define L:f(x)dx= LOoof(x)dx+ fooo f(x)dx,

if the improper integrals on the right-hand side are both convergent. If an improper integral is not convergent, it is called divergent.

Example 2

Solution

For which values of the exponent r is loo x' dx convergent? We have

I

lim (\'dx= lim x'+ll b = lim b,+I-I b--'>oo JI b--'>oo r + I b--'>oo r + 1

(r=l= -1).

If r + 1 > 0 (that is, r> -1), the limit limb_Hob,+1 does not exist and the integral is divergent. If r + 1 < 0 (that is, r< -1), we have limb-->oob,+1 = 0 and the integral is convergent-its value is -I/(r + I). Finally, if r = -I we have ftx -I dx = lnb, which does not converge as b ~ 00. We conclude that H'x' dx is convergent just for r < - I .... Example3

Solution

Findfoo ~. -00 1+ x 2 We write f':': 00 (dx/(I + x 2» = f~ 00 (dx/(l + x 2» + fg' (dx/(I + x 2». To evaluate these integrals, we use the formula f(dx/(l + x 2» = tan-Ix. Then

fo

-00

~= lim (tan-IO-tan-Ia) 1+ x 2

a->-oo

=0-

lim tan-Ia= 0 - ( -'TT) = J!... a--'>-oo 2 2

(See Fig. 5.4.5 for the horizontal asymptotes of y = tan -IX.) Similarly, we have ("" ~ = lim (tan-Ib - tan-IO) = J!.., b--'>oo 2

Jo I + x 2

so

oo dx -, f-ool+x

-2

'TT

= -

2

+ -'TT

2

=

'TT • ...

Sometimes we wish to know that an improper integral converges, even though we cannot find its value explicitly. The following test is quite effective for this situation.

530

Chapler 11 Limits, L'Hopital's Rule, and Numerical Methods

Comparison Test Suppose that fand gare functions such that Ci) If(x)1

< g(x) for all x

;;. a and

(ii) Lbf(x)dx and L bg(x)dx exist for every b

> a.

Then (I) If L"Og(x)dx is convergent, so is Loof(x)dx, and

(2) if Loof(x)dx is divergent, so is

Loo g(x)dx.

Similar statements hold for integrals of the type

f:oof(x)dx

and

L:f(x)dx.

Here we shall explain the idea behind the comparison test. A detailed proof is given at the end of this section. Ir fex) and g(x) are both positive functions (Fig. 11.3.3(a», then the region under the graph of fis contained in tht! region under the graph of g, so )'

x

X=a

=0

)' = [(.t)

.y = g(x)

Flgure 11.3.3. Illustrating the comparison test.

(b)

(a)

the integral f~( x) dx increases and remains bounded as b ~ 00. We expect, therefore, that it should converge to some limit. In the general case (Fig. 11.3.3(b», the sums of the plus areas and the minus areas are both bounded by f:'g(x)dx, and the cancellations can only help the integral to converge. Note that in the event of convergence, the comparison test only gives the inequality - f:'g(x)dx < f:'f(x)dx < f:'g(x)dx, but it does not give us the value of f:'f(x)dx.

Example 4 Solullon

Show that (00

Jo

dx

b + x8

is convergent, by comparison with

l1 x 4 •

We have I/b + x 8 < I/R = l1 x 4, so it is tempting to compare with fO'(dxlx4). Unfortunately, the latter integral is not defined because IIx 4 is unbounded near zero. However, we can break the original integral in two parts: 00 (00 dx (' dx dx Jo ~=JO~+I ,h+x8

1.

The first integral on the nght-hand side exists because I I ~8 is continuous on [0, I]. The second integral is convergent by the comparison test, taking g(x) = l1 x 4 and fex) = I I~I + x 8 • Thus fO' (dxlb + x 8 ) is convergent. ...

11.3 Improper Integrals

Example 5 Solution

Show that (00

Jo

(I

sinx

+ x)

2

531

dx converges (without attempting to evaluate).

We may apply the comparison test by choosingg(x) = 1/(1 + xi andj(x) = (sinx)/(I + xi, since Isinxl";; To show that fO'(dx/(1 + xi) is convergent, we can compare 1/( I + X)2 with 1/ x 2 on [I, 00), as in Example 4, or we can evaluate the integral explicitly:

1.

(00

Jo

(1

r

dx

(b

dx

r [

-I

+ x/ = b~~ Jo (I + x)2 = b~~ (I + x)

]I

b

0

=lim [1-_1_]=1.. 1+ b b~oo

Example 6 Solution

Show that (00

dx

J1 ~I + x 2

is divergent.

We use the comparison test in the reverse direction, comparing I/~I + x 2 with I/x. In fact, for x) I, we have I/b + x 2 ) 1/~X2 + x 2 = I/li x. But f~(dx/Ii x) = (I/Ii)lnb, and this diverges as b ~ 00. Therefore, by statement (2) in the comparison test, the given integral diverges. • We shall now discuss the second type of improper integral. If the graph of a function j has a vertical asymptote at one endpoint of the interval [a, b], then the integral f!t( x) dx is not defined in the usual sense, since the function j is not bounded on the interval [a,b]. As with integrals of the form r:j(x)dx, we are dealing with areas of unbounded regions in the plane-this time the unboundedness is in the vertical rather than the horizontal direction. Following our earlier procedure, we can define the integrals of unbounded functions as limits, which are again called improper integrals.

Integrals 01 Unbounded Functlons Suppose that the graph of f has X o = b as a vertical asymptote and that for a fixed, j is integrable on [a,q] for all q in [a,b). If the limit limq-->b_ f'!J(x)dx exists, we shall say that the improper integral f!t(x)dx is convergent, and we define

(bj(x)dx= lim (qj(x)dx.

Ja

Similarly, if x

q~b-

= a is a vertical asymptote, we define

(bj(x)dx =

Ja

Ja

lim

p~a+

(bj(x)dx,

Jp

If the limit exists. (See Fig. 11.3.4.)

If both x = a and x = bare vertical asymptotes, or if there are vertical asymptotes in the interior (a,b), we may break up [a~bJ into subintervals such that the integral of j on each subinterval is of the type considered in the preceding definition. If each part is convergent, we may add the results to get f!t(x)dx. The comparison test may be used to test each for convergence. (See Example 9 below.)

532

Chapter 11 Limits, L'Hopital's Aule, and Numerical Methods ! '

Figure 11.3.4. Improper integrals defined by (a) the limit limq->b_ f'!J(x)dx and (b) the limit limp->a+ f~f(x)dx.

Example 7 Solution

a

If h

(bI

(al

For which values of r is LI x' dx convergent? If r If ;

> 0, x' is continuous on [0, I] and the integral exists in the ordinary sense.

< 0, we have limx->o+ x' =

00, so we must take a limit. We have

1

lim fix' dx= lim X,+I 1 = _1_ p->O+ P p->O+ ,+ I p r + I

(I _ p->O+ lim p'+ I),

provided r =1= -1. If r + I > 0 (that is, , > - I), we have limp->o+ p'+ 1 = 0, so the integral is convergent and equals 1/(, + I). Ir r + I < 0 (that is, r < -I), limp->o+ pr+ 1 = 00, so the integral is divergent. Finally, if r + I = 0, we have limp->o+ J!x r dx = limp->o+ (0 -In p) = 00. Thus the integral Jbx r dx converges just for r > -1. (Compare with Example 2.) •

Example 8 Solution

Find

f

Inxdx.

We know that Jlnxdx

= xlnx - x +

C, so

(Ilnxdx= lim (llnl-I-plnp+p) p->o+

Jo

=0-1-0+0=-1 (limp->o+ p In p = 0 by Example 6, Section 11.2). •

Example 9 Solution

Show that the improper integral (00 e- x dx is convergent.

Jo {X

This integral is improper at both ends; we may write it as /1 + /2' where /1

= f(e-X/{X)dx

and

/2

= .J:oo(e-X/{X)dX

and then we apply the comparison test to each term. On [0, I], we have e -x .;;; I, so e -x / {X .;;; 1/ {X. Since JÖ{dx / -5) is convergent (Example 7), so is /1' On [1,00), we have 1/-5.;;; I, so e- x/-5 .;;; e- x ; but j'f'e-xdx is convergent because

(ooe-xdx= lim

J1

b->oo

(b e -

J1

xdx = lim (e- I - e- b ) b->oo

= e- I

Thus /2 is also convergent and so Jg'(e- X/-5)dx is convergent. •

11.3 Improper Integrals

Improper integrals arise tangents.

Example 10 Solution

533

arc length problems for graphs with vertical

in

=b -

Find the length of the curve y geometrically.

x 2 for x in [-I, I]. Interpret your result

By formula (I), Section 10.3, the arc length is

J~IJI + (dy/dxi dx= flYI + (-x/b =

J b dx- x

X 2 )2

dx

I

-I

2

•

The integral is improper at both ends, since

=

I

!im x-+-l+

b - x2

I

lim x-+I-

b - x2

= 00.

We break it up as

f o JI dx- x - 1

=

(I 2

lim

+ Jo (0

p-+-I+Jp

=

lim

p-+ - 1+

=0 -

(

dx

b - x2 dx

JI- x 2

+ lim (q q-+I-JO

dx

b - x2

(sin-10-sin-p)+ lim (sin-1q-sin-10)

-1) + 1- 0 =

q-+ 1-

7T.

Geometrically, the curve whose arc length we have just found is a semicircle of radius I, so we recover the fact that the circumference of a circle of radius I is 27T. Ä

Example 11

Luke Skyrunner has just been knocked out in his spaceship by his archenemy, Captain Tralfamadore. The evil captain has set the cpntrols to send the spaceship into the sun! His perverted mind insists on a slow death, so he sets the controls so that the ship makes a constant angle of 300 with the sun (Fig. 11.3.5). What path will Luke's ship follow? How long does Luke have to wake up if he is 10 million miles from the sun and his ship travels at a constant velocity of a million miles per hour?

Figure 11.3.5. Luke Skyrunner's ill-fated ship.

Solution

We use polar coordinates to describe a curve (r(l),9(t» such that the radius makes a constant angle a with the tangent (a = 30 0 in the problem). To find dr / d(}, we observe, from Fig. 11.3.6(a), that

tJ.r-;:::: rM

tana

so

dr

d(}

= _r_ tana

(1)

534

ehapler 11 Limits, L'Hopital's Rule, and Numerical Methods

Figure 11.3.6. The geometry of Luke's path.

(a)

(b)

We can derive formula (1) rigorously, but also more laboriously, by calculating the slope of the tangent line in polar coordinates and setting it equal to tan(O + a) as in Fig. 11.3.6(b). This approach gives

tanO(dr/dO) + r tanO + tana dr / dO - r tan 0 = tan( 0 + a) = I - tan 0 tan a ' so that again

dr = _r_ dO tana The solution of equation (I) is3 r(O) = r(0)e 9/ tana .

(2)

For this solution to be valid, we must regard 0 as a continuous variable ranging from - 00 to 00, not as being between zero and 2'11. As O~ 00, r(O)~ 00 and as O~ - 00, r(O)~O, so the curve spirals outward as 0 increases and inward as 0 decreases (if 0 < a < '11/2). This answers the first question: Luke follows the logarithmic spiral given by equation (2), where IJ = 0 is chosen as the starting point. From Section 10.6, the distance Luke has to travel is the arc length of equation (2) from 0 = 0 to 0 = - 00, namely, the improper integral

( dr dlJ

)2 + r2 d(J= JO-

00

=JO

-00

r(0)e9/tana_._I_dO sma

= r(O)

cosa

e9/tanalO -00

=

r(O) cosa

With velocity = 10 r(O) = 10 and cosa = cos30° =,f3 /2, the time needed to travel the distance is 6,

time = dista~ce = --.!!L velocIty ,f3 /2

7,

X

_1_ = 20 ~ 11.547 hours 106 ,f3

Thus Luke has less than 11.547 hours to wake up. '" 3

See Section 8.2. If you have not read Chapter 8, you may simply check direct1y that equation (2)

is a solution of (I).

~

~

•

11.3 Improper Integrals

535

The logarithmic spiral turns up in another interesting situation. Place four love bugs at the corners of a square (Fig. 11.3.7). Each bug, being in love, walks directly toward the bug in front of it, at constant top bug speed. The result is that the bugs all spiral in to the center of the square following logarithinic spirals. The time required for the bugs to reach the center can be calculated as in Example 11 (see Exercise 46). We conclude this section with a proof of the comparison test. The proof is based on the following principle: Let F be a function defined on [a, 00) such that (i) Fis nondecreasing; i.e., F(x l) < F(x2) whenever XI < x2; (ii) Fis bounded above: there is a number M such that F(x) < M for all x. Then limx-->ooF(x) exists and is at most M. The principle is quite plausible, since the graph of F never descends and never crosses the line y = M, so that we expect it to have a horizontal asymptote as X ~ 00. (See Fig. 11.3.8).

~

Figure 11.3.7. These love bugs follow logarithmic spirals.

-- -

)'=M

---------

Flgure 11.3.8. The graph of a nondecreasing function lying below the line y = M has a horizontal asymptote.

x

A rigorous proof of the principle requires a careful study of the real numbers,4 so we shall simply take the principle for gran ted, just as we did for some basic facts in Chapter 3. A similar principle holds for nonincreasing functions which are bounded be1ow. Now we are ready to prove statement (1) in the comparison test as stated in the box on p. 530. (Statement (2) follows from (I), for if f:'g(x)dx converged, so would f:'f(x)dx.) Let

y

x

~(X)

if if

fex) fex)

>0 00 j(x) exists.

{I7i

11.4 Limits of Sequences and Newton's Method Solutions of equations can ojten be found as the limits of sequences. This section begins with a discussion of sequences and their limits. The topic will be taken up again in Section 12.1 when we study infinite series. A sequence is just an "infinite list" of numbers: a l ,a2 ,a3 , ••• , with one an for each natural number n. A number I is called the limit of this sequence if, roughly speaking, an comes and remains arbitrarily elose to I as n increases.

538

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

Perhaps the most familiar example of a sequence with a limit is that of an infinite decimal expansion. Consider, for instance, the equation

t = 0.333 ...

(1)

in which the dots on the right-hand side are taken to stand for "infinitely many 3's." We can interpret equation (1) without recourse to any metaphysical notion of infinity: the finite decimals 0.3,0.33,0.333, and so on are approximations to t, and we can make the approximation as good as we wish by taking enough 3's. Our sequence a 1,a2, ••• is defined in this case by an = 0.33 ... 3, with n 3's (here the three dots stand for only finitely many 3's). In other words, _ 3

10 +

an -

3 100

+ ... +

3

Ion .

(2)

We can estimate the difference between an and Multiplying equation (2) by 10 gives

lOa

n

=

3 + -3

10

t

3+ ... + IOn-I'

by using some algebra.

(3)

and subtracting equation (2) from equation (3) gives 3 9an = 3 - Ion '

an

=

t -t (.~ ).

Finally,

t - = t(I~n an

(4)

).

As n is taken larger and larger, the denominator Ion becomes larger and larger, and so the difference t - an becomes smaller and smaller. In fact, if n is chosen large enough, we can make t - an as small as we please. (See Fig. ll.4.1.)

o

0.3

I

I!

"3

Figure 11.4.1. The decimal approximations to t form a sequence converging to t.

Example 1 Solution

How large must n be for the error

t - an

to be less than 1 part in 1 million?

By equation (4), we must have

1(_1 )< 10-6 3 Ion or IO- n < 3 . 10- 6 • It suffices to have n ~ 6, so the finite decimal 0.333333 approximates t to within I part in a million. So do the longer decimals 0.3333333, 0.33333333, and so on.... There is nothing special about the number 10- 6 in Example 1. Given any positive number €, we will always be able to make t - an = t (1/ Ion) less than € by letting n be sufficiently large. We express this fact by saying that t is the

11.4 Limits of Sequences and Newton's Method

539

limit of the numbers

a

n

3 3 3=-+-+ ... +Ion 10 100

as n becomes arbitrarily large, or

lim(l+~+ ... +~)=!. 10

n->oo

Ion

100

3

We may think of a sequence a"a 2,a3, ... as a function whose domain consists of the natural numbers 1,2,3, ... (Occasionally, we allow the domain to start at zero or some other integer.) Thus we may represent a sequence graphically in two ways-either by plotting the points a \, a2 , • •• on a number line or by plotting the pairs (n,an ) in the-plane. Example 2

(a) Write the first six terms of the sequence an = n/(n + I), n = 1,2,3, .... Represent the sequence graphically in two ways. Find the value of limn->oo[n/(n+ I)]. (b) Repeat for an = (_l)n/n. (c) Repeat for an = (-Itn/(n + I).

Solution

(a) We obtain the terms a\ through a6 by substituting n = 1,2,3, ... , 6 into the formula for an' giving 1, ~, ~, ~, ~. These values are plotted in Fig. 11.4.2. As n gets larger, the fraction n/(n + I) gets larger and larger hut never exceeds I; we may guess that the limit is equal to 1. To verify this guess, we look at the difference I - n/(n + I). We have

*,

1 _ _ n_ n+1

= n + 1-

n

n+1

= _1_

n+l'

which does indeed become arbitrarily small as n increases, so lim n( I) n->oo n+

=

1.

(b) The terms a, through a6 are -1,1, - 1.*, - !,i. They are plotted in Fig. 11.4.3. As n gets larger, the number (- It / n seems to get closer to zero. Therefore we guess that limn -> 00[( - I t / n] = O.

•

al

0.5

•

• 0.7 • 0.8 ••

I

1

0.6

•

a 3 as

al

U3 a4 a 5 a 6

a2

1•

0.9

-\

3 4

\ -~

-

\ 4

•• • •

ab "4

• I.

0

\ 4

a2

\ 2

an

1.\

an

1.0 0.9 0.8 0.7 0.6 0.5 0

•

• • ••

0

• 1

•

I

'2

3

•

I

"2 ~

3

4

5

6

Figure 11.4.2. The sequence an = n/(n + 1) represented graphically in two different ways.

7

11

--I

• 4

•

Figure 11.4.3. The sequence an = ( - 1)" / n plotted in two ways.

•

• 6

540

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

(c) We have, for a 1 through a6 , - hi, - i,~, - %,~. They are plotted in Fig. 11.4.4. In this case, the numbers an do not approach any particular number. (Some of them are approaching 1, others -1.) We guess that the sequence does not have a limit. •

•• • -4"

Q5 Q3

-I

al

J

3

0

I 4

2

.,

..

a2 a 4 a6

, J 4

J

2

I 3 4

an

•

I

2

0 J

Figure 11.4.4. The sequence an = (-I)"n/(n plotted in two ways.

•

-2

+ I)

-I

2

3

•

•

4

6

•

n

•

Just as with the e-8 definition for limits of functions, there is an e-lv definition for limits of sequences which makes the preceding ideas precise.

Limits 01 Sequences The sequence a 1,a2 ,a3' ... , an' ... approaches 1 as a limit if an gets elose to and remains arbitrarily elose to 1 as n becomes large. In this case, we write lim an = I.

n~oo

that

1- ;,

1 +;,

Figure 11.4.5. The relationship between an, I, and e in the definition of the limit of a sequence.

Example 3 Solution

In precise terms, limn-->o,ßn lan - I1 < e for all n ;;. N.

= 1 if, for every e > 0, there is an N such

It is useful to think of the number e in this definition as a tolerance, or allowable error. The definition specifies that if 1 is to be the limit of the

sequence an' then, given any tolerance, all the terms of the sequence beyond a certain point should be within that tolerance of I. Of course, as the tolerance is made smaller, it will usually be necessary to go farther out in the sequence to bring the terms within tolerance of the limit. (See Fig. 11.4.5.) Tbe purpose of the e-N definition is to lay a framework for a precise discussion of limits of sequences and their properties-just as the definitions in Sec ti on 11.1. Let us check the limit of a simple sequence using the e- N definition. Prove that limn->oo(lln)

= 0, using the e-N definition.

°

To show that the definition is satisfied, we must show that for any e > there is a number N such that 11/n - 01< e if n > N. If we choose N ;;. l/e, we get, for n > N,

..; 11.n - 01 = 1.n < ~ N

e.

Thus the assertion is proved. •

11.4 Limits of Sequences and Newton's Method

541

mCalculator Discussion Limits of sequences can sometimes be visualized on a calculator. Consider the sequence obtained by taking successive square roots of a given positive number a: a o = a,

and so forth. (See Fig. 11.4.6.) [ ~.?803SIJ8

Figure 11.4.6. For a recursively defined sequence an + 1 = j(an ), the next member in the sequence is obtained by depressing the "f' key.

I

I t5100S3IJ I v

Press V

0000

Herej=r.

Cl

0000 0000 0000 0000

key

For instance, if we start by entering a = 5.2, we get

ao = 5.2,

a. = fIT = 2.2803508, a2 = "2.2803508 = 1.5100830, a 3 = Vt.5100830 = 1.2288544,

r

and so on. After pressing the repeatedly you will see the numbers getting closer and closer to 1 until roundoff error causes the number 1 to appear and then stay forever. This sequence has 1 as a limit. (Of course, the calculation does not prove this fact, but does suggest it.) Observe that the sequence is defined recursivery-that is, each member of the sequence is obtained from the previous one by some specific process. The sequence 1,2,4,8,16,32, ... is another example; each term is twice the previous one: an + 1 = 2an • • Limits of sequences are closely re1ated to limits of functions. For exampie, if j(x) is defined for x ~ 0, then an = Jen) is a sequence. If limHooj(x) exists, then limn ..... ooan exists as weH and these limits are equal. This fact can sometimes be used to evaluate some limits. For instance, lim-x-=lim 1 = 1 =_1_=1 x ..... ool+l/x l+lim(l/x) 1+0 '

x ..... oox+1

x ..... 00

and so limn ..... oo[n/(n + I)] = I, confirming our calculations in Example 2(a). Limits of sequences also obey roles similar to those for functions. 5 We iHustrate: Example4

Find (a) lim ( n 22+ 1 ) n ..... oo 3n + n and (b) lim

n ..... oo

Solution

(I _1n + ~I ). n+

(a) Write

(n

2 · + 1) I1m n..... oo 3n 2 +n

5

=

n

I·1m (1 + 1/ 2 3+1/n

n ..... oo

)

These are written out formally in Section 12.1.

(dividing numerator and denominator by n 2 )

542

ehapler 11 Limits, L'Hopital's Rule, and Numerical Methods

1+ n-+oo Iim (1/n 2)

(quotient and sum ruIes)

3+ nI~(I/n)

=I+O=! 3 +0

3·

(b) Iim(l-l+_n-)=1-3(lim!)+lim( 1 ) n-+oo n n+I n-+oo n n-+oo 1 + l/n

=1-

3 . 0 + 1 = 2. •

Tbe connection with limits of functions allows us to use I'Hopital's rule to find limits of sequences. ~ Example 5

Solution

(a) Using numerical calculations, guess the value of limn --+ oo (ri. (b) Use I'Hopital's rule to verify the result in (a). (a) Using a calculator we find:

1ii

n I

5 10 50 100 500 1000 5000 10,000

I

1.37973 1.25893 1.08138 1.04713 1.01251 1.00693 1.00170 1.00092

Tbus it appears that limn --+ oo 'Vii = 1. (b) To verify this, we use l'Hopital's rule to show that limx--+oox l / x = 1. Tbe limit is in 00° form, so we use logarithms: Xl/x

=

e(lnx)/x.

Now limx--+oo(lnx/ x) is in li

x-+~

~

form, and l'Hopital's rule gives

l/x_ O xlnx_li - x-+~ -1- .

Hence lim xl/x= exp(

X~ 1, if , = 1, if 0 I, and so limn~oopn = 00. Therefore, limn~oo'" = limn--+oo(l I p n) = 0 (compare the reeiproeal test for limits of funetions in Seetion 11.1).

,n

Example 6 Solution

Evaluate (a) limn~oo3n, (b) lim"~OOe-n, (e) limn~oo(e + (ttt (a) Here, = 3 > I, so lim"--+oo3 n = 00. (b) e- n = (li et, and 11 e < 1, so limn-->ooe-" = O. (e) limn--+oo[e + (t)"t = [e + lim"~oooo(sinn)ln = O. (b) I( -lt Inl " I/n~O, so (-I)"ln~O by the eomparison test. Thus - - + Iim !!=O+I=I. ... lim ( -I)n+n) = lim (-I)")

n

,,~oo

n--+oo

n

n--+oo n

Many questions in mathematies and its applieations lead to the problem of solving an equation of the form J(x)

= 0,

(5)

where J is some funetion. The solutions of equation (5) are called the rools or ze,os of J. If J is a polynomial of degree at most 4, one can find the roots of J

by substituting the eoeffieients of j into a general formula (see pp. 17 and 173). On the other hand, if J is a polynomial of degree 5 or greater, or a funetion involving the trigonometrie or exponential functions, there may be no explicit formula for the roots of J, and one may have to search for the solution numerieally. Newton's method uses linear approximations to prorluee a sequenee X O,x\,x2 ' ••• whieh eonverges to a solution of j(x) = O. Let Xo be a first guess. We seek to eorreet this guess by an amount 6.x so that j(xo + 6.x) = O. Solving this equation for tu is no easier than solving the original equation (5), so we manufaeture an easier problem, replaeingjby its first-order approximation at x o; that is, we replaee j(x o + 6.x) by j(x o) + f'(x o)6.x. If j(xo) is not equal to zero, we ean solve the equation j(xo) + f'(x o)6.x = 0 to obtain 6.x = - j(xo)If'(xo), so that OUT new guess is XI = Xo + 6.x = Xo - J( xo)1 f'( x o)·

544

Chapter 11 LiI'nits, L'Hopital's Rule, and Numerical Methods

Geometrically, we have found XI by following the tangent line to the graph of f at (xo, f(xo)} until it meets the X axis; the point where it meets is (XI' 0) (see Fig. 11.4.7). y=f(xl

x

Figure 11.4.7. The geometry of Newton's method.

Now we find a new guess of xo; that is, x 2 = XI

-

f(x l ) f'(x l )

X2

= Xn -

XI

in place

•

In general, once we have found Xn+1

by repeating the procedure with

f(x n ) f'(x n ) '

Xn ,

we define x n + I by (6)

Let us see how the method works in a case where we know the answer in advance. (This iteration procedure is particularly easy to use on a programmable calculator.)

mExample 8 Solution

Use Newton's method to find the first few approximations to a solution of the equation x 2 = 4, taking X o = 1. To put the equation x 2 = 4 in the form f(x) = 0, we let f(x) = x 2 - 4. Then f'(x) = 2x, so the iteration rule (6) becomes Xn+1 = X n - (x; - 4)/2xn • which may be simplified to X n + 1 = !(xn + 4/xn ). Applying this formula repeatedly, with X o = 1, we get (to the limits of our calculator's accuracy) XI

=2.5

= 2.05 x3 = 2.000609756 x 4 = 2.000000093 X2

s =2 x6 = 2

X

and so on forever. The number 2 is, of course, precisely the positive root of our equation x 2 = 4....

mExample 9

Use Newton's method to locate a root of X S - x 4 - X + 2 = O. Compare what happens with various starting values of X o and attempt to explain the phenomenon.

11.4 Limits of Sequences and Newton's Method

Solution

545

The iteration formula is

x" + I

= x" -

x~ - x: - x"

+2

5x4 _ 4x3 _ I n

n

4x~ - 3x: - 2 I .

= 5x4n _ 4x3n -

For the purpose of convenient calculation, we may weite this as

x

,,+1

(4x" - 3)x: - 2 (5x" - 4)x! - I .

= -'-----'---:--

Starting at Xo = I, we find that the denominator is undefined, so we can go no further. (Can you interpret this difficulty geometrically?) Starting at X o = 2, we get XI

= 1.659574468,

x 2 = 1.372968569,

X3 = 1.068606737, X4 = - 0.5293374382, Xs = 169.5250382. The iteration process seems to have sent us out on a wild goose chase. To see what has gone wrong, we look at the graph of fex) = X S - x 4 - X + 2. (See Fig. 11.4.8.) There is a "bowl" near X o = 2; Newton's method attempts to take us down to a nonexistent root. (Only after many iterations does one converge to the root-see Exercise 59 and Example 10.)

20

\0

Flgure 11.4.8. Newton's method does not always work.

o Finally, we start with Xo =

Xo = -2,

= - 1.603603604, X2 = -1.323252501, X3 = - 1.162229582, X4 = -1.107866357, Xs = -1.102228599, X6 = -1.102172085, X7 = -1.102172080,

XI

- 2. The iteration gives !(xo) = -44; !(XI) = -13.61361361; !(X2) = -3.799819057; !(X3) = -0.782974790; !(X4) = -0.067490713; !(xs) = -0.000663267; ! (X6) = - 0.()()()()()()()61 ; !(X7) = -0.000000003.

546

Chapler 11 Limits, L'Hopital's Rule, and Numerical Methods

Since the numbers in the f(x) column appear to be converging to zero and those in the x column are converging, we obtain a root to be (approximately) -1.10217208. Since f(x) is negative at this value (where f(x) = -0.000000003) and positive at -1.10217207 (where f(x) = 0.000000115), we can conelude, by the intermediate value theorem, that the root is between these two values. • Example 9 illustrates several important features of Newton's method. First of all, it is important to start with an initial guess which is reasonably elose to a root-graphing is a help in making such a guess. Second, we notice that once we get near a root, then convergence becomes very rapid-in fact, the nu mb er of correct decimal pi aces is approximately doubled with each iteration. Finally, we notice that the process for passing from x n to x n + 1 is the same for each value of n; this feature makes Newton's method particularly attractive for use with a programmable calculator or a computer. Human intelligence still comes into play in the choice of the first guess, however.

Newton's Method To find a root of the equation f(x) = 0, where f is a differentiable function such that j' is continuous, start with a guess Xo which is reasonably elose to a root. Then produce the sequence XO'X I 'X 2 ' ••• by the iterative formula:

Xn+1

f(X n )

= Xn - j'(xn) .

If limn-->ooxn = X, then f(x)

= O.

To justify the last statement in the box above, we suppose that limn-->ooxn = x. Taking limits on both sides of the equation x n+1 = x n - f(xn)/j'(x n), we obtain x = x - limn-->oo[f(xn)/ j'(xn)), or Iim n-->oo[f(xn)/j'(xn)) = O. Now let an = f(x n ) /j'(xn). Then we have limn-->ooan = 0, while f(x n) = aJ'(xn). Taking limits as n ~ 00 and using the continuity of fand 1', we find lim f(x n) = !im an lim j'(xn), n---=»oo n---'::'oo

n~oo

so

f(x) = O· 1'(x) = 0. Newton's method, applied with care, can also be used to solve equations involving trigonometric or exponential functions. ~ Example 10

Solution

Use Newton's method to find a positive number x such that sinx

Withf(x)

x

= sinx -

x/2, the iteration formula becomes

sinx - x /2

n+L

= x/2.

2(xncosxn - sinxn) 2cosxn - 1

n n = xn - ----"--...;.:.:-,.:=_ cosxn - 1/2

Taking X o = 0 as our first guess, we get x I = 0, x 2 = 0, and so forth, since zero is already a root of our equation. To find a positive root, we try another guess, say Xo = 6. We get XI =

13.12652598

x 2 = 30.50101246 X3

= 176.5342378

x4

= 448.4888306

Xs =

x6

266.0803351

= 143.3754278

= -759.1194553 x lO = 3,572.554623 x9

11.4 Limits of Sequences and Newton's Method

547

We do not seem to be getting anywhere. To see what might be wrong, we drawa sketch (Fig. 11.4.9). The many places where the graph of sinx - x/2 y

x y=sinx

y

X2

Figure 11.4.9. Newton's method goes awry.

.

x

x

y=smx-"2

has a horizontal or nearly horizontal tangent causes the Newton sequence to make wild excursions. 6 We need to make a better first guess; we try Xo - 3. This gives

= 2.08799541, x 2 = 1.91222926,

s = 1.89549427,

XI

X

X3

= 1.89565263,

X4

= 1.89549428,

= 1.89549427, x 7 = 1.89549427, Xg = 1.89549427. X6

We conelude that our root is somewhere near 1.89549427. Substituting this value for X in sinx - x/2 gives 1.0 X 10- 11 • There may be further doubt about the last figure, due to internal roundoff errors in the calculator; we are probably safe to announce our result as 1.8954943.... You may find it amusing to try other starting values for X o in Example 10. For instance, the values 6.99, 7, and 7.01 seem to lead to totally different results. (This was on a HP 15C hand calculator. Numerical errors may be crucial in a calculation such as this.) Recently, the study of sequences defined by iteration has become important as a model for the long-time behavior of dynamical systems. For instance, sequences defined by simple rules of the form xn + I = axn(l - x n) display very different behavior according to the value of the constant a. (See the supplement to this section and Exercise 59.)

Supplement to Sectlon 11.4 Newton's Method and Chaos

The sequences generated by Newton's method may exhibit several types of strange behavior if the starting guess is not elose to a root: (a) the sequence XO'X I'X 2 ' ••• may wander back and fortb over the realline for some time before converging to a root; Try these ca1culations and those in Example 9 on your calculator and see if you converge to the root after many iterations. You will undoubtedly get different numbers from ours, probably due to roundoff errors, computer inaccuracies and the extreme sensitivity of the calculations. We gol four different sets of numbers with four calculators. (Tbe ones here were found on an HP ISC which also has a SOLVE algorithm which deverly avoids many difficulties.) ti

548

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

(b) slightly different choices of X o or the use of different calculators may lead to very different sequences; (c) the sequence x O'X 1'X 2 ' ••• may eventually cycle between two or more values, none of which is a root of the equation we are trying to solve; (d) the sequence XO,XI,X2' ••• may wander "forever" without ever settling into a regular pattern. Recent research in pure and applied mathematics has shown that the type of erratic behavior just described is the rule rather than the exception for many mathematical operations and the physical processes which they model (see Exercise 59 for a simple example). Indeed, "chaotic" behavior is observed in fluid flow, chemical reactions, and biological systems, and is responsible for the inherent unpredictability of the weather. Some references on this work on "chaos", aimed at the nonexpert reader, are: M. J. Feigenbaum, "Universal behavior in nonlinear systems," Los Alamos Science 1 (Summer 1980),4-27. D. R. Hofstadter, "Metamagical themas," Scientific American 245 (November 1981), 22-43. L. P. Kadanoff, "Roads to chaos," Physics Today 36 (December 1983),46-53. D. Ruelle, "Strange attractors," Math. Intelligencer 2 (1980), 126-137. D. G. Saari and J. B. Urenko, "Newton's method, circle maps, and chaotic motion," American Mathematical Monthly 91 (1984), 3-17; see also 92 (1985) 157-158.

Exerclses tor Sectlon 11.4 18. lim n 3 + 3n 2 + I " .... 00 n4 + 8n 2 + 2

1. If an = 1/10 + 1/100 + ... + I/Ion, how large must n be for ~ - an to be less than 1O- 6 ?

2 n --3n · 17 • I1m 2 " .... 00 n + 1

2. If an = 7/10 + 7/100 + ... + 7/Ion, how large must n be for ~ - an to be less than 1O- 8 ?

19. lim [ 3n 2 - 2n + I

Find the limits of the sequences in Exercises 3 and 4. 3. an = I + 1/2 + 1/4 + ... + 1/2n • 4. 0" = sin(mT /2). Write down the first six terms of the sequences in Exercises 5-10.

= nb,,_I/(1 + n); bo = t. 8. C,,+ I = - c,,/[2n(4n + I)]; CI = 2. 9. an + 1= [I/(n + 1)]L7=oaj; 00 = t·

12. lim

n~oo

E- N

(I - .!n ) = I

13. lim - 321 =0

14. lim _2_ =0 n+ ".... 00 2n + 5 Evaluate the limits in Exercises 15-24. " .... 00

15. lim ~ " .... 00

n+ I

.

(n 2

16. lim 8 2n I n---+oo n-

n(n + 2) + I)(n + 3)

]2

2 + I/n -

2)/(n 2 + I)

(sinn)2

21. lim - " .... 00 n+2 (-It· 2 23. !im 1

. (I + n)cos(n + I) 22. hm - - - ; : - - - ".... 00 n2 + 1

26. Yn/3

25. Yn/2

10. kn = {3n 2 + 2n; n = 1,2,3, .... Establish tbe limits in Exercises 11-14 using the definition.

1n =0

HOO

(n

" .... 00

7. bn

n~oo

20. tim

+ 1)

. cos('ITf,i) 24. hm ----=-n+ ".... 00 n2 Ii!! Using numerical calculations, guess tbe limit as n ~ 00 of tbe sequences in Exercises 25-28. Verify your answers using l'Höpital's rule.

5. k" = n 2 - 2f,i; n = 0,1,2, .... 6. an = (-lt+I[(n - I)/n!]; n = 0,1,2, ... (O! = I, n! = n(n - I) ... 3·2· 1.)

11. lim

n(n

" .... 00

28. yn(n + 3)/7 27. Yn(n + 1)/4 Find tbe limits in EKercises 29-34. 29. lim 1. 30. lim '17" n---'J>oo

31. lim

sn

n

" .... 00

33. lim [ " .... 00

+ (3/4t 32. n2 + 2 3b + (1/2)2" ]3 2

n - I

;

n-+oo

Find lim ('17

".... 00

b constant

2")2 . (2a+e-I 34. hm ; a constant n---+oo

n-

+ m")3

11.4 Limits of Sequences and Newton's Method 1ilI35. (a) Use Newton's method to find a solution of x 3 - 8x 2 + 2x + I = O. (b) Use division and the quadratic formula to find the other two roots.? 1ilI36. Use Newton's method to find all real roots of

x 3 -x+1ö. 1\137. Use Newton's method to locate a root of fex) = x 5 + x 2 - 3 with starting values Xo = 0, Xo= 2. 11138. Use Newton's method to locate a zero för fex) = x 4 - 2x 3 - 1. Use Xo = 2, 3, and - I as starting values and compare the results. 1ilI39. Use Newton's method to locate'a root of tanx = x in ["Ir /2,3"77"/2]. 11140. Use Newton's method to find the following numbers: (a) If; (b) 31f 1ilI41. The equation tan x = ax appears in heat conduction problems to determine values AI' A2' A3' ... that appear in the expression for the temperature distribution. The numbers A" A2' . .. are the positive solutions of tanx = ax, listed in increasing order. Find the numbers A" A2, A3 for a = 2, 3,5, by Newton's method. Display your answers in a table. 11142. (a) Use Newwn's method to solve the equation x 2 - 2 = 0 to 8 decimal places of accuracy, using the initial guess Xo = 2.

* (b) Find a constant C such that IXn -If I .;; CJXn-1 -If 12 for n = I, 2, 3, and 4. (See Review Exercise IOI for the theory of the rapid convergence of Newton's method.) 11143. Experiment with Newton's method for evaluation of the root 1/ e of the equation e - ex = 1/ e. 11144. Enter the display value 1.0000000 on your calculator and repeatedly press the "sin" key using the "radian mode". This process generates display. numbers al = 1.0000000, a2 = 0.84147, a3 = 0.74562, .... (a) Write a formula for an' using function notation. (b) Conjecture the value of limn-+ooan. Explain with a graph. 11145. Display the number 2 on your calculator. Repeatedly press the "x 2 " key. You should get the numbers 2, 4, 16, 256, 65536, .... Express the display value an after n repetitions by a formula. 1\146. Let fex) = I + I/x. Equipped with a calculator with a reciprocal function, complete the following: (a) Write out f(f(f(f(f(f(2)))))) as a division problem, and calculate the value. We abbreviate this as f(6) (2), meaning to display the value 2, press the "I/x" key and add I, successively six times. (b) Experiment to determine limn-+oo[l/j O. Prove that there is a positive integer N such that an > 0 for all n > N. 48. Let an = I if n is even and - I if n is odd. Does \imn-+ooan exist? 49. If a radioactive substance has a half-life of T, so that half of it decays after time T, write a sequence an showing the fraction remaining after time nT. What is limn-+ooan? 50. Evaluate:

+(-lr+I(.!.+~+J..)]; 3 n n2

(a) lim [(_I)n n-->oo 3

I . {(_I)n (b) hm - - +nIl+(-lt+ (3n+I)]} . 2 n-->oo 2 6n - 5n + 2

Find the limit or prove that the limit does not exist in Exercises 51-54. 51.

)im, ( ~ + ( -

l)n (

~-

*)).

52. lim ( (- Itsin(n"lr /2) )( n2 + I ). n-->oo. n n+1

. (3n

53. n-->oo hm -4--1 n+ 54. \im

(I

*56. *57.

*58.

(-I)n Sinn ) n+l'

+ n)cosn

. n (a) Give an A-N definition of what limn-->ooan = 00 means. (b) Prove, using your definition in part (a), that limn-->oo[(l + n 2 )/(1 + 8n)] = 00. If an~O and Ibni.;; janl, show that bn~O. Suppose that an' bn, and cn, n = 1,2,3, ... , are sequences of numbers such that for each n, we have an< bn < cn · (a) Ir Iim n-+ oo an = Land if limn-->oobn exists, show that limn-+oob.;;' L. [Hint: Suppose not!] (b) If limn-+ooan = L = lim n-+ 00 Cn, prove that limn-+oobn = L. A rubber baU is released from a height h. Each time it strikes the noor, it rebounds with twothirds of its previous velocity. (a) How far does the ball rise on each bounce? (Use the fact that the height y of the baU at time t from the beginning of each bounce is of the form y = vt - ~ gt 2 during the bounce. The constant g is the acceleration of gravity.) (b) How long does each bounce take? (c) Show that the ball stops bouncing after a finite time has passed. (d) How far has the ball travelled when it stops bouncing? (e) How would the resuIts differ if this experiment were done on the moon? n-+oo

*55.

+

For a computer, this method is preferable to using the formula for the TOotS of a cubic!

550

m*59.

Chapler 11 Limits, L'Hopital's Rule, and Numerical Methods (Research Problems) (a) Experiment with the recursion relation X n + I = aXn(l - x n) for various values of the parameter a where 0 < a .;; 4 and Xo is in [0, I]. How does the behavior of the sequences change when a varies?

(b) Study the bizarre behavior of Newton's method in Example 9 for various starting va lues xo. Can you see a pattern? Does X n always converge? (c) Study the bizarre behavior of Newton's method in Example 10.

11.5 Numerical Integration Integrals can be approximated by sequences wh ich can be computed numerically. The fundamental theorem of calculus does not solve all our integration problems. The antiderivative of a given integrand may not be easy or even possible to find. The integrand might be given, not by a formula, but by a table of values; for example, we can imagine being given power readings from an energy cell and asked to find the energy stored. In either case, it is necessary to use a method of numerical integration to find an approximate value for the integral. In using a numerical method, it is important to estimate errors so that the final answer can be said, with confidence, to be correct to so many significant figures. The possible errors include errors in the method, roundoff errors, and roundoff errors in arithmetic operations. The task of keeping careful track of possible errors is a complicated and fascinating one, of which we can give only some simple examples. 8 The simplest method of numerical integration is based upon the fact that the integral is a limit of Riemann sums (see Seetion 4.3). Suppose we are given j(x) on [a,b], and divide [a,b] into subintervals a = x o < XI < ... < Xn = b. Then ft;J(x)dx is approximated by 2:.7=d(c;)!lxi , where Ci lies in [Xi_I,X;]. Usually, the points Xi are taken to be equally spaced, so !lXi = (b - a)/n and Xi = a + i(b - a)/ n. Choosing Ci = Xi or xi+ I gives the method in the following box.

Riemann Sums To calculate an approximation to lbj(x)dx, let Xi

=

a + i(b - a)/n

and form the sum (la) or ( Ib)

8

For a further discussion of error analysis in numerical integration, see, for example, P. J. Davis,

Interpolation and Approximation, Wiley, New York (1963).

11.5 Numerical Integration

~ Example 1

Solution

551

Let f(x) = eosx. Evaluate J,(/eosxdx by the method of Riemann sums, taking 10 equally spaeed points: Xo = 0, XI = .,,/20, X2 = 2.,,/20, ... , x IO = 10.,,/20 = .,,/2, and Ci = Xi. Compare the answer with the aetual value. Formula (Ia) gives d ." ( ." 2." 9." ) J(,,/2 o eosx x~ 20 I + eos 20 + eos 20 + ... + eos 20

= ~ (I + 0.98769 + 0.95106 + ... + 0.15643) = ~ (6.85310) = 1.07648. Tbe aetual value is sin(." /2) - sin(O) = I, so our estimate is about 7.6% off. • Unfortunately, this method is ineffieient, because many points Xi are needed to get an aeeurate estimate of the integral. For this reason we will seek alternatives to the method of Riemann sums. To get a better method, we estimate the area in eaeh interval [Xi _ I' xJ more aeeurately by replaeing the reetangular approximation by a trapezoidal one. (See Fig. 11.5.1.) We join the points (Xi' f(x i» by straight line segments to obtain a set of approximating trapezoids. Tbe area of the trapezoid between Xi _ 1 and Xi is

Ai=Hf(xi-I)+ f(x i)].1xi sinee the area of a trapezoid is its average height times its width. y

J'

(a) Riemann sums method

(h) Trapezoid"1 mcthod

Figure 11.5.1. Comparing two methods of numerical integration.

The approximation to J~(x)dx given by the trapezoidal rule is + f(x;)).1x i· Tbis beeomes simpler if the points Xi are equally spaeed. Then .1xi = (b - a)/n, Xi = a + i(b - a)/n, and the sum is ~7=d-[J(Xi-l)

(b

~ a)

,± t [f(xi- I) + f(xi)]

.=1

whieh ean be rewritten as b

2n a [J(x o) + 2f(x l) + ... + 2f(x

n_

l) + f(x n )]

sinee every term oeeurs twiee exeept those from the endpoints. Although we used areas to obtain thisJormula, we may apply it even if j(x) takes negative values.

552

ehapler 11 Limits, L'Hopital's Rule, and Numerical Methods

Trapezoldal Rule To calculate an approximation to Lbj(X)dx, let Xi

=a+

i(b - a)/n

and form the sum (2)

Formula (2) turns out to be much more accurate than the method of Riemann sums, even though it is just the average of the Riemann sums (la) and (lb). Using results of Section 12.5, one can show that the error in the method (apart from other roundoff or cumulative errors) is < [(b - a)/12)M2(~X?, where M 2 is the maximum of 1f"(x)1 on [a,b). Of course, if we are given onIy numerical data, we have no way of estimating M 2 , but if a formula for j is given, M 2 can be determined. Note, however, that the error depends on (~xr, so if we divide [a, b) into k times as many divisions, the error goes down by a factor of 1/ k 2• The error in the Riemann sums method, on the other hand, is 0 on [2,4], so 11"(x)1 = 1"(x) « 1"(2) = 14e- 4 = M 2 , so the error is at most f'(x) = _2xe- x2 ,

and 1"(x) = _2e- x2 X2

x2

x2

•

X2

b- a M(A)2 1 14e-4A 2 /..lX ="6' (/..lx).2

~

To make this less than 10- 6, we should choose Ax so that

!. 14e-\Ax)2 < 10-6, (AX)2< e 4 1O- 6 • ~

= 0.0000234,

Ax < 0.0048. That is, we should take at least n = (b - a)j Ax = 416 divisions. For Simpson's rule, the error is at most [(b - a)jI80]MiAxt. Here f""(x) = 4(4x 4

-

12x 2 + 3)e- x2 •

On [2,4], we find that 4x 4 1f""(x)1

« 4(4· 44 -

-

12x 2 + 3 is increasing and e- x2 is decreasing, so

12.42 + 3)e- 4

=61.17=M4 · Thus [(b - a)jI80]MiAx)4 = fo . 61.17(Axt = 0.68(Axt Hence if we are to have error less than 10-6, it suffices to have

« 10- 6, Ax « 0.035.

0.68(Lh)4

Thus we should take at least n = (b - a)j Ax

= 57 divisions ....

Exercises tor Section 11.5 Use the indicated numerical method(s) to approximate the integrals in Exercises 1-4. ~ I. f~ 1(X 2 + I) dx. Use Riemann sums with n = 10 (that is, divide [- I, I] into 10 subintervals of equal length). Compare with the actual value. ~2. fo/ 2(x + sinx)dx. Use Riemann sums and the trapezoidal rule with n = 8. Compare these two approximate values with the actual value. ~3. ft[(sin'1Txj2)j(x 2 + 2x - l)]dx. Use the trapezoidal rule and Simpson's rule with n = 12.

g4. ~5.

f5(lj-R+T)dx. Use the trapezoidal rule and Simpson's rule with n = 20. Use Simpson's rule with n = 10 to find an approximate value for fÖ(xjrxr+2)dx.

1116. Estimate the value of fklx dx, using Simpson's rule with n = 4. Check your answer using x = u2, dx = 2udu. g7. Suppose you are given the following table of

data: 1(0) = 1.384 1(0.4) = 0.915 j(O.I) = 1.179 j(0.5) = 0.768 j(0.2) = 0.973 j(0.6) = 0.511 j(0.3) = 1.000 j(0.7) = 0.693

1(0.8) = 0.935 j(0.9) = 1.262 j(1.0) = 1.425

Numerically evaluate fA{x + j(x»dx by the trapezoidal rule. 18. Numerically evaluate fÖ2j(x)dx by Simpson's rule, where j(x) is the function in Exercise 7. g9. Suppose that you are given the following table of data:

j(O.O)

= 2.037 j(0.2) = 1.980 j(0.4) = 1.843 j(0.6) = 1.372 j(0.8) = 1.196 j(1.0) = 0.977 j( 1.2) = 0.685

j(1.3) = 0.819 j(1.4) = 1.026 j(1.5) = 0.799 j( 1.6) = 0.662 j( 1. 7) = 0.538 j(1.8) = 0.555

556

Chapter 11 Limits, L'Hopital's Rule, and Numerical Methods

~1O.

11.

12.

~ 13.

14.

~15.

Numerically evaluate J~8j(X) dx by using Simpson's role. [Hint: Watch out for the spacing of the points.] Numerically evaluate JA·s.!(x)dx by using the trapezoidal role, where j(x) is the function in Exercise 9. Evaluate J~(px2+qx+r)dx. Verify that Simpson's role with n = 2 gives the exact answer. What happens if you use the trapezoidal role? Discuss. Evaluate J~(px3 + qx 2 + rx + s)dx by Simpson's role with n = 2 and compare the result with the exact integral. How large must n be taken in the trapezoidal rule to guarantee an accuracy of 10- 5 in the evaluation of the integral in Exercise 2? Answer the same question for Simpson's role. Gaussian quadrature is an approximation method based on interpolation. The formula for integration on the interval [-I, I] is J~d(x)dx =j(l//3) + j( -1//3) + R, where the remainder R satisfies IR I ..;; M /135, M being the largest value of j{4>(X) on -I ..;; x ..;; 1. (a) The remainder R is zero for cubic polynomiais. Check it for x 3, x 3 - I, x 3 + X + 1. (b) Find J~ I[ x 2/ (I + x 4 )] dx to two places. (c) What is R for J~ IX 6 dx? Why is it so large? A tank 15 meters by 60 meters is filled to a depth of 3.2 meters above the bottom. The time T it takes to empty half the tank through an orifice 0.5 meters wide by 0.2 meters high placed 0.1 meters from the bottom is given by

(2.2)T

~

I

=

J.190

~19.6 130

*

P(x)

= yILI(x) + Y2L 2(x) + ... + YnLn(x),

dx

(x

+ 20)3/2 -

X 3/ 2 •

Compute T from Simpson's role with n = 6. * 16. A metropolitan sports and special events complex is circular in shape with an irregular roof that appears from a distance to be almost hemispherical (Fig. 11.5.3). x - axis

H _ __

x

ates the roof by revolution about the x axis (x and Y in feet, x vertical, y horizontal). (a) Find the square footage of the roof via a surface area formula. This number determines the amount of roofing material required. (b) To check against construction errors, a tape measure is tossed over the roof and the measurement recorded. Give a formula for this measurement using the are length formula. (c) Suppose the curve fis not given explicitly in the plans, but instead j(O), j(4), j(8), j(l2), ... ,j(H) are given (complex center-to-ceiling distances every 4 feet). Discuss how to use this information to numerically evaluate the integrals in (a), (b) above, using Example 3 as a guide. (d) Find an expression which approximates the surface area of the roof by assuming it is a conoid produced by a piecewise linear function constrocted from the numbers j(O), j(4), j(8), ... ,j(H). *17. How many digits in the approximate value A R:l2.3757 in Example 3 can be justified by an error analysis? 18. (Another numerical integration method) (a) Let (XI> YI), (X2' Y2), ... , (x n, Yn) be n points in the plane such that all the x;'s are different. Show that the polynomial of degree no more than n - I whose graph passes through the given points is

I--~"":"

where Lj(x) = Aj(x)j A'(xj),

= (x - XI)(X - X2) ... (x Aj(x) = A (x)j(x - x;), A(x)

i = 1,2, ... , n.

(P is called the Lagrange interpolation polynomial.) (b) Suppose that you are given the following data for an unknown functionj(x): j(O) = 0.01, j(O.I) = 0.12,

y

y - axis

Figure 11.5.3. The profile of the roof of asports complex. A summer storm severely damaged the roof, requiring a roof replacement to go out for bid. Responding contractors were supplied with plans of the complex from which to determine an estimate. Estimators had to find the roof profile Y = j(x), 0 ..;; x ..;; H, which gener-

x n ),

j(0.2)

j(O.3) = 1.18, j(O.4)

= 0.91.

= 0.82,

Estimate the value of j(0.16) by using the Lagrange interpolation formula. (c) Estimate Jg'~(x)dx (I) by using the trapezoidal rule, (2) by using Simpson's role, and (3) by integrating the Lagrange interpolation polynomial. (d) Estimate Jo/ 2cos x dx by using a Lagrange interpolation polynomial with n = 4. Compare your result with those obtained by the trapezoidal and Simpson's roles in Examples 2 and 4.

557

Review Exercises for Chapter 11

Review Exercises for Chapter 11 Verify the limits in Exercises 1-4 using the e-ll definition. 1. lim (x 2 + x - I) = I x--'>I 2. !im (x 3 + 3x + 2) = 2 x--'>O

3. !im(x 2 - 8x + 8) = -4 x--'>2 4. !im(x 2 - 25) = 0 x--,>5 Calculate the limits in Exercises 5-16. 5. lim tan( x + I ) x--,>O X - I 6.

!~ cos[ ( ~ :

7.

)~( ~ ~ ~)

8. li

(

9. !im (~ x~oo

11.

40. lim (sine-X)I/,'; x--'>oo 41. lim (I + sin 2X)I/x 42. !im (COS2X)I/x' x--'>o+ x~o+ (Inx)2 !im x 2 + x - 6 43. lun - 44. x--'>oo x x--+2 x 2 + 2x - 8

)

45. (00 1. dx.

-11 x) 10. 12.

Hm (~4X2 + I - 2x)

x~oo

lim x--'>2+

sin~ Jx - 2

13. lim [x sin(3/ x 2)]

14. limsin(R+2 -x)

15. lim x 4 + 8x x--'>o 3x 4 + 2

16. lim x + 3 x--'>o 3x + 8

x--'>O

x--'>o

17. Find the horizontal asymptotes of the graph y = tan -1(3x + 2). Sketch. 18. Find the vertical asymptotes for the graph of y = 1/(x 2 - 3x - 1Of, Sketch. Find the horizontal and vertical asymptotes of the functions in Exercises 19 and 20, and sketch.

=~ x 2+ I

Find the limits if they exist, Exercises 21-44. 3 21. tim x + 8x + 9 x--'>oo 4x 3 - 9x 2 + 10 23. Hm 1 - cosx x--,>O 3x - 2x (R+9 -3) 25. lim sinx x--'>o 27. 29. 30. 31. 32. 33.

lim x 3(lnxi

x--'>o+

Decide which improper integrals in Exercises 45-54 are convergent. Evaluate when possible.

!im _1_ x--'>I- JI - x

19. fex)

38.

x--'>O+

?T ]

x2 + 2

.!.) (_1___ .!.)

lim (_._1_ smx x 36. !im 1 __ x--'>I Inx x- I 2 x--'>o+

37. lim x 2e- x x--'>oo 39. lim x sinx

i )32

x--,>~ 3x 2 + 2x + I

35.

20. fex)

= 2x + 3

3x + 5

using l'Hdpital's rule, in 22. lim _x_ X--'>oo X + 2 XX - -1 24. ru n x--'>I X - I

Vx 3 + 27 - 3 x x--'>O 2x Hm sin5x !im tan 28. X x--,>O x--,>O x 2 . sin(x - 2) - x + 2 hm - - - - - : - - x--'>2 (x - 2)3 lim 24cosx - 24 + 12x 2 - x 4 x--,>O x5 tan(x + 3) - tan3 !im - - - - - x--'>O x lim cos,fX + I X--,>,,2 X - ?T 2 34. lim cotx !im x cotx x--+o x--'>o+ Inx 26. lim

J1

46. Joo

x2

-

sinx dx.

00 x 2 + 3

47. (00 Idx . [Hint: Prove Inx " x for x;;. 2.] J2 nx 48. (00

JI

dx

~X2 + 8x + 12

49. (2 _1_ dx.

50.

JI ,;x=T

51.

Lo (I - dxx) I

JO

x+I

-I~

dx.

2/5 .

54. fooo(x + 2)e-(x 2+4X) dx.

53. folxlnxdx.

Evaluate the limits in Exercises 55 and 56. 55. lim (x x--,>ooJo

dt

t2 + t

+I

56.

lim (I dt

x--'>o+ Jx

(t

57. The region under the curve y = xe- x on [0,00) is revolved about the x axis. Find the volume of the resulting solid. 58. The curve y = sin x / x 2 on [I, 00) is revolved around the x axis. Determine whether the resulting surface has finite area. Evaluate the limits of the sequences in Exercises 59-72. 59.

)~ ( 8 + ( i

60. !im n-->oo 61. lim

n---+oo

3

rr

Sn 3 - 2n + 1 n3 + I

(1 + ~)n n

62. lim ( n - 3 n--ioOQ

n

),-2n I

63. lim (n 2 + 3n + l)e- n n--'>oo 64. !im 2n 65. lim _n_ n--'>oo n2 n--'>oo n + 2 66. lim n2 + 2n n-->oo 3n 2 + I 68. !im [ln(n 2 + n--'>oo

67. !im tan[ ~ ] n--+oo n+8 2 1) - ln(3n + 5)]

558

ehapler 11 Limits, L'Hopital's Rule, and Numerical Methods , sin( 71n /2) 69. 11m ------::-11-->00 n-3

70. lim n cos 4nw 11-->00 2n + I

11. lim 8 - 2n

72.

5n

11-->00

lim

11-->00

(I _3n2 ++nI )

Using l'Hdpital's rule if necessary, evaluate the limits of the sequenees in Exereises 73-76. 73. lim "".J3n 74. lim mloglo(2- m ) n-+oo

75. lim

11-->00

m-tooo

[

1..2 n

I

(n + 1)2

]

3'

76. lim ~ j-->oo e3j

+l

Use Newton's method for Exereises 77-80. Locate the roots of x 3 - 3x 2 + 8 = O. Find the eube root of 21. Solve the equation e X = 2 + x. Find two numbers, eaeh of whose square is ten times its naturallogarithm. 181. Evaluate ß(x 2 dx/rxr+T) by the trapezoidal rule with n = 10. 1182. Evaluate the integral in Exercise 81 by Simpson's rule with n = 10. 183. Use Simpson's rule with n = 10 to ealeulate the volume obtained by revolving the eurve y = fex) on [1,3] about the x axis, given the data:

177. 178. 179. 180.

f(l) = f(1.2) = f(1.4) = f( 1.6) = f(1.8) = f(2) =

2.03 2.08 2.16 2.34 2.82 3.01

f(2.2) f(2.4) f(2.6) f(2.8) f(3)

88. Show that fg"[(sinx)/(I + x)]dx is eonver. gent. [Hint: Integrate by parts.] 89. (a) Show that

+ f(xo - h) f "( Xo ) = I'1m f(xo + h) - 2f(xo) 2 h-->O h if f" is continuous at xo. [Hint: Use l'Hdpi. tal's rule.] *(b) Find a similar formula for f"'(xo). 90. Show that

f "(Xo ) -- lim f(xo Ax~o

+ 2Ax) + f(xo)2 -

2f(xo + Ax)

(Ax)

if f" is continuous at xo. 91. Use Riemann sums to evaluate 11

lim ~ (Inn -Ini)/n.

n-+oo i-I

92. Let

Prove that lim"-+ooS,, = 4 using Riemann sums. 93. Let

= 3.16 =

3.01

= 2.87

= 2.i5 = 1.96

184. (a) Evaluate (2/ fi>JAe- il dt by using Simpson's rule with 10 subdivisions. (b) Given an upper bound for the error in part (a). (See Example 6 of Section 11.5.) (c) What does Simpson's rule with 10 subdivisions give for (2/fi)!AOe- t2 dt? (d) The funetion (2/fi)!'Oe- il dt is denoted erf( x) and is called the error funetion. Its values are tabulated. (For example: HaTJdbook of Mathematicol Functions, National Bureau of Standards, Applied Mathematics Series 55, June 1964, pp. 31O-311.) Compare your results with the tabulated results. Note: limx-+ooerf(x) = I, and erf( 10) is so close to I that it probably won't be listed in the tables. Explain your result in part (e). 85. Let fex) = COS,X for x ;> 0 and fex) = I for x < O. Decide whether or not f is continuous or differentiable or both. 86. Let fex) = xl/sin(x-I). How should f(l) be defined in order to make f continuous? 87. Find a function on [0, I] which is integrable (as an improper integral) but whose square is not.

Prove that SII ~ i as n ~ 00 by using Riemann sums. 94. Expressing the following sums as Riemann sums, show that: 11 (a) lim ~

11-->00 i-I

[{f (' )3/2] i-i n

n

-I = -4 ; n

15

3n 2 - -21 . (2n + i) 95. P dollars is deposited in an account eaeh day for a year. The account earns interest at an annual rate r (e.g., r = 0.05 means 5%) eompounded continuously. Use Riemann sums to show that theamount in the account at the end of the year is approximately (365P / r) (er - I). *96. Evaluate: (b) lim:± 11-->00

I_I

I·Im x-->,,2

IX + tanvx . [Sin r:: 1 (IX -71)( IX + 71)

*97. Limits can sometimes be evaluated by geometrie techniques. An important instance oceurs when the curve y = fex) is trapped between the two intersecting lines through (a, L) with slopes m and - m, 0 < Ix - 0\ 0, m > O. The resulting algebraie eondition is ealled a Hölder condition:

a" m,

If(x) - LI

0
o[Az / Ax) = f'(yo)g'(xo). (This proof avoids the problem of division by zero mentioned on p. 113.)

=

0 is the

f(x n ) x n + I = Xn - f'(xo)'

This is ealled a Lipschitz condition. (e) Argue that a Lipsehitz eondition implies limx->J(x) = L, by appeal to the definition of limit. *98. Another geometrie teehnique for evaluation of limits is obtained by requiring that y = fex) be trapped on 0 " Ix - al ..; h between two power eurves

y

559

*

sometimes known as Picard's method. Notiee that this method requires evaluating f' only at the initial guess Xo and so requires less eomputation at eaeh step. (a) Show that, if the sequenee xo, XI' x2, ... eonverges, then limn-4ooxn is a solution of fex) = o. Iil (b) Compare Pieard's method and Newton's method on the problem x 5 = x + I, using the initial guess Xo = I in eaeh ease and iterating until the solution is found to six deeimal plaees of aecuraey. (e) Suppose that f(q) = 0 and in addition that 0 0.] (d) How many iterations are needed to solve x 2 - 2 = 0 to within 20 decimal plaees, assuming an initial guess in the interval [1.4. 1.5)1

Chapter12

Infinite Series

Infinite sums can be used to represent numbers and functions.

The decimal expansion 1 = 0.3333 . .. is a representation of 1 as an infinite sum fö + lÖo + l~ + lO.k + .... In this chapter, we wql see how to represent numbers as infinite sums and to represent functions of x by infinite sums whose terms are monomials in x. For example, we will see that

In2

= I _1 + 1_1 + 2

3

4

and .

smx

=

x3 x - ~

x5

+ 1.2.3.4.5

Later in the chapter we shall use our knowledge of infinite series to study complex numbers and some differential equations. There are other important uses of series that are encountered in later cour~es. One of these is the topic of Fourier series; this enables one, for example, to decompose a complex sound into an infinite series of pure tones.

12.1 The Sum of an Infinite Se ries The sum of infinitely ma'9' numbers may be finite.

An infinite series is a sequence of numbers whose terms are to be added up. If the resulting sum is finite, the series is said to be convergent. In this section, we define convergence in terms of limits, give the simplest examples, and present some basic tests. Along the way we discuss some further properties of the limits of sequences, but the reader should also review the basic facts about sequences from Section 11.4. Our first example of the limit of a sequence was an expression for the number 1: 1 . (3 -3 3=J.!.~ 10+ 100

3 )

+ ... + IOn·

This expression suggests that we may consider

1 as the sum

562

Chapter 12 Infinite Series

3 3 3 10 + 100 + ... + 10" + ... of infinitely many terms. Of course, not every sum of infinitely many terms gives rise to a number (consider 1 + 1 + 1 + ... ), so we must be precise about what we mean by adding together infinitely many numbers. Following the idea used in the theory of improper integrals (in Section 11.3), we will define the sum of an infinite series by taking finite sums and then passing to the limit as the sum includes more and more terms.

Convergence of Se ries Let a"a 2 ,

be a sequence of numbers. The number Sn =

•••

L~=,ai

= a, + a2 + ... + an is called the nth partial sum of the a;'s. If the

sequence S" S2' ... of partial sums approaches a limit S as n -:; 00, we say that the series a, + a2 + ... = L:"'= ,ai converges, and we write 00

,

'" ..c.. a·= S· ;~

I

,

that is, n

00

~ a;

is defined as

;=,

lim ~ a;

n-4oo ;=1

and is called the sum of the series. If the series L:"'-,a; does not converge, we say that it diverges. In this case, the series has no sumo In summary: n

a senes

converges if

lim ~ ai exists (and is finite);

n~oo i=

1

n

a senes

Example 1

n-4oo i = I

Write down the first four partial sums for each of the following series:

(b) 1 -

1 '21 + '41 + '81 + 16 + ... ,. 1. + 1. - 1. + 1. - 1. + ....

(c) 1 +

1. + ~ + ~ + ~ + ....

(a) 1 +

00

2

5

345

3

(d) ~ i+t. ;-0 2

Solution

lim ~ ai does not exist (or is infinite).

diverges if

52

53

6

54

'

'

12.1 The Sum of an Infinite Series

(d)

~

_3_ = J. + 1. + 1. + 1. + . .. so SI 2;+' 2 22 23 24 ' 21 1 1 1 1 1 S3 = 3("2 + "4 + 8" ) = ""8' and S4 = 3("2 + "4 ;=0

= J.

2' 1

S2

563

= J. + J. = 2.

1

2

4

4'

45

+ 8" + 16 ) = 16' ...

00 not confuse a sequence with aseries. A sequence is simply an infinite list of numbers (separated by commas): a.,a 2 ,a3, .... Aseries is an infinite list of numbers (separated by plus signs) which are meant to be added together: a. + a2 + a3 + .... Of course, the terms in an infinite series may themselves be considered as a sequence, but the most important sequence associated with the series a. + a2 + ... is its sequence of partial sums: S"S2,S3"" -that is, the sequence

o

SI S3

S2 S4

Figure 12.1.1. The term aj of aseries represents the "move" from the partial sum Sj_1 to Sj. Sn is the cumulative result of the first n moves.

a. ,al + a2 ,a. + a2 + a3 , · · · • We may illustrate the difference between the a;'s and the Sn's pictorially. Think of a.,a 2 ,a3' ... as describing a sequence of "moves" on the real number line, starting at O. Then S" = a. + ... + a" is the position reached after the nth move. (See Fig. 12.1.1.) Note that the term a; can be recovered as the difference Si - Si _ •. To study the limits of partial sums, we will need to use some general properties of limits of sequences. The definition of convergence of a sequence was given in Section 11.4. The basic properties we need are proved and used in a manner similar to those for limits of functions (Section 11.1) and are summarized in the following display.

Properties 01 Limits 01 Sequences Suppose that the sequences a"a2' ... and b l ,b2 , and that c is a constant. Then: 1. 2. 3. 4.

•••

are convergent,

lim" .... oo(a" + b,,) = lim"->ooa,, + lim" .... oob". lim,,->oo( ca,,) = c lim".... ooa". lim" .... oo(anb,,) = (lim,,->ooa,,) . (limn->oob,,). If lim".... oob" =1= 0 and b" =1= 0 for all n, then

. -a" Ilßl b"

"->00

lim" .... a" = ".,----::00

lim" .... oob,,·

5. If fis continuous at lim" .... ooa", then

li~f(a,,) = fCli~a,,). 6. 7. 8. 9.

lim" .... ooc = c. limn->oo(l/n) = O. If limx->oof(x) = /, then lim" .... oof(n) = /. If Irl < I, then limn->oor" = 0, and if Irl > 1 or r = -I, lim".... oor" does not exist.

Here are a couple of examples of how the limit properties are used. We will see many more examples as we work with series. Example2

Solution

Find (a) lim 3 + n and (b) lim sin( 2 'Im 1 ). "->00 2n + 1 n-+oo n+ (a)

. 3+n . 3/n + 1 nl~~ 2n + 1 = J~ 2 + l/n =

3Iim" .... ool/n + lim".... ool _ 3·0 + 1 1 lim,,->oo2 + lim,,->ool/n - 2 + 0 ="2'

564

ehapter 12 Infinite Series

This solution used properties I, 2, and 4 above, together with the facts that limn-+ool/n = (property 7) and limn-+ooc = c (property 6). (b) Since sinx is a continuous funetion, we can use property 5 to get

°

nli~ sin( 2n'IT: I ) = sin[ Ji~ ( 2n'IT: I )] =

sin[ Ji~( 2 +'ITI / n )]

= sin( I) = 1. • We return now to infinite series. A simple but basic example is the geometrie series

a+ar+ar 2 + ... in whieh the ratio between each two suecessive terms is the same. To write a geometrie series in summation notation, it is convenient to allow the index j to start at zero, so that ao = a, a l = ar, a2 = ar 2, and so on. The general term is then ai = ari, and the series is compactly expressed as ~f=oarl. In our notation ~f=lai for a general series, the index i will start at 1, but in special examples we may start it wherever we wish. Also, we may replace the index j by any other letter; ~f-Iai = ~j_laj = ~~_Ian' and so forth. To find the sum of a geometrie series, we must first evaluate the partial sums Sn = ~7_oari. We write

= a + ar + ar 2 + ... + ar n, ar + ar2 + ... + ar n + ar n+ l. rSn = Sn

Subtracting the second equation from the first and solving for Sn' we find

Sn

a(1 - r n + l )

= --=--:-1--~ -r

(I)

The sum of the entire series is the limit

a(1 - r n + l ) = _a_ lim (1 - r n + I) n--+oo 1- r 1 - r n--+oo n I = _a_ (1 - n--+oo lim r + ). 1- r

= lim

°

(We used limit properties 1, 2, and 6.) If Irl < 1, then limn-+oor n+ 1 = (property 9), so in this case, ~f-Iari is convergent, and its sum is a/(1 - r). If Irl > 1 or r = -1, limn-+oor n+ 1 does not exist (property 9), so if a 0, the series diverges. Finally, if r = 1, then Sn = a + ar + ... + ar n = a(n + 1), so if a 0, the series diverges.

*"

*"

Geometrie Series If Irl < 1 and a is any number, then a + ar + ar2 + converges and the sum is a/(l - r). If

Irl ; ;.

1 and a

*" 0, then ~f-oari diverges.

... = ~f..oari

12.1 The Sum of an Infinite Series

Example 3 Solution

1

565

1

tr + ... , (b) ~ n/2' and (e) ~ i · n=O 6 i=1 5 (a) This is a geometrie series with , = t and a = 1. (Note that a is the first Surn the series: (a) 1 +

t+~+i

00

+

00

term and , is the ratio of any term to the preceding one.) Thus 1+

1+1+ 3

9

...

= ~ ( 1 )i = i=O 3

I

1 - 1/3

= 1.

2·

(b) L:'..o[l/(6 n/2 )] = 1 + (1/';6)+ (l/';6)2 + ... =a/(1-,), where a = land, = 1//6, so the sum is 1/(1 - 1//6) = (6 + /6)/5. (Note that the index here is n instead of i.) (e) L~=ll/5i = 1/5 + I/5 2 + ... = (1/5)/{1 - 1/5) = 1/4. (We mayaiso think of this as the series L~=ol/5i with the first term removed. The sum is thus 1/[1 - (1/5)] - 1 = 1/4.) .& The following example shows how a geometrie series may arise in a· physical problem.

Example 4

A bouneing ball loses half of its energy on eaeh bounee. The height reached on each bounce is proportional to the energy. Suppose that the ball is dropped vertieally from a height of one meter. How far does it travel? (Fig. 12.1.2.)

Figure 12.1.2. Find the total distance travelled by the bouncing ball.

Solution

Eaeh bounce is 1/2 as high as the previous one. After the ball falls from a height of 1 meter, it rises to 1/2 meter on the first bounce, (1/2)2 = 1/4 meter on the second, and so forth. The total distance travelled, in meters, is 1 + 2(1/2) + 2(1/2)2 + 2(1/2)3 + ... , whieh is 1+2

Jl (1/2)i = 1 + 2 00

(

1/2 ) 1 _ 1/2

= 3 meters. A

Two useful general rules for summing series are presented in the box on the following page. To prove the validity of these rules, one simply notes that the identities n

n

n

~(ai+bi)= ~ai+ ~bi and

i-I

i-I

i-I

n

n

~ cai= c~ai

i-I

i-I

are satisfied by the partial sums. Taking limits as n ~ 00 and applying the sum and eonstant multiple rules for limits of sequenees results in the rules in the box.

566

Chapter 12 Infinite Series

Aigebraic Rules For Serles If

Lf'= lai and Lf'= lbi converge, 00

~ (ai

i=l

00

00

i=l

i=l

Sum rule then Lf'..l(ai + b;) converges and

+ b;) = ~ ai + ~ bi •

Constant multiple rule If Lf'= lai converges and c is any real number, then and 00

00

i=l

i=l

Lf'= lcai converges

~ cai = c ~ ai •

Example 5 Solution

3i _ t . i=O 6' 00

Sum the series ~

We may write the ith term as

Since the series Lf'=o(l/2)i and Lf'=o(l/3Y are convergent, with sums 2 and ~ respectively, the algebraic rules imply that

.~ 3i ~ 2i = .~ [( l)i + (-1)( 1 )i]

,=0

6

,=0

2

1

3

i

= i~O ( 2" ) + ( 00

Example 6 Solution

1

)i

3

1

i~O"3 = 2 - 2" = 2" . • 00

1)

(

Show that the series q + 3* + 7i + 15* + ... diverges. [Hint: Write it as the difference of a divergent and a convergent series.] The series is Lf'..J2i - (!YJ. If it were convergent, we could add to it the convergent series Lf'..o(1-Y, and the result would have to converge by the sum rule; but the resulting series is Lf'.,0[2i - (1-Y + (!)i] = Lf'..02i, which diverges because 2 > 1, so the original series must itself be divergent. • The sum rule implies that we may change (or remove-that is, change to zero) finitely many terms of aseries without affecting its convergence. In fact, changing finitely many terms of the series Lf'= lai is equivalent to adding to it aseries whose terms are an zero beyond a certain point. Such a finite series is always convergent, so adding it to the convergent series produces a convergent result. Of course, the sum of the new series is not the same as that of the old one, but rather is the sum of the finite number of added terms plus the sum of the original series.

Example 7

Show that

1+2+3+4+1+~+~+~+ ... 4

42

43

44

is convergent and find its sumo Solution

The series 1/4 + 1/42 + 1/43 + 1/44 + . .. is a geometrie series with sum (1/4)/(1 - 1/4) = 1/3; thus the given series is convergent with sum 1+ 2 +

12.1 The Sum of an Infinite Series

3 + 4 + 1/3 = 10

567

t. To use the sum rule as stated, one can write

1+2+3+4+!+.1+.1+ ... 4 42 43 = (I + 2 + 3 + 4 + 0 + 0 + 0 + ... ) + (0 + 0 + 0 + 0 + ! + .12 + .13 + ... ) .• 4 4 4 We can obtain a simple necessary condition for convergence by recalling that a; = S; - S;-I' If limHOOS; exists, then limHooS;_1 has the same value. Hence, using properties land 2 of limits of sequences, we find limHooa; = limHooS; - limHOOS;_1 = O. In other words, if the series L~ la; converges, then, the "move" from one partial sum to the next must approach zero (see Fig. 12.1.1).

The ith Term Test If L~ la; converges, then limHOOa; = O. If limHOOa; =f= 0, then Lr'= la; diverges. If limHOOa; = 0, the test is inconclusive: the series could converge or diverge, and further analysis is necessary.

The ith-term test can be used to show that aseries diverges, such as the one in Example 6, but it cannot be used to establish convergence.

Example 8

Solution

Test for convergence: (a) (a) Here a;

00 . 00(1) L00. - I '.; (b) L (-IY-'-. ; (c) L -;- .

;=1

+,

= I ~ i = 1//+ I ~ I

;=1

..fl+i

;=1'

as i ~ 00. Since a; does not tend to zero,

the series must diverge. (b) Here la;1 = i/..ff+i = {i / i + I ~ 00 as i ~ 00. Thus aj does not tend to zero, so the series diverges. (c) Here aj = 1/ i, which tends to zero as i ~ 00, so our test is inconclusive. •

b/

As an example of the "further analysis" necessary when limHOOaj consider the series

I+!+!+!+ .. 2

3

4

= 0,

we

·=f~ ;= 1 '

from part (c) of Example 8, called the harmonie series. We show that the series diverges by noticing a pattern: 1

2

t+t

>t+t=t >i+i+i+i=t ~+ '" +16>16+'" +16=t n+'" +-i2 >-i2+'" +11=t t+i+~+i

and so on. Thus the partial sum S4 is greater than I + t + t = I +~, S8 > I + t + t + t = I + t and, in general S2n > 1+ n/2, which becomes arbitrarily large as n becomes large. Therefore, the harmonie series diverges.

568

Chapter 12 Infinite Se ries

Example 9 Solution

Show that the series (a)

!

+;\ + i + i + ... and (b)

L~=ll/(l

+ i) diverge.

(a) This series is L~= 1(1 /2i). If it converged, so would twice the series Lf'_12· (1/2i), by the constant multiple rule; but Lf'=1(2· 1/2i) = Lf'_11/i, which we have shown to diverge. (b) This series is ! + t + ;\ + ... , which is the harmonie series with the first term missing; therefore this series diverges too. A

Supplement 10 Sectlon 12.1: Zeno's Paradox Zeno's paradox concerns a race between Achilles and a tortoise. The tortoise begins with a head start of 10 meters, and Achilles ought to overtake it. After a certain elapsed time from the start, Achilles reaches the point A where the tortoise started, but the tortoise has moved ahead to point B (Fig. 12.1.3).

Figure 12.1.3. Will the runner overtake the tortoise?

After a certain further interval of time, Achilles reaches point B, but the tortoise has moved ahead to a point C, and so on forever. Zeno conc1udes from this argument that Achilles can never pass the tortoise. Where is the fallacy? The resolution of the paradox is that although the number of time intervals being considered is infinite, the sum of their lengths is finite, so Achilles can overtake the tortoise in a finite time. The word forever in the sense of infinitely many terms is confused with "forever" in the sense of the time in the problem, resulting in the apparent paradox.

Exerclses tor Sectlon 12.1 Write down the first four partial sums for the series in Exercises 1-4. 1. !+!+*+~+ ...

! + ! - ! + -k - ... ~ (~);

2. I 3.

;=1 3 00

4.

L

~

n=1 3n

Sum the series in Exercises 5-8.

5. I + 1. 7

+ -.L + -.L + ...

6. 2 + ~ 9

+ 2 + 2 + ...

L

7..

00 (

,=1

72

7

"8

92

)i

73

93

8.

00

(13)ft

ft~1 15

12.1 The Sum of an Infinite Series 9. You wish to draw $10,000 out of a Swiss bank account at age 65, and thereafter you want to draw i as much each year as the preceding one. Assuming that the account earns no interest, how much money must you start with to be prepared for an arbitrarily large life span? 10. A decaying radioactive source emits 1'0 as much radiation each year as the previous one. Assuming that 2000 roentgens are given off in the first year, what is the total emission over all time? Sum the series (if they converge) in Exercises 11-20. 11.

~

j=l

13. 15.

00

_1 13 j 23i+4

12.

00

L-

14.

j~3( i Y

16.

;=0 32i + 5

r +3 L n= I 6n 00

17.

00

19.

L n=5

(±)k

5 44/ + 2

I~O 53/ + ~ 5( l3

i=4 00

n

18.

2n + 1

3n -

~

k=l

r

80

l

/

2

3 + 1 L k=l 27k 2k

569

nth term an can be expressed as an = b n + I - b n for some sequence bn • (a) Verify that al + a2 + a3 + ... + an = bn + I-bi; therefore the series converges exactly when limn--+oobn + I exists, and 2:;;'= lan = limn-->oobn+ I-bi· (b) Use partial fraction methods to write an = I/[n(n + I)] as bn + 1 - bn for some sequence bn • Then evaluate the sum of the series 2:;;'= II/[n(n + 1)]. 36. An experiment is performed, during which time successive excursions of a deflected plate are recorded. Initially, the plate has amplitude bo. The plate then deflects downward to form a "dish" of depth b l , then a "dome" of height b2 , and so on. (See Fig. 12.1.4.) The a's and b's are related by al = b o - b l ,a2 = b l - b 2 ,a3 = b 2 b3 • • • • The value an measures the amplitude "lost" at the nth oscillation (due to friction, say).

~ __ V

t?_!~l

2

~~~

~

21. Show that 2: 1= 1(1 + 1/2;) diverges. 22. Show that 2:1=0(3; + 1/3 i) diverges. 23. Sum 2 + 4 + ! + ~ + i + .... 24. Sum 1+ 1/2 + 1/3 + 1/32 + 1/3 3 + .... Test the series in Exercises 25-30 for convergence. 25.

L 00

;= I

26.

00

28.

..;r+T

~ II + I ;= 1

27.

•

_1_

II + 8 3

L -5 5· ;=1 + / 6

L --. 00

;=17 + 7/

2~ I+t+~+~+i+i+i+i+~+ '-----'

.-,----'

2

4

30. I + ~ + ~ + ~ + ~ + ~ + ~ + '-.-'

2

~

4

~

t. + i;- +

...

'----,------'

8

31. Show that the series 2:)= 1(1 - 2 - j) / j diverges. 32. Show that the series t + ! + ~ + ! + . .. diverges. 33. Give an example to show that 2:1= I(ai + b;) may converge while both 2:r:.la; and 2:1= Ib; diverge. 34. Comment on the formula I + 2 + 4 + 8 + ... =1/(1-2)=-1. 35. A lelescoping series, like a geometric series, can be summed. Aseries 2:;;'= lan is telescoping if its

Figure 12.1.4. The deflecting plate in Exercise 36.

(a) Find 2:;;'= lan · Explain why b o - 2:;;'= lan is the "average height" of the oscillating plate after a large number of oscillations. (b) Suppose the "dishes" and "domes" decay to zero, that is, limn--+oobn + 1 = O. Show that 2:;;'= lan = bo, and explain why this is physically obvious. 37. The joining of the transcontinental railroads occurred as folIows. The East and West crews were setting track 12 miles apart, the East crew working at 5 miles per hour, the West crew working at 7 miles per hour. The official with the Golden Spike travelled feverishly by carriage back and forth between the crews until the rails joined. His speed was 20 miles per hour, and he started from the East. (a) Assurne the carriage reversed direction with no waiting time at each encounter with an East or West crew. Let Ik be the carriage transit time for trip k. Verify that 12n + 2 = r n + l . (12/13), and 12n+1 = r n . (12/27), where r = (13/27)· (15/25), n = 0, 1,2, 3, ....

(b) Since the crews met in one hour, the total time for the carriage travel was one hour, i.e., limn-->oo(tl + 12 + 13 + 14 + ... + In) = 1. Verify this formula using a geometric series.

570

Chapler 12 Infinite Series

12.2 The Comparison Test and Alternating Serles Aseries with positive terms converges if its terms approach zero quickly enough. Most series, unlike the geometrie series, cannot be summed explicitly. If we can prove that a given series converges, we can approximate its sum to any desired accuracy by adding up enough terms. One way to tell whether aseries converges or diverges is to compare it with aseries which we already know to converge or diverge. As a fringe benefit of such a "comparison test," we sometimes get an estimate of the difference between the nth partial sum and the exact sumo Thus if we want to find the sum with a given accuracy, we know how many terms to take. The comparison test for series is similar to that for integrals (Section 11.3). The test is simplest to understand for series with non-negative terms. Suppose that we are given series ~f=laj and ~f=lbj such that 0 ~ aj O and ao=l; an+I=Han+B/an). Show that an ~[Ji. *64. Let an + I = 3 - (I/an); ao = 1. Prove that the sequenee is inereasing and bounded above. What is lim n--> 00 an ? *65. Let an+1 =!an +IQ;,; ao= I. Prove that an is inereasing and bounded above. What is lim n--> 00 an ? *66. Let an+I = ! (I + an)' and ao = 1. Show that limn-->ooan = I. *67. Give an alternative proof that limn-->oor n = 0 if o < r < I as folIows. Show that r n deereases and is bounded below by zero. If the limit is I, show that rI = 1 and eonclude that 1 = O. Why does the limit exist? *68. Suppose that ao = I, an+1 = I + 1/(1 + an). Show that an eonverges and find the limit. *69. The eelebrated example due to Karl Weierstrass of a nowhere differentiable continuous function fex) in - 00 < x < 00 is given by

a 12 = 2.30278 ...

The sequenee appears to be eonverging to a number 1R; 2.30278 ... , but the numerieal evidenee only suggests that the sequenee eonverges. The inereasing sequenee test enables us to prove this. 57. Let the sequenee an be defined induetively by the mies ao = 0, an = ';4 + an_I. (a) Write out al, a2, and a3 in terms of square roots. llil (b) Calculate al through a\2 and guess the value of limn-->ooan to four signifieant figures. *58. (a) Prove by induetion on n that for the sequenee in Exercise 57, we have an > an-I and an < 5. (b) Conclude that the limit 1 = limn-->o,,an exists. (e) Show that 1 must satisfy the equation

*59. an

*61. an

where (x + 2) = (x), and (x) on 0.;; x .;; 2 is the "triangle" through (0, 0), (1, I), (2, 0). By eonstmetion, 0 .;; (4nx) .;; 1. Verify by means of the eomparison test that the series eonverges for any value of x. [See Counterexamples in Analysis by B. R. Gelbaum and J. M. H. Olmsted, HoldenDay, San Franeiseo (1964), p. 38 for the proof that f is nowhere differentiable.] * 70. Prove tha t an = (I + 1/ n)n is inereasing and bounded above as folIows: (a) If 0 .;; a .;; b, prove that n+1 b n+1 b=: «n+l)b n. That is, prove bn[(n + I)a - nb] < a n+l • (b) Let a = 1 + [l/(n + 1)] and b = 1 + (l/n) and deduee that an is inereasing. (e) Let a = 1 and b = 1 + (I/2n) and deduee that (l + 1/2n)2n < 4. (d) Use parts (b) and (e) to show that an < 4. Conclude that an eonverges to some number (the number is e-see Seetion 6.3).

12.3 The Integral and Ratio Tests

579

12.3 The Integral and Ratio Tests The integral test establishes a connection between infinite series and improper integrals.

The sum of any infinite series may be thought of as an improper integral. Namely, given a series L~ la;, we define a step function g(x) on [1,00) by the formulas: g(x) = a 2

x< 2) (2, x< 3)

g(x) = a;

(i,x 0 for all x in [1,00); 2. j( x) is decreasing on [I, 00).

For example, if a; = I/i, the harmonie series, we may take j(x) = I/x. We may now compare j(x) with the step function g(x). When x satisfies i , x < i + I, we have

0, j(x) , j(i)

= a; = g(x).

Hence 0, j(x) , g(x). (See Fig. 12.3.1.) J'

Figure 12.3.1. The area under the graph of fis less than the shaded area, so 0.;; JJ+~(x)dx.;; L7=IQj'

3

4

6

x

It follows that, for any n,

0,

J.n

+I

I

j(x)dx,

J.n + I g(x)dx= ~a;. n I

;=1

(I)

We conclude that if the series L~lai converges, then the integrals f7+Y(x)dx are bounded above by the sum Li': 1 aj, so that the improper integral ffJ(x)dx converges (see Section 11.3). In other words, if the integral ffj(x)dx diverges, then so does the series

580

Chapter 12 Infinite Series

Example 1

Show that I+

! + . . . + ~ ~ In( n + I)

and so obtain a new proof that the harmonie series diverges. Solution

We take our funetionj(x) to be I/x. Then, from formula (I) above, we get 1+ -I + ... + -I 2 n

= ~n -;-I ~ Jn+ I -I dx= ln(n + i-l/'I

X

I).

Sinee limHooln(n + I) = 00, the integral Jf(l/ x)dx diverges; henee the series L~I(l/i) diverges, too . • We would like to turn around the preeeding argument to show that if J'i'f(x)dx eonverges, then Lf'=lai eonverges as well. To do so, we draw the reetangles with height ai to the tejt of x = i rather than the right; see Fig. 12.3.2. This proeedure defines a step funetion h(x) on [I, 00) defined by

h(x)=aj+1

(i..;x so L7=2aj = J7h(x)dx. j

j(i + I) = a j + 1 = h(x)

Henee j(x)

~

h(x)

~

~

If x satisfies i..; x

O.

O. (See Fig. 12.3.2.) Thus n

J nj(x)dx~Jnh(x)dx= ~ aj~ O. 1

(2)

i=2

I

y

Figure 12.3.2. The area under the graph of fis greater than the shaded area, so

y. /(x)

2

0..: 2:7=2aj ..: f'if(x)dx.

3

4

6

x

If the integral ffj(x)dx eonverges, then the partial sums L7_ la j = a l + L7=2aj are bounded above by a l + ffj(x)dx, and therefore the series Lf'=laj is eonve~gent (see the Supplement to Seetion 12.2).

Integral Test 00

To test the eonvergenee of a series ~ aj of positive deereasing terms, i= 1

find a positive, deereasing funetion j( x) on [l, 00) such that j( i) If iOOj(x)dx converges, so does

00

i~1 a

j •

00

If iOOj(X)dx diverges, so does

Example 2 Solution

Show that 1 +

i+~+k+

i~1 a

j •

. .. converges.

This series is Lf'=I(l/P). We letj(x) = l/x 2 ; then (00

)1

~ dx= lim (b ~ dx= lim x2

b~OO)1 x 2

b~oo

(1 - -bI) = 1.

The indefinite integral converges, so the series does, too. •

=a

j •

12.3 The Integral and Ratio Tests

Example 3

Show that

~

m=2

Solution

I

m=2

Note that the series start at m (00

)2

~

diverges, but

m~lnm

581

I 2 converges. m(lnm)

= 2 rather than m = 1. We consider the integral

I p dx= lim (\lnxfP 1 dx x(lnx) b~00)2 X . (Inx) -p+ I = b~oo hm - p + 1

I

b

2

I lim [(Inb)-p+1 - (In2)-P+I]. -p + 1 b~oo

=

The limit is finite if p = 2 and infinite if p = !' so the integral converges if p = 2 and diverges if p =!. It follows that L::=2[1/(m~lnm)] diverges and L::=2[l/m(lnmf] converges . • Examples land 2 are special cases of a result called the p-series test, which arises from the integral test with fex) = 1/ x p • We recall that ffx n dx converges if n < - 1 and diverges if n ;;. - 1 (see Example 2, Section 11.3). Thus we arrive at the test in the following box.

p-Series If P < 1, then If P > 1, then

~ .~

diverges.

~ .~

converges.

i= I i= I

1

1

The p-series are often useful in conjunction with the comparison test.

Example 4

Test for convergence: (a)

Solution

00

,,_1_. i~1 1 + P'

(b)

·2

00

"

LJ I j=

}

}

·4 -

+ 2':J

. 3'1·2 + 10 '

() c

~ 3n +[,1. n=1 2n3/2+2'

(a) We compare the given series with the convergentp series L::"'=ll/P. Let ai = 1/(1 + p) and bi = I/P. Then 0< ai < bi and L:~lbi converges, so L:~ lai does, too. (b) Let aj = C/ + 2j)/(/ - 3/ + 10) and bj = /// = 1//. Then

I·1m

- = 1m 1 lajll'

j~oo bj

j~oo

I+2/j

1- 3// + 10//

1=

1.

Since L:i= tbj converges, so does L:i= taj' by the ratio comparison test. (c) Take an = (3n + [,1.)/(2n 3/ 2 + 2) and bn = n/(n 3/ 2) = I/rn. Then . 3 + (I/rn) 3 hm =2'

n~oo 2 + (2/ n 3!2)

Since L:~=lbn diverges, so does L:~=lan' •

582

Chapter 12 Infinite Series

What is the error in approximating a p-series by a partial sum? Let us show that L~=I(l/nP) approximates L~=I(l/nP) with error whieh does not exeeed 1/[(p - I)NP-I].

Indeed, just as in the proof of formula (2), we have

00

1

~

001

p< ( pdx. n=N+I n JN x

The left-hand side is the error:

00

~ n=N+I Th us, error

I!iIExample5 Solution

1

00

nP =


o(

-~-

x smx

-

1-) x2

(use a eommon de-

nomina tor). · 1 + cosx 30• IImX~'f1 2 (x - '11') Expand eaeh of the functions in Exercises 31-36 as a Maclaurin series and determine for what x it is valid. 31.

1- x

33. - - - - I-x I+x 35.

1- x 2

32.

1+ x

1 + 1 ) I-x l+x 1 1 36. I+x 1- x2

34. 1(

2"

37. Find the Maclaurin series for fex) = (I + X 2 )2 in two ways: (a) by multiplying out the polynomial; (b) by taking sueeessive derivatives and evaluating them at x = 0 (without multiplying out). 38. Write down the Taylor series for Inx at Xo = 2. 39. Find apower scries expansion for Hin t dt. Compare this with the expansion for x In x. Wh at is your conclusion? 40. Using the Taylor series for sin x and eos x, find the terms through x 6 in the series for (sin X)2 + (eosxl

606

Chapter 12 Infinite Series

Let fex) = ao + alx + a2x2 + .... Find ao, al' a2' and for each of the functions in Exercises 41-44. 41. secx 42. ~

a3

43. (d/dx)~ 44. e l + x Find Maclaurin expansions through the term in X S for each of the functions in Exercises 45-48. 45. (1- cosx)/x 2 46. x - sin3x x3 2 47. 1- x 48. d I+x dx 2 ..[l+?

49. Find the Taylor polynomial of degree 4 for lnx at: (a) Xo = I; (b) Xo = e; (c) Xo = 2. 50. (a) Find apower series expansion for a function fex) such that f(O) = 0 and f'(x) - fex) = x. (Write fex) = ao + alx + az x 2 + ... and solve for the a;'s one after another.) (b) Find a formula for the function whose series you found in part (a). Find the first four nonvanishing terms in the power series expansion for the functions in Exercises 51-54. 51. ln(l + e X) 52. eX'+x

liI 55. An engineer is about to compute sin(36°), when the batteries in her hand caJculator give out. She quickly grabs a backup unit, only to find it is made for statistics and does not have a "sin" key. Unperturbed, she enters 3.1415926, divides by 5, and enters the result into the memory, called "x" hereafter. Then she computes x(l - x 2 /6) and uses it for the value of sin(36°). (a) What was her answer? (b) How good was it? (c) Explain what she did in the language of Taylor series expansions. (d) Describe a similar method for computing tan(IOO). 56. An automobile travels on a straight highway. At noon it is 20 miles from the next town, travelling at 50 miles per hour, with its acceleration kept between 20 miles per hour per hour and - 10

miles per hour per hour. Use the formula x(t) = x(O) + x'(O)t + fb(t - s)x"(s) ds to estimate the auto's distance from the town 15 minutes later. *57. (a) Let fex) = {(SinX)/x,

I,

x~O,

x =0.

Find 1'(0),1"(0), and 1"'(0). (b) Find the Maclaurin expansion for (sinx)/x. *58. Using Taylor's formula, prove the following inequali ties: (a) e X - I ~ x for x ~ O. (b) 6x - x 3 + X S /20 ~ 6 sin x ~ 6x - x 3 for x ~ O. (c) x 2 - x 4 /l2.;;; 2 - 2cosx .;;; x 2 for x ~ O. *59. Prove that In2 = I -! + t - ~ + .... *60. (a) Write the MacJaurin series for the functions I/~ and sin-Ix. Where do they converge? (b) Find the terms through x 3 in the series for sin -I(sin x) by substituting the series for sinx in the series for sin -IX; that is, if sin-Ix = ao + alx + a2x2 + ... , then

X3 XS + a2 ( x - 3T + sr -

...

)2

+ .... (c) Use thc substitution method of part (b) to obtain the first five terms of the series for sin -IX by using the relation sin -I(sin x) = x and solving for ao through as. (d) Find the terms through X S of the Maclaurin series for the inverse function g(s) of fex) = x 3 + x. (Use the relation g(j(x» = x and solve for the coefficients in the series for g.)

12.6 Complex Numbers

607

12.6 Complex Numbers Comp/ex numbers provide a square root Jor - 1.

This section is abrief introduction to the algebra and geometry of complex numbers; i.e., numbers of the form a + bH. We show the utility of complex numbers by comparing the series expansions for sinx, cosx, and e X derived in the preceding seetion. This leads directly to Euler's formula relating the numbers 0, I, e, 'TT, and H: e'llN + 1=0. Applications of complex numbers to second-order differential equations are given in the next section. Section 12.8, on series solutions, can, however, be read before this one. If we compare the three power series

. x sm

= x - -

cosx

=

e

x 3 + -X S - ... 3! 5!

2 1- -x 2!

+ -x 4 - ... 4!

(I) (2)

'

= I + x + -x 2 + -x 3 + ...

(3) 2! 3! ' it looks as if sin x and cos x are almost the "odd and even parts" of e X • If we write the series X

x x e -x =I-x+---+'" 2

3

(4)

2! 3! subtract equation (4) from equation (3) and divide by 2, we get eX-e- x

2

x3

XS

=x+3T+5T+ ....

(5)

Similarly, adding equations (3) and (4) and dividing by 2, gives eX+e- X x4 x2 2 = I + 2T + 4! + . . . .

(6)

These are the Maclaurin series of the hyperbolic functions sinhx and coshx; they are just missing the alternating signs in the series for sin x and cos x. Can we get the right signs by an appropriate substitution other than changing x to - x? Let us try changing x to ax, where ais some constant. We have, for example, ax

+

-ax

2

4

6

e = I + a 2 ~ + a 4 ~ + a 6 ~ + ... 2 2! 4! 6! . This would become the series for cosx if we had a 2 = a 6 = a lO = ... = -I and a 4 = a 8 = a 12 = ... = I. In fact, all these equations would follow from the one relation a 2 = - 1. We know that the square of any real number is positive, so that the equation a 2 = - I has no real solutions. Nevertheless, let us pretend that there is a solution, which we will denote by the letter i, for "imaginary." Then we would have cosh ix = cos x.

cosh ax

Example 1 Solution

= e

What is the relation between sinh ix and sin x? Since i 2 = - 1, we have i 3 = - i, i 4 substituting ix for x in (5) gives

= (-

i) . i

=

1, i S = i, i 6 =

-

I, etc., so

608

Chapter 12 Infinite Series .

.

smhlx=

eix_e- ix

2

. .x 3 .x 5 .x7 = I X - I - + I - - I - + •••

3!

5!

7!

Comparing this with equation (1), we find that sinh ix = i sin x . ... The sum of the two series (5) and (6) is the series (3), i.e., e X = coshx Substituting ix for x, we find

+ sinhx.

e ix = cosh ix + sinh ix or

e ix = cosx + isinx.

(7)

Formula (7) is called Euler' s formula. Substituting 7T for x, we find that

e i7T = -1, and adding 1 to both sides gives

e i7T + 1=0,

(8)

a formula composed of seven of the most important symbols in mathematics:

0, I, +, =, e, i, and 7T. Example 2

Using formula (7), express the sine and eosine funetions in terms of exponentials.

Solution

Substituting - x for x in equation (7) and using the symmetry properties of eosine and sine, we obtain

e- ix = eosx - isinx. Adding this equation to (7) and dividing by 2 gives eosx=

e ix

+ e- ix 2

'

while subtracting the equations and dividing by 2; gives

sinx Example 3

Solutio"

=

e ix _ e- ix

2;

.... -

Find e i( 7T /2) and e 27Ti • Using formula (7), we have

e i('1T/2) = cos.!. +; sin.!. = ; 2

2

and

e27Ti = eos27T + isin27T = 1. ... Since there is no real number having the property i 2 = - 1, all of the calculations above belong so far to mathematieal "science fiction." In the following paragraphs, we will see how to construet a number system in whieh - 1 does have a square root; in this new system, all the ealculations which we have done above will be eompletely justified. When they were first introdueed, square roots of negative numbers were deemed merely to be symbols on paper with no real existenee (whatever that means) and therefore "imaginary." These imaginary numbers were not taken seriously until the eubie and quartic equations were solved in the sixteenth eentury (in the formula in the Supplement to Seetion 3.4 for the roots of a eubie equation, the symbol {- 3 appears and must be contended with, even if

12.6 Complex Numbers

609

all the roots of the equation are real.) A proper way to define square roots of negative numbers was finally obtained through the work of Girolamo Cardano around 1545 and Bombelli in 1572, but it was only with the work of L. Euler, around 1747, that their importance was realized. A way to understand imaginaries in terms of real numbers was discovered by Wallis, Wessel, Argand, Gauss, Hamilton, and others in the early nineteenth century. To define a number system which contains i = Fl, we note that such a system ought to contain all expressions of the form a + bFl = a + bi, where a and bare ordinary real numbers. Such expressions should obey the laws (a

+ bi) + (c + di) = (a + c) + (b + d)i

(a

+ bi)(c + di) =

and ac

+ adi + bei + bdi 2 =

(ac - bd)

+ (ad + bc)i.

Thus the sum and product of two of these expressions are expressions of the same type. All the data in the "number" a + bi is carried by the pair (a, b) of real numbers, which may be considered a point in the xy plane. Thus we define our new nu mb er system, the complex numbers, by imposing the desired operations on pairs of real numbers.

Complex Numbers A complex number is a point (a, b) in the xy plane. Complex numbers are added and multiplied as follows:

(a.b)

b

imaginary axis

- ... I I real axis

+ (c,d) = (a + c,b + d), (a, b)( c, d) = (ac - bd, ad + bc).

(a,b)

I I

a

x

Figure 12.6.1. A complex number isjust a point (a,b) in the plane.

The point (0, I) is denoted by the symbol i, so that i 2 = (- 1,0) (using a = 0, c = 0, b = I, d = 1 in the definition of muliplication). The x axis is called the real axis and the y axis is the imaginary axis. (See Fig. 12.6.1.)

It is convenient to denote the point (a, 0) just by a since we are thinking of points on the real axis as ordinary real numbers. Thus, in this notation, i 2 = -I. Also, (a,b)

= (a,O) + (O,b) = (a,O) + (b,O)(O, I)

as is seen from the definition of multiplication. Replacing (a,O) and (b,O) by a and b, and (0, I) by i, we see that (a,b) = a + bio

Since two points in the plane are equal if and only if their coordinates are equal, we see that a

-+ ib = c + id

if and only if a = c and b = d.

Thus, if a + ib = 0, both a and b must be zero. We now see that sense can indeed be made of the symbol a + ib, where i 2 = - I. The notation a + ib is much easier to work with than ordered pairs, so we now revert to the old notation a + ib and dispense with ordered pairs in our calculations. However, the geometrie picture of plotting a + ib as the point (a, b) in the plane is very useful and will be retained.

610

Chapter 12 Infinite Series

It can he verified, although we shall not do it, that the usual laws of algebra

hold for complex numbers. For example, if we denote complex numbers by single letters such as z = a + ib, W = c + id, and u = e + iJ, we have z(w + u) = zw + zu,

z(wu) = (zw)u, etc. Example 4 Solution imaginaryaxis

(a) Plot the complex number 8 - 6i. (h) Simplify (3 + 4i)(8 + 2i). (c) Factor x 2 + x + 3. (d) Find {[. (a) 8 - 6i corresponds to the point (8, - 6), plotted in Fig. 12.6.2. (b)

(3 + 4i)(8 + 2i)

= 3·8 + 3· 2i + 4· 8i + 2· 4F =

= 16 + 38i.

8

---I+-'I-+-I-H~f---.. real axis

-6i

• 8-6i

Figure 12.6.2. The point 8-6; plotted in the xy plane.

Example 5

24 + 6i + 32i - 8

(c) By the quadratic formula, the roots of x 2 + X + 3 = 0 are given hy ( - I ± ~) /2 = ( - 1/2) ± ({IT /2)i. We may factor using these two roots: x 2 + x + 3 = [x + (1/2) - ({IT /2)i][x + (1/2) + ({IT /2)i]. (You may check by multiplying out.) (d) We seek a number z = a + ib such that Z2 = i; now Z2 = a 2 - b 2 + 2abi, so we must solve a 2 - b 2 = 0 and 2ab = l. Hence a = ±b, so b = ±(I/fi). Thus there are two numbers whose square is i, namely, ±[(I/fi) + (i/fi)], i.e., {[ = ±(I/fi)(1 + i) = ±(fi /2)(1 + i). Although for positive real numbers, there is a "preferred" square root (the positive one), this is not the case for a general complex number .... (a) Show that if z

= a + ib 1= 0,

then

is a complex number whose product with z equals I; thus, 1/ z is the inverse of z, and we can divide by nonzero complex numbers. (b) Write 1/(3 + 4i) in the form a + bio Solution

(a)

(a2 - ib2 )(a a+b

+ ib) =

1 (-2a+b

-2

)(a - ib)(a

= ( __1_ )(a 2 + aib a 2 + b2

=

(_I _ )(a a +b 2

2

2

+ ib) iba - b 2i 2)

+ b2) = l.

Hence z( a2 - ib2 ) = I, so (a - ib)/(a 2 + b 2) can be denoted I/z. Note that a

+b

z 1= 0 means that not both a and b are zero, so a2 + b2 1= 0 and division by the real number a 2 + b2 is legitimate. (b) 1/(3 + 4i) = (3 - 4i)/(3 2 + 42 ) = (3/25) - (4/25)i by the formula in (a) .

...

12.6 Complex Numbers

611

Terminology for Complex Numbers If z

= a + ib is a complex number,

then:

(i) a is called the real part of z; (ii) b is called the imaginary part of z (note that the imaginary part is itself areal number); (iii) a - ib is called the complex conjugate of z and is denoted z; (iv) r = Ja 2 + b 2 is called the length or absolute value of z and IS denoted Izl; (v) 0 defined by a = rcosO and b = rsinO is called the argument of z.

The notions in the box above are illustrated in Fig. 12.6.3. Note that the real and imaginary parts are simply the x and y coordinates, the complex conjugate is the reflection in the x axis, and the absolute value is (by Pythagoras' theorem) the length of the line joining the origin and z. The argument of z is the angle this line makes with the x axis. Thus, (r,O) are simply the polar coordinates of the point (a, b). y

r = absolute value

--~---

z = a + ib

imaginary part

x

Figure 12.6.3. IIlustrating various quantities attached to a complex number.

z = a - ib = complex

conjugate of z

The terminology and notation above simplify manipulations with complex numbers. For example, notice that z·

z = (a + ib)(a -

ib)

= a 2 + b 2 = Iz1 2,

so that 1/ z = Z/lzl2 which reproduces the result of Example 5(a). Notice that we can remember this by: 1

z

z

z

-; = -;""i = zz = ~. Example 6

Solution

(a) Find the absolute value and argument of I + i. (b) Find the real parts of I/i, 1/(1 + i), and (8 + 2i)/(I - i). (a) The real part is I, and the imaginary part is 1. Thus the absolute value is

f12+i2

= fi, and the argument is tan - I( I / I) = 7T / 4. (b) I/i = (1/ i)( - i / - i) = - i/I = - i, so the real part of I/i = - i is zero. 1/(1 + i) = (1 - i)/(1 + i)(1 - i) = (1 - i)/2, so the real part of 1/(1 + i) is 1/2. Finally, 8 + 2i I-i

=

(8 + 2i) 1 + i (I - i) I + i

= 8 + lOi 2

so the real part of (8 + 2;)/(1 - i) is 3....

2

= 6 + lOi = 3 + 5i 2

'

612

Chapter 12 Infinite Series

Properties 01 Complex Numbers (i) z lz2=Zj'Z;, ZI/Z2=Zj/Z;; (ii) z is real if and only if z = z; (iii) IZ 1Z21 = Izd ·IZ21, IZI/Z21 = Izd/lz21; and (iv) Iz 1 + z21 ..; Iz d + IZ21 (triangle inequality).

The proofs of these properties are left to the examples and exercises. Example 7 Solution

Example 8 Solution

(a) Prove property (i) of complex numbers. (b) Express (I + i)IOO without a bar. (a) Let Zl = a + ib and z2 = c + id, so Zl = a - ib, z2 = c - id. From ZlZ2 = (ac - bd) + {ad + bc)i, we get z lZ2 = (ac - bd) - (ad + bc)i; we also have ZJ'Z; = (a - ib)(c - id) = (ac - bd) - ibc - aid =Z lz2' For the quotient, write Z2 . z 1/ z2 = z 1so by the rule just proved, Z; . (z 1/ Z2) = ZI' Dividing by Z2 gives the result. (b) Since the complex conjugate of a product is the product of the complex conjugates (proved in (a)), we similarly have ZlZ2Z3 =Zlz2Z3 = ZIZ2Z3 and so on for any number of factors. Thus zn = zn, and hence (I + i)IOO = (1 + i)100 = (1 - i)100 . .& Given z

=

a + ib, construct iz geometrically and discuss.

If z = a + ib, iz = ai - b = - b + ia. Thus in the plane, z = (a, b) and iz = ( - b, a). This pOInt (- b, a) is on the line perpendicular to the line Oz since the slopes are negative reciprocals. See Fig. 12.6.4. Since iz has the same length as z, we can say that iz is obtained from z by a rotation through 90°. .& y iz

= -b + ia

Figure 12.6.4. The number iz is obtained from z by a 90° rotation about the origin.

x

Using the algebra of complex numbers, we can define fez) when fis a rational function and z is a complex number. Example 9 Solution

If fez) = (l + z)/(l - z) and z = 1 + i, express fez) in the form a + bio

Substituting 1 + i for z, we have f(l+i)=

1+1+~ = 2+.i = -1-4= -1+2i . .& 1-(1+,) -/ /

How can we define more general functions of complex numbers, like er? One way is to use power series, writing Z2 Z3 00 zn e Z = 1 + z + ,. + ,. + ... = ~ ,.. 2. 3. n=O n.

12.6 Complex Numbers

613

To make sense of this, we would have to define the limit of a sequence of complex numbers so that the sum of the infinite series could be taken as the limit of its sequence of partial sums. Fortunately, this is possible, and in fact the whole theory of infinite series carries over to the complex numbers. This approach would take us too far afield,6 though, and we prefer to take the approach of defining the particular function eiX, for x: real, by Euler's formula e ix =

Since

Example 10

cosx + isinx.

e X+ Y

(9)

= eXeY , we expect a similar law to hold for

e ix •

(a) Show that

(10) (b) Give adefinition of e Z for z

Solution

=x +

iy.

(a) The right-hand side of equation (10) is (cosx + isinx)(cosy + isiny)

= cosx cos y -

sinx sin y + i(sinx cos y + sin ycosx)

= cos(x + y) + isin(x + y) =

ei(x+y)

by equation (9) and the addition formulae for sin and cos. (b) We would like to have e Z,+Z2 = e Z'e Z2 for any complex numbers, so we should define eX+ry = e X • e(", i.e., e X +(1' = eX(cos y + j sin y). [With this defi'. nition, the law e Z ,+Z2 = e Z'e Z2 can then be proved for all ZI and Z2'] .& Equation (10) contains all the information in the trigonometrie addition formulas. This is why the use of e ix is so convenient: the laws of exponents are easier to manipulate than the trigonometrie identities.

Example 11

(a) Calculate e i9 and le i9 1. (b) Calculate I + cos9 + cos29 + ... + cosn9 by considering I +

Solution

e i9

+

e 2i9

+ ... +

e hr / 2

= -I

2

and

ehr.

(c) Prove that

I ( cosn9 - cos(n + 1)9) + 2 1- cos9

e ni9 •

(a) e i9 = cos9 + isin9, so by definition of the complex conjugate we should change the sign of the imaginary part: e ilJ

= cos (J -

i sin 9 = cos( -

(J)

+ i sin( -

since cos( - 9) = cosO and sin( - 0) =

(J)

- sinO.

= e - i9,

Thus leilli

= Jcos2(J + sin2(J = I

using the general definition Izi = Ja 2 + b 2 , where z = a + ib. (b) e hr / 2 = cos('1J' /2) + i sin('1J' /2) = i and ei'IT = COS'1J' + isin'1J' = -1. (c) Since cosnO is the real part of e in9 , we are led to consider 1+ e ill + e i211 + ... + einII. Recalling that I + r + ... + r n = (1 - r n + 1)/(1 - r), we get 1+

e ill

+

e i211

+ ... +

e in9

=

1- e i(n+I)1I 1- e ill 1- e i (n+I)9 l-e iIJ

1- e- ill I-e- ill

See a text on complex variables such as J. Marsden, Basic Comp/ex Analysis, Freeman, New York (1972) for a thorough treatment of complex series.

6

614

ehapler 12 Infinite Series

1- e- ilJ _ ei(n+I)/J + e inIJ 2 - (ei/J + e-i/J) 1- e-i/J _ ei(n+I)/J 2(1 - cosO)

+ einIJ

Taking the real part of both sides gives the result. • Let us push our analysis of e ix a little further. Notice that ei/J = cosO + isinO represents a point on the unit circle with argument O. As 0 ranges from 0 to 271, this point moves once around the circle (Fig. 12.6.5). (This is the same basic geometrie picture we used to introduce the trigonometrie functions in Section 5.1). x Recall that if z = a + ib, and r,O are the polar coordinates of (a,b), then a = rcos9 and b = rsin9. Thus

y

-I

z = rcosO + ir sin 9 = r(cosO + isinO) = rei/J.

-;

Figure 12.6.5. As (J goes from 0 to 2'IT, the point e ilJ goes once around the unit circ1e in the complex plane.

Hence we arrive at the following.

Polar Representatlon

0' Complex Numbers

If z = a + ib and if (r,9) are the polar coordinates of (a,b), i.e., the absolute value and argument of z, then z = rei/J.

This representation is very convenient for algebraic manipulations. For exampie, which shows how the absolute value and arguments behave when we take products; i.e., it shows that IZIZ21 = IzJ! IZ21 and that the argument of Zl Z2 is the sum of the arguments of Zl and Z2' Let us also note that if z = re ilJ , then zn = rne inlJ . Thus if we wish to solve zn = w where w = pe i, we must have r n = p, i.e., r = nlP (remember that r,p are non-negative) and ein/J = e i, i.e., ei(nIJ- 0 is a constant. (Can you see why there is a minus sign before y?). If we rewrite equation (6) as (7)

where ß = y / m and w 2 = k / m, it has the form of equation (1) with a = 1, b = ß, and C = w2• To solve it, we look at the characteristic equation which has roots

,=

12.7 Second Order Linear Differential Equations

621

> 4w 2 (i.e., ß > 2w), then there are two real roots and so the solution is = cle"t + C2e'2t, where 'I and '2 are the two roots H- ß ± Vß2 - 4w 2 ).

If ß2 x

Note that 'I and '2 are both negative, so the solution tends to zero as t~ 00, although it will cross the taxis once if CI and c2 have opposite signs; this case is called the ove,damped case. A possible solution is sketched in Fig. 12.7.2. x

Figure 12.7.2. Damped harmonie motion.

= 4w 2, there is a repeated root

'I = - ß/2, so the solution is This case is called critically damped. Here the solution also tends to zero as t ~ 00, although it may cross the taxis once if CI and c2 have opposite signs (this depends on the initial conditions). A possible trajectory is given in Figure 12.7.2. Finally, if ß2 < 4w 2, then the roots are complex. If we let w= !V4w 2 - ß2 = ß2/4w 2 , then the solution is

If ß2

x

= (CI + c zt)e- ßt / 2 •

wVI -

x

= e-ßt/ 2(clcoswt + c2 sinwt)

which represents unde,damped oscillations with frequency w. (Air resistance slows down the motion so the frequency w is lower than w.) These solutions may be graphed by utilizing the techniques of Section 8.1; write x = Ae-ßt/ 2cos(wt - B), where (A,B) are the polar coordinates of CI and c2 • A typical graph is shown in Fig. 12.7.2.At t = 0, 0 = x = CI' Example 4

Consider aspring withß = 7T / 4 and

w

= 7T / 6.

(a) Is it over, under, or critically damped? lilJ (b) Find and sketch the solution with x(O) = 0 and x'(O) = I, for t ;;. O. lilJ (c) Find and sketch the solution with the same initial conditions but with ß = 7T/2. Solution

(a) Here ß2 - 4w 2 = 7T 2 /16 - 47T 2/36 = -77T 2/36 damped. (b) The effective frequency is

w= wVi -

= 7T17 /24, so the general solution is

At t = 0, 0 = x = CI' Thus, x

= c2e-"t/Ssin( 7T1:

t).

< 0, so the spring Is under-

ß2/4w 2

= (7T/6)/1 - 9/16

622

ehapter 12 Infinite Series

Hence

x' = C [(- !!.)e- 7Tt / 8sin( 71{f 8

2

24

t) + 71{f e-7T1/ cos( 71{f t)] 24 24' 8

At t = 0, x' = 1, so 1= C2[71{f /24], and hence C2 = 24/71{f. Thus the solution is

x

t).

= ~ e- 7T1 / 8sin( 71{f 71{f

24

Tbis is a sine wave multiplied by the decaying factor e- 7T1 / 8 ; it is sketched in Fig. 12.7.3. Tbe first maximum occurs when x' = 0; i.e., when tan«71{f /24)t)

= 8{f /24, or t ~ 2.09, at which point x R:: 0.84. x

V

1.0 .9

\

.8

24

y = 11:.J7 e-ltt/8

\

.7 .6

.5 .4 .3 .2

Figure 12.7.3. Graph of the solution to Example 4.

.I

10

(c) For ß = 71/2, we have ß2 - 4 0, so the spring is overdamped. Tbe roots 'I and '2 are

1(- ß

±

~ ß2 -

4w 2 ) =

1(-t

±

~)=

*(-

1±

1),

so the solution is of the form x = c le(7T/4)(-I+JS/3)1 + c 2e(7T/4)(-I-JS/3)1. At t = 0, x = 0, so CI + c2 = or CI = - c2 • Also, at t = 0,

°

1 = x' = CI!!. (-1

4

so CI

+ 15 ) + c2 !!. (-1 3 4

- 15 ) = CI 7115 ,

3

6

= 6/7115 and c2 = -6/7115. Tbus our solution is x = cl(eT,1 - e T21 ) = ~e(7T/4)(-I+JS/3)1 _ _ 6_ e (7T/4)(-I-JS/3)1 7115

= (0.8541)(e-o.200t _

e-1.375t)

Tbe derivative is x' = CI('leT,1 ishes when

-'I -_ e(T2-T,)1

'2

or

7115

'2eT21)

-1 +15 /3

-1-15 /3

=

CI('I -

= e-(7TJS/6)1

at this point, x R:: 0.731. See Fig. 12.7.3. •

'2e(T2-T,)I)eT,1

'

or t R:: 1.64;

which van-

12.7 Second Order Linear Differential Equations

623

In the preeeding diseussion we have seen how to solve the equation (I): ay" + by' + cy = O. Let us now study the problem of solving

ay" + by' + cy = F(x),

(8)

where F(x) is a given funetion of x. We eall equation (1) the homogeneous equation, while equation (8) is ealled the nonhomogeneous equation. Using what we know about equation (1), we ean find the general solution to equation (8) provided we ean find just one partieular solution.

Nonhomogeneous Equations: Particular and General Solutions = c,y, + c2Y2 is the general solution to the homogeneous equation (I) and if yp is a partieular solution to the nonhomogeneous equation (8), then

If Yh

y = yp + c,y, + c2 Y2 = yp + Yh

(9)

is the general solution to the nonhomogeneous equation.

To see that equation (9) is a solution of equation (8), note that

a(yp + Yh)" + b(yp + Yh)' + c(Yp + Yh)

= (ay; + by; + cYp) + (ay; + by~ + CYh) = F(x) + 0 = F(x). To see that equation (9) is the general solution of equation (8), note that if j is any solution to equation (9), then j - yp solves equation (1) by a ealculation similar to the one just given. Henee j - yp must equal Yh for suitable c, and C2 sinee Yh is the general solution to equation (I). Thus j has the form of equation (9), so equation (9) is the general solution. Sometimes equation (8) ean be solved by inspeetion; for example, if F(x) = Fo is a eonstant and if c 0, then y = Fo/ c is a partieular solution.

'*

Example 5 Solution

(a) Solve 2y" - 3y' + Y = 10. (b) Solve 2x" - 3x' a funetion of t). (e) Solve 2y" - 3y' + Y = 2e- x .

+ x = 8eos(t/2) (where xis

(a) Here a partieular solution is y = 10. From Example 1,

Thus the general solution, given by equation (9), is

y = c,e x + c2 e x / 2 + 10. (b) When the right-hand side is a trigonometrie funetion, we ean try to find a partieular solution whieh is a eombination of sines and eosines of the same frequeney, sinee they reproduee linear eombinations of eaeh other when differentiated. In this ease, 8 eos( t /2) appears, so we try xp = A eos(

-i )+

B sin(

-i ),

where A and Bare eonstants, ealled undetermined coefficients. Then

624

Chapter 12 Infinite Series

i) -

i)]

2x" - 3x' + xP = 2[ - A cos(2 4 !i sin( PP 4 2

- 3[

-1 f )+ ~ f )] sin(

cos(

f )+ B sin( f )] = ( ! A - ~ B )cos( f )+ ( ~ A + ! B )sin( f ). + [ A cos(

For this to equal 8 cos(t /2), we choose A and B such that

!A

-

tB = 8,

and

tA +! B = O.

The second equation gives B = - 3A which, upon substitution in the first, gives ! A + ~ A = 8. Thus A =! and B = so xp

= ~ [cos(

f )- 3 sin( f )l

and the general solution is x

= cle t + C2 et / 2 + ~ [ cos(

-zr,

f )- 3 sin( f )J-

A good way to check your arithmetic is to substitute this solution into the original differential equation. (c) Here we try Yp = Ae- x since e- x reproduces itself, up to a factor, when differentiated. Then 2y; - 3y; + Yp = 2Ae- x

+ 3Ae- x + Ae- x • require 6A = 2 or

For this to match 2e x , we A particular solution, and so the general solution is Y

= cle x + c2e x/ 2 + te-x.

= t.

Thus Yp

= te-x

is a

Ä

The technique used in parts (b) and (c) of this example is called the method of undetermined coefficients. This method works whenever the right-hand side of equation (8) is a polynomial, an exponential, sums of sines and cosines (of the same frequency), or products of these functions. There is another method called variation of parameters or variation of constants which always enables us to find a particular solution of equation (8) in terms of integrals. This method proceeds as folIows. We seek a solution of the form

(10) where YI and Yz are solutions of the homogeneous equation (1) and VI and v 2 are functions of x to be found. Note that equation (10) is obtained by replacing the constants (or parameters) CI and c2 in the general solution to the homogeneous equation by Junctions. This is the reason for the name "variation of parameters." (Note that a similar procedure was used in the method of reduction of order-see equation (4).) Differentiating VIYI using the product rule gives (VIY.)'

= V;YI + VIY;'

and

"+ 2VIYI ,. '+ VIYI' " (VIYI )" = VIYI and similarly for V2Yz. Substituting these expressions into equation (8) gives

12.7 Second Order Linear Differential Equations

625

+ 2v;y; + VIY;') + (v2'Yl + 2VZY2 + V2Y2')] +b[(v;YI + VIY;) + (V2Yl + V2Y2)] + C(VIYI + v2 Yl) = F.

a[ (vi'YI

Simplifying, using (1) for YI and Yl, we get

a[ V;'YI + 2v;y; + v2'Yl + 2V2Y;)] + b[ V;YI + V;Yl] = F.

(11)

This is only one condition on the two functions VI and V2' so we are free to impose a second condition; we shall do so to make things as simple as possible. As our second condition, we require that the coefficient of b vanish (identically, as a function of x):

(12) This implies, on differentiation, that vi'yl tion (11) simplifies to

+ v;Y; + v2'Yl + V;Y2 = 0,

so equa(13)

Equations (12) and (13) can now be solved algebraically for v; and V2 and the resulting expressions integrated to give VI and V2' (Even if the resulting integrals cannot be evaluated, we have succeeded in expressing our solution in terms of integrals; the problem is then generally regarded as "solved").

Variation

0' Parameters

A particular solution of the nonhomogeneous equation (8) is given by

Yp = VIYI

+ V2Yl,

where YI and Yl are solutions of the homogeneous equation and where and V2 are found by solving equations (12) and (13) algebraically for and v2 and then integrating.

VI

v;

Combining the two preceding boxes, one has a recipe for finding the general solution to the nonhomogeneous equation.

Example 6

(a) Find the general solution of 2y" - 3y' + Y = e2x + e- 2x using variation of parameters. (b) Find the general solution of 2y" - 3y' + Y = 1/(1 + x 2 ) (express your answer in terms of integrals).

Solution

(a) Here YI = e X and Yl = e X / 2 from Example 1. Thus, equations (12) and (13) become

v;e + v;e x/ 2 = 0, X

v' eX I

2x + -2x + _I v' ex/ 2 = e e 2

2'

2

respectively. Subtracting,

I , x/2 _ ( e2x + e- 2x ) - 2 '

'2 v2e and so

V2

=

Similarly,

_e- X / 2(e 2x

+ e- 2x ) = _e 3x / 2 _ e- 5x / 2.

626

Chapler 12 Infinite Series

and so integrating, dropping the constants of integration, V2

3x / 2 + le- 5x / 2 = _le 3 S '

VI

= eX

-

t e - 3x .

Hence a particular solution is 2x _le 2x + le- 2x = le 2x + .le- 2x Yp = VIYI + V2Y2 = e 2x _le3 3 S 3 IS ,

and so the general solution is Y = cIe X

2x + .le2x + c2 e x/ 2 + le 3 IS •

The reader can check that the method of undetermined coefficients gives the same answer. (b) Here equations (12) and (13) become

v;e + v;e x / 2 = 0, X

'x

vle

I,

+2: v2e

x/2 _

I

I

-2:1+x2 '

respectively. Solving,

, V2

=

so

and

, Vl

e -x = 1 + x2 '

f+

-x/2

e -x/2 -I + x 2 '

V2

= -

le

x 2 dx

l= f I ~:2 dx. V

so

Thus the general solution is

Let us now apply the above method to the problem of forced oscillations. Imagine that our weight on aspring is subject to a periodic external force Focos~h; the spring equation (6) then becomes m

d2~ dt

= - kx - 'Y ddx

t

+ Focos fl.t.

(14)

A periodic force can be directly applied to our bobbing weight by, for example, an oscillating magneiic field. In many engineering situations, equation (14) is used to model the phenomenon of resonance; the response of a ship to a periodic swell in the ocean and the response of a bridge to the periodic steps of a marching army are examples of this phenomenon. When the forcing frequency is elose to the natural frequency, large oscillations can set in-this is resonance. 9 We shall see this emerge in the subsequent development. Let us first study the case in which there is no damping: 'Y = 0, so that m(d 2x/dt 2) + kx.= Focos~k This is called a forced oscillator equation. A particular solution can be found by trying xp = C cos Q t and solving for C. We find xp=[Fo/m(w 2 _Q2)]cosQt, where w=Jk/m is the frequency of the unforced oscillator. Thus the general solution is

For further information on resonance and how it was involved in the Tacoma bridge disaster of 1940, see M. Braun, Differential Equations and their Applications, Second Edition, 1981, SpringerVerlag, New York, Section 2.6.1.

9

827

12.7 Second Order Linear Differential Equations

x = A coswt

+ Bsinwt +

Fo

2

2

m(w - 0)

cosOt,

(15)

where A and Bare determined by the initial conditions.

Example 7

Solution

Find the solution of d 2x / dt 2 + 9x sketch its graph.

= 5 cos 2t

with x(O) = 0, x'(O) = 0, and

We try a particular solution of the form x = Ccos2t; substituting into the equation gives -4Ccos2t + 9Ccos2t = 5cos2t, so C must be 1. On the other hand, the solution of the homogeneous equation d 2x / dt 2 + 9x = 0 is A cos 3t + B sin 3t, and therefore the general solution of the given equation is x(t) = A cos3t + Bsin3t + cos2t. For this solution, x(O) = A + 1 and x'(O) = 3B, so if x(O) = 0 and x'(O) = ~ we must have A = -1 and B = O. Thus, our solution is x(t) = -cos3t + cos2t. To graph this function, we will use the product formula

= Hcos(R To recover -cos3t + cos2t, R = 1 and S = t. Thus x(t) = 2sin(1t)sin(!t). sinRtsinSt

+ S)t]. we must have R + S = 3 and

S)t - cos(R

R- S

= 2,

so

We may think of this as a rapid oscillation, sin 1t, with variable amplitude 2 sin t t, as illustrated in Fig. 12.7.4. The function is periodic with period 2'11", with a big peak coming at '11",3'11",5'11", etc., when -cos3t and cos2t are simultaneously equal to 1. .& x

x

x

Figure 12.7.4. x(t) = -cos3t + cos2t = 2sin!tsin~t.

If in equation (15), x(O) = 0 and x'(O) = 0, then we find, as in Example 7, that x(t)

= =

Fo

2

2

m(w - 0)

(cosOt - coswt)

2Fo [. (w+O)t)] sm (W-O)t)][. 2 2 sm 2· m(w 2 - 0)

628

ehapler 12 Infinite Series

n is smalI, then this is the product of a relatively rapidly oscillating function [sin«w + n)t/2») with a slowly oscillating one [sin«w - n)t/2»). The slowly oscillating function "modulates" the rapidly oscillating one as shown in Fig. 12.7.5. The slow rise and fall in the amplitude of the fast oscillation is the phenomenon of beats. It occurs, for example, when two musical instruments are played slightly out of tune with one another.

If w -

2FO sm. (W-U)t) --

x(t)

m(w 2 -U 2 )

2

Figure 12.7.5. Beats.

The function (15) is the solution to equation (14) in the case where 'I = 0 (no damping). The general case ('I =1= 0) is solved similarly. The method of undetermined coefficients yields a particular solution of the form xp = a cos nt + ß sin nt, which is then added to the general solution of the homogeneous equation found by the method of Example 4. We state the result of such a calculation in the following box and ask the reader to verify it in Review Exercise 110.

Damped Forced Oscillations The solution of

m d2~ dt

= -

x(t)

cle,,1 + c2e,,1 +

kx - 'I ddx + Focosnt t

is

where

J

CI

=

Fo

~m2(w2 _

cos(nt - c5),

(16)

g})2 + 'Y2w2

and c2 are constants determined by the initial conditions,

= k / m , r land r2 are roots of the characteristic equation mr2 + 'Ir + k = 0 [if r l is a repeated root, replace cle,,1 + c2e,,1 by (CI + c2t)e"I), and

w

In equation (16), as t~ 00, the solution cle,,1 + C2e,,1 tends to zero (if 'I > 0) as we have seen. This is called the transient part; the solution thus approaches

12.7 Second Order Linear Differential Equations

629

the oscillatory part,

Fo

---;.======- cos(Ot ~m2(w2 _ 0 2)2 + y 202

8),

which oseillates with a modified amplitude at the foreing frequeney 0 and with the phase shift 8. If we vary w, the amplitude is largest when w = 0; this is the resonance phenomenon. Example 8

Consider the equation

d 2x + 8 dx + 25x dt 2

dt

=

2 eos t.

(a) Write down the solution with x(O) = 0, x'(O) = O. (b) Diseuss the behavior of the solution for large t.

Solution

(a) The characteristic equation is

r2 + 8r + 25

=

0

which has roots r = (-8 ± ~64 - 100 )/2 = -4 ± 3i. Also, m = I, w = 5, 0= I, Fo = 2, and y = 8; so 2

~;:::::=::=7

= -2- = -

1

~ 0.079

~m2(w2 _ 0 2)2 + y 202 J576 + 64 J640 4{IO and 8 = tan - l( -!4) = tan - l( t ) ~ 0.322. The general solution is given by equation (16); writing sines and eosines in place of the eomplex exponentials, we get

x(t) = e- 41 (A cos3t + Bsin3t) + _1_ eos(t - 8). 4{IO

At t

= 0 we get 0= x(O) = A + _1-cos 8 4,fW

=A+_I_.~ 4{IO J640 -A -

so A = -

+ 3 . 40'

10. Computing x'(f) and substituting t =

0= x'(O) = -4A =

0 gives

+ 3B + _1- sin8 4,fW

-4A +3B+ _1_. _8_ 4{IO J640 1

12

1

= -4A + 3B + 40 = 40 + 3B + 40 . Thus B = -

No, and our solution beeomes -41

x(t) = - ~20 (9cos3t + 13sin3t) + 0.079cos(t - 0.322). (b) As t ~ 00 the transient part disappears and we get the oseillatory part 0.07geos(t - 0.322) ....

630

ehapler 12 Infinite Series

Supplement to Sectlon 12.7:

Wronsklans

In this section we have shown how to find solutions to equation (I): ay" + by' + cY = 0; whether the roots of the characteristic equation ar 2 + br + C = 0 are real, complex, or coincident, we found two solutions YI and Yz. We then asserted that the linear combination CIYI + c2Yz represents the general solution. In this supplement we shall prove this. Suppose that YI and Yz are solutions of equation (I); our goal is to show that every solutiony of equation (1) can be written asy = CIYI + c2 Yz. To do so, we try to find CI and c2 by matching initial conditions at x o : y(xo) = CIYI(XO) + c2Yz(xO)' y'(x o) = cIY;(XO) + c2Y;(XO)·

We can solve these equations for CI by multiplying the first equation by y;(xo), the second by Yz(xo), and subtracting: CI =

y(xo)y;(x o) - Yz(xo)y'(x o) YI(XO)Y;(x o) - Yz(xo)y;(x o) .

(I7a)

Y(Xo)Y~ (xo) - Y~ (xo)y(xo) Yl (xo)y;(x o) - Y2(xo)y~ (x o) .

(I7b)

Similarly, C2

=

These are valid as long as YI(XO)Y;(x o) - Yz(xo)y;(x o) W(x)

= YI(X)Y;(x) -

* O. The expression (18)

Yz(x)y;(x).

is called the Wronskian of YI and Yz [named after the Polish mathematician ·Count Hoene Wronski (1778-1853)]. (The expression (18) is a determinantsee Exercise 43, Section 13.6). Two solutions YI and Yz are said to be a fundamental set if their Wronskian does not vanish. It is an important fact that W(x) is either everywhere zero or nowhere zero. To see this, we compute the derivative of W: W'(x)

= [y;(x)y;(x) + YI(X)Y;(x)]

- [y;(x)y;(x)

+ Yz(x)y;'(x)]

=YI(X)Y;(x) - Yz(x)y;'(x). IfYI andYz are solutions, we can substitute -(bja)y; -(cja)YI fory;' and to get similarly for

Y;

W'(x)

= YI(X)[ - ~ Y2(X) - ~ Yz(X)] - Yz(X) [ - ~ y;(x) - ~ YI(X)] = -

!z.a [YI( x)Y2( x) -

Yz( x)y;( x)].

Thus W'(X) = -

!z.a W(x).

Therefore, from Section 8.2, W(x) = Ke-(b/a)x for some constant K. We note that W is nowhere zero unless K = 0, in which case it is identicallY zero. If YI and Yz are a fundamental set, then equation (17) makes sense, and so we can find CI and c2 such that CIYI + c2 Yz attains any given initial conditions. Such a specification of initial conditions gives a unique solution and determines Y uniquely; therefore Y = CIYI + c2Yz. In fact, the proof of uniqueness of a solution given its initial conditions also follows fairly easily from what we have done; see Exercise 46 for a special case and Exercise 47 for the general case. Thus, in summary, we have proved:

12.7 Second Order Linear Differential Equations

631

If YI and h are a fundamental set of solutions for ay" + by' + cy = 0, then y = cIYI + c2h is the general solution. To complete the justification of the claims about general solutions made earlier in this section, we need only check that in each specific case, the solutions form a fundamental set. For example, suppose that r l and r2 are distinct roots of ar 2 + br + c = O. We know that YI = er,x and h = e r2X are solutions. To check that they form a fundamental set, we compute W(x) = YI(X)Y;(x) - h(x)y;(x)

This is nonzero since r 2 =I=- r l , so we have a fundamental set. One can similarly check the case of a repeated root (Exercise 45).

Exerclses tor Section 12.7 Find the general solution of the differential equations in Exercises 1-4. 1. y" - 4y' + 3y = O. 2. 2y" - Y = O. 3. 3y" - 4y' + Y = O. 4. y" - y' - 2y = O. Find the particular solutions of the stated equations in Exercises 5-8 satisfying the given conditions. 5. y" - 4y' + 3y = O,y(O) = O,y'(O) = I. 6. 2y" - Y = O,y(1) = O,y'(1) = 1. 7. 3y" - 4y' + Y = 0, y(O) = I, y'(O) = 1. 8. y" - y' - 2y = O,y(l) = O,y'(I) = 2. Find the general solution of the differential equations in Exercises 9-12. 9. y" - 4y' + 5y = O. 10. y" + 2y' + 5y = O. 11. y" - 6y' + 13y = O. 12. y" + 2y' + 26y = O. Find the solution of the equations in Exercises 13-16 satisfying the given conditions. 13. y" - 6y' + 9y = 0, y(O) = 0, y'(O) = 1. 14. y" - 8y' + 16y = O,y(O) = -3,y'(0) = O. 15. y" - 212y' + 2y = O,y(1) = O,y'(1) = 1. 16. y" - 2/3 y' + 3y = 0, y(O) = 0, y'(O) = -1. In Exercises 17-20 consider aspring with ß, w, x(O), and x'(O) as given. (a) Determine if the spring is over, under, or critically damped. (b) Find and sketch the solution. 17. ß = 'IT/16, '" = 'IT/2, x(O) = 0, x'(O) = 1. 18. ß = I, '" = 'IT/8, x(O) = I, x'(O) = o. 19. ß = 'IT /3, w = 'IT / 6, x(O) = 0, x'(O) = 1. 20. ß = 0.03, '" = 'IT /2, x(O) = I, x'(O) = 1. In Exercises 21-28, find the general solution to the given equation (y is a function of x or x is a function of 1 as appropriate). 21. y" - 4y' + 3y = 6x + 10. 22. y" - 4y' + 3y = 2e x • 23. 3x" - 4x' + x = 2 sin I. 24.3x"-4x'+x=e'+e-'. 25. y" - 4y' + 5y = x + x 2 . 26. y" - 4y' + 5y = 10 + e- x • 27. y" - 212y' + 2y = cosx + sinx. 28. y" - 212y' + 2y = cosx - e- x .

In Exercises 29-32 find the general solution to the given equation using the method of variation of parameters. 29. y" - 4y' + 3y = 6x + 10. 30. y" - 4y' + 3y = 2e x • 31. 3~" - 4x' + x = 2sin/. 32.3x"-4x'+x=e'+e-'. In Exercises 33-36 find the general solution to the given equation. Express your answer in terms of integrals if necessary. 33. y" - 4y' + 3y = tanx. 34. y" - 4y'

+ 3y = _1_. x2 + 2

35.y"-4y'+5y=

36.

I 1+ cos2x

Y"-4Y'+5Y=~.

1+ x 2 In Exercises 37-40, find the solution of the given forced oscillator equation satisfying the given initial conditions. 37. x" + 4x = 3 cos I, x(O) = 0, x'(O) = O. 38. x" + 9x = 4 sin 4/, x(O) = 0, x'(O) = O. 39. x" + 25x = COS/, x(O) = 0, x'(O) = 1. 40. x" + 25x = cos6t, x(O) = I, x'(O) = o. In Exercises 41-44, (a) write down the solution of the given equation with the stated initial conditions and (b) discuss the behavior of the solution for large I. d 2x dx 41. dl 2 + 4 dt + 25x = 2cos2t, x(O) = 0, x'(O) = O. 42.

dx 2 dx dt 2 + 2 dt + 36x = 4 cos 3/, x(O) = 0, x'(O)= O.

d 2x dx 43. + - + 4x dl 2 dl

x(O) = I x'(O) = O. ,

dx + 2 -d + 9x = cos4t, x(O) = I, x'(O) = O. dt t 45. If rl is a repeated root of the characteristic equation, use the Supplement to this section to show that YI = er,x and Y2 = xer,x form a fundamental set and hence conc\ude that y = clYI + C2Y2 is the general solution to equation (I).

44.

d 2x -2

= cos 1'

632

Chapter 12 Infinite Series

*46. Suppose that in (I), a > 0 and b 2 - 4ac < O. If Y satisfies equation (1), prove w(x) = e(b/2a)xy (x) satisfies w" + [(4ac - b2 )/4a 2 ]W = 0, which is a spring equation. Use this observation to do the following. (a) Derive the general form of the solution to equation (I) if the roots are complex. (b) Use existence and uniqueness results for the spring equation proved in Seetion 8.1 to prove corresponding results for equation (1) if the roots are complex. *47. If we know that equation (I) admits a fundamental set Y1' J2, show uniqueness of solutions to equation (I) with given values of y(xo) and y'(xo) as folIows. (a) Demonstrate that it is enough to show that if y(xo) = 0 and y'(xo) = 0, then y(x) == o. (b) Use facts above the Wronskian proved in the Supplement in order to show that Y(X)y'1(X) - Y'(X)Y1(X) = 0 and that y(x)Yz(x) - y'(x)J2(x)

*49. (a) Show that the basic facts about Wronskians, fundamental sets, and general solutions proved in the Supplement also apply to the equation in Exercise 48(a). (b) Show that the solutions of Euler's equations found in Exercises 48(b) and 48(c) form a fundamental set. Write down the general solution in each case. *50. (a) Generalize the method of variation of parameters to the equation a(x)y" + b(x)y' + c(x)y

= F(x). (b) Find the general solution to the equation xJI" + 5xy' + 3y = xe X (see Exercise 48; express your answer in terms of integrals if necessary). *51. In Fig. 12.7.1, consider the motion relative to an arbitrarily placed x axis pointing downward. (a) Taking all forces, including the constant force g of gravity into account, show that the equation of motion is

= o.

(c) Solve the equations in (b) to show thaty(x) =0. *48. (a) Generalize the method of reduction of order so it applies to the differential equation a(x)y" + b(x)y' + c(x)y = 0, a(x) O. Thus, given one solution, develop a method for finding a second one. (b) Show that x r is a solution of Eu/er's equation xJI" + axy' + ßy = 0 if r 2 + (a - I)r + ß=O. (c) Use (a) to show that if (a - 1)2 = 4ß, then (lnx)x(l-a)/2 is a second solution.

*

m dy dt2

*52. *53. *54. *55. *56.

= _ k(y _ Y ) e

+g_

y dy

dt '

where Ye is the equilibrium position of the spring in the absence of the mass. (b) Make a change of variables x = y + c to reduce this equation to equation (6). Show that solutions of equation (15) exhibit beats, without assuming that x(O) = 0 and x'(O) =0. Find the general solution of y"" + Y = o. Find the general solution of y"" - Y = o. Find the general solution of y"" + Y = e Find the general solution of y"" - Y = cosx. X •

12.8 Series Solutions of Differential Equations Power series solutions of differential equations can often be found by the method of undetermined coefficients.

Many differential equations cannot be solved by means of explicit formulas. One way of attacking such equations is by the numerical methods discussed in Section 8.5. In this section, we show how to use infinite series in a systematic way for solving differential equations. Many equations arising in engineering and mathematical physics can be treated by this method. We shall concentrate on equations of the form a(x)y" + b(x)y' + c(x)y = f(x), which are similar to equation (1) in Seetion 12.7, except that a, b, and c are now functions of x rather than constants. The basic idea in the power series method is to consider the a;'s in a sum y = 2~Oaixi as undetermined coefficients and to solve for them in successive order.

12.8 Series Solutions of Differential Equa1ions

Example 1 Solution

Find apower series solution of y"

633

+ xy' + Y = O.

If a solution has a convergent series of the form y = ao + alx = 2:~~oaixi, we may use the results of Section 12.4 to write

y' = a l + 2a2x + 3a3x2 + ...

00

=

2: iaix

+ a2x 2 + ...

I and

i-

i= I

y" = 2a 2 + 6a 2x + 12a4x 2 + ... =

00

2: i(i -

l)ai x i - 2 .

i=2

Therefore

y" + xy' + y =

00

2: i(i -

i=2

l)aix i - 2 +

00

00

i=1

i=O

2: iaix i+ 2: aix i= o.

In perforrning manipulations with series, it is important to keep careful track of the summation index; writing out the first few terms explicitly usually helps. Thus,

y" + xy' + y (2a 2 + 6a3x + 12a4x2 + ... ) + (alx + 2a 2x 2 + ... )

=

+ (a o + alx + a2 x z + ... ). To write this in summation notation, we shift the summation index so all x's appear with the same exponent:

y" + xy' + Y =

00

00

00

i=O

i= I

i=O

2: (i + 2)(i + l)ai+Zx i+ 2: iaix i+ 2: aix i=

O.

(Check the first few terms from the explicit expression.) Now we set the coefficient of each Xi equal to zero in an effort to determine the ai' The first two conditions are

2a z + ao = 0 6a3 + 2a l

=

(constant term), (coefficient of x).

0

Note that this deterrnines az and a3 in terms of ao and a l : az = a3 = - tal' For i ~ I, equating the coefficient of Xi to zero gives

(i + 2)(i + 1)ai+2 + (i + I)ai = 0 or _

1

a,+z - - (i + 2) ai · Thus,

az =

-

a4 = a6

1 lao 1

1 4· 2 ao

1

1

4" a2 =

a3 =

-

31 al

a5 =

-

5"1 a3 =

= - 6" a4 = - 6. 4 . 2 ao

Hence

aZ n

= 2n . (2n -

(-1( a 2) . (2n - 4) ... 4 . 2 o

1

Da I

- tao,

634

Chapter 12 Infinite Series

=

(-lf

(_1)"

a = -n-aO 2 ... 2·n(n-l) ... 2·1 o 2n!

and

(-I)"2nn! a2"+' = (2n + 1)(2n - 1)(2n - 3) ... 5.3 a, = (2n + I)! a,. (-1)"

Thus, we get (using O! = I), 00 i X 2i ) ( 00 i 2ii! X 2i +' ) y=ao( .~(-I)j. +a, .~(-1) (2.+1)' . 1=0 2z! 1=0 I.

(1)

What we have shown so far is that any eonvergent series solution must be of the form of equation (I). To show that equation (1) really is a solution, we must show that the given series eonverges; but this eonvergenee follows from the ratio test. • The eonstants ao and a, found in Example 1 are arbitrary and play the same role as the two arbitrary eonstants we found for the solutions of eonstant eoeffieient equations in the preeeding seetion.

Example 2 Solution

Find the first four nonzero terms in the power series solution of y" satisfying y (0) = 0, y' (0) = 1.

+ xy = 0

Let y = ao + a,x + a2x2 + .... The initial eonditions y(O) = 0 and y'(O) = I ean be put in immediately if we set ao = 0 and a, = 1, so that y = x + a2x2 + .... Then

y"

= 2a2 + 3· 2a3x + 4· 3a4x2 + 5· 4asx 3 +

... + (i + l)iai+,x i-' + ...

and so

= x 3 + a2x4 + a3xs + Settingy" + xy = 0 gives xy

... + ai_3xi-' + .... ( constant term), (eoefficient of x), (eoeffieient of x 2 ), (eoefficient of x\

as = - 15·4

(coefficients of x 4,X S,X 6), (eoefficient of x \ ete. Thus, the first four nonzero terms are

y

1

1

5

1

9

= x - 5. 4 x + 9. 8 . 5 . 4 x - 13· 12 . 9 . 8 . 5 ·4

[The reeursion relation is ai

+, = -

1

i(i + 1) ai -

3

and the general term is ( -l)i

1

(4j + 1)(4j) ... 9·8·5·4

x4j+'

.

The ratio test shows that this series converges.] •

X'3

+ ...

.

12.8 Series Solutions of Differential Equations

Example 3 Solution

635

(Legendre's equatlon)1O Find the recursion relation and the first few terms for the solution of (1 - x 2)y" - 2xy' + AY = 0 as apower series. We write y = Lr:oajX j and get y

= ao + a.x + a2x2 +

... + ajx j + ... ,

= a. + 2a2x + 3a3x 2 + -2xy' = -2a.x - 2· 2a2x 2 y'

y" - xy"

... + iajx j-· + ... , 2· 3a3x 3 - ... - 2iajx j - ... ,

= 2a 2 + 3· 2a3x + 4· 3a4 x 2 + ... + i(i - l)ajx

= -2a2x2 -

3· 2a3x3 - ... - i(i - I)ajx j

-

j

-

2

+ ... ,

••••

= 0 gives (constant term), 2a2 + Aao = 0 (coefficient of x), 3 . 2a3 - 2a. + Aa. = 0 (coefficient of x 2 ), 4· 3a4 - 2a2 - 4a 2 + Aa 2 = 0 (coefficient of x 3 ), 5 . 4a s - 3 . 2a3 - 2 . 3a3 + Aa3 = 0

Tbus, setting (I - x 2)y" - 2xy' + AY

Solving, A a2 = - iao, a4 = as

a3

6-A

~a2=

-

2-A a ., = T2

6-A A i ao ,

~

12 - A 12 - A 2 - A = """"5T a3 = """"5T . T2 a. , etc.

Thus, the solution is (6-A)x4+ Y =ao(I-~x24·3·2 2 +a. ( x +

2- A

T2 x

3

+

... )

(12 - A)(2 - A) s ) 5.4.3.2 x + . . . .

The recursion relation comes from setting the coefficient of x j equal to zero: (i + 2)(i + l)ai+2 - i(i - I)aj

-

2iaj + Aa;

= 0,

so i(i+l)-A a j +2 = (i + 2)(i + I) a j

•

From the ratio test one sees that the series solution has a radius of convergence of at least 1. It is exactly I unless there is a nonnegative integer n such that A = n(n + I), in which case the series can terminate: if n is even, set a. = 0; if n is odd, set ao = O. Tben the solution is a polynomial of degree n called Legendre's polynomial; it is denoted Pn(x). Tbe constant is fixed by demanding Pn(l) = 1. Ä

10 This equation occurs in the study of wave phenomena and quantum mechanics using spherical coordinates (see Seetion 14.5).

636

Chapter 12 Infinite Series

Example 4 Solution

(Hermite's equation)lI Find the recursion relation and the first few terms for the solution of y" - 2xy' + Ay = 0 as apower series.

Again write y = ao + alx + a2x2 + ... + ajx i + ... , so

Ay = Mo + MIX + M2X2 + ... + Aa;x i + ... , -2xy' = -2a lx - 4a2x2 - 2· 3a3x3 - ... - 2ia;x; - ... , and

y"=2a 2 +3·2a3x+4·3a4 x 2 + ... +i(i-l)a;x;-2+ .... Setting the coefficients of powers of X to zero in y" - 2xy' + Y = 0, we get

2a2 + Aao = 0

(constant term),

3 . 2a3 - 2a l + MI = 0

(coefficient of x),

4 . 3a4 - 4a2 + M 2 = 0

(coefficient of x 2 ),

and in general

(i + 2)(i + 1)ai +2 - 2iai + Mi

O.

=

Thus

A a2 = - 2"ao, a4 =

4-A

~ a2

2-A a3 = 3:-r al ' A(4-A)

= - 4. 3 . 2 ao,

and in general,

2i - A a;+2= (i + 2)(i + l)a;. Thus

a5 =

6 -A ~

a3 =

(6 - A)(2 - A) 5!

al

,

etc., and so

Y= a

(1 _~X2 _ (4 - A)A 2 4!

o

x4 _

(8 - X)(4 - X)X x6 _

6!

... )

(2 - A) 3 (6 - X)(2 - X) 5 (10 - X)(6 - X)(2 - X) ) +a l( x+-3-!-x + 5! x + 7! + ... This series converges for all x. If A is an even integer, one of the series, depending on whether or not A is a multiple of 4, terminates, and so we get a polynomial solution (called a Hermite polynomial). Ä Sometimes the power series method runs into trouble-it may lead to only one solution, or the solution may not converge (see below and Exercise 23 for examples). To motivate the method that folIows, which is due to Georg Frobenius (1849-1917), we consider Euler's equation:

xy" + axy' + ßy = O. Here we could try y = ao + alx + a2 x 2 + ... as before, but as we will now show, this leads nowhere. To be specific, we choose a = ß = 1. Write

11

This equation arises in the quantum meehanies of a harmonie oscillator.

12.8 Se ries Solutions of Differential Equations

637

= a o,+ alx + a2x 2 + a3x 3 + ... , xy' = alx + 2a 2x 2 + 3a3x3 + ... , y

and

xy"

2a2x 2 + 3 . 2a3x3 + ....

=

Setting xy"

+ xy' + Y = 0,

ao = 0\ 2al

=0

we get

(constant term), (coefficient of x), (coefficient of x 2 ), (coefficient of x 3), etc.,

and so all of the a;'s are zero and we get only a trivial solution. The difficulty can be traced to the fact that the coefficient of y" vanishes at the point x = 0 about which we are expanding our solution. One can, however, try to find a solution of the form x'. Letting y = x', where r need not be an integer, we get

y' = rx,-I

so axy'

= arx'

and

y"

= r(r -

so xy"

I)X'-2

= r(r -

I)x'.

Thus, Euler's equation is satisfied if

r(r - I) + ar + ß = 0 which is a quadratic equation for r with, in general, two solutions. (See Exercise 48 of Section 12.7 for the case when the roots are coincident.) Frobenius' idea is that, by analogy with the Euler equation, we should look for solutions of the form y = x'~~=Oaixi whenever the coefficient of y" in a second-order equation vanishes at x = O. Of course, r is generally not an integer; otherwise we would be dealing with ordinary power series.

Example 5 Solution

Find the first few terms in the general solution of 4xy" - 2y' the Frobenius method.

+Y =0

using

We write

so

- 2y'

= - 2raoX,-1 - 2(r + 1)alx' - 2(r + 2)a2x'+ I -

4xy"

= 4r(r -

l)aoX'-1

=0

and

+ 4(r + l)ra lx' + 4(r + 2)(r + l)a2x'+1 + ....

Thus to make 4xy" - 2y' + Y

ao[ 4r(r - 1) - 2r]

• ..

= 0, we set

(coefficient of X'-I).

If ao is to be allowed to be nonzero (which we desire, to avoid the difficulty encountered in our discussion of Euler's equation), we set 4r(r - 1) - 2r = O. Thus r(4r - 6) = 0, so r = 0 or r = First, we take the case r = 0:

t.

y

= a o + alx + a2x2 + a3 x3 + ... ,

-2y'= -2al-2·2a2x-3·2a3x2_ ... , 4xy"

= 4 . 2a2x + 4 . 6a3x2 + 4 . 12a4x3 + ....

638

Chapter 12 Infinite Series

Then 4xy" - 2y' + Y

= 0 gives

a o - 2a l

=0

(constant term),

4 . 2a2 - 2 . 2a2 + a l

=0

(coefficient of x),

4 . 6a3 - 3 . 2a3 + a2 = 0

(coefficient of x 2);

so a l = tao. a2 = - ~al = - iao. and a3 = - -ha2 = Iltao. Thus

y

= ao{1 + tx - ix2 + Iltx3 -

... ).

For the case r =!' we have

= aoX 3/ 2 + a lx 5/ 2 + a2x 7/ 2 + a~/2 + ... , -2y' = -3aox l / 2 - 5a lx 3/ 2 -7a2xs/2 - 9a3x7/2 - ... , y

and

4xy"

= 3aoXl/2 + 5· 3a lx 3/ 2 + 7· 5a2x 5 / 2 + 9· 7a3x7/2 + ....

Equating coefficients of 4x" - 2y' + Y = 0 to zero gives

3ao - 3ao = 0

+ ao =

(coefficient of X I / 2).

0

(coefficient of x 3 / 2 ).

=0 9 . 7 a3 - 9a3 + a2 = 0

(coefficient of X 5 / 2 ),

5 . 3a l

5a l

-

7· 5a2 -7a2 + a l

(coefficient of x 7/2);

so

al

_ -

-

I 10 ao,

-

54

a2 =

-

al

28

I

= 280 ao,

and

a3 =

a2

I 280· 54 ao ·

=-

Thus Y

= a (X 3 / 2 _ ...L x 5 / 2 + _1_ X 7/2 _ o

10

280

I x 9/ 2 + 280 . 54

... )

.

The general solution is a linear combination of the two we have found: 1 X2 +_I-x 3 - ... ) Y =C(I+lx_ I 2 8 144

+C (x 3/ 2 _ 2

...L x 5 / 2 + _1_ x 7/2_ 10

280

••• )

.

....

The equation that determines r, obtained by setting the coefficient of the lowest power of x in the equation to zero, is called the indicial equation. The Frobenius method requires modification in two cases. First of all. if the indicial equation has a repeated root r I' then there is one solution of the form YI(x) = aoX" + alx" + I + . .. and there is a second of the form Y2(X) = YI(x)lnx + box" + blx,,+1 + .... This second solution can also be found hy the method of reduction of order. (See Exercise 48, Section 12.7.) Second, if the roots of the indicial equation differ by an integer, the method may again lead to problems: one may or may not be ahle to find a genuinely new solution. If r2 = r l + N, then the second series boX" + blx,,+1 + ... is of the same form, aoX" + alx,,+1 + ... , with the first N coefficients set equal to

12.8 Series Solutions of Differential Equations

639

zero. Thus it would require very special circumstances to obtain a second solution this way. (If the method fails, one can use reduction of order, but this may lead to a complicated computation). We conclude with an example where the roots of the indicial equation differ by an integer. Example 6 Solution

Find the general solution of Bessel's equation '2 x)" + xy' + (x 2 - k 2)y with k =!. We try y

y

=0

= XrL~=oa;x;. Then 00

= ~ a;x;+r, ;=0

x) =

00

~ a;x;+r+2 ;=0

00

= ~ a;_2 X;+r, ;=2

00

00

;=0

;=-)

y' = ~ (i + r)a;xi+r-' = ~ (i + r + I)a;+,x;+r, xy'

00

= ~

(i + r)a;x;+r,

;=0 00

y" = ~ (i + r + l)(i + r + 2)a;+ 2Xi+r, ;= -2

and 00

x)"

= ~ (i + r - l)(i + r)a;x;+r. ;=0

Setting the coefficient of x r in x)" get

+ xy' + (x 2 - k 2)y = 0 equal to zero, we

0= (r - l)ra o + rao - iao, so the indicial equation is 0 = (r - l)r + r - i = r2 - i, the roots of which are r, = -! and r2 =!' which differ by the integer 1. Setting the coefficient of x r+' equal to zero gives

0= r(r + I)a, + (r + I)a, - ia, = [(r + 1)2 - i ]a" and the general recursion relation arising from the coefficient of x;+r, i :> 2, is

0= r(r + i)a; + (r + i)a; - ia; + a;_2 = (r + i)2 - i )a; + a;_2. Let us work first with the root r, = -!. Since -! and +! are both roots of the indicial equation, the coefficients a o and a, are arbitrary. The recursion relation is a;-2 a;_2 a;-2 a· = = - - - = - -'---'--01/2 + i)2 - 1/4 ;2 - i i (i - 1) I

( _

for i ~ 2. Thus ao

= - 2-1 '

a6

= - 6.5 = - 6! ' ... ,

a4

ao

a2

a2

a4

= - 4. 3 = 4· 3 . 2 . I ' ao( _I)k

ao

a 2k

= (2k)!

12 This equation was extensively studied by F. W. Bessel (1784-1846), who inaugurated modern practical astronomy at Königsberg Observatory.

640

Chapler 12 Infinite Series

Similarly a l ( _I)k a 2k + 1 = (2k

+ I)!

Our general solution is then X- I / 2 [

ao( 1 - ~!2 + ~!4 - ~!6 + . ..

)

x3 x5 + a l ( .x - 3T + 5T - ... )]

which we recognize to be ao(cosx)/IX + al(sinx)/IX. Notice that in this case we have found the general solution from just one root of the indicial equation....

Exercises tor Section 12.8 In Exercises 1-4, find solutions of the given equation in the fonn of power series: y = L~Oajxj. I. y" - xy' - y = O. . 2. y" - 2xy' - 2y = O. 3. y" + 2xy' = O. 4. y" + xy' = O. In Exercises 5-8, find the first three nonzero terms in the power series solution satisfying the given equation and initial conditions. 5. y" + 2xy' = 0, y(O) = 0, y'(O) = I. 6. y" + 2xy = 0, y(O) = I, y'(O) = O. 7. y" + 2xy' + Y = 0, y(O) = 0, y'(O) = 2. 8. y" - 2xy' + Y = 0, y(O) = 0, y'(O) = I. 9. Airy's equation is y" = xy. Find the first few tenns and the recursion relation for apower series solution. 10. Tchebycheff's equation is (l - x 2 )y" - xy' + 1X2y = O. Find the first few tenns and the recursion relation for apower series solution. What happens if a = n is an integer? In Exercises 11-14, use the Frobenius method to find the first few tenns in the general solution of the given equation. 11. 3xy" - y' + y = O. 12. 2xy" + (2 - x)y' - y = O. 13. 3xy" + 2xy' + Y = O. 14. 2xy" - 2xy' + Y = O. 15. Consider Bessel's equation of order k, namely, xy" + xy' + (x 2 - k 2 )y = O. (a) Find the first few terms of a solution of the fonn Jk(x) = aoX k + alx k + 1 + .... (b) Find a second solution if k is not an integer.

16. Laguerre functions are solutions of the equation xy" + (I - x)y' + ;\y = O.

17.

* 18.

*19. *20. *21. *22.

(a) Find apower series solution by the Frobenius method. (b) Show that there is a polynomial solution if ;\ is an integer. Verify that the power series solutions of y" + = O' are justy = A coswx + B sinwx. Find the first few terms of the general solution for Bessel's equation of order ~. (a) Verify that the solution of Legendre's equation does not converge for all x unless ;\ = n(n + I) for some nonnegative integer n. (b) Compute P1(x), P2(x), and P3(x). Use Wronskians and Exercise 49 of Section 12.7 to show that the solution found in Example I is the general solution. Use Wronskians and Exercise 49 of Seetion 12.7 to show that the solution found in Example 3 is the general solution. Prove that the Legendre polynomials are given by

wy

Rodrigues' formula: P (x)

n

I -d n (x 2 - I )n. = -2nn! dx n

*23. (a) Solve xY' + (x - I)y - I = 0, y(O) = I as a power series to obtain y = Ln! x n , which converges only at x = O. (b) Show that the solution is

e--I /-x y= x

J--dx. e x l/ x

Review Exercises tor Chapter 12

641

Review Exercises for Chapter 12 In Exereises 1-8, test the given series for convergenee. If it ean be summed using a geometrie series, do so.

~

_I_i (12) 00 3i+1 3. '" - i 1 i7:1 5 I.

2.

;= 1

~

;= 1 00

+ .! + ..!. + ..!. + ...

6. 100 +

.! + ..!.2 + ..!. + ...

7

+ 1)

4. '" - i ;7:1 9

5. I + ~

3

9

~

100(, 8

32

9

33

~

~ 9 . ;=1 10+ Ili

00

6

8. ~­ i=1 7 + 8;

In Exercises 9-24, test the given series for eonvergenee. 00

9. ~ 5- n n=1

00 k 11. '" k~1 3k 00 (-I)nn

13

~--

. n=1 00

3n ( _ 1)2n

15. ~-n=1 n 00

2n2

00 4n 10. ~ n=1 (2n+ I)! 00 2n 12. '" - n~1 n + 3

14.

--f:!!-

16. ~-­ j=O / + 8 00

•

18. ~ - ' i=1 ;3 + 8

17. ~ n=1 n! 00

rn

00

19. ~ ne- n2 n=1 00 I 21. ~-n=2 (Inninn

20. ~ - : - - : - n=1 n 2 - sin299n

23. ~ n=1 (n + I).

24. ~ _ n _ n=1 (n+I)!

n,

00

11!1

~

n=1 n + 3 00 (-1)1

rnI)

00(1 22. ~ --n=1 n

2

00

Sum the series in Exercises 25-32 to within 0.05. 25. I 26. t -

*+ +. - 12 + ... ~

+ i - .i + -fi - ... ~ (-I)nn 28. Ln=1 6n 2 - I 00

2

30. ~ !!... n=1 n! 32.

38. If a senes converges, it must also eonverge absolutely. 39. The error made in approximating a eonvergent series by a partial sum is no greater than the first term omitted. 40. eosx = ~k=O< _1)kx 2k /(2k)!. 41. If ~;:'Iaj and ~k=obk are both eonvergent, then ~j= laj + ~k=obk = bo + ~r:.I(aj + b;). 42. ~r:.I( -IY[3/(; + 2)] eonverges conditionally. 43. The eonvergenee of ~;;'= lan implies the eonvergenee of ~;;'=I(an + an+I). 44. The eonvergenee of ~;;'_I(an + an+l) implies the eonvergenee of ~;;'=Ian. 45. The eonvergenee of ~;;'= IO x5 x4 . (I + X)3/2 _ (1_ x)3/2 73. hm 2

1109. The current I in the electric circuit shown in Figure 12.R.1 satisfies L d 21 dl 2

X

x->O

74. 1~[1

Find the first few terms in an appropriate series for at least one solution of the equations in Exercises 105108. 105. 5xy" + y' + y = O. 106. xy" + y' - 4y = O. 107. Bessel's equation with k = I. 108. Legendre's equation with A = 3.

+ cosx -

(x - 'IT)2

2

+

(x - 'IT)4 24 ].

+ 'lTi = O. z8 =.;s + 3i.

83. Solve for z: z2 - 2z

84. Solve for z: Find the general solution of the differential equations in Exercises 85-96. 86. y" -4y=0 85. y" +4y =0

+ 6y' + 5y = 0 88. y" - 6y' - 2y = 0 + 3y' - lOy = e + cosx y" - 2y' - 3y = x 2 + sinx y" - 6y' + 9y = cos( y" - lOy + 25 = cos(2x) y" + 4y = x . (Express your answer in

87. y"

89. y" 90. 91. 92. 93.

X

I)

rxr+T

terms of integrals.) 94. y" - 3y' - 3y =

sinx . (Express your an";cos2x + I swer in terms of integrals.) 95. y'" + 2y" + 2y' = 0 96. y'" - 3y" + 3y' - Y = e X In Exercises 97-100, identify the equation as aspring equation and describe the limiting behavior as I ~ 00. 97. x" + 9x + x' = cos2/. 98. x" + 9x + 0.001 x' = sin(50/). 99. x" + 25x + 6x' = COS('lT/). 100. x" + 25x + O.OOlx' = cos(60'IT/). In Exercises 101-104, find the first few terms of the general solution as apower series in x. 101. y" + 2xy = 0 102. y" - (4sinx)y = 0 103. y" - 2x)/ + 2y = 0 104. y" + y' + xy = 0

dl

C

= dE

dl'

where E is the applied voltage and L, R, C are constants.

In Exercises 75-78 find the real part, the imaginary part, the complex conjugate, and the absolute value of the given complex number. 76. 2 - lOi 75. 3 + 7i 78. (2 + i)/(2 - i) 77. J2 - i In Exercises 79-82, plot the given complex numbers, indicating rand (J on your diagram, and write them in polar form z = re i9 • 80. I + ~ 79. I - I I - I 81. ie"i/2 82. (I + i)e i"/4

+ R dl + !...

L

R

c Figure 12.R.l. An electric circuit. (a) Find the values of m, k, y that make this equation a damped spring equation. (b) Find 1(/) if 1(0) = 0, 1'(0) = 0 and L = 5, C = 0.1, R = 100, and E = 2cos(60'IT/). 110. Verify formula (16) in Section 12.7. 111. Verify that L~_oX2(1 + x 2)-n is a convergent geometric series for x*-O with sum I + x 2 • It also converges to 0 when x = O. (This shows that the sum of an infinite series of continuous terms need not be continuous.) 112. A beam of length L feet supported at its ends carries a concentrated load of P Ibs at its center. The maximum deflection D of the beam from equilibrium is

D= 2L 3p ~ Isin(mr/2)1. EI'lT 4

n=\

n4

(a) Use the formula L~=.(I / n 4 ) = 'lT 4 /90 to show that

[Hinl: Factor out 2- 4 .] (b) Show that

k~O (2k ~ 1)4

= (

!~ )( ~ );

hence D = (1/48XL3p / EI). [Hinl: A series is the sum of its even and odd terms.] (c) Use the first two nonzero terms in the series for D to obtain a simpler formula for D. Show that this result differs at most by 0.23% from the theoretical value.

Review Exercises tor Chapter 12

113. The deflection y(x, t) of astring from its straight profile at time t, measured vertically at location x along the string, 0 ,.; x ,.; L, is y(x, t) =

n~l Ansin( nlx )cos( n7t ).

where An, Land c are constants. (a) Explain what this equation means in terms of limits of partial sums for x, t fixed. (b) Initially (at t = 0), the deflection of the string is

118. In eaeh of the following, evaluate the indicated derivative: (a) /12)(0), wheref(x) = x/(l + x 2); (b) /10), where fex) = x 6e x + I. 119. Let Z B2 2 B3 3 --=I+Blz+-z +-z

2Po y(L) = 3L I/ 3

where a

=

and subtract. (b) Differentiate 2:~=oan = 1/(1 - a) with respect to a, and then subtraet your answer from 2:~=oan = 1/(1 - a). 121. In highway engineering, a transitional spiral is defined to be a eurve whose eurvature varies directly as the are length. Assurne this curve starts at (0,0) as the continuation of a road coincident with the negative x axis. Then the parametrie equations of the spiral are x

*

10

L3/2

u(az)dz u(aL 3/ 2)

00

ß74

('" cosU dU,

)0 /0

y

=k

('" sin U dU.

)0 /0

(a) By means of infinite series methods, find the ratio x / y for cf> = 'Ir / 4. (b) Try to graph the transitional spiral for k = I, using accurate graphs of (cosU)//O, (sin U)/ /0 and the area interpretation of the integral. 122. The free vibrations of an elastic eircular membrane can be described by infinite series, the terms of which involve trigonometrie funetions and Bessel functions. The series 00

i=O

k

The values of the gamma function r may be found in a mathematical table or on some calculators asf(x) = (x - I)! [r(D = 1.3541, ren = 0.9027, rm = 1.5046, r( 1f) = 4.0122]. The function u is the Bessel function of order - t. (a) Find the smallest positive root of u(az) = 0 by using the first four terms of the series. (b) Evaluate y(L) approximately by using the first four terms of the series. 1iI1l7. (a) Use apower series for {f+x to calculate eorrect to 0.01. (b) Use the result of part (a) to caIculate ß. How aceurate is your answer?

=k

(_I)i . , (x/2r+ 2 ; I. (n + I).

~.,

a 2kz 2k u(az) = - 2 - kL;:o (-I) 22kk! r(k + 2/3) 1/3

= a + 2a 2 + 3a 3 + ... + nan,

aSn =a 2 +2a 3 + ... +(n-l)an+na n+ 1

~JW/ EI, and

az -

+ ....

eZ - I 2! 3! Determine the numbers BI' B 2 , and B 3 . (The Bi are known as the Bernoulli numbers.) 120. Show in the following two ways that 2:~= Inan = a/(l - a)2 for lai< I. (a) Consider Sn

Find the defleetion value as an infinite series at the midpoint x = L/2. 114. In the study of saturation of a two-phase motor servo, an engineer starts with a transfer function equation V(s)/ E(s) = K/(l + ST), then goes to the first-order approximation V(s)/ E(s) = K(I - ST), from which he obtains an approximate equation for the saturation dividing line. (a) Show that 1/(1 + ST) = 2:~=0( - STY, by appeal to the theory of geometrie series. Which values of ST are allowed? (b) Discuss the replacement of 1/(1 + ST) by I - ST; include an error estimate in terms of the value of ST. 115. Find the area bounded by the curves xy = sinx, x = I, x = 2, Y = O. Make use of the Taylor expansion of sin x. ; 116. A wire of length L inches and weight w Ibs/inch, clamped at its lower end at a small angle tan-Ipo to the vertical, defleets y(x) inches due to bending. The displacement y(L) at the upper end is given by

643

*

is called the Bessel Function Jn(x); n is an integer;;. O. (a) Establish eonvergenee by the ratio test. (b) The frequencies of oseillation of the circular membrane are essentially solutions of the equation Jn(x) = 0, x> O. Examine the equation Jo(x) = 0, and see if you ean explain why J o(2A04) = 0 is possible. (e) Check that Jn satisfies Bessel's equation (Example 6, Seetion 12.8). 123. Show that g defined by g(x) = e - '/x 2 if x 0 and g(O) = 0 is infinitely differentiable and g(il(O) = 0 for all i. [Hint: Use the definition of the derivative and the following lemma provable by I'H6pital's rule: if P(x) is any polynomiai, then limx~oP(x)g(x) = 0.)

'*

644

Chapter 12 Infinite Series

* 124. Let fex) = (I + x)a, where a is areal number.

Show by an induetion argument that j. is analytie for lxi< 1. *125. True or false: The eonvergenee of L~=.a; and L~= .b~ implies absolute eonvergenee of L~=.anbn·

* 126. (a) Show that if the radius of eonvergenee of

*

L~= .anx n is R, then the radius of convergenee of L~= .an x 2n is IR. (b) Find the radius of eonvergenee of the series L~-O('IT14 2n • 127. Let fex) = Lf';.oaiXi and g(x) = f(x)/(1 - x). (a) By multiplying the power series for fex) and 1/(1 - x), show that g(x) = Lf=obix i, where bi = ao + . . . + ai is the ith partial sum of the series L;":.Oai. (b) Suppose that the radius of eonvergenee of fex) is greater than land that J(I) 'i= O. Show that lim;-.oobi exists and is not equal to zero. What does this tell you about the

rx

radius of eonvergenee of g(x)? (e) Let e X 1(1 - x) = Lf=obix i. What is lim;-.oobi ? 128. (a) Find the seeond-order approximation at T = 0 to the day-Iength funetion S (see the supplement to Chapter 5) for latitude 38° and your own latitude. (b) How many minutes earlier (compared with T = 0) does the sun set when T = I, 2, 10, 30? (e) Compare the results in part (b) with those obtained from the exaet formula and with listings in your loeal newspaper. (d) For how many days before and after lune 21 is the seeond-order approximation correet to within I minute? Within 5 minutes? *129. Prove that e is irrational, as folIows: if e = alb for some integers a and b, let k> band let a = klee - 2 - ~ - ~ - ... - ir). Show that a is an integer and that a < I I k to derive a contradietion.

*

Chapter 7 Answers

A.43

Chapter 7 Answers 7.1 Calculating Integrals 1. x 3 + x 2

-

1/2x 2 + C

3. e X + x 2 + C

-(eos2x)/2 + 3x 2 /2 + C -e - x +2sinx+Sx 3 /3+C 1084/9 tt. lOS /2 844/S 15. 1/12 19. 6 3w/4 23. w/12 I 27. (e 6 - e 3)/3 + 3(25/ 3 - I)/S 29. InS 31. 41n2 + 61/24 33.400 35. 116/15 37. (b) e(e 2) - e + 3 39. (a) 11 5. 7. 9. 13. 17. 21. 25.

°

(b) -8 (e) N ote that f~ f(t) dt is negative

41. -2tJe 12 + sinSt 4

= lO00t -

(2

7.2 Integration by Substitution 1. Hx 2 + 4)5/ 2 + C 3. -1/4(y8+4y-I)+C 5. - 1/2 tan20 + C 7. sin(x 2 + 2x)/2 + C 9. (x 4 + 2)1 / 2/2 + C tt. -3(t 4 / 3 + 1)-1/2/2+ C 13. -cos 4(r 2)/4 + C 15. tan - l(x 4)/4+ C 17. -cos(O + 4) + C 19. (x 5 + x)lol/101 + C 21. ';t 2 + 2t + 3 + C 23. (t 2 + 1)3/ 2/3 + C

°

43. 3 45. (a) (b) 5/6 -eosx { (e) eos(2)-2eosx

65. (a) Evaluate the integral. (b) A = $45,231.46 67. (a) R(t) = 2000e l / 2 - 2000, C(t) (b) $S7,279.90 69. 1 + In(2) - In(l + e) ~ 0.380

if 0 < x .;; 2 if 2 < x';; w

y

1.584

0.416

25. 27. 29. 31. 33. 35. 37. 39.

sin 0 - sin 30/3 + C Inllnxl + C 2sin - l(x/2) + x~ /2 + C In(l + sin 0) + C - cos(ln t) + C -3(3+ l/x)4/ 3/4+ C (sin 2x)/2 + C m a non-negative integer and n an odd positive integer, or n a non-negative integer and m an odd positive integer.

x - I

47. 2 + tan- 12 - ilnS 49. 16.4 51. (1/2)(e 2 - 1) 53. 16/3 - w 55. y

- I

57. w/4 59. (a) Differentiate the right-hand side. (b) Integrate both sides of the identity. (c) 1/8 61. Use the fact that lan -la and tan -Ib lie in the interval ( - w/2, w/2) 63. 16,000,014 meters

7.3 Changing Variables in the Definite Integral 1. 2(313 - 1)/3 5. 2[(25)9/ 4 - (9)9/ 4]/9 9. (e - 1)/2

°

3. (5/5 - 1)/3 7. 1/7 tt. -1/3 15. 1 19. 1/2

13. 17. In( y1 cos( w/8» 21. 4 - tan- I (3) + w/4 23. (a) w /2 (b) w/4 (c) w/8 25. The substitution is not helpful in evaluating the integral. 27. (y1 /2)[tan -12y1 - tan -I( y1 /2)] 29. (1/13)ln[(4 + 3y1)/(1 +13)] 31. Let u = x - t. 33. (Sy1 - 2/5)/10 35. (w/27)(I45[i45 - IOfiO) 37. (a) 1/3 (b) Yes.

Chapter 7 Answers

A.44

7.4 Integration By Parts 1. (x + I)sinx + cosx + C 3. x sin 5x/5 + cos5x/25 + C 5. (x 2 - 2)sinx + 2xcosx + C 7. (x + l)e X + C 9. x In(lOx) - x + C 11. (x 3 /9)(3Inx - I) + C 13. e 3s (9s 2 - 6s + 2) /27 + C 15. (x 3 - 4)1/3(x 3 + 12)/4 + C 17. t 2sin t 2 + cos t 2 + C 19. -(I/ x)sin(1/ x) - cos(1/ x) + C 21. -[ln(cosx)f /2 + C

23. xcos-I(2x) -~ /2 + C 25. yb/y - I - tan-1b/y - I + C 27. sin 2x/2 + C 29. Tbe integral becomes more complicated. 31. (16 + '17)/5 33. 3(3 In 3 - 2)

35. y2[('17/4)2+3'17/4-2]/2-1. 37. .f3 /8 - '17/24 39. e - 2 41. _(e 2". - e- 2")/4 43. H22/ 3(22/ 3 + li/ 2 - 25 / 2 + H2 7/ 2 - (22/ 3 + If/2))~4.025 45. ('17 - 4)/8y2 - 1/2 47.

f Fxr

dx -

10,12 Fxr dx =

- ~,12 Fxr dx 49. 51. 53. 55. 57.

is - 1/8 the area of a circ\e of

radius y2 corrected by the area of a triangle (draw a graph). (-2'17cos2'170)/0+(sin2'170)/02. (Tbis tends to zero as 0 tends to 00. Neighboring oscillations tend to cancel one another.) (b) (5e 3"./10 - 3)/34 (a) Use integration by parts, writing cosnx = cosn-Ix X cosx. 2'17 2 (a) Q = f EC(a 2 /w + w)e-O/sin(wt)dt (b) Q(t)= EC{I- eO/[cos(wt) + asin(wt)/wn

59. y

y

Review Exerclses for Chapter 7 1. x 2/2 - cosx + C

3. x 4 /4 + sinx + C 5. e x 3 /3 - Inlxl + sin x + C 7. eH + 0 3/3 + C 9. -cos(x 3 )/3 + C 11. e(x 3) /3 + C 13. (x + 2)6/6 + C X

-

17. -!cos 3 2x + C - x/2 + tan- lx/2 + C sin- l (t/2)+t 3/3+C xe 4x /4 - e 4x /16 + C x 2sinx + 2xcosx - 2sinx + C (e-Xsinx - e-Xcosx)/2 + C x 31n3x/3 - x 3 /9 + C (2/5)(x - 2)(x + 3)3/2 + C x sin3x/3 + cos3x/9 + C 3x sin2x/2 + 3 cos2x/4 + C x 2e x'/2 - e X ' /2 + C x 2(lnxf/2 - x 2(lnx)/2 + x 2 /4 + C

15. e 4x '/12

19. 21.

23. 25. 27. 29. 31.

33. 35. 37. 39.

+C

x 2tan- lx/2

41. 2e'Ix(/i - 1) + C 43. sin x Inlsin xl - sin x + C 45. x tan -Ix ~ In(1 + x 2)/2 + C 47. -I 49. '17/25 51. sin(1) - sin(I/2) 53. ('17 2 /32 + 1/2)tan- I('I7/4) - '17/8

55. (4y2 - 2)/3 + (2y2 - 2)a 57. 3J3/5 59. 399/4 y

40 30

20 13

.
0

21. Increasing 23. 27. 25. 33,000 years 31. 29. 1.5 X 109 years 35. 33. 49 minutes 37. 18.5 years 39. The annual percentage ~ 18.53%. 41. (a) 300 e- O.3t (b) 2000; 2000 books will

1.2 m( dx)2 dt

and V(x) must remain

smalI, so both dx / dt and x - Xo remain smalI.

8.2 Growth and Deeay 1. dT/dt - -O.ll(T - 20) 3. dQ/dt - -(0.00028)Q 5. 2e- 3t 7. e3t 9. 2e St - S 11. 2e6 -2.s 13. 7.86 minutes

15. 2,476 years

Decreasing 173,000 years 2,880 years 4.3 minutes rate is 1000e rilOO - I)

eventually be sold.

(c)

29. There is no restriction on b. 31. Multiply (9)by ",sin",t and (10) by cos",t and add. 33. (a) V"(xo) > 0, so the second derivative test applies. (b) ComputedE/dt using the sum and chain rules. (c) Since E is constant, if it is initially smalI, the

sum of

2

S(I)

2000

----

-----

1000

o

10

20

1

43. K is the distance the water must rise to fill the tank. 45. (a) Verify by differentiation. (b) a(t) = t(e- I / t + 1 - e- I ) 47. (2m/8)ln2

8.3 The Hyperbolle Funetlons 1. Divide (3) by cosh~. 3. Proceed as in Example 2.

5.

fx (coshx) = i fx

(eX

+ e- X) =

i

(eX - e- X)

= sinhx. 7. Use the reciprocal rule and Exercise 5.

Chapler 8 Answers 9. 11. 13. 15. 17. 19. 21. 23.

(3x 2 + 2x)cosh(x 3 + x 2 + 2) cosh x sinh 5x + 5 cosh 5x sinh x - 8 sin 8x cosh(cos 8x) 4 sinh x cosh x - 3 csch 23x (2 sech 22x)exp(tanh 2x) [sinhx(l + tanhx) - sechx]/(I + tanhx)2 (sinhxXj[dx/(l + tanh 2x)]) + coshx/(1 + tanh 2x) 25. (sinh 3t)/3

11. -3sin3x;";cos23x + I 13. 0.55 15. 1.87 17. Lety = COSh-IX, so X = He Y + e- Y ). Multiply by 2eY , solve the resulting quadratic equation for eY and take logs. 19. Let y = sech -Ix so x = 2/(eY + e-Y). Invert and proceed as in Exercise 17. 21.

27. 2coshß 1 29. cosh 3t + (sinh 3t)/3 31. 2 cosh 61 33.

~tanh-Ix= dx

~ sech y

2

35.

d I =--z:: = I _ tanh y sech'y I - tanh dy

25. 27. 29. 31. 33. 35. 37.

x

- sech y tanh y

-I

-I

x,.jI - sechy

x~

Differentiate the right hand side. Differentiate the right hand side. (l/4)lnl(l + 2x)/(1 - 2x)1 + C (I/2)ln(2x +R+!)+ C In(sin x + ~sin2x + I ) + C (1/2)lnl(l + e X )/(1 - eX)1 + C No

8.5 Separable DiHerential Equatlons

y ------- 1

Y

I

y

-3

A.47

-------

1. Y = sinx + I

2

x

(sinh 3x)/3 + C Inlsinh xl + C (sinh 2x)/4 - x/2 + C e 2x /4 - x/2 + C cosh 3x/3 + C [y - cosh(x + y)]/[cosh(x + y) - x) - 3y sech 23xy /(cosh y + 3x sech 23xy) (a) Xo = 1 (b) d 2x/dl 2 = 2(x - I) 53. Use the definitions of sinhx and coshx. (Don't expand the nth power!)

37. 39. 41. 43. 45. 47. 49. 51.

8.4 The Inverse Hyperbolle Functions 1. 2x;";x 4 +4x 2 + 3 3. (3 - sinx) / Y'(3-x-+-co-s-x-)2=-+-1 5. tanh- 1 (x2

-

I) + 2/(2 - x 2 )

7. [(I + I/y?"=l)(sinh-Ix + x)-

(x + cosh-Ix)(1 + I/v?+l)]/(sinh-Ix + xi 9. [exp(l + sinh - IX »)/ v?+l

3. y = exp(x 2 - 2x + I) - I 5. Y = -2x 7. eY(y - I) = (I/2)ln(x 2 + I) 9. y = 2x + I 11. y=exp(-sinx)+ I 13.

2.1

1.25

3

4

6

15. (a) Q = EC(I - exp( - t / RC» (b) 1 = RC In(lOO) 17. Verify that the equations hold with dx/dt = 0 and dy/dt = O. 19. P = PoA exp(Pokl)/[1 + A exp(Pokt»)

o

21. As T increases, COSh(

mfox ) ~ I, so Y ~ h, which

represents a straight cable.

ehapter 8 Answers

A.48

23. (a)y'= -y/x (b)y' = X/y;y2

35. !im y(x) = I x-->oo

= x 2 + C.

y 6

I I

5

I I I I I I I I I I I I

~}J~J!t~}

4 3

""""

:/""///////

\\\\\\\\

-I

\ \ \

-2

I I I I

37. 61 39. h(y)dy = -

J

25. (a)

x

\

J(1/

g(x»dx

8.6 Linear First-Order Equatlons 1. y = 2 + ( - 31nll - xl + C)(I - x) 3. Y = I + Cexp(x 4 /4) 5. y = -2 + 2 exp(sin x) 7. y=(eX-e)/x 9. The equation is L

~ + RI = Eoeos wl + EI

has solution (b)y' = 3cx 2 (e) y' = -1/3cx 2 ; Y 27. (a) \

\

,

\ \

I

E

+ Ce- tR / L + ~

, I /

I I I / I 11 /

I 1/ / I" / ,-'~ ..... '-

\ ,"'~~\

(!!.

1= E o I sinwl _ weoSWI) L (R/L)2+ W 2 L

= 1/3cx + C

y

,, \ "\ \ ' ..... .....' ~\

,

and

x

R 11. I = EoC - EoC exp( - 1/ RC); 1-+ EoCas1-+ +00. 13. Set y = .9 X 2.51 X Hr and verify the value of I. 15. 6.28 X 105 -(8.28 X 10 5 )exp(-2.67X 10- 11) - (2.01 X 1OS)exp( - 1.07 X 10 - 6/ ) 17. 15 seconds; 951 meters. 19. Use separation of variables to get

v =/mgjy tanh(hg/m I) FMo g 2 2

21. -

2

\

I

y

3

+ MI)

y I

33. }i,IIJ.,y(x) = 3

I -

2Mf

23. Y = -2(x + I) + Ce x

(b)y = kx 2 29. y(l) ~ 2.2469 31. y(l) ~ 0.4683

4

- --(Mo

Mf

-

-

-

------

-4,

I I I

/

/

/

/

/

/

I

I

I

I

I

I

I

I

I

I

I

I

I

I

I

I

I

I

I

I

,, ,,, I

2

3 x

\

,

,

-

,

/4

I

I

/

\

I

\ \

I

\ \ \ \ -4

/

I

,

/

-/

I I

\- I

4, x

,

25. If YI and Y2 are solutions, prove, using methods of Section 8.2, uniqueness for y' = P(x)y and apply it to y = YI - Yl. (This is one of several possible procedures.)

ehapter 8 Answers

27. (a) w' = (l - n)[ Q + Pw] y

(b) Y = ± 1/(xFJC2) 29. (a) v =

~ y- r

-

g(Mo - rt) y - 2r

+ C(Mo -

where C = MJ-y/r( gMo y - 2r

rl)y/r-I

~) y- r

------- 3

and

where the air resistance force is yv. (b) At burnout, v =

~

y- r

_

gM I

y - 2r

-----

-4

4

x

+ cMy/r-l. I

Review Exerclses for Chapter 8 31. 33. 35. 37. 39.

1. y = e3t 5. y = (4e 3t - 1)/3 3. y=(I/ß)sinßt 7. y=4/(4-t4 ) 9. j(x) = e4x 11. j(/) = cosh2t + sinh2t/2 13. X(/) = cos t - sin t 15. x(t) = (sinh 3t)/3 17. Y = -In(l/e + 1- e X ) 19. x(t) = e- 4t 21. y = - t

41. cosh 3x /

23. g(/) = cos(fi73 t - (2/fi73)sin(fi73 I); amplitude is ~19/7 ; phase is -

x = et y = x 2 /2 - x - 2e- x y(x)=sinh5x/sinh5 6x cosh(3x 2) 2x / ~,....(x-2-+-I-/""'--1

/3fi tan -1(2b/7 )

+2

{T+T + 3 sinh 3x sinh -IX

43. (-3/~)exp(l- cosh-I(3x». 45. tan-I(sinhx) + C 47. (l/3)tanh- l (x/3) + C if lxi< 3 (I/3)coth- l (x/3) + C if lxi> 3 49. xcoshx - sinhx + C 51. x(t) = cosJ2.1/5 t 53. (a) k = 640 (b) - 6400 newtons 55. (a) y" + (w 2 - ß)y = 0 (c) x(t) = e-t(coS(ß t) + (l/ß sinß t» ,

x

-1.648

1----271/w =4.1--~ I

571

3v3

25. y

/ I I'':> .:-:; ' , l'

/

,

,

x

, h

r "I

57. 59. 61. 63.

27. y

7

5

3

5

x

66.4 years 54,150 years 27 minutes (a) 73 years (b) S(t) = ke- at where k = S(O)

A.49

A.50

ehapter 9 Answers

65.

EI R

-------------

69. 15.2 minutes, no. [Tbe "no" could be "yes" if you allow a faster addition of fresh water after draining.] 71. 1= 2(3 sin '/TI - '/T cos'/T/)/(9 + '/T2) + [1 + 2'/T /(9 + '/T 2 )]e- 3t

73. y = -4/3 + Ce 3x y

14

13 1~

11

67. (a) y2/9 + x 2 = k, k = 2C /9 y

,.3

I" 1,,

.

"\ \ \ \

"

\ \

'I ,\,, ,I ,, ,' '", , I I ,,

\

I I I I

111 -I

\

11 11

, 1, , , I I , I

x

, ,\.' ,,'/,./""

, , I

\1\ \ ,

I , ,

\ \ \ \ \\

·- 2

\

\

\

-3

\

\ \ \

-4

\

/

/

x

/

""" \

\

\

\

\

\

75. 1 77. Y

~I ,

I I I

f--1

"

-3

= e X is the exact solution; y(1) = e ~ 2.71828. 79. Y = -I/(x - 1) is the exact solution, it is not defined at x = 1. 81. Y = Ce Qt - (b/a); the answers are all the same;. 83. (a) Verify using the chain rule (b) Integrate the relation in (a) (e) Solve for T = I; the period is twiee the time to go from (J = 0 to (J = (Jo . 85. (a) y = eosh(x + a) or y = 1. (b) Area under curve equals are length.

Chapter 9 Answers f!.1 Volumes by the Slice Method 1. 3'/T 3. Ah/3

15. 38'/T }'

5. 2125/54

7. 4,ß /3 9.

XI

= (1-

3ff74)h,

(1-

Vf/4)h

X3 =

X2

= (1- Vfli)h, 3

11. 0.022 m3 13. 1487.5 em 3 x

x

Chapter 9 Answers

A.51

5. 71(17 + 4v'2 - 6/3)/3

19. 7171/105

y

)'

6

4

2

Je

23. 1371 \'

8

o

6

9. 971 (See tbe Figure for Exercise 23, in tbe left-hand column.) 11. 971 )' .

4

8

6 4

6

,..

4

25. 1371 (See Exercise 11, Section 9,2 for tbe figure,) 27. 5 cm 3 29. V

= 71 2(R + r)(R

- r)2/4

31. For tbe two solids, A I(X) = A 2(x). Now use tbe

4

slice metbod',

9.2 Volumes by Shell Method

6

x

13.471/5 (You get a cylinder when this volume is added to that of Example 5, Sec ti on 9,1.) y

"

Je

Je

15. /3 ",/2 17. 24",2 19. (a) V = 4",r 2h + "'h 3/3 (b) 4",,2, it is the surface area of a sphere, 21. (a) 271 2a2b (b) 2",2b(2ah (c) 471 2ab

23. ",3/4 -

",2

+ h 2)

+ 2",

ehapler 9 Answers

A.S2

9.3 Average Values and the Mean Value Theorem tor Integrals 1. 1/4

3.ln..ß72

2 7. '/T/4 11. ~2/3'/T '/T/2 - 1 9 +ß 15. 1/2 55° F 19. (a) x 2/3 + 3x/2 + 2 (b) Tbe function approaches 2, which is the value of f(x) at x = O.

5. 9. 13. 17.

17. x = 3 (ln 3)/2, y = 26/27 19. x = 4/(3'/T), Y = 4/(3'/T) 21. x. = 4/3,y = 2/3 mx + mx + m3 x 3 •• 2 2 23. Smce Xj " b, x = + + m. m2 m3 m.b + m2b + m3b " m. + m2 + m3 = b. Similarly a " x. Tbe center of mass does not lie outside the group of masses. x to get the velocity of the center of mass and use the definitions of P and M.

25. Differentiate 27.

x = -4/21, Y = 0 y

21. Use the fundamental theorem of calculus and the definition of average value. 23. Tbe average of [J(x) + k) is k + [the averge of f(x»). x

\'

-

-

[(x) +k

"" '"

(x)

a

25. f(b) - f(a)

=

i

b

b j'(x)

dx

[(x)

+k

29. 31.

[(x)

x

= j'(c)' (b

- a), for

some c such that a < c < b.

x=

(Jin/4 - 1)/(Ji - 1), Y = 1/[4(Ji - 1») (YI + Y2 + Y3)/3

x = (XI + x2 + xN3,y =

Supplement to 9.5: Integrating Sunshine

27. exp [ .eIn fex) dx/(b - a) ]

1. The arctic circle receives 1.25 times as much energy as the equator.

29. Write F(x) - F(xo) = ff(s)ds. If If(s)1 " M on

3. (a) F = ~ ,{ Jcos 21- sin2D

Xo

[a,b) (extreme value theorem), IF(x) - F(xo)1 " Mix - xol, so given E> 0, let 6 = E/ M.

9.4 Center ot Mass

3. 5.

7.

9.

m.x. + m2x2 + m3x3 m. + m2+ m3 Let M. = m. + m2 + m3 and M 2 = m4' x=3 x = 67 x = I, Y = 4/3

11. x = 29/23, Y = 21/23 13. (a) x = 1/2,y =ß /6 (b)

x = 3/8,y =ß /8

15. m.x.

+ (m2 + m3 + m4) [

364

T-O

+

sin I sin D cos-·( - tan ltan D)} (b) Expressing sinD in terms of T, the sum in (a) yie1ds

.J;'"{J'''~

-"n'.

,,,'(2.T/365)

+ sin I sin a cos(2'/TT/365) X cos -

I[

-

tan I sin a cos(2'/TT/365) ]}dT.

)1 - sin a cos (2'/TT /365) 2

2

This is an "elliptic integral" which you cannot evaluate.

5. nsinlsinD m2x2 + m3 x 3 + m4 x 4 ] m2 + m3+ m4

m. + (m2 + m3 + m4) m.x. + m2x2 + m3x3 + m4x4 m. + m2 + m3 + m4

7. 0.294; it is consistent with the graph (T = 16.5; about July 7).

Chapter 10 Answers

33. 120/ 7T joules

9.5 Energy, Power, and Work 1,890,000 joules 360 + 96/7T watt-hours 3/2 7. 0.232 98 watts 11. (a) 18t 2 joules (b) 360 watts 13. 1.5 joules 15. (a) 45,000 joules (b) 69.3 meters/seeond 19. 125,685,000 joules 17. 41,895,000 joules 23. 1.48 X 108 joules 21. 0.15 joules 1. 3. 5. 9.

Review Exereises for Chapter 9 1. (a) n2 /2 (b) 2n 2

under the graph y = fex), 0.;;; x .;;; a, revolved about the y-axis. 37. (a) The force on a slab of height fex) and width dx is dx fo!(X)pgy dy

=

t

pg[f(x)f dx. Now in-

tegrate. (b) If the graph of fis revolved about the x axis, the total force is pg/27T times the volume of the solid. (e)

"32 Pg X

106

= 6.53

X

10 9 Newtons.

(e) Show that if the standard deviation is 0, k; - /L = 0, whieh implies k; = /L.

17. 1/3,4/45,215 /15 19. I, (e 2 - 5)/4, ~ /2 21. 3/2, 1/4, 1/2

(d)

23. (a) 7Tfp(X)[f(x)fdx

27. 29. 31.

35. pgw fx2[f(a) - f(x)]j'(x)dx; the region is that

3. (a) 3n/2 (b) 2n(21n2 - 1)

5. 64.j2 7T /81 7. 72n/5 9. 5/4 11. I 13. 6 15. Apply the mean value theorem for integrals.

25.

A.53

(b) (l47T /45) grams x = 5/3,y = 40/9 x = I / 4(2 In 2 - I), Y = 2(ln 2 x = 27/35, Y = -12/245 (a) 7500 - 21OOe- 6 joules (bH(l25 - 35e- 6 ) watts

IN (2 In 2 -

I)

{l ±[ai -l ±a;]2} 1/2 n

n

j=1

;=1

(e) All numbers in the list are equal. 41. Let g(x) = f(ax) - c. Adjust a so g has zero integral. Apply the mean value theorem for integrals to g. (There may be other solutions as weil.) 43. The average value of the logarithmie derivative is ln[f(b)/f(a)l/(b - a).

Chapter 10 Answers 10.1 Trigonometrie Integrals 1. (eos 6x)/6 - (eos 4x)/4 + C 3. h/4 5. (sin 2x)/4 - x/2 + C 7.1/4- 7T/16 9. (sin 2x)/4 - (sin 6x)/ 12 + C 11. 0 13. -1/(3eos3x)+ 1/(5eos5x)+ C 15. The answers are both tan -IX + C 17.

Jx

2 -

29. 1,0,1/2,0,3/8,0,5/16.

+ 1»-1 y = (tan- 12 - 7T/4)/[2In«ß + 2)/(.j2 +

31. x=(ß -.j2)/ln«ß +2)/(.j2

35. ,[3,9.j2 /4

37. (a) Differentiate [S(t)]3 and integrate the new expression. (b) [3(-teost + sint + t/8 - (l/32)sin4t)]1/3 (e) Zeros at t = nw, n a positive integer. Maxima oeeur when n is odd.

4 - 2eos-1(2/x) + C

19. (l /2)(sin -lU + u[f"7) + C 21.

,f4+? +C

10.2 Partial Fraetions

23. (-1/3)F? (x 2 + 8) + C

25. (1/2)sinh- I«8x + 1)/[i5) + C 27. __1_ Inl6X + I +

6,[3

[i3

(6x

+ 1)2

13

I))]

33. 125

I -I +C

1. (l / l25){ 41n[(x 2 + 1)/(x 2 - 4x + 4)] + (37 /2)tan -Ix + (15x - 20)/2(1 + x 2 )_ 5/(x-2)}+C 3. 5/4 - 37T/8 5. (I/5){ln(x + (3/2)ln(x 2 + 2x + 2)tan-I(x + I)} + C

2i

A.54

ehapler 10 Answers

7. 2 + (l/3)ln3 + (2/ß)(tan- I (5/ß)tan -1 9. (I/S)ln«x 2 - 1)/(x 2 + 3» + C 11. (I /2)ln(5/2) 13. 2..[X - 2 tan -I..[X + C 15. i(x 2 + 1)4/3 + C 17. -2/(1 + tan(x/2» + C 19. w/16 - (1/4)lnl(l + tan(w /S»(I + 2 tan(w /S)tan2(w/S»I~ -0.017 21. wIn(225/ 176)

(3J3»

23. 3(1 + Xf/3/4 + (3/4 V4 )lnl 3 l4 (I + x)2/3 + (2 + 2X)I/3 + II - (1/2 \1432)tan -1[(2(4 + 4X)I/3

Vi /6"fiOS] + C

+

Illx-SOI 1 1n"3 4 25. (a) 20 n x _ 60 = kt + 20 (b) x =

SO(I - e- 20kl ) 4

]-e

-20kl

31. (a)

f

JI + (I - 1/ X 2)2 dx

(b) 2w f(l/x + x)~ri-+-(I-_-I/-x-2)~2 dx 33. Dividing the curve into I mm segments and revolving these, we get about 16 cm2.

35. Use Isinß xl < I to get L 0), horizontal tangents at 1 = n7T, n a nonzero integer. The slope is - 1/2 at 1 = 0 although the eurve ends.

37. (a) (h) (e) (d)

y

x

We estimated about 338 miles. We estimated about 688 mi1es. It would probably he longer. The measurement would depend on the definition and seale of the map used. (e) From the World Almanac and Book 0/ Facls (1974), Newspaper Enterprise Assoe., New York, 1973, p. 744, we have eoastline: 228 miles, shoreline: 3,478 miles.

10.5 Length and Area in Polar Coordinates

-I

21.

y2 = (l - x)/2, vertieal tangents at 1 = n7T, n an integer

y

1. 12/2

3. (4/3)(13 3/ 2 - 8) 5.9'TT/4 Ji

3

x

23. (133/ 2 - 8)/27 25. (1/2)(.;5 + (l/2)ln(2 +.;5)] 27. (a) Calculate the speed direetly to show it equals

lai-

lal(/1 -

(h) Calculate direetly to get 29. (a) y = -x/2 + 'TT/2 - I

10)

y

1f

x

9. 33'TT/2

2 x

Ji

(e) 31. 5 33. (a)

f'{5 -

3 eos2(J - 2eos (J

d(J

-4

4

x=

k(eos wl - wl sin w/); j = k(sin wl + wl eos w/). (h) k{1 + w 2/ 2

(e) 2mkw 35. (a) x = 1 + (l + 4/ 2)-1/2, Y = 12 + 2/(1 + 4/ 2)-1 / 2 (b) x = ±(1/2)...j1/(x 2

-

y) - 1 + ";x 2 - y.

A.55

3

11. 2'TTr

x

A.56

Chapter 10 Answers

13. S =J"'/2

-",/2

23. A

~sec\0/2)/4 + tan2(0/2) dO

L

= ",/2 = 2", + 8

A =2 - ",/2

y

y

x

r

15. s

= 10.,,/4 ~sec2fj tan2fj + sec2fj + 4sec 0 + 4 dO

A = 1/2 + ",/2 + In(3 + 2/2)

25. /2(e 2(n+I)'1T _ e211'lT) 27. (a) Use x = acost,y = bsint, where T= 2",. (b) Substitute into the given formula.

y

/

/

/

/

r

;

Review Exerclses for Chapter 10

= 3.414

1. sin3x + C 3. (cos2x)/4 - (cos8x)/16 + C 5. (1- X 2)3/2 +C

2

3

-F?

x

17. s =

1o'1t/2~(l + cosO -

0 sin 0)2 + 0 2(1 + 2 cosO +

cos~.)

A = (1/2)[",3/16 + ",2/2 - 4 - ",/8]

19. s = 1o"'/2J(5 + 4 sin 20)/(1 + 2 sin 20) dO A

21. A L

= ",/2 = =

(1/4X5",/6-ß) (2 +ß)",/6 y

dO

7.4(x/4-tan- l (x/4»+C 9. (2tr /7)tan- I[(2x + I)/tr] + C 11. Inl(x + I)/xl-I/x + C 13. (1/2)[lnlx 2 + I1 + 1/(x 2 + I)] + C 15. tan-I(x + 2) + C 17. - 2jX cosjX + 2 sinjX + C 19. -(1/2a)cot(ax/2) - (1/6a)cot3(ax/2) + C 21. Inlsecx + tanxl- sinx + C 23. (tan-lxi /2 + C . 25. (l/3W)[lnlx-

WI-ln~x2+ Wx+3 3ß

+ßtan- I«2x/W + I)/ß)]+ C.

27. 2jX e./x - e./x + C 29. x - In(e X + I) + C

31. (-1/4)[(2x 2 - 1)/(x 2 - If] + C 33. -(1/1O)cos5x - (l/2)cosx + C

35. In~

+C

37. 2e'1x + C 39. tln2 41. tln(x 2 + 3) + C 43. x 4lnx/4 - x 4 /16 + C 45. H(ln6 + 5)4 - (ln 3 + 5)4]R:: 186.12 47. (1/4)sinh2 - 1/2

49.0 51. (733 3/ 2 - 43 / 2)/243 53. 59/24 55. ",(53 / 2 - 1)/6 57. R:: 31103

Chapter 10 Answers 59. x

=

(y

A.57

89. (a) b - a + (b n+1 - an+I)/(n + I) if n -:!= -I. If n = -I, we have b - a + In(b/a).

+ 1)2

y

(b) n = 0: L = b - a; n = 1: L = {2 (b - a); n = 2: see Example 3 of Section 10.3; for n = (2k + 3)j(2k + 2), k = 0, 1,2,3, ... L

-1

={

I/(I-n)

_n._ _ (I

n- I

+ n 2x 2n - 2)3/2

-2

n =t :L

2:k

(_I)k- j

(~). (I + n 2x 2n - 2'/ j=o] 2] + 3

= ir[(4 + 9b)3/2 -

}IX=b x=a

(4 + 9a)3/2].

(e) Around the x-axis, we have

61. Y

=

2x /3

w[ b - a +

+1

b2n+ 1 _ a2n+ 1 2(bn+ 1 - a n+l ) n+ 1 + 2n + 1

]

if n -:!= - I or - 1/2. For n = - 1 we have

w[ b - a + 2In(b/a) -

(a- I - b-I)J.

For n = -1/2 we have

'17 [ b - a + 4yb - 4{ti +In(b/a)]Around the y-axis we have

2(bn+2 _ a n+2) ]

w[ b 2 - a 2 + 63. x

= O,y > 0

----,--;;--n+2

if n -:!= -2. For n = -2, we have 'I7[b 2 - a 2 + 2ln(b / a)]. (d) A", = 2wL (from 89(b» + A x (from 88(d» A y = Ay (from 88(d» Some answers from 88(d) needed here are: 88(d).

y

n = 0; Ax

x

= 2'17(b -

a)

n = I; A x ={2 w(b 2 - a 2)

n = 2; Ax = ;;

65. Y = 3x/4 + 5/4 67. (1/8)(,;B7 . 16 + Inl,;B7 + 161) 69. L=(1/3)[(w 2 /4+4)3/2_8] A = wS /320 71. L = rJ'(5-/-4-)-+-e-os-2-0-+-3-si-n2:-2-0 dO

x=b [(I + 8x 2)2xv'1+'47 -ln(2x +v'1+'47) JL=a n = 3; A x

=

w

4

27 (1 + 9x)

3/2I x -

b x=a

n = (2k + 3)j(2k + 1); k = 0, 1,2,3, ... A x = ~n(l+n)/(I-n)(1 n -

A = 3'17/8

73. L = 5{2 A = 315w/256 + 9/4

75. b 2 = I, an others are zero. 77. a3 = I, all others are zero. 79. a4 = 3, an others are zero. 81. ao = I, a2 = -1/2, an others are zero. 83. (a) (1/ k 2)In[NoCk I N(t) - k 2 )/ N(t)(kINo - k 2)] (b) N(t) = k, N o/[k I N o(1 - e k2') + k 2e k2'] (e) The limit exists if k 2 > 0 and it equals k 2 /k l • 85. Use (eOScf»dcf> = (sincf>m)(eos ß)dß and substitute. 87. a- I/ 2InI2ax + b + 2 {ti ~ax2 + bx + c 1+ C, a>O (- a)-1/2sin- I[( -2ax - b)!Vb 2 - 4ac] + C, a 0 in each case). 51. (a) Add alI three forces (b) Substitute and differentiate. 53. cle A + C2eiA + C3e - A + c"e'- jA where A= (l + i)/,f2 or

(_I)nx2n+1

00

[

I

9. Y = ao ( I + - + - + ... 6 180

[e 2Z sin2x/21J (e 2X [(1 - tan2x)' (cos2x) +

(I + tan2x)(sin2x)](1 + cos2x)} -It/x 37. x= -cos2t + cost=2sin(3t/2)sin(t/2) 39. x = (-1/24)cos5t + (1/5)sin5t + (l/24)cost 41. (a)

I

5. Y = x - x 3 /3 + x 5/ 10 - ... 7. y=2x-x 3 +7x 5/20-'"

+ rot 2x)(sin 2x)] . (l + ros2x)} -I dx +

(I

55.

y

60

x 7/ 3 + X '0/ 3 7 140

or XI / 6 [ blCOS( 11 ~nx )

)

1920 _

+

x 13 / 3 + ... ) 5460

+ b2sin( 11 ~nx )]

(no

further terms). k x k +2 X k +4 15. (a) x + 4k + 4 + (4k + 4)(8k + 16) + ...

+

x k + 2j

+ 1)(2k +4)··' (jk + /) -k+2 (b) x- k + ~4k + 4 -k+4

+

+

4

j (k

(-4k

,

X

+ 4)( -8k + 16)

+ ...

+ ...

X- k + 2j

4 J( - k + 1)( - 2k + 4) ... ( - jk + /)

+ ...

17. Solve recursively for roefficients, then rerognize the series for sine and cosine. 19. (a) Use the ratio test (b) x, - t + ~X2, - ~x + !X 3 21. Show that the Wronskian is non-zero 23. (a) Solve recursively (b) Substitute the given function in the equation. (To discover the solution, use the methods for solving first order linear equations given in Section 8.6).

X

53.

12.8 Serles Solutlons of Differential Equatlons 1. y = ao [

a I

[

~ 00

1-0

L 00

n-O

2:2n] + n. 1

2nl n. x 2n + 1 ] (2n + I)!

Review Exercises for Chapter 12 1. 3. 7. 11. 15. 19.

Converges to 1/11. Converges to 45/2 5. Converges to 7/2 9. Converges Diverges 13. Converges Converges 17. Diverges Diverges 21. Converges Converges

ehapter 12 Answers 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47.

Jxsin(2x) dx R+T + sin(2x) J dx

Converges 0.78 -0.12 -0.24 0.25 False False False False True True True Use the comparison test.

93. -cos(2x)

xcos(2x)

R+T

95. CI + e-X(c2cOsx + C3sinx) 97. m = I, k = 9, y = I, F o = I, g = 2, '" = 3, II = tan -10). As t ~ 00, the solution approaches

focos[2t - tan-l(~)].

99. m = I, k = 25, y = 6, F= I, g = '!T, '" = 5, II = tan- I [6'!T1(25 - '!T 2)]. As t~ 00, the solution apo proaches I COS['!Tt - tan- I( ~)] J625 - 14'!T2 - '!T 4 25 - 77 2

49. 1/8 51. I 53.R=00 55.R=00 57. R = 2

101. ao( 1-

~3

103. ao( I _ x2

+al ( x 61. 63.

65.

105. I - x +

Li"=I[(-I)i+lx 4i li]

Lf. 1[( - I )iX 2i I (2i)!] Li"= I[X i I[i(i!)]]

107. x +

x "8 3

+ ... ) + a l ( x _

~4

+ ... )

+ x64 _ ~5 + ... ) 4

x 3x3 +"6 -

x T 2

x5

20

+ .. , )

11x 3

-6- + ...

x5 + 192 + ...

l1 c, y = R (b) O.OI998e-19.90t - O.02020e-O.l00St + 0.002099 sin(60xt) + 0.0002227 cos(60xt)

109. (a) m = L, k =

67. Lf_ole 2(x - 2Y/i!], R = 00 ($J Hf-I)"'O-i+l) . 69. ~ ., (x - I)' i=O

A.67

I.

71. '!T 2 /2 73. 3 75. 3,7,3 -7i,/58 77. ±(I+!ß)I/2~±1.46, +(-I+!l5)1/2~ ;:0.344, ~2 + i ~ ± 1.46 ± 0.344i, 4,j5 ~ 1.50 79. z =,fIexp(-'!Ti/4) y

111. Factor out x 2• 113. (a) The partial sums converge to y(x, t) for each (x, t). (b) Lk-O( -llA 2k + I 115. ~0.659178 117. (a) ~ 1.12 (b) ~ 2.24. It is accurate to within 0.02. 119. -1/2,1/6,0 121. (a) ~ 3.68 (b) y

x

2

-1

81. z = exp('!Ti) y

x

2 x

83. I ±~I - '!Ti ~2.4658 - 1.0717i,

m

- 0.4658 + 1.07 85. clcos2x + c2sin2x 87. y = clexp( - 5x) + C2exp( - x) 89. y = - eX 16 - 11 cos xl 130 + 33 sin xl 130 + clexp( - 5x) + c2exp(2x) 140 (x) 48. 91 • cle + C2 xe ]x + 1369 cos 2" - 1369 sm 2"

3x

(X)

123. Show by induction that g(n)(x) is a polynomial times g(x). 125. True 127, (a) Collect terms (b) The radius of convergence is at most I. (c) e 129. Show that a < I1 k by using a Maclaurin series with remainder.

Index Includes Volumes 1and 11 Note: Pages 1-336 refer to Volume I; pages 337-644 refer to Volume 11.

Abel, Nils Hendrik 172 absolute vaIue 22 funetion 42, 72 properties of 23 absolutely eonvergent 574 accelerating 160 aeeeieration 102, 131 gravitationaI 446 Aehilles and tortoise 568 addition fonnulas, 259 air resistanee 136 Airy's equation 640 aIgebraie operations on power series 591 aIgebraie rules 16 aIternating series test 573 amplitude 372 analytie 600 angular frequeney 373 rnomentum 506 annuaI percentage rate 382 antiderivative 128 of Ir 323, 342 of eonstant mutiple 130, 338 of exponentiaI 342 of hyperbolie funetions 389 of inverse trigonometrie funetion 341 of lIx 323, 342 of polynomiaI 130 of power 130, 338 mies 337, 338 of sum 130, 338 of trigonometrie funetion 340

of trigonometrie funetions 269 Apostol, Tom M. 582 applieations of the integral, 240 approaches 58 approximation, first-order (see linear approximation) approximation, linear (see linear approximation) arc length 477 in polar coordinates 500 Archimedes 3,5,6 area 4,251 between graphs 211, 241 between intersecting graphs 242 in polar coordinates 502 of a sector 252 signed 215 of a surface 482 of a surface of revolution 483 under graph 208, 212, 229 under graph of step funetion 210 argument 40 arithmetie mean 188 arithmetic-geometrie mean inequality 436 astroid 198 astronomy 9 asymptote 165 horizontal 165, 513, 535 vertieal 164, 518, 531 asymptotie 164 average 3 power 464, 465

1.2

Index average (cont.) rate of change 100 value 434 velocity 50 weighted 437 axes 29 axial symmetry 423 axis of symmetry 440

8·1) definition of limit 516 ball 421 Bascom, Willard 306(fn) base of logarithm 313 beats 628 Beckman, P. 251, fn Berkeley, Bishop George 6(fn) Bemoulli, J. 252(fn), 521 equation 414 numbers 643 Bessel, F.W. 639 equation 639 functions 643 binomial series 600 bisection, method of 142, 145 blows up 399 bouncing ball 549 bounded above 575 Boyce, William 401 Boyer, C. 7(fn), 252(fn) Braun, Martin 380,401, 414, 426 Burton, Robert 8 bus, motion of 49, 202, 207, 225

Calculator discussion 49, 112, 166, 255, 257, 265, 277, 309, 327, 330, 541 calculator symbol 29 calculus differential 1 fundamental theorem of 4, 225, 237 integral 1,3 Calculus Unlirnited iii, 7(fn) capacitor equation 406 carbon-14 383 ClU'dano, Girolamo 172 cardioid 298 cartesian coordinates 255 catastrophe cusp 176 theory 176 catenary 402 Cauchy, Augustin-Louis 6, 521 mean value theorem 526 Cavalieri, Bonaventura 8,425 center of mass 437 in the plane 439 of region under graph 441

of triangular region 445 chain rule 112 physical model 116 change average rate of 100 instantaneous rate of 10 linear or proportional 100 proportional 95 rate of 2, 100, 101, 247 of sign 146 total 244 chaos, in Newton's method 547 characteristic equation 617 chemical reaction rates 407 circle 34,44, 120,251,421 equations of 37 parametric equations of 490 circuit, electric 413 circular functions 385 circumference 251 city Fat 116 Thin 115 climate 180 closed interval 21 test 181 College, George 383 common sense 61, 193 comparison test 570 for improper integrals 530 for limits 518 for sequences 543 completing the square 16, 17, 463 complex number 607, 609 argument of 611 conjugate of 611 imaginary part of 611 length or absolute value of 611 polar representation of 614 properties of 612 real part of 611 composition of functions 112, 113 derivative of 113 concave downward 158 upward 158 concavity, second derivative test for 159 conditionally convergent 574 conoid 486 conservation of energy 372 consolidation principle 438 constant function 41, 192 derivative of 54 rule for limits 62, 511 constant multiple rule for antiderivatives 130 for derivatives 77 for limits 62

Index for series 566 eonsumer's surplus 248 eontinuity 63, 72 of rational funetions 140 eontinuous 139 eontinuous flmetion 63 integrability of 219 continuously compounded interest 331, 382, 416 convergence, absolute 574 conditional 574 of series 562 of Taylor series 597 radius of 587 eonvergent integral 529 convex funetion 199 eooling, Newton's law of 378 coordinates 29 eartesian 255 polar 253, 255 eoriolis force 499 eosecant 256 inverse 285 eosine 254 derivative of 266 hyperbolic 385 inverse 283 series for 600 eosines, law of 258 cost, marginal 106 cotangent 256 inverse 285 Creese, T.M. 401 critical points 151 critically damped 621 cubic function 168 general, roots and graphing 172 curve 31(fn) parametrie 124, 298, 489 cusp 170 catastrophe 176 eycloid 497

dam 454 damped force oseillations 628 damping 377 in simple harmonie motion 415 Davis, Phillip 550 day length of 30, 302 shortening of 303 decay 378 decimal approximations 538 declerating 160 decrease, rate of 101 deereasing function 146 definite integral 232

1.3

by substitution 355 eonstant multiple ruIe for 339 endpoint additivity rule for 339 inequality rule 339 power rule 339 properties of 234, 339 sum rule 339 wrong-way 339 degree as angular measure 252 of polynomial and rational functions 97 delicatessen, Cavalieri's 425 delta 50(fn) demand eurve 248 Demoivre, Abraham 614 formula 614 density 440 uniform 440 depreciation 109 derivative 3, 53, 70 of Ir 318 of eomposition of funetions 113 of constant multiple 77 of eosine 266 formal definition of 70 of hyperbolie funetions 388 of implicitly defined funetion 122 of integer power 87 of integral with respect to endpoint 236 of integral whose endpoint is a given function 236 of inverse hyperbolic funetions 396 of inverse funetion 278 of inverse trigonometrie funetions 285 Leibniz notation for 73 as a limit 69 of linear funetion 54 logarithmie 117, 322, 329 of logarithmie funetion 321 of lIx 71 of polynomial 75, 79 of power 75, 119 of power of a function 110, 119 of produet 82 of quadratie function 54 of quotient 85 of rational power 119 of rational power of a funetion 119 of reciprocal 85 of sum 78 second 99, 104, 157 summary of rules 88 of Vx 71 Dido 182 differenee quotient 53 differentiable 70 differential algebra 356

1.4

Index differential (cont.) ealeulus 1 equation 369 Airy's 369 Bessel' s 639 first order 369 harmonie oseillator 370 Hermite' s 636 Legendre's 635 linear first order 369 numerieal methods for 405 of growth and decay 379 of motion 369 seeond order linear 617 separable 398, 399 series solutions 632 solution of 39 spring 370 Tehebyehefrs, 640 differential notation 351, 359, 374, 398 differentiation 3, 53, 122, 201 implicit 120, 398 logarithmie 117, 322, 329 of power series 590 diminishing returns, law of 106 Diprima, Riehard 390, 401 direetion field 403 diseriminant 17 disk 421 method 423 displaeement 230 distanee formula, in the plane 30 on the line 23 divergent integral 529 domain 41 double-angle formulas 259 drag 136 resistanee 414 dummy index, 203

e, 319, 325 as a limit, 330 E-A definition of limit 513 E-8 definition of limit 509 ear popping 116 earth's axis, inelination of 301 economies 105 electrie eireuit 399, 413 element 21 elliptie integral 417, 506, 507 endpoints 181 of integration 17 energy 201,445 conservation of 372 potential 446 equation of circle and parabola 37

differential (see also differential equation) pararnetric 124, 298 simultaneous, 37 spring 376 of straight line 32 of tangent line 90 error function 558 Eudoxus 4 Euler, Leonhard 251(fn), 252(fn), 369 method 404 equation 636 formula 608 evaluating 40 even function 164, 175 exhaustion method of 5, 7 existence theorem 180, 219 exponent, zero 23 exponential function 307 derivative of 320 graphing problems 326 limiting behavior 328 growth 332 series 600 spiral 310, 333 exponentiation 23 exponents integer 23 laws of 25 negative 26 rational 27, 118 real, 308 extended product rule for limits 62 extended sum rule for limits .62, 69 extensive quantity 445 extreme value theorem 180

factoring 16 falling object 412, 414 Feigenbaum, Mitchell J. 548 Ferguson, Helaman 602 Fermat, Pierre de 8 Fine, H.B. 468 first derivative test 153 first-order aproximation (see linear approximation) ftying saucer 430 focusing property of parabolas 36, 95, 97 football 453 force 448 on a darn 454 forced oscillations 415, 626 Fourier coefficients 506 fractals 499 fractional (see rational) frequeney 259

Index friction 377 Frobenius, George 636 frustum 485 funetion 1, 39 absolute value 42, 72, 73 average value of 434 eomposition of 112, 113 eonstant 41, 192 continuous 63 eonvex 199 eubie 168 definition of 41 differentiation of 268 even 164, 175 exponential 307 graph of 41, 44 greatest integer 224 hyperbolic 384, 385 identity 40, 277, 384, 385 inverse 272, 274 inverse hyperbolie 392 inverse trigonometrie 281, 285 linear 192 odd 164,175 piecewise linear 480 power 307 rational 63 squaring 41 step 140, 209, 210 trigonometrie, antiderivative of 269 trigonometrie, graph of 260 zero 41 fundamental integration method 226 set 630 fundamental theorem of ealeulus 4, 225, 237 alternative version 236 Galileo 8 gamma funetion 643 Gauss, Carl Friedrieh 205, 615 Gear, Charles W. 405 Gelbaum, Bernard R. 576, 600 general solution 618, 623 geometrie mean 188, 436 series 564,600 global 141, 177 Goldstein Larry 172 Gould, S.H. 6(fn) graphing in polar coordinates 296 graphing problems exponential and logarithmie funetions 326 trigonometrie funetions 292 graphing procedure 163 graphs 41, 163 area between 241

area under 212, 229 of funetions 41, 44 gravitational acceleration 446 greatest integer funetion 224 growth 378 and deeay equation, solution 379 exponential 332 half-life 381, 383 hanging eable 401 Haraliek, R.M. 401 Hardin, Garrett 416 harmonie series 567 Hermite polynomial 636 Hermite's equation 636 herring 156 Hipparchus 256(fn) Hofstadter, Douglas 548 Hölder eondition 559 homogenous equation 623 Hooke's Law 99,295 horizontal asymptote 165, 513, 535 horizontal tangent 193 horsepower 446 horseraee theorem 193 hyperbolie eosine 385 hyperbolie funetions 384, 385 derivatives 388 antiderivatives 389 inverse 392 hyperbolie sine 385 inverse 393 I method 361

identity funetion 40, 277 rule for limits 60 identity, trigonometrie 257 illumination 183 imaginary axis 609 imaginary numbers 18 implicit differentiation 120, 122, 398 improper integrals 528, 529 comparison test for 530 inelination of the earth' s axis 30 1 inerease, rate of 101 increasing funetion 146 test 148 theorem 195 sequence property 575 inereasing on an interval 149 indefinite integral (see antiderivative) indefinite integral test 233 independent variable 40 indeterminate form 521 index dummy 203

1.5

1.6

Index index (cont.) substitution of 205 indicial equation 638 induetion, principle of 69 inequality 18 arithmetie-geometrie mean 188, 436 Minkowski' s 365 properties of 19 infinite limit 66 infinite series 561 infinite sum 561 infinitesimals 73 method of 6, 8 infinity 21 infleetion point 159 test for 160 initial eonditions 371, 398 instantaneous quantity 445 instantaneous velocity 50, 51 integer power rule for derivatives 87 integers 15 sum of the first n 204 mtegrability of eontinuous funetion 219 intehable 217 integral 217 ealeulated "by hand" 212 ealeulus 1 eonvergent 529 definite 232 definition of 217 divergent 529 elliptie 417 of hyperbolie funetion indefinite 129 (see also antiderivative) improper 528, 529 of inverse funetion 362 Leibniz notation for 132 mean value theorem 239 mean value theorem 435 of rational funetions 469 of rational expression in sin x and eos x 475 Riemann 220 sign 129, 132, 217 tables 356 trigonometrie 457, 458 of unbounded funetions 531 wrong way 235 integrand 129 integration 33, 129, 201 applieations of 420 by parts 358 by substitution 347, 348, 352 endpoint of 217 limit of 217 method, fundamental 226 methods of 337 numerieal 550

of power series 590 intensity of sunshine 451 interest, compound 244, 331 intermediate value theorem 141, 142 intersecting graphs, area between 242 intersection points 39 interval 21 elosed 21 open 19 inverse eosecant 285 eosine 283 eotangent 285 funetion 272, 274 integral of 362 rule 278 test 276 hyperbolie funetions 392 integrals 396 derivatives 396 hyperbolie sine 393 secant 285 sine 281 tangent 283 trigonometrie funetions 281, 285 invertibility, test for 275, 276 irrational numbers 16 ith term test 567

joule 445

Kadanoff, Leo 548 Keisler, H. Jerome 7(fn), 73(fn) Kelvin, Lord 594 Kendrew, W.G. 180 Kepler, Johannes 8 second law 506 Kilowatt-hour 446 kinetie energy 446 Kline, Morris 182

I'Höpital, Guillaume 521 rule 522, 523, 525 labor 106 ladder 190 Lagrange's interpolation polynomial 556 Laguerre funetions 640 Lambert, Johann Heinrich 251(fn) latitude 300 law of mass action 476 law of refleetion 290 Legendre, Adrien Marie 251(fn) equation 635 polynomial 635 Leibniz, Gottfried 3, 73, 193(fn), 594

Index notation 73, 104, 132, 217 for derivative 73 for integral 132 lemniscate 136 length of curves 477 of days 300, 302 of parametrie eurve 495 librations 506 limac;on 298 limit 6, 57, 59 at infinity 65, 512 eomparison test 518 of (eos x - 1)/x 265 derivative as a 69 derived properties of 62 e-& definition of 509 of function '509 infinite 66 of integration 217 method 6 one-sided 65, 517 of powers 542 product rule 511 properties of 60, 511 reeiproeal rule 511 of sequence 537, 540 properties 563 of (sin x)/x 265 line 31(fn) equation of 32 perpendicular 33 point-point form 32 point-slope form 32 real number 18 secant 51, 191 slope of 52 slope-intereept form 32 straight 31(fn), 125 tangent 2, 191 linear approximation 90,91,92, 158, 159, 601 linear funetion 192 derivative of 54 linear or proportional change 100 linearized oseillations 375 Lipshitz condition 559 Lissajous figure 507 loeal 141, 151, 177 maximum point 151, 157 minimum point 151, 157 logarithm 313 base of 313 defined as integral 326 function, derivative of 321 laws of 314 limiting behavior 328 natural 319

1.7

properties of 314 series for 600 word problems for 326 logarithmie differentiation 117,322,329 logarithmie spiral 534, 535 logistic equation 506 logistic law 407 logistie model for population 335 Lotka-Voltera model 400 love bugs 535 lower sum 210 Luean 8(fn) Maclaurin, Colin 594 polynomial for sin x 602 series 594, 596 MACSYMA 465 majorize 199 Mandelbrot, Benoit 499 marginal cost, 106 produetivity 106 profit 106 revenue 106 Marsden, Jerrold 582, 615 Matsuoka, Y. 582 maxima and minima, tests for 153, 157 181 ' maximum global 177 point 151 value 177 maximum-minimum problems 177 mean value theorem 191 Cauchy's 526 eonsequences of 192 for integrals 239, 435, 455 Meech, L.W. 9 midnight SUD 30l(fn) minimum points 177 value 177 Minkowski's inequality 365 mixing problem 413, 414 modulates 628 motion, simple harmonic 373 with damping 415 natural growth or decay 380 logarithms 319 numbers 15 Newton, Isaac, 3(fn), 8(fn), 193(fn), 253(fn), 594 iteration 559 law of eooling 378 method 537, 546

1.8

Index Newton, Isaac (cant.) accuracy of 559 and chaos 547 second law of motion 369 nonhomogenous equation 623 noon 301(fn) northem hemisphere 301 notation differential 351,359,374,398 Leibniz 73, 104, 132, 217 summation 203, 204 nowhere differentiable continuous function 578 number complex 607, 609 imaginary 18 irrational 16 natural 15 rational 15 real 15, 16 numerical integration 550

odd function 164, 175 Olmsted, John M. H. 578, 600 one-sided limit 65, 517 open interval 21 optical focusing property of parabolas 36, 95, 97 order 18 orientation quizzes 13 origin 29 orthogonal trajectories 402 oscillations 294, 369 damped forced 628 forced 415 harmonic 373 linearized 375 overdamped 621 underdamped 621 oscillator, forced 626 oscillatory part 629 Osgood, W. 521 overdamped case 621 pH 317 Papprus theorem 454 parabola 34 equations of 37 focusing property of 36, 95, 97 vertex of 55 parameter 489 parametric curve 124,287,489 length of 495 tangent line to 491,492 parametric equations of circle 490 of line 490

partial fractions 465, 469, 591 partial integration (see integration by parts) particular solution 371, 623 partition 209 parts, integration by 358, 359 pendulum 376,391,417 period 259 periodic 259 perpendicular lines 33 Perverse, Arthur 367 pharaohs 416 phase shift 372, 629 Picard's method 559 plotting 29, 43, 163 point critical 151 inflection 159 interseetion 39 local maximum 151, 157 local minimum 151, 157 point-point form 32 polar coordinates 253, 255 arc length in 500 area in 502 graphing in 296 tangents in 299 polar representation of complex numbers 611 Polya, George 182 polynomial antiderivative of 130 derivative of 75, 79 pond, 74 population 117, 175, 189, 195, 335, 344, 382, 400, 407, 416 position 131 Poston, Tim 176 potential energy 446 power 445 function 307 integer 23 negative 26 of function rule for derivatives 110 rational 18, 27, 169 real 308 rule for antiderivatives 130 for derivatives 76, 119 for limits, 62 series 586 algebraic operations on 591 differentiation and integration of 590 root test 589 predator-prey equations 400 producer's surplus 248 product ruIe for derivatives 82

Index product rule (cont.) for limits 60 E-8 proof 520 productivity of labor 106 marginal 106 profit 329 marginal 106 program 40 projectile 295 proportional change 95 Ptolemy 256(fn) pursuit curve 499 Pythagoras, theorem of 30

quadratic formula 16, 17 function derivative of 54 general, graphing of 176 quizzes, orientation 13 quotient derivative of 85 difference 53 rule, for limits 62

radian 252 radius 34 of convergence 587 rate of change 2, 101, 247 decrease 101 increase 10 1 relative 329 rates, related 124 ratio comparison test for series 571 ratio test for power series 587 for series 582 rational exponents 118 expressions 475 function, continuity of 63, 140 numbers 15 power rule, for derivatives of a function 119 powers 118, 119 rationalizing 28 substitution 474 real axis 609 real

exponents 308 number line 18 numbers 15, 16 powers 308 reciprocal rule for derivatives 86

1.9

for limits 60 test for infinite limit 517 recursively 541 reduction formula 365 reduction of order 619 reflection, law of 290 region between graphs 240 related rates 124 word problems 125 relative rate of change 329 relativity 80(fn) repeated roots 620 replacement rule, for limits 60 resisting medium 412 resonance 415, 626, 629 revenue, marginal 106 revolution, surface of 482 Riccati equation 414 Richter scale 317 Riemann, Bernhard 220, fn integral 220 sums 220, 221, 551 Rivlin's equation 199 Robinson, Abraham 7, 73(fn) rocket propulsion 412 Rodrigues' formula 640 Rolle, Michel 193(fn) theorem 193 root splitting 619 root test for power series 589 for series 584 rose 297 Ruelle, David 548 Ruffini, Paolo 172

Saari, Donald G. 548 scaling rule, for integral 350 Schelin, Charles W. 257 (fn) school year 303 secant, 256 inverse 285 line 52, 191 second derivative 99, 104, 157 test for maxirna and minirna 157 test for concavity, 159 second-order approximation 601 second-order linear differential equations 617 sector, area of 252 separable differential equations 398, 399 sequence 537 comparison test 543 limit of 537, 540 properties 563 series altemating 572

1.10

Index series (cant.) comparison test for 570 constant multiple rule for 566 convergence of 562 divergent 562 geometrie 564 harmonie 567 infinite 561 integral test 580

P 581 power (see power series) ratio comparison test for 571 ratio test for 582 ooot test for 584 solutions 632 sum of 562 sum rule for 566 set 21 shell method 429 shifting rule for derivatives 115 for integral 350 sigma 203 sign, change of 146 signed area 215 similar triangles 254 Simmons, George F. 401 simple juumonic motion 373 damped 415 Simpson's rule 554 simultaneous equations 37 sine 254 derivative of 266 hyperbolic 385 inverse 281 series 600 sines, law of 263 slice method 420 slope 2, 31 of tangent line 52 slope-intercept form 32 Smith, D.E. 193(fn) Snell's law 305 solar energy 8, 107, 179, 180,221,449 solids of revolution 423,429 Spearman-Brown formula 520 speed 103, 497 speedometer 95 sphere 421 bands on 483 spiral exponential 310, 333 logarithmic 534, 535 Spivak, Mike 251(fn) spring constant 370 equation 370, 376 square, completing the 16, 17, 463

square ooot function, continuity of 64 squaring function 41 stable equilibrium 376 standard deviation 453 steady-state current 520 step function 5, 140, 209, 210 area under graph 210 straight line 31(fn), 125 (see also line) stretching rule, for derivatives 117 strict local minimum 151 Stuart, Ian 176 substitution definite integral by 355 integration by 347, 348, 352 of index 205 rationalizing 474 trigonometrie 461 sum rule 8-8 proof 520 for derivatives 78 for limits 60 physical model for 80 sum 203 collapsing 206 infinite 561 lower 210 of the first n integers 204 Riemann 220,221,551 rule for antiderivatives 130 telescoping 206 upper 210 summation notation 201, 203, 204 properties of 204, 208 sun 300 sunshine intensity 451 superposition 371 supply curve 248 surface of revolution 482 area of 483 suspension bridge 407 symmetries 163, 296 symmetry axis of 440 principle 440

tables of integrals 356, endpapers Tacoma bridge disaster 626 tangent hyperbolic 386 inverse 284 line 2, 191,491 horizontal 193 slope of 52 to parametrie curve 492 vertical 169 tangentfunction 256

Index Tartaglia, Nieeolo 172 Taylor series 594 test 59 eonvergenee of 597 Taylor, Brook 594 Tehebyeheff's equation 640 teleseoping sum 206 terminal speed 412 third derivative test 160 Thompson, D'Arcy 423 time of day 301 of year 301 torus 431 total ehange 244 traetrix 499 train 55, 80, 291 transeontinental railroad 569 transient 411, 628 transitional spiral 643 trapezoidal rule 552 triangles , similar 254 trigonometrie funetions antiderivatives of 269 derivatives of 264, 268 graphing problems 282 inverse 281, 285 word problems 289 trigonometrie identity 257 trigonometrie integrals 457, 458 trigonometrie substitution 461 triseeting angles 172

unbounded region 528 underdamped oseillations 621 undetermined eoefficients 623 unieellular organisms 423 uniform density 440 uniform growth or decay 381 unstable equilibrium 376, 390, 406 upper sum 210 Urenko, John B. 548

value absolute (see absolute value) maximum 177 minimum 177 variable ehanging 354

1.11

independent 40 varianee 453 variation of eonstants (or parameters) 378, 624 velocity 102, 131, 230 average 50 field 404 instantaneous 50, 51 positive 149 vertex 55 vertieal asymptote 164, 518, 531 vertieal tangent 169 Viete, Fran~ois 251(fn) Volterra, Vito 401 volume of bologna 426 disk method for 423 shell method for 429 sliee method for 419 of a solid region 419 washer method for 424

washer method 424 water 178, 247 fiowing 131, 144,343 in tank 126 watt 446 wavelength 263 waves, water 306 Weber-Feehner law 33 Weierstrass, Karl 6, 578 weighted average 437 window seat 291 word problems integration 247 logarithmie and exponential funetions 326 maximum-minimum 177 related rates 125 trigonometrie funetions 289 wrong-way integrals 235 Wronskians 630

yogurt 279

zero exponent 23 funetion 41

Undergraduate Texts in Mathematics (continued from page ii)

HiitonIHoltonIPedersen: Mathematieal Refleetions: In a Room with Many Mirrors. Iooss/Joseph: Elementary Stability and Bifureation Theory. Seeond edition. Isaac: The Pleasures ofProbability. Readings in Mathematics. James: Topologieal and Uniform Spaees. Jänich: Linear Algebra. Jänich: Topology. Kemeny/Snell: Finite Markov Chains. Kinsey: Topology of Surfaees. KIambauer: Aspeets ofCalculus. Lang: A First Course in Calculus. Fifth edition. Lang: Caleulus ofSeveral Variables. Third edition. Lang: Introduetion to Linear Algebra. Seeond edition. Lang: Linear Algebra. Third edition. Lang: Undergraduate Algebra. Seeond edition. Lang: Undergraduate Analysis. Lax/BursteinlLax: Calculus with Applieations and Computing. Volume 1. LeCuyer: College Mathematics with APL. LidIlPilz: Applied Abstract Algebra. Second edition. Logan: Applied Partial Differential Equations. Macki-Strauss: Introduction to Optimal Control Theory. Malitz: Introduction to Mathematical Logic. Marsden/Weinstein: Calculus I, 11, III. Second edition. Martin: The Foundations of Geometry and the Non-Euclidean Plane. Martin: Geometrie Constructions. Martin: Transformation Geometry: An Introduction to Symmetry. Millman/Parker: Geometry: AMetrie Approach with Models. Second edition. Moschovakis: Notes on Set Theory.

Owen: A First Course in the Mathematical Foundations of Thermodynamies. Palka: An Introduction to Complex Function Theory. Pedrick: A First Course in Analysis. PeressinilSullivanlUhl: Tbe Mathematics ofNonlinear Programming. PrenowitzlJantosciak: Join Geometries. Priestley: Calculus: A Liberal Art. Second edition. ProtterlMorrey: A First Course in Real Analysis. Second edition. ProtterlMorrey: Intermediate Calculus. Second edition. Roman: An Introduction to Coding and Information Theory. Ross: Elementary Analysis: The Theory of Calculus. Samuel: Projective Geometry. Readings in Mathematics. ScharlaulOpolka: From Fermat to Minkowski. Schiff: The Laplace Transform: Theory and Applications. Sethuraman: Rings, Fields, and Vector Spaces: An Approach to Geometrie Constructability. Sigler: Algebra. SilvermanlTate: Rational Points on Elliptic Curves. Simmonds: A Brief on Tensor Analysis. Second edition. Singer: Geometry: Plane and Faney. Singerrrhorpe: Lecture Notes on Elementary Topology and Geometry. Smith: Linear Algebra. Third edition. Smith: Primer ofModem Analysis. Seeond edition. Stanton/White: Construetive Combinatories. StillwrlI: Elements of Algebra: Geometry, Numbers, Equations. StilIweIl: Mathematies and lts History. StilIweIl: Numbers and Geometry. Readings in Mathematics. Strayer: Linear Programming and Its Applieations.

Undergraduate Texts in Mathematics Thorpe: Elementary Topics in Differential Geometry. Toth: Glimpses of Algebra and Geometry. Readings in Mathematics.

Troutman: Variational Calculus and Optimal Contro!. Second edition.

Valenza: Linear Algebra: An Introduction to Abstract Mathematics. Whyburn/Duda: Dynamic Topology. Wilson: Much Ado About Calculus.

ABrief Table of Integrals, continued. 29.

J

11 i ± i =± +i

cschx dx = Inltanh

= -

30. J sinh 2x dx =

sinh 2x -

x

31. J cosh 2x dx

sinh 2x

x

In

~~:~~ ~

!

32. J sech2x dx = tanh x

33. JSinh - I ~ dx a 34.

J

cosh - 1 ~ dx a

=

x sinh - I ~

=

{XCOSh-l~ -~

~a2

I

+ x2

+ ~ Inla 2 -

[ cosh

1_

= In(x + ~a2 + x 2 ) = sinh - 1 ~

~a2

> 0)

!

=

I dx=sin-I~ _ x2 a

42. J

1 3/2 dx (a 2 - x 2)

43. J

~X2 ± a 2 dx =

(a>O)

i (5a 2 - 2X2)~ + 3t sin

=

-I

~

(a

> 0)

(a>O)

41. J __1_ dx = ~ Inl a + x a2 _ x 2 2a a- x

1

x

a2~a2 - x 2

I ~x2 ± a2 ± ;2 Inlx + ~x2 ± a 2

J I dx = Inlx +~I 45. J I dx ! Inl-x-I x (a + bx) a a + bx 44.

-I( ~ ) < 0, a > 0] (a

a

39. J (a 2 - x 2)3/2 dx = 40.J

~ ) > 0, a > 0 J

1(

x21

tan - I ~ (a > 0) a a ~ x~ a2 x 38. J va -x dX=Iva -x +Tsin-'a

37.

> 0)

[ cosh -

a

dx

J__ dx a2 + x 2

(a

a xcoSh-l~ +~

35. J tanh - I ~ dx = x tanh -I ~ 36. J

~

-

a

~

= cosh- I

~

1

(a

a

> 0)

=

r;::-;-;:::: 2(3bx - 2a)(a + bx)3/2 46. J xya + bx dx = - - - - - - = - - - 15b 2

J7 48. J x ,Ja + bx 47.

49. ( .J

x,Ja

1

+ bx

aJ x,Ja + bx

dx

= 2..fa + bx +

dx

= -------;:----

dx

1

2(bx - 2a),Ja + bx 3b 2

= ~ Inl ,;a+iiX -..;a

..;a,Ja + bx +..;a

=

51.

~

dx

f x~ dx

I

_2_ tan -I ~ a + bx -a

~

50. J

dx

=~a2 = -

x2

j (a 2 -

(a

> 0)

(a

< 0)

-alnl a +~ I x 2)3/2 (a

> 0) Continued on overleaf.

ABrief Table of Integrals, continued. 53.

J

54.J 55.

I

+~

dx = - .!.lnj a a

1

x~

x

x dx=-~ Ja 2 - x 2

J

X2

a x a2 - x + 2 sin - I a 2'X V~ 2

dx = -

~

(a

> 0)

56.J~ dx=R+?-alnla+~1 57. J

~ dx=~ -acos-

I: I

I

=Jx 2 -a 2 -asec-I(~) 58. J xJx 2 ± a 2 dx = 59. J 60.J 61 • J 62. J

j (x 2 ± a 2)3/2

1 dx = .!.lnl x xR+? a a+R+?

dx=.!.COS-I~

1

x~

x dx Jx 2 ± a 2

63 J 1 . ax 2 + bx

+C

~ 2

+-

1 dx= ± a2

X 2 Jx 2

ax

= Jx 2 ± a 2 dx =

1 In!2ax Jb 2 - 4ac 2ax

2

tan-I

x ax 2 + bx

65. J

1 vax 2 + bx

+c

1 Inlax 2 + bx dx = -2 a

+c

dx =

2

68. J

r:::a

1

+c

xvax 2 + bx

2ax - b Vb 2 - 4ac

dx

vax 2 + bx + c a

70. J

-..!!...2a

J

+ c..[c

x4

+- ~(x2 ±

dx (a

> 0)

< 0)

4acS- b 2 a 1

Jvax

Jax 2 + bx

+c

1

2+ b

+c

dx

dx

(c > 0)

x bx+2c

Fe Ixlvb 2 - 4ac x3,~ dx =( ~x2 - -lsa2)~(a2 + X 2)3 Jx2 ± a 2

+c

= ..::...!.lnI 2..[c Jax 2 + bx + c + bx + 2c I

=_l-sin- I

69. J

(a

1 -

a

=

1 ax 2 + bx

a

2a~ + b vax 2 + bx + c +

=

Jax 2 + bx

+ cl - 2b J

~ Inl2ax + b + 2/ii vax 2 + bx + cl

/ii = _I_ sin -

Jvax + bx + c dx 67. J x dx 66.

+ b - Jb 2 - 4ac! (b 2 > 4ac) + b + Vb2 - 4ac 2ax + b (b 2 < 4ac)

v4ac - b 2

J4ac - b 2 64. J

I

(a>O)

lxi

a

(a>O)

(c