Smith R., Minton R. Calculus

Подождите немного. Документ загружается.

P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch05 MHDQ256-Smith-v1.cls December 14, 2010 20:15

LT (Late Transcendental)

CONFIRMING PAGES

5-59 CHAPTER 5

Review Exercises 373

Review Exercises

12. A swimming pool viewed from above has an outline given by

y =±(5 + x) for 0 ≤ x ≤ 2. The depth is given by 4 + x (all

measurements in feet). Compute the volume.

13. The cross-sectional areas of an underwater object are given.

Estimate the volume.

x 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2

A(x) 0.4 1.4 1.8 2.0 2.1 1.8 1.1 0.4 0

............................................................

In exercises 14–18, ﬁnd the volume of the indicated solid

of revolution.

14. The region bounded by y = x

, y = 0 and x = 1 revolved

about (a) the x-axis; (b) the y-axis; (c) x = 2; (d) y =−2

15. Theregionboundedby y = x

and y = 4revolvedabout(a)the

x-axis; (b) the y-axis; (c) x = 2; (d) y =−2

16. The region bounded by y = x, y = 2x and x = 2 revolved

about (a) the x-axis; (b) the y-axis; (c) x =−1; (d) y = 4

17. The region bounded by y = x, y = 2 − x and y = 0 revolved

about (a) the x-axis; (b) the y-axis; (c) x =−1; (d) y = 4

18. The region bounded by x = 4 − y

and x = y

− 4 revolved

about (a) the x-axis; (b) the y-axis; (c) x = 4; (d) y = 4

............................................................

In exercises 19–22, set up an integral for the arc length and

numerically estimate the integral.

19. The portion of y = x

for −1 ≤ x ≤ 1

20. The portion of y = x

+ x for −1 ≤ x ≤ 0

21. The portion of y =

√

x + 1 for 0 ≤ x ≤ 3

22. The portion of y = sin 2x for 0 ≤ x ≤ π

............................................................

In exercises 23 and 24, set up an integral for the surface area

and numerically estimate the integral.

23. The surface generated by revolving y = 1 − x

, 0 ≤ x ≤ 1,

about the x-axis

24. The surface generated by revolving y = x

, 0 ≤ x ≤ 1, about

the x-axis

............................................................

In exercises 25–32, ignore air resistance.

25. A diver drops from a height of 64 feet. What is the velocity at

impact?

26. If the diver in exercise 25 has an initial upward velocity of

4 ft/s, what will be the impact velocity?

27. An object is launched from the ground at an angle of 20

◦

with

an initial speed of 48 ft/s. Find the time of ﬂight and the hori-

zontal range.

28. Repeat exercise 27 for an object launched from a height of

6 feet.

29. A football is thrown from a height of 6 feet with initial speed

80 ft/s at an angle of 8

◦

. A person stands 40 yards downﬁeld

in the direction of the throw. Is it possible to catch the ball?

30. Repeat exercise 29 with a launch angle of 24

◦

. By trial and

error, ﬁnd the range of angles (rounded to the nearest degree)

that produce a catchable throw.

31. Find the initial velocity needed to propel an object to a height

of 128 feet. Find the object’s velocity at impact.

32. A plane at an altitude of 120 ft drops supplies to a location on

the ground. If the plane has a horizontal velocity of 100 ft/s,

how far from the target should the supplies be released?

............................................................

33. A force of 60 pounds stretches a spring 1 foot. Find the work

done to stretch the spring 8 inches beyond its natural length.

34. A car engine exerts a forceof 800 +2x pounds whenthe car is

at position x miles. Find the work done as the car moves from

x = 0tox = 8.

35. Compute the mass and center of mass of an object with density

ρ(x) = x

− 2x + 8 for 0 ≤ x ≤ 4. Explain why the center of

mass is not at x = 2.

36. Compute the mass and center of mass of an object with density

ρ(x) = x

− 2x + 8 for 0 ≤ x ≤ 2. Explain why the center of

mass is at x = 1.

37. A dam has the shape of a trapezoid with height 80 feet. The

width at the top of the dam is 60 feet and the width at the

bottom of the dam is 140 feet. Find the maximum hydrostatic

force that the dam will need to withstand.

38. An underwater viewing window is a rectangle with width

20 feet extending from 5 feet below the surface to 10 feet

below the surface. Find the maximum hydrostatic force that

the window will need to withstand.

39. The force exerted by a bat on a ball over time is shown in the

table. Use the data to estimate the impulse. If the ball (mass

m = 0.01slug)hadspeed120ft/sbeforethecollision, estimate

its speed after the collision.

t (s) 0 0.0001 0.0002 0.0003 0.0004

F(t) (lb) 0 800 1600 2400 3000

t (s) 0.0005 0.0006 0.0007 0.0008

F(t) (lb) 3600 2200 1200 0

40. If a wall applies a force of f (t) = 3000t(2 − t) pounds to a car

for0 ≤ t ≤ 2,ﬁndtheimpulse. Ifthecar(massm = 100slugs)

is motionless after the collision, compute its speed before the

collision.

P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch05 MHDQ256-Smith-v1.cls December 14, 2010 20:15

LT (Late Transcendental)

CONFIRMING PAGES

374 CHAPTER 5

Applications of the Definite Integral 5-60

Review Exercises

EXPLORATORY EXERCISES

1. As indicated in section 5.5, general formulas can be derived

for many important quantities in projectile motion. For an ob-

ject launched from the ground at angle θ

with initial speed

ft/s, ﬁnd the horizontal range R ft and use the trig iden-

tity sin(2θ

) = 2sinθ

cosθ

to show that R =

sin(2θ

)

Conclude that the maximum range is achieved with angle

= π/4 (45

◦

2. Tofollowuponexploratoryexercise1,supposethattheground

makes an angle of A

◦

with the horizontal. If A > 0 (i.e., the

projectile is being launched uphill), explain why the maximum

rangewouldbeachievedwithananglelargerthan45

◦

.IfA < 0

(launching downhill), explain why the maximum range would

be achieved with an angle less than 45

◦

. To determine the ex-

act value of the optimal angle, ﬁrst argue that the ground can

be represented by the line y = (tan A)x. Show that the pro-

jectile reaches the ground at time t = v

sinθ

− tan A cosθ

Compute x(t) for this value of t and use a trig identity to re-

place the quantity sin θ

cos A − sin A cos θ

with sin(θ

− A).

Thenuse anothertrigidentitytoreplacecos θ

sin(θ

− A)with

sin(2θ

− A) −sin A. At this stage, the only term involving θ

will be sin(2θ

− A). To maximize the range, maximize this

term by taking θ

P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch06 MHDQ256-Smith-v1.cls December 15, 2010 19:5

LT (Late Transcendental)

CONFIRMING PAGES

CHAPTER

Exponentials, Logarithms and

Other Transcendental Functions

The International Space Station is one of the most ambitious engineering

projects ever undertaken. A construction project in space offers a few

advantages over construction on Earth. For instance, the weightlessness

andlackofatmosphericconditionsreducethe need for structural strength.

In fact, the International Space Station could not support its ownweight if

it were constructed on Earth! The lightweight beamsholding the station’s

solar panels are long, yet relatively thin and ﬂexible. Unfortunately, these

characteristics can allow a minor tremor to magnify into a dangerous

vibration, due to the phenomenon of resonance. Consequently, this design

requires a system to maintain the stability of the structure. This is one area where

calculus plays a critical role in the design of the space station.

You can think of this stability problem

by imagining yourself operating a joystick,

where moving the joystick applies a force at

one of the beam’s joints. The goal is to apply

the appropriate forces to keep the beam from

vibrating.Forinstance,ifthebeamstartsmov-

ingtotheleft,youmightmoveyourjoystickto

the right, applying an opposing force. Mathe-

matically, think of this as a function; you sup-

ply the input (the force) that determines the

output (the motion of the beam). Your task

is then to solve an inverse problem. That

is, given the desired output (stability), you

must determine the correct input (force) that

produces it. We discuss inverse functions in section 6.2. In the remainder of this

chapter, we deﬁne several new functions that are essential to engineers investigat-

ing the stability of structures.

International Space Station

6.1 THE NATURAL LOGARITHM

At some point or other prior to your study of calculus, you likely encountered the

natural logarithm. The standard precalculus deﬁnition is that thenatural logarithm

is the ordinary logarithm with base e. That is,

ln x = log

where e is a (so far) mysterious transcendental number, whose approximate value

is givenas e ≈ 2.71828....So, why would a logarithm with a transcendental base

375

P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch06 MHDQ256-Smith-v1.cls December 15, 2010 19:5

LT (Late Transcendental)

CONFIRMING PAGES

376 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-2

be called natural? Further, why would anyone be interested in such a seemingly unusual

function? We will resolve both of these questions in this section.

First, recall the power rule for integrals,



dx =

n+1

n + 1

+ c, for n =−1.

Of course, this rule doesn’t hold for n =−1, since this would result in division by zero.

Then, what can we say about



dx?

From Theorem 4.1 in Chapter 4, we know that since f (x) =

is continuous for x = 0,

it must be integrable on any interval not including x = 0. (But, how do we ﬁnd its an-

tiderivative?) Notice, that by Part II of the Fundamental Theorem of Calculus,



is an antiderivative of

for x > 0. We give this new (and naturally arising) function a name

in Deﬁnition 1.1.

DEFINITION 1.1

For x > 0, we deﬁne the natural logarithm function, denoted ln x,by

ln x =



dt.

y  Q

12x 3 4

FIGURE 6.1a

ln x (x > 1)

y 

3 4

FIGURE 6.1b

ln x(0 < x < 1)

We’ll see later in this chapter that this deﬁnition is, in fact, consistent with your previous

understanding of ln x. First, let’s interpret this new function graphically. Notice that for

x > 1, this deﬁnite integral corresponds to the area A under the curve y =

from 1 to x,as

indicated in Figure 6.1a. That is,

ln x =



dt = A > 0.

Similarly, for 0 < x < 1, notice from Figure 6.1b that for the area A under the curve y =

from x to 1, we have

ln x =



dt =−



dt =−A < 0.

Using Deﬁnition 1.1, we get by Part II of the Fundamental Theorem of Calculus that

ln x =



dt =

, for x > 0. (1.1)

We illustrate this new derivative formula in example 1.1.

P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch06 MHDQ256-Smith-v1.cls December 15, 2010 19:5

LT (Late Transcendental)

CONFIRMING PAGES

6-3 SECTION 6.1

The Natural Logarithm 377

EXAMPLE 1.1 Differentiating a Logarithm

Find the derivative of ln(x

+ 7x

Solution Using the chain rule and (1.1), we have

ln(x

+ 7x

) =



+ 7x



+ 7x

)



+ 7x



(3x

+ 14x).



Now, note that ln |x| is deﬁned for x = 0. For x > 0, ln|x|=ln x and hence,

ln|x|=

ln x =

Similarly, for x < 0, ln|x|=ln(−x), and hence,

ln|x|=

ln(−x)

−x

(−x)

By the chain rule.

−x

(−1) =

Notice that we got the same derivative in either case. This proves Theorem 1.1.

THEOREM 1.1

For x = 0,

ln|x|=

EXAMPLE 1.2 The Derivative of the Log of an Absolute Value

For any x for which tan x = 0, evaluate

ln|tan x|.

Solution From Theorem 1.1 and the chain rule, we have

ln|tan x|=

tan x

sec



Of course, with the new differentiation rule in Theorem 1.1, we get a new integration

rule.

COROLLARY 1.1

In any interval not containing 0,



dx = ln |x|+c.

Corollary 1.1 is a very common integration rule. We illustrate this in example 1.3.

P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch06 MHDQ256-Smith-v1.cls December 15, 2010 19:5

LT (Late Transcendental)

CONFIRMING PAGES

378 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-4

EXAMPLE 1.3 Two Integrals Involving Logarithms

Evaluate the integrals (a)



+ 7

dx and (b)



tan xdx.

Solution For (a), notice that the numerator is nearly the derivative of the denominator:

+ 7) = 3x

This suggests the substitution u = x

+ 7, so that du = 3x

dx.Wenowhave



+ 7

dx =



+ 7



du =

ln|u|+c

ln|x

+ 7|+c.

For (b), we must ﬁrst rewrite the integrand in terms of sinx and cos x,toreveal



tan xdx=



sin x

cos x

dx.

Here, you should quickly observe that setting u = cos x gives us du =−sin xdx, so that



tan xdx =−



−sin x

cos x

dx =−



=−ln|u|+c =−ln|cos x|+c.



HISTORICAL

NOTE

John Napier (1550–1617)

A Scottish nobleman and inventor

who developed the concept of

the logarithm and constructed the

first table of logarithm values. He

invented a number of useful

techniques for working with large

numbers, including logarithms and

a calculation device sometimes

called “Napier’s bones” that can

be used for multiplication and the

estimation of square roots and

cube roots. Napier also

popularized the use of the decimal

point for representing fractions.

Notice that since ln x is deﬁned by a deﬁnite integral, we can use Simpson’s Rule (or

any other convenient numerical integration method) to compute approximate values of the

function. For instance,

ln2 =



dt ≈ 0.693147

and ln3 =



dt ≈ 1.09861.

We leave these approximations as exercises. (You should also check the values with the ln

key on your calculator.)

We now brieﬂy sketch a graph of y = ln x. As we’ve already observed, the domain of

f (x) = ln x is (0, ∞). Further, recall that

ln x

⎧

⎨

⎩

< 0 for 0 < x < 1

= 0 for x = 1

> 0 for x > 1

and that f



(x) =

> 0, for x > 0,

so that f is increasing throughout its domain. Next,



(x) =−

< 0, for x > 0,

P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch06 MHDQ256-Smith-v1.cls December 15, 2010 19:5

LT (Late Transcendental)

CONFIRMING PAGES

6-5 SECTION 6.1

The Natural Logarithm 379

and hence, the graph is concave down everywhere. You can easily use Simpson’s Rule or

the Trapezoidal Rule (this is left as an exercise) to make the conjectures

lim

x→∞

ln x =∞ (1.2)

and lim

x→0

ln x =−∞. (1.3)

We postpone the proof of (1.2) until after Theorem 1.2. The proof of (1.3) is left as an

exercise. We now obtain the graph shown in Figure 6.2.

1234 5

1

2

3

FIGURE 6.2

y = ln x

Now, it remains for us to explain why this function should be called a logarithm. The

answer is simple: it satisﬁes all of the properties satisﬁed by other logarithms. Since ln x

behaves like any other logarithm, we call it (what else?) a logarithm. We summarize these

properties in Theorem 1.2.

THEOREM 1.2

For any real numbers a, b > 0 and any rational number r,

(i) ln 1 = 0

(ii) ln(ab) = lna + lnb

(iii) ln





= lna − ln b and

(iv) ln(a

) = r lna.

PROOF

(i) By deﬁnition,

ln1 =



dt = 0.

(ii) Also from the deﬁnition, we have

ln(ab) =



dt =



dt +



dt,

from part (ii) of Theorem 4.2 in section 4.4. Make the substitution u =

in the last integral

only. This gives us du =

dt. Finally, the limits of integration must be changed to reﬂect

the new variable (when t = a,wehaveu =

= 1 and when t = ab,wehaveu =

= b),

to yield

ln(ab) =



dt +







dt +



du = lna + lnb.

From Deﬁnition 1.1.

(iv) Note that

ln(x

) =

From (1.1) and the chain rule.

r−1

From the power rule.

P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch06 MHDQ256-Smith-v1.cls December 15, 2010 19:5

LT (Late Transcendental)

CONFIRMING PAGES

380 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-6

Likewise,

[r ln x] = r

(ln x) =

Now, since ln(x

) and r ln x have the same derivative, it follows from Corollary 8.1 in

section 2.8 that for all x > 0,

ln(x

) = r ln x + k,

for some constant, k. In particular, taking x = 1, we ﬁnd that

ln(1

) = r ln1 +k,

where since 1

= 1 and ln1 = 0, we have

0 = r(0) + k.

So, k = 0 and ln(x

) = r ln x, for all x > 0.

Part (iii) follows from (ii) and (iv) and is left as an exercise.

EXAMPLE 1.4 Rewriting a Logarithmic Expression

Simplify ln

√

, where n is a positive integer.

Solution For any integer n > 0,

√

= ln(a

−1/n

) =−

lna.



CAUTION

√

= (lna)

−1/n

(What is wrong with this?)

Using the properties of logarithms will often simplify the calculation of certain deriva-

tives. We illustrate this in example 1.5.

EXAMPLE 1.5 Using Properties of Logarithms to Simplify Differentiation

Find the derivative of ln



(x − 2)

+ 5

Solution Rather than directly differentiating this expression by applying the chain

rule and the quotient rule, notice that we can considerably simplify our work by ﬁrst

using the properties of logarithms. We have



(x − 2)

+ 5



(x − 2)

+ 5



1/2



(x − 2)

+ 5



From Theorem 1.2 (iv).

[ln(x − 2)

− ln(x

+ 5)] From Theorem 1.2 (iii).

[3ln(x − 2) −ln(x

+ 5)] From Theorem 1.2 (iv).





x −2



(x − 2) −



+ 5



+ 5)



From (1.1)

and the

chain rule.



x −2

−

+ 5



P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch06 MHDQ256-Smith-v1.cls December 15, 2010 19:5

LT (Late Transcendental)

CONFIRMING PAGES

6-7 SECTION 6.1

The Natural Logarithm 381

Of course, you could simply compute the derivative directly using the original

expression. Try this for yourself. It’s not pretty (there are multiple chain rules and a

quotient rule; don’t say we didn’t warn you), but it’s entirely equivalent. The bottom

line is that the rules of logarithms can save you untold complication; so use them where

appropriate.



EXAMPLE 1.6 Examining the Limiting Behavior of ln x

Use the properties of logarithms in Theorem 1.2 to prove that

lim

x→∞

ln x =∞.

Solution We can verify this as follows. First, recall that ln 3 ≈ 1.0986 > 1. Taking

x = 3

, we have by the rules of logarithms that for any integer n

ln3

= n ln3.

Since x = 3

→∞,asn →∞, it now follows that

lim

x→∞

ln x = lim

n→∞

ln3

= lim

n→∞

(n ln3) =+∞,

where the ﬁrst equality depends on the fact that ln x is a strictly increasing function.



Logarithmic Differentiation

A clever technique called logarithmic differentiation uses the rules of logarithms to help

ﬁnd derivatives of certain functions for which we don’t presently have derivative formulas.

For instance, note that the function f (x) = x

is not a power function because the exponent

is not a constant. In example 1.7, we show how to take advantage of the properties of

logarithms to ﬁnd the derivative of such a function.

EXAMPLE 1.7 Logarithmic Differentiation

Find the derivative of f (x) = x

, for x > 0.

Solution As already noted, none of our existing derivative rules apply. We begin by

taking the natural logarithm of both sides of the equation f (x) = x

.Wehave

ln[ f (x)] = ln (x

)

= x ln x,

from the usual properties of logarithms. We now differentiate both sides of this last

equation. Using the chain rule on the left side and the product rule on the right side,

we get

f (x)



(x) = (1)ln x + x



(x)

f (x)

= ln x + 1.

Solving for f



(x), we get



(x) = (ln x +1) f (x) = (ln x + 1)x



P1: PIC/OVY P2: OSO/OVY QC: OSO/OVY T1: OSO

MHDQ256-Ch06 MHDQ256-Smith-v1.cls December 15, 2010 19:5

LT (Late Transcendental)

CONFIRMING PAGES

382 CHAPTER 6

Exponentials, Logarithms and Other Transcendental Functions 6-8

BEYOND FORMULAS

In Deﬁnition 1.1, we deﬁned a new function that ﬁlls a gap in our previously known

integration rules, namely, what to do with



dx when n =−1. Surprisingly, we can

then develop properties of this unusually deﬁned function and discover that it is, in

fact, a logarithmic function. Logarithms, in turn, trace their history back to a need

for a practical method of computing by hand with large numbers. Such unexpected

connections are common in mathematics. However, these remarkable connections can

only be fully appreciated with a sound understanding of the underlying mathematical

theory, such as that developed in this section.

EXERCISES 6.1

WRITING EXERCISES

1. Explain why it is (mathematically) legal to deﬁne ln x as



dt. For some, this type of deﬁnition is not very satis-

fying. Even though this is probably the ﬁrst function you have

seen deﬁned this way, try the following comparison. Clearly,

it is easier to compute function values for x

than for ln x, and

therefore x

is easier to understand. However, compare how

you would compute (without a calculator) function values for

sin x versus function values for ln x. Describe which is more

“natural” and easy to understand.

2. In this section, we gave two different “deﬁnitions” of ln x.

Explain why it is logically invalid to give different deﬁnitions

unless you can show that they deﬁne the same thing. If they

deﬁne the same object, either deﬁnition is equally valid and

you should use whichever deﬁnition is clearer for the task at

hand. Explain why, in this section, the integral deﬁnition is

more convenient than the base e logarithm.

3. Usethe integraldeﬁnitionoflnx (interpretedasarea)toexplain

why it is reasonable that lim

x→0

ln x =−∞and lim

x→∞

ln x =∞.

4. The graph of f (x) = ln x appears to get ﬂatter as x gets larger.

Interpret the derivative f



(x) =

as the slopes of tangent

lines to determine whether this is correct or just an optical

illusion.

In exercises 1–4, express the number as an integral and sketch

the corresponding area.

1. ln4 2. ln5 3. ln 8.2 4. ln 24

............................................................

5. Use Simpson’s Rule with n = 4 to estimate ln 4.

6. Use Simpson’s Rule with n = 4 to estimate ln 5.

............................................................

In exercises 7–12, ﬁnd the derivative of the function.

7. ln4x

8. ln(sec x) 9. ln(cos x)

10.

ln x

11. sin(ln(cos x

)) 12. ln(sec x +tan x)

............................................................

In exercises 13–20, evaluate the limit or derivative using prop-

erties of logarithms where needed.

13.





+ 1



14.

[ln(x

sin x cos x)]

15.



+ 1



16.

⎛

⎝



+ 1

⎞

⎠

17. lim

x→1



x − 1



18. lim

x→0

ln[(sin x)

1/x

]

19. lim

x→∞

ln[(x + 1)

4/ ln(x+1)

] 20. lim

x→0

ln[(cos x)

]

............................................................

In exercises 21–30, evaluate the integral.

21.



+ 1

dx 22.



+ 5

23.



tan2xdx 24.



x + 1

+ 2x − 1

25.



x ln x

dx 26.



√

x + 1)

27.



(ln x + 1)

dx 28.



cos(ln x)

29.



√

ln x

dx 30.



+ 1

............................................................

In exercises 31–34, use the properties of logarithms to rewrite

the expression as a single logarithm.

31. ln

√

2 + 3ln2 32. ln8 −2ln2

33. 2ln3− ln9 +ln

√

3 34. 2ln





− ln3 +ln





............................................................

35. Use properties (ii) and (iv) of Theorem 1.2 to prove property

(iii) that ln





= lna − ln b.

36. Prove equation (1.3).

............................................................

In exercises 37–40, use logarithmic differentiation to ﬁnd the

derivative of the given function.

37. f (x) = x

sin x

38. f (x) = x

4−x

39. f (x) = (sinx)

40. f (x) = x

ln x

............................................................