chapter 6 thomas 9 ed f 2008

Crnprpn

TranscendentalFunctions

OvERvBw Many of the functions in mathematics and science are inverses of oneanother. The functions lnx and e' are probably the best-known function-inversepair, but others are nearly as important. The trigonometric functions, when suitablyrestricted, have important inverses, and there are other useful pairs of logarithmicand exponential functions. Less widely known are the hyperbolic functions andtheir inverses, functions that arise in the study of hanging cables, heat flow, andthe friction encountered by objects falling through the air. We describe all of thesefunctions in this chapter and look at the kinds of problems they solve.

Inverse Functions and Their DerivativesIn this section, we define what it means for functions to be inverses of one anotherand look at what this says about the formulas, graphs, and derivatives of function-inverse pairs.

One-to-One FunctionsA function is a rule that assigns a value from its range to each point in its domain.Some functions assign the same value to more than one point. The squares of - Iand 1 are both 1; the sines of tr /3 and 2r 13 arc both J5 /2. Other functions neverassume a given value more than once. The square roots and cubes of differentnumbers are always different. A function that has distinct yalues at distinct pointsis called one-to-one.

Definition. A function /(r) is one-to-one on a domain D if f (x) I /(;2) wheneverxt * rz .

EXAMPLE I /(r) : .r4 is one-to-one on any domain of nonnegative numbersbecawe n$ I nEj whenever xl f x2. tr

M9

450 Chapter 6: Transcendental Functions

One-lo one: GraPh meels eacllhorizontal linc at most once.

Not one-to one: GlaPh Ineets one or

nlore hofizontal lines more than once

6. t Using the horlzontal l ine test, we seethat y : x3 and y = \tq arc one-to-one,but y = x2 and y : sinx are not.

EXAMPLE 2 g(x) = sinx ir /?ol one-to-one on the interval [0'r] because

sin (T /6) : sin (52/6). The sine ls one-to one on [0, z/2], however, beceuse sines

of angles in the first quadrant are distinct.

The graph of a one-to-one function -r : .f(r) can inte$ect a given horizontal

line at most once. If it intersects the line more than once it assumes the same

1-value more than once, and is therefore not one-to-one (Fig 6l) '

The Horizontal Line Test

A function } : /(r) is one-to-one if and only if its graph intersects each

horizontal line at most once.

lnversesSince each output of a one-to-one function comes liom just one input' a one-to-

one function can be reversed to send the outputs back to the inputs from which

they came. The function defined by reversing a one-to-one function .l is called the

inyerse of I The symbol for the inverse of f is/ t' read "/ inverse" (Fig 6 2)'

The I in / I is nol an exponent: /

r(r) does not mean l//({)

As Fig. 6.2 suggests, the result of composing I and / | in either order is the

identity function, the function that assigns each number to itself This gives a wry

to test whether two functions / and g are inverses of one another' Compute f o I

and g o /. If (/ o g)(x) : (g " "f)('t)

: 'r' then l and g are inverse ofone another;

otherwise they are not. If / cubes every number in its domain, 3 had hetter take

cube roots or it isn't the inverse of f

Functions / and g are an inverse pair if and only if

Same I value

511T /

"f (g(r)) :;r and

l n t h i s c a s e , g : J l a n d f - g ' \ .

g( l ( r ) ) = ; r .

,r. = Jirl- r : / ' 0 )

5.2 The inverse of a function f sends

each output back to the input fromwhich i t came.

A function has an inverse if urLd onLy if it is one-to-one This means, for

example, that increasing functions have inverses and decreasing functions haYe

inverses (Exercise 39). Functions with positive dedvatives have inverses because

they increase throughout their domains (Corollary 3 of the Mean Value Theorem.

Section 3.2). Similarly, because they decrease thtoughout their domairs, functions

with negative derivatives have inverses

Finding InversesHow is the graph of the inverse of a function related to the graph of the function?

If the function is increasing, say, its graph rises from left to right, like the graph in

Fig. 6.3(a). To read the graph, we start at the point r on the -t axis' go up to the

graph, and then move over to the y-axis to read the value of ] If we start with )

and want to find the r from which it came, we reverse the process (Fig 6 3b)'

The graph of / is the graph of f I with the input-output pails reversed' To

display the graph in the usual way, we have to reve$e the pairs by reflecting the

Doma;n of.l Range of/

Dorn3in of / L

(a) To find the value offatr, we stafi at,and go up to the curve and over to the ) axis.

6.1 lnverse Functions and Their Derivatives 45'l

(b) The graph of/can also serve as a graph of/-r.To find the r that gave y, we staft at ), and go overto the curve and down to the -{-axis. The domalnof /-r is the range of I The mnge of /

' is thedomain oft

I (b, a)

(c) To draw the graph ofl-r in theusual way, we rcflect it in the line ) = r.

(d) Then we interchange the letters i and :|.We noq have a gaph of/ | as a function of r.The graph of f 1(x).

How to Express f-t as a Functiono l x

Step ,.' Solve the equation f: f@)for x in terms of y.

Sfep 2: Interchange .r and y. Theresulting formula will be y: /-1(x).

graph in the 45' line y: ,r (Fig. 6.3c) and interchanging the letters r and y (Fig'

6.3d). This puts the independent variable, now called.r, on the horizontal axis and

the dependent variable, now called y, on the vertical axis. The graphs of /(-r) and

f | (x) are symmetric about the line / : t.The pictures in Fig. 6.3 tell us how to express / t as a function of .r, which

is stated at the left.

EXAMPLE 3 Find the inverse of n: ]t * 1, expressed as a function of .x.' 2

Solution

Step 1: Solve for x in terms ofY: I : ),

+ t

2 Y : x l 2

y : 2 y - 2 .

Step 2: Interchange x and y: Y :2x - 2.

The inverse of the function f (x):(l/2)x *I is the function /-1(t):Lt -2.

z v


6.4 Graphing fkl : ( lztx + 1 andf-l(xl =)2q 2 together shows thegraphs' symmetry with respect to the l i

lx1f t )=tl y r ( t ) = t + c o s t

l 4 ( t ) : tI v'(t) = t

I xr(t) = 116es 1I v,(t) : t

To check, we verify that both composites give the identity function:

r ' ( r ( ' ) ) : ' ( ) ' * t ) - ' : ' *2 - 2 : x

f ( f ' ( * ) ) : )e ,

- r ) * t : , r r1 : r .

See Fig. 6.4.

Find the inverse of the function y: x2,x > 0, expressed as aEXAMPLE 4function of ;.

Solution

Step 1: Solve for x in terms of y:

Y : x 2

J y : J 7 : l x l : x i : , b i , , u \ e r : o

Step 2: Interchange x ancl y: y : 1E.

The inverse of the function y : x2, x; 0, is the function I : "nuf. See Fig. 6.5.

Notice that, unlike the restricted function ) : tz, x 2 0, the unrestricted func-tion y: rr is not one-to-one ald therefore has no inverse. t

Technofogy tJsing Parametric Equations to Graph hyerses (See the Tech-nology Notes in Section 2.3 for a discussion of parametric mode.) It is easy tograph the inverse of the function y : f(x), using the parametric form

x( t ) : f ( t ) , y ( t ) : t .

You can graph the function and its inverse together, using

xt(t) : t, y1G): f(t) (the tunction)

x2(t) : f(t), yz(t) : t (its inverse)

Even better, gaph the function, its inverce, and the identity function y : "r,expressed parametrically as

x3(t) : t, !z(t): t (the identity function)

The graphing is particularly effective if done simultaneously.Try it on the functions y : x5 /G2 + 1) and y : I 1se5x. You will see

the symmetry best if you use a square window (one in which the r- and y-axesare identically scaled).

Derivatives of Inverses of Differentiable FunctionsIf we calculate the derivatives of /(;r) : (I/2)x + 1 and irs inverse .f ,(x) =2x - 2 from Example 3. we see that

ftrot: *(:..,)::frro: fi<2"-rt:2.

{

/

. : . l i The functions y =.1Q and y: x2,x > 0, are inverses of one another.

6.6 The slopes of nonvertlcal l inesreflected across the l ine y = x arereciprocals of one another.

6.1 Inverse Functions and Their Derivatives 453

The slopes afe rccilnrcrt,TL,,,= +R , ,

6.7 The graphs of inverse functions have reciprocal slopes at correspondlngpoints.

The dedvatives are reciprocals of one another' The grtph of / is the line 'i' =

(1/2).r + l, and the graph of / I is the l ine -r :2,t 2 (Fig 6 4). Theit sJopes

are reciprocals of one another.This is not a special case. Reflecting any nonholizontal or nonveltical line

across the l ine ,l :,t always inverts the l ine's slope. If the original l ine has slope

n * 0 (Fig. 6.6), the reflected l ine has slope I /rr (Exercise 36).

The leciplocal relation between the slopes of graphs of inverses holds tbr other

functions as well. If the slope of I = /(,r) at the point (4. l (. i)) is f '(.r) I 0. then

t h e s l o p e o f t , : . l r f r ) a t t h e c o r r e s p o n d i n g p o i n t ( / ( c r ) . a ) i s 1 / . f ' ( u l ( F i g . 6 7 )

Thus, the derivative of f I at /(a) equals the rcciprocal of the dei'ivative of f at

c. As you might imagine, we have to impose some mathematical conditions on .lto be sure this conclusion holds. The usual conditions. t iom ldvanced calculus. are

stated in Theorem 1.

Theorem 1The Derivative Rule for lnverseslf / is differentiable at every point of an interval 1 and d//dr ir never

zero on 1, then f I is differentiable at every point of the inter val / ('f ) The

valte of df-t ldx at any particular point ./(a) is the reciprocal of the value

of df /dx at a:

/ d f ' \ I

\ rt- ). r,",=eL ( 1 )

l

I

lJ ld) .4)

ln short,

( l ' ) ' : (2 )

454 ChaDter 6: Transcendental Functions

D

EXAMPLE 5 For f (x): x2, 'r :0, and its inverse f-t(x) = "/7

(Fig. 6.8),we have

) t r d [ - 1 d - |" L : L 6 2 1 = 2 x a n d : ! . : + J i : ^ F . r > 0 .dx dx ax qx z\/x

The point (4,2) is the mirror image of the point (2,4) across the line y :1.

{ : r , :z(z) :c.dx

4

3

2

I

Slope 4

At the point (2, 4):, , _ J ;

6.8 The derivative of f 1(x) =..f at thepoint (4 2) is the reciprocal of thederivative of fkl = x2 at (2, 4).

6.9 The derivative of f(i = x3 - 2 atx = 2 tells us the derivative of f 1 at

d t - t I I I IAr rhe poinr (4.2);

d_ :

,G: ,A: 4: dI /d_

Equation (1) sometimes enables us to find specific values of df t/dx

knowing a formula for /-1.

^ l:3x '1 , - , :12

: 1

Eq ' ( i )L2

trwithout

EXAMPLE 6 Let f (x) : x3 - 2. Find flrc value of df-1 ldx at x :6: f (2)

withour linding a formula for / I{x).

Solution

See Fig. 6.9.

Another Way to Look at Theorem 1If y : /(x) is differentiable at x : a and'we change "r by a small amount dr, the

conesponding change in y is approximately

dy : f'(a) clx.

This means that y changes about /'(a) times as fast as x and that.r changes about

1/f'@) nmes as fast as ).

-r4

slope 3r2 = 3(2)'z = 12

Exercises 6.1ldentifying One-to-One Functions Graphicallywhich of the functions graphed in Exercises 1 5 are one-to-one, and which are not?

Exercises 6.'1 455

Formulas for Inverse FunctionsEach of Exercises l3-18 gives a formula for a function ]l : /(r) andshows the graphs of / and /-1. Find a formula for ./-r in each case.

13, f (x): x2 + I, x > 0

Graphing Inverse FunctionsEach of Exercises 7-10 shows the graph of a functio! ), : /(r).Copy the gaph and draw in the line y =.r. Then use synmetry withrespect to the line y : a 16 add the graph of / ' to your sk€tch. (Itis not necessary to frnd a formula for /

t.) Ideltify the domain andrange of ./-r.

9.

- ! - , - T I2 - - - )

a) Graph the function f (x) : ^rrl=7, O: r : 1. What sym-metry does the gaph have?

b) Show that / is its own inve$e. (Remember that r/7: .ri f r > 0 . )

a) Graph the function /(.x) : 1/r. What symmetry does thegraph have?

b) Show that jf is its own inverse.

14 , f ( x ) : x2 , a <Q

)

16.tf ( x ) = x 2 - 2 t ! 1 , t c 7 _ l

J

11.

y = f ( x ) = | - i , , t 0

12.

456 chaDter 6: Transcendental Functions

1 7 . f ( r ) - ( x * l ) 2 , . r > I 1 8 . / ( j r ) : t 2 l 3 , t > 0

)

Each of Exercises 19 24 gives a formula for a function 1, : /(r). Ineach case, find /-r(.r) and identify the domain and range of / 1. Asa check, show that ./(./

'(r)) : / '(/( ir)) : r.

Let f(x): x2 4x - 5, x > 2. Find the yalw of df-t ldx att h e p o i n t . r : 0 : " f ( 5 ) .

Suppose that the differentiable function ) : /(r) has an inverseand that the graph of ./ passes through the point (2, 4) and hasa slope of 1/3 there. Find the value of d/ | /dx at x = 4.

Suppose that the differentiable function ], : g(r) has an inverseand that the graph of g passes through the odgin with slope 2.Find the slope of the graph of g I at the origin.

35. a) Find the iriverse of the function /(-r) : m-r, where m is aconstant different from zerc.

b) What can you conclude about the inverse of a function

): /(r) whose graph is a line thrcugh the origin witha nonzerc slope ,r?

Show that the graph of the inverse of f (tt) : rpx I b, where mand , are constants and m +0, is a line with slope l/m andy-intercept -b/m.

a) Find the inverse of /(:r) :.t + 1. Graph jr and its inversetogether. Add the line ) : r to your sketch, drawing it withdashes or dots for contast.

b) Find the inverse of /(.t) : x * b (b constant). How is thegraph of /-r related to the gaph of /?

c) What can you conclude about the inverses of functionswhose graphs are lines parallel to the line l, : J?

Find the inverse of f(x): i + l. Graph the l ine y:-r + 1 together with the line,y : r. At what angle do thelines intersect?Find the inverse of /(r) : r + , (, constant). What angledoes the line y: -x + b make with the line y : r?What can you conclude about the inverses of functionswhose gaphs are lines perpendicular to the line ) : ir?

Increasing and Decreasing Functions39. lncreasing functions and decreasing functlons. As in Section

3.2, a function f(i) increases on an interval 1 if for any twopoints rr and 12 in 1,

x2 > xt :+ fQt2). f(nl).

Similarly, a function decreases on 1 if for any two points xt andx2 rn I,

x z > x r + f G ) . f o t ) .Show that increasing functions and decreasing functions are one,to-one. That is, show that for any rl and xz in I, h I jrl implies

f (x) + f (xt).

Use the results of Exercise 39 to show that the functions in Exercises40-44 have inverses over their domains. Find a fomula for d/-l/dtusing Theorem l.

40. /(.r) : (r/3)x + (5/6)

4r. f (x) : 2'7 xr

4 2 . f ( x ) : l - $ a 3

36.

19. / ( : r ) :15

21. f ( r ) : x3 + |

23 . f (x ) : l / x2 , x > 0

20. f ( t ) : va , a > 6

2 2 . f ( x ) : ( 1 / 2 ) x - 7 / 2

2 4 . f ( x ) : 1 / x ' , x l 0

Derivatives of Inverse FunctionsIn Exercises 25 28:

a) Find /-r(r).b) Graph / and /-r togetherc) E\alu..te df ldr at x : a and df-t /d t at r = /(a) to show that

at these points d/ | ldx:11(.df/ux1.

2 5 , f ( x ) : 2 x - 1 3 , a : - l

26. f(x) - (r/5)x t7, a - |

2 7 . f ( t c ) : 5 - 4 x . a : 1 / 2

2 8 . f ( t c ) : 1 a 2 , r > 0 , a - , 5

29. a) Show that /(,r) : 'r3 and g(,r) : ]T are inverses of oneanother.

b) Graph / and g over an r-interval large enough to showthe graphs intersecting at (1, 1) and (-1, 1). Be sure thepicture shows the required symmetry in the line y :.r.

c) Find the slopes of the tangents to the graphs of / and I at(1, 1) and ( l, 1) (four tangents in all).

d) What lines are tangent to the curves at the origin?

30. a) Show thar h(x) : x3 14 and t(r) : (4r)t/:' arc inverses ofone anomer.

b) Graph ft and fr over an .r-interr'al large enough to showthe graphs intersecting at (2,2) ^nd ( 2, 2). Be sure thepicture shows the required symmehy about the line ) - r.

c) Find the slopes of the tangents to the gaphs at t and,t at(2, 2) and (-2, -2).

d) What lines are tangent to the cuNes at the origin?

31. I-et /(r) : x3 -3x2 - l,x-2. Find the yalue of df-t/hc atthe point r: -l : /(3).

38. a)

b)

c)

Exercises 6.1 457

43.

u,/ ( r ) : ( l r ) .

Theory and Applications45. If / (r) is oneto-one, can anything be said about I (,r) = - f (x)?

Give reasons for your answer

46. lf f (x) is one-to-one and /(r) is never 0, can anything be saidabout r(t) :1lf Q)1 Give reasons for your answer.

47. Suppose that the range of g lies in the domain of / so thatthe composite / o g is dellned. If / ard g are one-to-one, can

anything be said about / o g? Give reasons for your answer.

48. If a composite / o g is one-to-one, must 8 be one-to-one? Give

reasons for your answer.

49. Suppose /(:r) is positive. continuous, and increasing over theinterval [.r, r]. By interpreting the graph of / show rhat

fb | fth\

I f t x t d x + | . [ ' ( y \ d ; - b l ( b \ a l h \ .

J t J f @)

50. Determine conditions on the constants @, b, c, and d so that the

rational function

Then show that the l'unctions W and 5 agree at a point of [,?, bland have identical derivatives on [d, ,]. As you saw in Section,1.2, Exe.cise 56, this will guarantee w(t) - s(/) fbr all r in

fa, Dl. In particular, w(b) = S(b). (Source: "Disks and ShellsRevisited," by Walter Cadip, American Mathetuaticd[ Monthll,Vol. 98, No. 2, February 1991, pp. 154-156.)

S CAS Explorations and ProjectsIn Exercises 53-60, you willexplore some functions and their inversestogether with their derivatives and linear approximating funclions at

specified points. Perform the following steps using your CAS:

a) Plot the function 1 : /(r) together with its derivative ovef thegiven inteNal. Explain why you know that / is one to one overthe interval.

b) Solve the equation r - /(x) for -r as a function of l, and name

the resulting inverse function g.

c) Find the equation tbr the tangent line to / at the specified point(.rn,

"f(,.0)).d) Find the equation for the tangent line to 8 at the point (J (,ro), .ro)

located symmetrically across the 45' line 1 :1 (which is thegraph of the identity function). Use Theorem I to find the slopeof this tangent line.

e) Plot the functions / and.9, the identity, the two tangent lines, andthe line segment joining the points (.r0.

"f(,ro)) and (l (,16), rc6).Discuss the symmetries you see across the main diagonal.

_ 25 3 . 1 : / 3 x - 2 .

1 = x ! 4 , * o - 3

l v - t t5 4 . 1 : j _ ] . -

2 . t 2 . ) o = t / 2l x - l l

s s . r : j . . L \ ' 1 . t , = t , 2r j + I

y l

5 6 . y L r - 1 . r , = 1 , 2- . \ r + I

w(r)

s(/)

)1\ ' : r - J r - l - I ^ ' : . . 1 , , :

t 0

) : 2 - x - r r , 2 = t = 2 , t o : 3 -z

: I ,,,,',' "("

'(r))' - o') 1r'

- l "

znx{Jt t) - : f (x))dx.

a x + bc x + d

has an 1nvelse.

Sl. st i l i anather t , /ay ta r ie' t t Theatem 1.If we write 8().) tbr

,f ' (r) , Eq. (1) can be wdtten as

1g r f t a t l : - ; : - . o r R t [ t a \ t , I t , r t = L

I \4)

If we then write ,r for d, we get

s ' ( / ( r ) ) ' / ( ' ) : 1 .The latter equation may remind you ofthe Chain Rule, and indeedthere is a connection.

Assume that / and g are differentiable functions that areinverses ofone another so that (8 o /)(r) - jr. Differentiate both

sides of this equation with respect to r, using the Chain Rule to

express (g o /)'(r) as a product ol derivatives of g and /. What

do you find? (This is not a proofofTheorem 1 because we assume

herc the theorem's conclusion that g: /-l is differentiable.)

52. EotJivalence of the v' tasher and shel l methods fat f indlng,tolumc. Let f be differentiable on the interval 4 : -r : b, with

a > 0, and suppose that I has a differentiable inverse, / I. Re-

volve about the )-axis the region bounded by the graph of /and the lines a:4 and y = /(r) to generate a solid. Then thevalues of the integrals given by the washer and shell methods forthe volume have identical values:

f ' b ' , ^ . t ^I r l r f ' r v ) ) - d - ) r r i | 2 n t t l r b t - f t ' ( t d ( .

J l \ , ' ) J d

To prove this equality, define

59. 1 : gr,

60. 1 - 5 i11,

In Exercises 6l and 62, repeat the steps above to solve for the functions

): /(r) and r = / '(t) deflned implicitly by the given equations

over the interval.

6 1 . ; 1 / 3 - l : ( x f 2 ) r , 5 : j : 5 , t i o : - 3 1 2

62. cosl : yl ls, 0 = jr : l , -ro : l /2

56.

3 = r = 5 , i o = I

- - < - r < - . L , : l) - - 2

r-


:,-r The graph of y: In x and itsrelation to the function Y : 1lx, x > O.The graph of the logarithm rises abovethe x-axis as x moves from I to the right,and it falls below the axis as x movesfrom 1 to the left,

Natural LogarithmsThe most important function-inverse pair in mathematics and science is the pair

consisting of the natural logarithm function ln.r and the exponential function e''

The key to understanding e' is ln;, so we introduce ln.x first. The importance

of logarithms came at first from the improvement they brought to arithmetic. The

revolutionary propefiies of logarithms.made possible the calculations of the great

seventeenth-century advances in offshore navigation and celestial mechanics Nowa-

days we do complicated arithmetic with calculators, but the propefties of logarithms

remain as important as ever.

The Natural Logarithm FunctionThe natural logarithm of a positive number r, written as ln x, is the value of an

intesral.

If 'r > 1' then ln { is the area under the curye y : l/t from t : 1 to t : x

(Fig. 6.10). For 0 < x < 1, ln r gives the negative of the area under the curve from

I

rro < r < r, thenrnr = l,' + "

= -1,' + "gives ihe negative of this area.

r r ,> r ,o"nrn '=J lar

t ,**", : f t j r ,=oT"

gives this arca.

,, '\

Y = l n ,

lypical 2-place values of ln x

r l n t

6.2 Natural Logarithms 459

r to 1. The function is not defined for -r < 0. We also have, , ,

lnl : I :, dt = 0. l I ,., .t| i i ,. . l imir\ equ.,lJ t r

Notice that we show the graph of y : 1/"t in Fig. 6.10 but use y : l/t r theintegral. Using r for everything would have us wdting

n,: [ ' !a, ,, J t x

with r meaning two different things. So we change the variable of integrationto t.

The Derivative of y = In xBy the first part of the Fundamental Theorem of Calculus (in Section 4.6),

) d f ' l II l n x : : I : d t : : '

a x a x J t t x

For every positive value of "{. therefore.

i 1

d x x

If a is a differentiable function of -r whose values are positive, so that ln ,, isdefined, then applying the Chain Rule

dy dy dudx du dx

to the function y : ln, givesl d uu d r

l d u- - - ' u > Uu dJc

EXAMPLE 1d ^ | d . ^ | . ^ . I

d x " ' ^ 2 x d x ' ' ^ ' - 2 x ' ' ' -

r U

Notice the remarkable occunence in Example 1. The function :y : ln 2.x hasthe same derivative as the function :y : ln.r. This is true of l: lna-r for anynumber a:

d t d t l- - - t t t u ^ - - . - \ s i , , - - ( s / -

dx alx alx ax x

00.050.5I234

l0

undef,ned-3.00-0.69

00.691,101.392.30

d

dx

d . du- t n u . -du dx

d- - l n a :dx

(2)


In the late 1500s, a Scottish baron, JohnNapier, invented a device called theIogarithm t|-:at simplified aithmetic byreplacing multiplication by addition. Theequation that accomplished this was

l n a r : l n 1 7 + l n t .

To multiply two positive numben 4 and r,you looked up their logarithms in a table,added the logarithms, found the sum in thebody of the table, and read the tablebackward to lind the product a.r.

Having the table was the key, of course,and Napier spent the last 20 years of his lifeworking on a table he never finished (whilelhe asrronomer Tycho Brahe uaited in vainfor the infomation he needed to speed hiscalculations). The table was completed afterNapier's death (and Brahe's) by Napier'sfriend Henry Briggs in London. Base 10logarithms subsequently became known asBriggs's logarithms (what else?) and somebooks on navigation still refer to them thisway.

Napier also invented an artillery piece thatcould hit a cow a mile away. Hodfied by theueapon'\ accuracy. he stopped productionand suppressed the cannon's design.

EXAMPLE 2 Equat ion (1) wi thu:x2 *3gives

l rn ' ' * r ) :dx x2 +3

d , ^ l ^ z x. - . ( . r - +JJ

d x x 2 + 3 x t t 3 tr

The properties that made logarithms the single most important improvement in

arithmetic before the advent of mode.m computers are listed in Table 61' The

propefiies made it possible to replace multiplication of positive numbers by addition,

and division of positive numbers by subhaction. They also made it possible to

replace exponentiation by multiplication. For the moment, we add the restriction

that the exponent n in Rule 4 be a rational number You will see why when we

orove the rule.

Properties of Logarithms

EXAMPLE 3

a ) l n 6 : l n ( 2 . 3 ) : l n 2 * l n 34

b) ln4 - ln5 : ln - :1n0.8

Ic ) l n g : - l n 8 R.crUocal

tr

EXAMPLE 4

a) ln4 * ln sin.t : ln (4 sin;r) PfodL,cI

b ) l n ^ : = l n { x i - l t - l n ( 2 x - 3 ) o " r i r '2 x - 5

1c) ln sec.r : ln - : - ln cos.r Reciproc'rl

cos.t1

d) In Vx * I : ln tx + I ) ' /r : ,

ln ("r * 1) Po$cf E

Proof that In ax = In a + In x The argument is unusual-and elegant. It starts

by observing that lnar and lnx have the same derivative (Eq. 2). According to

Corollary 1 of the Mean Value Theorem, then, the functions must differ by a

Table 6.1 Properties of natural logarithms

For any numben a > 0 and "r > 0,

l n a x = l n a * l n x

L r 1 : h a - l n , r

. l

l r l . rn : , ln.r

1. Product Rule:

2. Quotient Rule:

3. Reciprocal Rule:

4. Power Rule:

R u l c l \ i r h d - 1


constant, which means that

ln ax : lnx lC (3)

for some C. With this much accomplished, it remains only to show that C equalslnq.

Equation (3) holds for all positive values of .r, so it must hold for x : 1. Hence,

l n ( a ' 1 ) : l n 1 * C

l n a : 0 i C t n l : o

C: lna . Reananged

Substituting C :bte in Eq. (3) gives the equation we wanted to prove:

l n a x : l n a l l n x . (4)

D

Proof that ln (a/x) = ln a - In x We get this from Eq. (4) in two stages. Equation(4) with a replaced by I/x. gr'ves

I / l \l n ' * l n x : l n l - . . x 1

x \ { /: l n l : 0 ,

so that

m1: - ln r .x

Equation (4) with x replaced by i/.x then gives

In l : rn f , .1 ) : l na r rn l- { \ x / x

: lna - In.r.

Proof that ln x" = n In x (assuming n rational) We useargument again. For al1 positive values of .r,

d - | d . "- : - I n r " : - , ( r ' )dx x ' ax

: ! , , , ,

l d= n - - - - , ( | 1 l n x ] .

x a x

Since ln-rn and nlnx have the same derivative,

l n x n : n T n x t C

for some constant C. Taking x to be I identifies C as zeto, and we're done. D

As for using the rule lnx' : nlnx for irrational values of n, go right ahead

and do so. It does hold for all n, and there is no need to pretend otherwise. From

the point of view of mathematical develoPment, however, we want you to be aware

that the rule is far from proved,

tr

the same-derivative

E q . ( l ) w i l h , r : t "

Herc is u,here $,e need ,to be rational. al least fofnow. We have prored lhcPolver Rule only forralional expoDents.


and

The Graph and Range of In xThe derivative d(lnx)/dx:1/;r is positive for r > 0, so ln.r is an increasingfunction of .r. The second derivative, -l/x2, is negative. so the graph of ln,r isconcave down.

We can estimate ln 2 by numerical inte$ation to be about 0.69. We thereforeknow that

ln2-"n

2

It follows that

lim lnx : oo and lim ln-x : *oo.x+0+

The domain of ln.{ is the set of positive real numbersl the range is the entire realline.

Logarithmic DifferentiationThe derivatives of positive functions given by formulas that involve products, quo-tients, and powers can often be found more quickly if we take the natural logarithmof both sides before differentiating. This enables us to use the rules in Table 6.1to simplify the fomulas before differentiating. The process, called logarithmicdifferentiation, is illustrated in the next example.

EXAMPLE 5 Find dy/dx if y :( a 2 + r ) ( x + 3 ) 1 / 2

, x > l .x - 1

So/ufion We take the natural logarithm of both sides and simplify the result withthe rules in Table 6.1:

. . ( " '+ l ) ( . r + 3)r / ' ]||I - || ----------------

x - l

1)(- lr + 3) ' / '?) - ln (x - 1) Quotient Rulc: r n ( t x - + l

: ln(x2 t 1) +ln(.r*3)r/2 -ln(.r - 1) Producr Rulc

: ln ( . r2 * t ) + lm(. r +: ) - ln(x - l ) . powcr Rule)

We then take derivatives of both sides with r€spect to x, using Eq. (1) on the left:

t d y t ^ I I Iy d x

: r , + l '

- * - r ' x - i

- r - I

Next we solve for dy/dx:

rn2 ' : n1 "2 , " ( ) ) : ;

: nln2. -, (;) :

d t / 2 x Id x - r \ x z ] - 1 . ' 2 x ] - 6

1 \. - 1 ) '


Finally. we substitute fbr r:

d l _ ( r ' ] + t ) ( r + 3 ) 1 i : ( 2 x *

|d x r I \ . r r + 1 2 x i 6 * ) fl

How to Differentiate y = f(x) > 0 by Logarithmic Differentiation

1. ln y : ln /(y) T . l r 1 ' , ' i l L h r , t s

nn, , j ^ ^ , -

d - r ( l n / ( r ) , I r r ' . . . , , . . , . I

3 . I d , ' -

- d ,

r l n 1 , t ) , ' . , r ' r . . r Il d r o r

d v d4 . , = y . ( l n / ( x ) )

ax ax

i r , t

5. " , ' = /1x;a 1 ln 71.r ;1qx dx

The Integral I AO a,Equation (1) leads to the integral formula

t lI d u : l n u l C

.t u

when r/ is a positive differentiable flnction, bur whar if a is negative? If irnegative. then rr is positive and

f 1 f 1l - d u : I - - d ( ' )

.l u J Gu) (6)

: l n ( r ' i ) * C . F . q . 1 r ) \ r l h r f r t r i r r ! L i t , l

We can combine Eqs. (5) and (6) into a single formula by noticing rhat in each casethe expression on the right is ln lrrl * C. In Eq. (5). ln I : ln lul because a > u:in Eq. (6), ln (-a) = ln lu because a < 0 . Whether r is positive or negative, thein tegra l o f ( l /u )du is ln la *C.

If r is a nonzero differentiable function,

r1. l ;0"

: ln ) tL t C. \7)

We know that

I t r " d u - : ' + C ' t l - 1 '. l l t + l

Equation (7) explains what to do when ir equtrls l.

(s)

i s


Equation (7) says that integrals of a certain form lead to logarithms. That is,I f t t - \

I #a r : l n r / ( r ) l *CJ J I X )

whenever /(r) is a differentiable function that maintains a constant sign on thedomain given for it.

EXAMPLE 5t 2 ' r -t -

Jn " , _ 5o*

:a t 2 ) : I

*ln 5

r ' : 3 + 2 s i n r . d : 2 t ( \ s 0 l Hu l n / 2 ) : I . t n / 1 ) = s

Eq. (7)

4 - l t r l

I ! ! :n tu ' IJ - s u ' l - . / / r r i r = ) '

l n i - l l - l n l - 5 1 : l n I - l n 5 : D

EXAMPLE 7

f"/2 4coso fsI -". ^de = 1

J "/z J + zstna Jt

: ? l n

: 2 l n :.12 l n l 1 l : 2 l n 5

2- d uu

l 5l u l I

J '

15 l -

The Integrals of tan x and cot xEquation (7) tells us at last how to integrate the tangent and cotangent functions.For the tangent,

I t ^ *a* : [ " i " * a r : [

-o 'J J C O S X J U

f d u: - J ; : - ln la l rc

: -ln lcosrl * c : ln -f * a Reciprocal Rrr.l c o s r l

: ln I secxl * C.

F^r rhA ^^ ra- -^ - r

t , I c o s x d x I d u / , _ , i , , .Jco lxd r : J ; , " r : J ; ' ; ' - ' " . ' 1 '

: l n l z l f C : l n l s i n . r l * C : - l n l c s c x l * C .

| ,un* a,

| "o tu ao

: - l n l c o s a l i C : l n l s e c a l - C

: ln ls ina l t C : - ln lcsc. r l + C

Exercises 6.2 455

EXAMPLE 8

fu"'' ,^rr o* :S L L b \ ( i l L r l c r : l ! .

tar lu du , r - r i i / I i .i r { l ) r ( lr / r r r h r : r r

I- l n 1 ) : ^ l n 2

z

lo"'t ,un, +:: Ir ' ' 'jr" r*",r]"" : la' ' z

Exercises 6.2

Using the Properties of Logarithms1. Express the following logarithms in terms of ln 2 and ln 3.

ln ,t-t t . , - _- 1 + l n . r

23. y : ln (ln x)

25. y : 0(sin (ln 6) * cos (h0))I

t t . , - l n -' x.,,/ x + |

l + l n tt o - -I t n r

31. 1 : 11 (sec(ln 9))

/ ( x 2 + 1 ) 5 \3 3 . ] : l n | , . I

\ v L ) 4 , /

35 . ! : | . l n ^ / t d t

37. y: JiQT]r]f t3e. i :V r+ t

41. y: JE + 3 sin?

43. 1 : ti11 1111-1- 2.1

0 +545. ],,:

a) ln 0.75

d ) h i 6b) ln (4/9)

e) lrL 3xD

c) ln ( l /2)

f) h ^"43J

ir ln ,rt t - - _- l + ] n r

24. ): ln (ln (ln r))

26. y :111(sec 9 + tan d)

28. r, = i h j--l-12 1 - x

30. y: Jni/ , / \ in 4 cos P \

32 . 1 : l n t , ' r r - I\ ' - " " , /

34. r.. : ln

x . y : [ # n ta rJ ! t

38. ) : \,{" +lx,r l)'?

42. y - (tan| JE +1

I4 4 , ) : -' t \ t + I ) 1 1 + 2 )

9s ind

v sec u

2. Express the following logarithms in terms of In 5 and ln 7.

a:) tn (r 1125) b) ln 9.8 Q tt1rt

d) ln 1225 e) In 0.056

f) (ln 35 * ln (l/1))/(rn25)

Use the properties of logarithms to simplify the expressions in Exer-

cises 3 and 4./ s i n 6 \

J . a ) ln s ind - t " l ; /

- / l \b ) ln (3x ' / - 9x ) + ln

\3 rJ

o 1 ln t4 ror - ln 22

4 , a ) l n s e c d + l n c o s ab) ln (8:r + 4) - 2 ln2c ) g t n & t l - t n ( r + t )

Derivatives of Logarithmsln Exercises 5 36, find the derivative of ,v with respect to,r, r' or 6'

as appropriate.

5 . y : ln 3x

7 . y = ln ( t2 )

39 . 1 : l n :

1 1 . ) - l n ( 0 + 1 )

1 3 . y : l n : r 3

15. ) : t(]n t)2

rc4 l*1 7 . y - ; 1 n x . ;

l n t1 9 ' v :

Logarithmic DifferentiationIn Exercises 37 50, use logarithmic differentiation to find the deiva-

tive of ] with respect to the given independent variable

6. y: ln,tr, k constant

8 . y : ln i t3 / '? ;

l01 0 . 1 : l n -

12. | : ln Qe + 2)

1 4 . y - ( l n x ) 3

16. y : t ̂ ,G7

tJ r4 t4 7 . v : -

(-[ + I )a'

, - , - 1 ,

4 e . J = : l . 'I L ' f l

0 cos9

4 8 . v :

. 3 . 3.o : l l n r L' 3 9

l + l n t

t

(x * 1)5

( r * 2 ) r0

(x * l )10

@+rf

r ( ) r + 1 ) ( t - 2 )(-r': + 1)(2' + 3)

18.

50 . 1 ' :


IntegrationEvaluate the integrals in Exercises 51 68.

77. Find the lengths of the following curves.

a ) y = ( x 2 1 8 ) l n r , 4 S r : 8b ) x : ( y / 4 ) 2 - 2 h ( y / 4 : ) , 4 3 y = 1 2

78. Find a curve through the point (1, 0) whose length from x:1t o t : 2 i s

, ' [ ,

ur. J" t^)a'

l 2 d x51. '

f 2.r d,t< 1 ' -

f " s i n ls s . l ^ d t

Jo Z - COS t

5L l - ax

se. , ---:'J ) x ( l n . I , , '

61. I ."':- '

clrJ 6 + J tanr

es. |,","2*torao

-'l#E

s2. I =:=

s4. I ::::

f"tt 4sines6. | _d0

J 0 I - 4 C O S y

s8. / 1"J2 x \ n I

60. I -+Jz 2x.,/ln x

- ^ / s e c l t a n y ,oz. | -:--aY

J Z + S e C ]

64, I cott dt

66. I 6tan3x dx

,- I secx 1.t6 X t -

J Vln (sec x + tan.r)

, l r r " dx .

Theory and Applications69. Locate and identify the absolute exteme values of

a) ln (cosx) on l-r /4, r /3),b) cos (h .t) on l l/2,21.

70. a) Prove that /(r) = r - ln r is increasing for x > 1.b) Using part (a), show that ln r < r ifx > l.

71. Find the area between the curves ),: ln r and y : ln 2.r from, r : 1 t o r : 5 .

72. Find the area between the curve y : tan.r and thejr-axis fromx : - n / 4 t o x : t t / 3 .

73. The region in the first quadrant boutrded by the coordinate axes,the line y:3, and the curve x:2/Jy11 is revolved aboutthe )-axis to generate a solid. Find the volume of the solid.

74. The region between the curve y:."/iot.r and the r-axis fromx : tt /6 to x = n/2 is revolved about the r-axis to generate asolid. Find the volume of the solid.

75. Th€ rcgion between the curve ) = l/* and the r-axis fromx =7/2 to x =2 is revolved about the y-axis to genente asolid. Find the volume of the solid.

76. In Section 5.4, Exercise 6, we revolved about the y-axis the regionbetween the curve y - 9x 1Jt\ 9 and the .r-axis from x = 0to r - 3 to generate a solid of volume 362. What volume doyou get if you revolve the region about the ,r-axis instead? (SeeSection 5.4, Exercise 6, for a graph.)

E 29. carcurarona) Find the centroid of the region between the curve y : 1/,t

and the r-axis from.x - I to.r - 2. Give the coordinatesto 2 decimal places.

b) Sketch the region and show the centoid in your sketch.

80. a) Find the center of mass of a thin plate of constant densitycovering the region between the curye ) : llJi ar'd thel-axis from -:r : I to .r = 16.

b) Find the center of mass if, instead of being constant, thedensity function is t(x) : +laF.

Solve the initial value problems in Exercises 81 and 82.d\i l

8 1 . : - . : I + - . y ( t ) : J

82 . - = sec2x . y t0 ) :0 and l ' (0 ) : It lx '

83. The linearization of ln(1 +x)atx:0. Instead of approximat-ing r near.r : 1, we appro mateln(1 + r) near r : 0. We geta simpler formula this way.

a) Dedve the linearization ln (1 + t) x a at x :0.

E f1 cnfcufnfoR Esrimate ro 5 decimal places the eror in-volved in replacing ln (1 *,r) by .r on the interval [0, 0.1].

EE c) GRAPHER Graph ln (l *l) and r together for 0 5 j{ s0.5. Use different colors, if available. At what points doesthe approximation of ln (1 + r) seem best? least good? Byreading coordinates from the graphs, find as good an upperbound for the error as your grapher will allow.

84. Estimating values of ln x with Simpson's rule. Although lin-earizations are good for replacing the logarithmic function overshort intervals, Simpson's rule is bettet fot estimating particulorvalues of ln ,r.

As a case in point, the values of ln (1.2) aird ln (0.8) to 5places are

ln (1.2) : 0.18232, ln (0.8) : -0.22314.

Estimate ln (1.2) and ln (0.8) nrst with the formula ln (1 + r) ^, xand then use Simpson's rule with', :2. (Impressive, isn't it?)

85. Find

I. ln (r ').

JJ6 ln j!

Generalize this result.

86. The derivative of In kx. Could ] - ln 2jr and )l : In 3, pos_.ibly ha\e rhe same derirarjve al each porntl lDrfferentrare rhemto nnd out.) What about ) : h ft.r, lor other positive values ofthe constant i? Give reasons for your answer

I Grapher Explorations87. Graph ln x, ln 2x, ln 4r, ln 8.r, and ln 16.r (as many as you can)

togerher for 0 <.r: 10. Whar is going on? Explain.

6.3 The Exponential Function 467

88. Graph ) : ln sinrl in the window O = x 522, -2: ) : 0.Explain what you see. How could you change the formula totum the arches upside down?

89. a) Graph l = sini and the curves ) : ln (a i sin,r) for a :2, 4, 8,20, and 50 together for 0 : r : 23.

b) Why do the cuNes flaften as c increases? (I1irr..Find and-dependent upper bound for ly,l.)

90. Does the graph "of y : uE - ln t, :r > 0, have an inflectionpoint? Try ro answer the quesrion (a) by graphing, (b) by us-ing calculus.

6 . t t T h e g r a p h s o f y : I n x a n dy : ln ' x . The number e i s ln r i .

9:n,a l

If, in addition, ) : _y0 when I :0, the function is the exponential funcrron y :yoer'. This section defines the exponential function (it is the inverse of ln .r) andexplores the properties that account for the amazing frequency with which thefunction appears in mathematics and its applications. We wiil loot at some of theseapplications in Section 6.5.

The Inverse of In x and the Number eThe function ln .r, being an increasing function of _r with domain (0, oo) and range(-m,oo), has an inverse ln-'x with domain (_oo, oo) and range (0,co). Thegraph of ln-r ,r is the graph of ln x reflected across the line y : 'r i, yuo .un ,"",

lim ln-r .r : oo and l im ln - r r :0 .

The number ln | 1 is denoted by the letter e (Fig. 6.11).

Definition

Although e is nor a rational number, we will see in Chapter g that it is possible tofind its value with a computer to as many places as we want with the formula

. . / | i l \e : l i m I I I I + . _ t . . . + _ ' I, , . @ \ 2 6 n , . /To 15 places,

The Exponential FunctionWhenever we have a quantity ) whose rate of change over time is proportional tothe amount of ) present, we have a function that satisfies the differential equation

e:2.7 1828 1828 45 90 45.


er (rounded)

TheFunct iony=exWe can raise the number e to a rational power -r in the usual way:

e 2 : e . e , - ' I"

- : a . e , - : \ /e ,

and so on. Since e is positive, e' is positive too. This means that e' has a logarithm.When we take the logarithm we find that

l n e " : r l n e : , r . 1 : r . ( 1 )

Since ln r is one-to-one and ln (ln-r,t) :.t, Eq. (1) tells us that

e' : ln-t x for x rational. (2)

Equation (2) provides a way to extend the definition of €' to irrational valuesof -r. The function ln-l;r is defined for all x, so we can use it to assign a value toe' at every point where e' had no previous value.

ln - ir.

1012

10100

0.3712.'127.39220262.6881 x l0a3

DefinitionFor every real number .r,

EXAMPLE 1

a) ln ez =2b ) l n e r : - 1

_ lC , l n . / e = -' 2

d) ln €'in* : sinr

e) et"2 :2

f) ern("+r) - 12 + 1g) e3tnz : etn2' - etns - Ih ) s : rn : - @t"2)3 :23 - 8

Equations Involving In x and e"Since ln,r ar'd. e" are inverses of one another. we have

Invers€ Equations for e' and ln x

e t n ' : x ( a l l r > 0 )

ln (e') : r (all 'r)

(3)

(4)

You rnight want to do parts of the next example on your calculator.

Tlpical Values of e'

Table 5.2 Laws of exponents for e'

For all numben x, .rl, and t2,

l. ett . e'x - e't+t2

I

4. (.ett)4 : e'\" : (e")al


EXAMPLE 2 Find y if ln y :3t + 5.

sorution Exponen,ru," ro* ,ru".;,,, _

".,*.y : e3'+5. Eq (3)

EXAMPLE 3 ttn4 1 i1 s2* = 10..

so/utron Take the natural logarithm of both sides:

ezk : l0

l n e x : l n I O

2k : ln 10 Eq. (4)

1k : -ln 10.

Laws of ExponentsEven though e' is defined in a seemingly roundabout way as ln-l.x, it obeys the

familiar laws of exponents from algebra (Table 6.2).

Proof of Law 1 Let

Useful Operating Rules

1. To Gmove logadthms from anequation, exponentiate both sides.

2. To remove exponentials, take theloeadthm of both sides. tr

tr

\ : e ' ' a n d Y z : e " (s)

Then

= lnlr and xz: lrr Y2

: lnyr + ln)2

: lnyly2

: Y1Y2

The proof of Law 4 is similar. Laws 2 and 3 follow from Law 1 (Exercise 78).

X l

\ + ) e 2

Take logs of bothsides of Eqs. (5) .

Producl Rule

Exponentlate.

a

EXAMPLE 4

t) s" +t"z : e' . eta 2 : 2e'

, 1 1b) e " -- -';:- : -

X

e

1 e 3 1 , : e 3 , : ( e ' ) 3

c)

d)

Law l

La$, j

La* 4 D

470 Chaoter 6: Transcendental Functions

Transcendental numbers andtranscendental f unctions

Numbers that are solutions of polynomialequations with rational coeflicients are calledalgebraic: 2 is algebruic because it satisfiesthe equation x + 2 : 0, and r/5 is algebraicbecause it satisfies the equation .I2 - 3 : 0.Numbers that ar:e not algebraic are calledtraNcendental, a term coined by Euler todescribe numbers, like e and z, that appearedto "tuanscend the power of algebraicmethods." But it was not until a hundredyears after Euler's death (1873) that CharlesHemite proved the transcendence of e in thesense that we describe. A few years later(1882), C. L. F. Lindemann proved thetranscendence of r.

Today we call a function ) : /(r)algebraic if it satisfies an equation of theform

& y ' * . * & r * P o : 0in which the P's are polynomials in r withrational coefficients. The functionf :1lJx +1ls algebraic because itsatisfies the equation (r + l))'2 I :0.Here the polynomials arePz = x I 1 ,4 :0 , and Po : -1 .Polynomials and rational ftmctions withrational coefncients are algebraic, as are allsums, products, quotients, rational powers,and rational roots of algebraic functions.

Functions that are not algebraic arecalled transcendental. The six basictrigonometric functions are tanscendental, asare the inverses of the t gonometricfunctions and the exponential and logarithmicfunctions that are the main subject of thePresent chapter

The Derivative and Integral of e'The exponential function is differentiable because it is the inverse of a differentiablefunction whose derivative is never zelo. Starting with ! : e' , we have, in order,

r t r r - ^ Logenrhm\ of borh s ides

Derivalives of both sides wilh respccr

: e' . ] replaced by e'

The startling conclusion we draw from this sequence of equations is that €' is itsown derivative.

As we will see in Section 6.5, the only functions that behave this way areconstant multiDles of er.

d "

d x -(6)

1 /l\

y d.x

EXAMPLE 5, l i

a l ] e ' = e ^ + t - x ) : e ' \ - 1 ) : - e - x r q , 7 ' $ i , h , - - \dx dx/t ,1

by a " t in

' : " ' ' "

' l ( s in x ) : e ' i " * . cosrax a)(

The integral equivalent of Eq. r7t is

dydxdydx

EXAMPLE 5

:5e ' t rThe Chain Rule extends Eq. (6) in the usual way to a more general form.

If a is any differentiable function of x, then

d . . . . d u

4 X A X(7)

d . - " . _ d ". ( ) e ' ) = J - e "ax a)t

Eq. (7) wi th r : s in l

le 'au:e"+c.


, = * . \au : a ' . , rot : o.

I l ( l n 2 ) = 3 l n 2 : l n 2 r : l n 8

EXAMPLE 7

EXAMPLE 8

th2

J" e3'dx

. r l n S I= I e " ; du

J O

I f l n 8: ; l e "du

I . l r n8

- , d ! |

r -lo

1 7: ; t8 -11 : ;

J J tr

f " t ' ' l -

Anl iderrvat i \e f iom

/ et'", cos x dx : e' '"t I r\xmpte r,

J o l o

: e l - e o : e - \ t r

EXAMPLE 9 Solving an initial value problem

Solve the initial value problem

" d y€ --. = "t .

, . J3, Yl2t :0.a)c

so/ution We integrate both sides of the differential equation with respect to .x to

obtain

e , : x 2 + C .

We use the initial condition to determine C:

c : e o - ( 2 ) 2

: t _ 4 : _ 3 .

This completes the formula for e):

. ' " : x2 - 3 '

To find y, we take logarithms of both sides:

h e r : h ( r 2 - 3 )

' : ln (;r '?- 3)'

Notice that the solution is valid for -r > J5.

(8)

(e)

472 Chaoter 6: Transcendental Fundions

It is always a good idea to check a solution in the original equation. From Eqs.(8) and (9), we have

) , , )4 2 : " t L I n ( x 2 - 3 ) l o , a ,

dx dx

x ' - J

- / . 2 _ l \ _* \ 4

x ' * ) F q 1 8 )

The solution checks. tr

Exercises 6.3

Algebraic Calculations with the Exponential andLogarithmFind simpler expressions for the quantities in Exercises 1+.

l , a\ e1"12 b) "

t t c) stn '- tn Y

2, a) etn ("+r') b) e tn o: c) stn "' t" z

3 . a t 2 l n , / i b ) l n { l n e " l c ) l n ( e ' ' " )

4, a) ln (e,* d) b) In (e(")) c) ln (e2r" ')

Solving Equations with Logarithmic or Exponen-tial TermsIn Exercises 5-10, solve for ), in terms of t or r, as appropriate.

DerivativesIn Exercises 17-36, find the derivative of y with respect to,r, t, or6, as appropriate.

17, y -- e-5'

19, y = ss-t'

21 , y : vp ' - " '

2 3 , y = ( x z - 2 x + 2 ) e '

25 . } , :ed(s ind+cosd)

2'1. y : cos (e e')

29 , y : In (3 te - ' )

31. ":h f-4- ' \\ l + e b /

33. ) : d(cosr+'n ' )

at. ,:,/. sin e'dt

In Exercises 3140, find dy/dtt.

37. ln y : ar sina

39. e" = 5i16 1 3y;

IntegralsEvaluate the inte$als in Exercises 41-62.

5 . l t \ y =2 t +4

7. ln (y - 40) :5t

9 . l n ( y - l ) - l n 2 = i + l n t

10. ln (1'?- 1) - ln (), + 1) : ln (sinr)

In Exercises 11 and 12, solve for ft.

18. y -- eb/3

20. y = ewE+")

22. y: (l + 2x)e-2'

?A. y = (9x2 6x + Z)e3'

26 . Y :h (3ee-o)

28. y : 03 e-ze cos 50

30. y : ln (2e-' sin r)

| - / A \3 2 . y : 1 q t ; ' j ^ l

\ r + v t l

34 . ) : es i n ' on I + 1 )

N. t :1" " n ta t

38, ln ry = g*+r

4 0 . t a n y : e ' + l n - r

42. I@

- 3e-2') d'

u' lo ,^""-' a'

6 . l n ) = * t + 5

8 . l n ( l - 2 y ) : t

l l . a) elk :4

12 . a ) esk : !

In Exercises 13-16, solve for t.

13, a) e-.o3t :2't b) "r '

:; c) eono.2)t :O.4

14. at e o.o,!: rooo br e,, : fr ", " ^r, : !

lS. el : x2 16. et") e(r+1) - el

b) 100eroe :200 c) ek/tt]@ - a

b) 80?r : I c.) errno s){ - 0.8

a. | {"" +s" \a,

43' J,",

e' dx

4s. J

8"on\ d'

n . l'"n" "', o'

4e. J fdr

sr. lz," " a,

t . [ " I 0 ,

tt, fo"''

r, + e^o) sec2 o do

s6. 1,"'o' {t * """) """t

o ae

Sl. | ""*"'

r"" 1rt tan rt dt

5 8 . I " ' " " " c s c ( z + t ) c o t ( n + t ) d l

Theory and Applications67. Find the absolute maximum and minimum values of /(t):

e' 2.t on [0, 1].

68. Where does the periodic function /(t) = 2esin {x/2) take on itsextreme values and what are these values?

Exercises 6.3 473

69. Find the absolute maximum value of /(t) : t'?ln (l/r) and saywhere it is assumed.

fE zO. cnlpgea Graph /(r) : (t -3)2e' and its first derivative to-gether Comrnent on the behavior of / in rclation to the signsand values of /'. Identify significant points on the graphs withcalculus, as necessary.

71. Find the area of the ltdangular" region in the fiIst quadrant that isbou[ded above by the curve ] : e], below by the curve 1' : e',and on the dght by the line I :ln 3.

72. Find the area of the "triangular" region in the first quadrant thatis bounded above by the a)Ne ! = e'/2, below by the curvey : e '/2, and on the right by the litre t : 2 ln 2

73. Find a cuwe through the odgin in the,r1-plane whose lengthf r o m , r : 0 t o 1 : 1 i s

74. Find the area of the surface generated by revolving the curve

x = kr + e-r)12,0 : ) : ln 2, about the )-axis.

75. a) Show that / ln xdt: xln t - x * C.b) Find the average value of ln.r over [1, e].

76. Find the average value of /(r) : I lx on ll, 21.

77, The linearization of e' at x: o

- a) Dedve the linear approximation e' ^, I + x at r - 0.

E 11 CalCumfoR Estimate to 5 decimal places the magnitudeof the enor involved in replacing e' by 1 + 'I on the interval

t0 ,0 .21 .EE c) GRAPHER Graph e' and 1 +,r together for -2 S x = 2.

Use different colors, if available. On what intervals does theapproximation appear to overestimate ei? underestimate er?

78. Laws of Exponents.

a) Starting with the equation e',e' - ett+rl, derived in thetext, show that e-': l le'for any real number .I. Thenshow that e', /ex, : €'1 " for any numbers :r1 and ,r2.

b) Show that (er )D - etlr' : (e')r\ for any numbers .r1 and.r:2

@7g. a decimal representation of e. Find e to as many decimalplaces as your calculator allows by solving the equation ln ,r = 1.

E 80. rh" lru"rru ." lation between e" and ln x. Find out how goodyour calculator is at evaluating the composites

er" and ln (e').

u. lzee ' ta t

48. f'ntu "'ro

dtJo

5 0 . l - - - - d r

s2. I

feo at

sa. J ,d '

59, | 2eu cos e'du

. [ " , d ,. l l + e "

lnitial Value ProblemsSolve the initial value problems in Exercises 63 66.

6 3 . 1 - e ' s i n ( e ' - 2 ) . ) f l n 2 ) : od t

,hM.

-'- : e-' sec'lne ' 1. ylln 4l - 2ln

d t

65 . -= 2e - , , . y (0 ) = L and

6 6 . - : I - e ' z ' , ) , ( l ) : - l

l d x6 2 . ' -

. l | + e '

60. loffi

z" "" "0"{"")a,

:r'(0) : 0

and y'(l) = 0


81, Show that for any number 4 > I

(See accompanying ligure.)

The geometric, logarithmic, and arithmetic mean inequality

a) Show that the graph of e" is concave up over every intervalof r-values.

b) Show, by reference to the accompanying figure, that if 0 <a < b t h e n

? i l 1 - l r i ' . ( l n b - l n a ) - l ' " , , a ^ .J " "

- . ( l n , - t n d ) .

c)

1," n ,a ,+

lo ' ' " " ,d r -a tna.

82.

NOTTO SCALE

a, and log"xWhile we have not yet deyised a way to raise positive numbers to any but rationalpowers, we have an exception in the number e. The definition e' : ln l_r definese' for every real value of;, irrational as well as rational. In this section. we showhow this enables us to raise any other positive number to an arbifiary power andthus to define an exponential function 1: a' for any positive number a. We alsoprove the Power Rule for differentiation in its final form (good for all exponents)and define functions like r' and (sinx)bn' that involve raising the values of onefunction to powers given by another

Just as s' js but one of many exponential functions, ln;r is one of many loga-dthmic functions, the others being the inverses ofthe function a'. These loqarithmicfunctions have important applications in science and engineering.

The Function axSince a : eln" for any positive number n, we can think of aJ as (er. a)i : e'ln a.

We therefore make the following definition.

DefinitionFor any numbers a > 0 and -r,

Use the inequality in (b) to conclude that

. , b - a c t + b\ /AD < -

l n b - l n a 2

This inequality says that the geometdc mean of two positivenumben is less than their logarithmic mean, which in turnis less than their arithmetic mean.

(For more about this inequality, see "The Geometdc.Logarithmic, and Arithmetic Mean Inequality" by FrankBirk, American Mathematical Monthly, YoI. 94, No. 6.June-July 1987, pp. 527 528.)

( 1 )

Table 6.3 Laws of exponents

For a > 0, and any x ald 1:

l. a' . a' : a'+'

I

a i

- r ) _ r ,y \ r

6.4 a' and /og,x 475

EXAMPLE 1

a) 2& : eJ1t"2b1 2" :

" "^ t D

The function a' obeys the usual laws of exponents (Table 6.3). We omit theproofs.

The Power Rule (Final Form)We can now define -rn for any .r > 0 and any real number z as xn : e' 'n'. Therefore,

the n in the equation ln xn : nlnx no longer needs to be rational-it can be any

number as long as .r > 0:

ln r' : ln (e't") : nln x ' ln e ln etl : u' ar,y u

: n h ) t .

Together, the law a'fav:a"-r and the definition xn - ertn' enable us to

establish the Power Rule for differentiation in its final form. Differentiating .x' with

respect to .{ SivesL

,^ : d

,^"' Dennidon of .r,'. .ir > odx ' - dx -

= p'r" ' . :( l? In x) Chain Rule for P'tlx

-- x' ' \ The delinition againx

dx

The Chain Rule extends this equation to the Power Rule's final form.

Power Rule (Final Form)

If a is a positive differentiable function of .r and n is any real number, then

a' is a differentiable function of .r and

d " . - , du- - u ' : n u ' - .ax q r

EXAMPLE 2n-

a ) - xu t : \ / 2x r t - t ( x > 0 )tlx

,]b ) i G i n x ) " : 7 r { s i n . r ) " ' c o s , r

dx

Table 6.3, Law 3

In short, as long as .r > 0,

o(sin x > 0)


J('tl

. . (

The Derivative of a'We start with the definition a' = et tn i .

d. _ ( r l n d )

d x '( h iL i r r l t L : l f

If 4 > 0, then

d- i ' = a ' l n a .dx

With the Chain Rule, we get a more general folm.

If a > 0 and u is a differentiable function of .r, then a' is a differentiablefunction of r and

!o , = o ,n n ! . e )d t "

- - " d x '

Equation (2) shows why e* is the exponential function preferred in calculus. Ifa : e, then ln c : 1 and Eq. (2) simplifies to

L" ' - " , rne :e^ .d x -

EXAMPLE 3

a) :3 ' : 3 ' ln 3A I

,l ,lb ) : 3 ' = 3 ' I n 3 ; ( - - r ) : - 3 ' , l n 3

tlx dx

n nc ) + 3 ' " ' i : 3 \ i n r l n 3 ] 1 s i n x ) - 3 ' i n ' r l n 3 ; c o s . r

d x d x J

From Eq. (2), we see that the derivative of a' is positive if ln a > 0, or a > 1.and negative if ln a < 0, or 0 < a < 1. Thus, a' is an increasing function of r ifa > I and a decreasing function of ,r if 0 < a < 1. In each case, a' is one-to-one.The second derivative

,12 ,1

2@\ = |@' tn a) : ( \ r a)za 'dx. dx

is positive for all -r, so the graph of a' is concave up on every interval of the reall ine (Fig. 6.12).

Other Power FunctionsThe ability to raise positive numbers to arbitrary real powers nakes it possibleto define functions like r' and,rrn'for x > 0. We lind the dedvatives of suchfunctions by rewriting the functions as powers of e.

) , 1" - r l n d _ - r i n a---u - -: <

dx dx

- a ' l n a .

- t 0

6.72 Exponential functions decrease i f0 <a < 1 and increase i f d > 1. Asx --+ co,w e h a v e a ' + 0 i f 0 < a < 1 a n d a r - - + c o i fa > l. As x + -cc, we have ax -+ co if0 < a < l a n d a x - + 0 i f a > ' 1 .

6.4 a' and log"x 477

EXAMPLE 4 Find dy ld.x if y : a' ,

so/ution Write r' as a power of a:

Then differentiate as usual:

x > 0 ,

: e ' t n ' . E q . ( l ) w i t b a : r

dv d , n ,dx dx

: 'e- t" , i@ tn x)ax

/ 1 \: x ' { x . : + l n x l

\ x /: r ' (1 * ln x) . tr

The Integral of auIf a I I, so that ln a + 0, we can divide both sides of Eq (2) by ln a to obtain

. . d u I d" d x - I n a d x ' -

' '

Integrating with respect to x then givesJ 1 f ,l l

Io ' ! !a r : I=L ! -k " tdx= . I i 1a 'ydx= - a '+c .. l d r J l n a d x l n a l d x t n a

Writing the first integral in differential form gives

| '"a": h*' (3)

EXAMPLE 5

ul I

r a* : #*,

Eq (3) with 'r :2, tr : ,

u) |

z"h' cos tc ax

: | , '0,: h*,: ' # * , u : s i n r i n E q . ( 3 )

Logarithms with Base aAs we saw earlier, if a is any positive number other than 1, the function a' is one-

to-one and has a nonzero derivative at every point. It therefore has a differentiable

inverse. We call the inverse the logarithm of x with base 4 and denote it by

log,.r.

tr


6.73 The graph of 2, and its inverse,log, x.

DefinitionFor any positive number a I 1,

logo _:r : inverse ofa*.

The graph of y = log,,y can be obtained by reflecting the graph of _r=a,c ros \ the l ine r - x {F ig . 6 .13) .

Since log,,.r and a' are inverses of one another, composing them in either ordergives the identity function.

Inverse Equations for a' and logo.r

log,(a') : r

(-r > o)

(all r)

(4)

(s)

EXAMPLE 6

a) log, (25) = 5

c) 2rocr (r) - 3

b)

d)

logro (10-7) = -7

l0 ro80(4) :4 J

The Evaluation of loga xThe evaluation of log,, "r is simplified by the obseNarion that log,, x is a numericalmultiDle of ln r.

, I l n rr o g a r : - . l n r : : -t n a l n a

We can derive Eq. (6) from Eq. (4):

4log,,(r) _ r

ln 4los, '(r) - ln r

l o g , ( r ) . l n a : J n r

l n rlog, r : -_"

ln r-r

h q ( l )

Trkc lhe n:r t rLrLr l logar i rhnr

The Po$ef RrLle in hLrte 6.1

Solvc lbr los,, r.

ln 2 0.69315l o 8 ' n l - , n -

* : _ r O X n ' 0 . 3 0 1 0 1

EXAMPLE 7

!

Table 5.4Properties of base a logarithms

For any numbers x > 0 and y > 0,

l. Product Rule:log. ry : 1og, ; * 1og. 1

2. Quotient Rule:, . xrog, - ro& ir roga )

3. Reciprocal Rule:

. 1log , : - log"y

4. Power Rule:log, .r) = y log, .r

6.4 a" and log,x 479

The arithmetic properties of logo -r are the same as the ones for ln r (Table 6.4).These rules can be proved by dividing the coresponding rules for the naturallogarithm function by ln a. For example,

ln"ry : ln x -f ln y

l n J y l n r l n l

1 " t -

1 " t - 1 " t

log, xy : 1og, x * 1og, y.

The Derivative of logauTo find the derivative of a base a logarithm, we first convert it to a natural logarithm.If i.r is a positive differentiable function of x, then

d d l l n u \ | d I l d u

d r ( r o 9 " t : d r \ r " "

) = r n a d x t t n ' ) -

r n o ' ; d r

d . 1 l d u, (log- r) : -- ' ---

d x - "

l n a u d x(7)

EXAMPLE 8

ftbg,o{2,+ r) : n roa . }-fto, + u : (ln 10) (3r + 1)

Integrals Involving log, xTo evaluate integrals involving base a logarithms, we convefi them to natural log-arithms.

EXAMPLE 9l ' l o e , , r I l l n r t n f|

- . : - r t , : , - . | . . dx r , , ; . , - rn lJ X t n z J . : r

I t , \ n ' t " ' , .

- :--= t uatl) n t J

I u ) ^ I t l n , r ) 2 ^ r l n x ) 2- 1 "21 *c - , - - , +c - 2 rn2+c - l

*Base 10 LogarithmsBase 10 logadthms, often called common logarithms, appear in many scientificformulas. For example, earthquake intensity is often reported on the logarithmicRichter scale. Here the fomula is

M a g n i r u d e R - l o g , n ( i ) * t

where a is the amplitude of the ground motion in microns at the receiving station,Z is the period of the seismic wave in seconds, and B is an empiical factor that

Rulc I 1br nr tLrra ll o g a i h n s . . .

. . . d iv idcd b) ln . r . . .

. . . g jves Rulc 1 101 LrLrse dlog.rrithnr\.

D


Most foods are acidic (pH < 7).

Food pH Value

Bananas 4.54.7Grapefi'Lrit 3.0-3.3Oranges 3.0-4.0Limes 1.8-2.0Mitk 6.3_6.6Soft drinks 2.04.0Spinach 5.1-5.7

allows for the weakening of the seisrnic wave with increasing distance from theepicenter of the earthquake.

ffiUpte tO For an earthquake 10,000 km ftom the receiving statron, I =6.8. If the recorded vertical ground motion is a: 10 microns and the period is7: 1 sec, the earthquake's magnitude is

_ / 1 0 \f f = l o g r o l ; l + 6 . 4 : I t 6 . 8 : 7 . 8 .

\ r . / "

An earthquake of this magnitude does great damage near its epicenter. JThe pH scale for measuring the acidity of a solution is a base 10 logarithmic

scale. The pH value (hydrogen potential) of the solution is the common logadthmof the reciprocal of the solution,s hydronium ion concentration, [H3O+]:

pH : logro =+= : - logrn [H3O+].[ t t : U , l

The hydronium ion concentration is measured in moles per liter. Vinegar has a pHof 3, distilled water a pH of 7, seawater a pH of g.15, and householj ammoma apH of 12. The total scale ranges from about 0.1 for normal hydrochloric acid to 14for a normal (1 N) solution of sodium hydroxide.

Another example of the use of common logarithms is the decibel or db (,deebee") scale for measuring loudness. If 1 is the intensity of sound in wans persquare meter, the decibel level of the sound is

Sound level: l0 log,o (1 x 1012) db.

Sound level with 1 doubled : l0 log (21 x l0r2)

: 10 log (2 . I x 1012)

(8)If you ever wondered why doubling the power of your audio amplifier increases thesound level by only a few decibels, Eq. (g) provides the answei. As the followingexample shows, doubling 1 adds only about 3 db.

EXAMPLE 11 Doubling 1 in Eq. (8) adds about 3 db. Writing log for lo916 (acomrnon practice), we have

Typical sound levelsThreshold of hearingRusde of leavesAverage whisper

Quiet automobileOrdinary conversationPneumatic drill 10 feet awayThrcshold of pain

0 d bl0 db20 db50 db65 db90 db

120 db

: 10 1og 2 + 10 log(1 x 1012)

: original sound level + l0 log 2

I original sound level + 3.

E q . ( 8 ) $ i ( h21 fot I

l

Exercises 6.4

Algebraic CalculationsSimplify the expressions in Exercis€s l-4. d)

2. a)

loga 16

2tae,3

e)

b)

log rE

lOlosrc 0/2)

,) ,*.(i)c) fttog 7

1. a) 5i"c.7 b) 8r"s""' c) 1.3roc,,75

Exercises 6.4 481

46. ) : (ln .r)r"d) loglr l2l

3. a) 2to&'

4. a) 25r'e'(31)

r) bc, (;)

c) 1og2(e(h 2)G- *))

c) log4(2d' i " ' )

c)

e) logr2r 11

b) 9t's''

b) log"(e')

s. al P4log3,r

6. a) loge rrogr t

t ^ n -

b ) :logs'tlog l ;xb)+tog!5 x

I t3t' ax

J 15" de

/ )J1

J, Jro"I"" r'1 \""' ,.. ',.i,

Jo \ 3,/

I -_

dx

I tog..x

f ' 2 l n l 0 l o e , , , *

l, - ;-"/ro

log,o(l0r) r,

[3 2 rog2(r' t)

dx

f d t. / . r ( log* . r )2

Express the ratios in Exercises 5 and 6 as ratios of natuml logarithmsand simplify.

log,a u. I s'a'I '

2 3 d 0 5 0 .

a5xztr)) dx 52.

ss. lu'''

t""' ' "in , a, s4.

ss. lro ""lt+

ln r)djr 56.

a. I t:el' a, 62.

63. fo ln2logrt

d* 64.

ar. ['t:3{:2ox 66.J o x + z

st. [" 2-be'ot:l \ ,, 68.

J 0 . { I _ t

es. I d' io.

J j{ rogrn x

Evaluale rhe inregrals in Exerciies 7l 7

n . [ ' " ' ! a t . , . t 72 .l t t

n. | , " ' la , '

, .0 74.

log; alog.blogba

Solve the equations in Exercises 7 l0 for r.

7. 3losr (?) + 2loc1(5) : 5roci (r)

8. 8ros! (3) _ eli 5 : 12

'ltoe13r)

9. 3 locr G'?) - 5ern r 3, ldoe. t r i )

110, lne +4 ,2ro&6) : : loq , . (100)

DerivativesIn Exercises 1l-38, find the derivative of ) with respect to the givenindepeldent variable.

l l. y :21

13. y : 5"4

15. y : an

17. y: (cos?)&

19, y :7 ' *e ln '1

21, y :2"i 3'

23. Y:log250

25. y:logo16 .r1oto tz

27. Y : logtT ' 1oto,

/ i - r r r h r \2e . y : r og , { (= ) I

\ \ r - ' I l /

31. y=9s in ( logr9)

12.

14.

16.

18.

20.

24.

26.

28.

40, Y: t (J+' l

& ' Y : t J i

1, : [n 0)"

1 : 3 ' " " l n 3

) : log3(1 + 6ln3)

y - logzs e' -logs JV

) : 1 o g : r ' l o g s rf - . h 5

ro. ): loe.,/(;+; )\ \ 5 x i 2 , /

/ s i n d c o s 9 \32. y : rog.,

\ ", ,, )

/ x2e2 \3 4 . y = l o s , l - l' --

\2r,zx * I /

36. y:3 legu11or";

38. ) = t 1og3 (e(' i" ')(t"r))

33 . y=1og re '

37 . ) : l og2 (8 t t n2 )

3 9 . y = ( n * 1 ) '

41. y : QI)'

Logarithmic DifferentiationIn Exercises 39-46, use logarithmic differentiation to flnd the dedva-tive of )r with respect to the given independent variable.

Theory and Applications75. Find the area of the region between the cuNe r -2x/(l + ''2)

and the interval 2 5 r 5 2 of the .{ aris.

76. Find rhe area of lhe region belween the curve y - 2l ' and theinterval I =r 5l of the i-axis.

!a ,t

r ^ 1J' , o'' "o

4.

I,I

1 " "

43. y: (sin:r) '

45. y : atn'

IntegrationEvaluate the integrals in Exercises 47-56

I,I,

Evaluate the integrals in Exercises 57 60.

s t . I

t *La" s t . |

* ' ' z ta^

se. lo'trt+ rt, ' 'a* m. l, ' ru^ u 'a*

Evaluate the integrals in Exercises 61-70.


77. Blood pH. The pH of human blood nomally lalls between 7.37and 7.44. Find the corresponding bounds for [H3O +].

78. Brain fluid pH. The cerebrospinal fluid in the brain has a hy-dronium ion concentration of about [H3O +] : 4.8 x 10 3 molesper liter. What is the pH?

79. Audio amplifiers. By what factor & do you have to multiply rheintensity of 1 of the sound from your audio amplifier to add 10db to the sound level?

80. Audio amplifiers. You multiplied the intensity of the sound ofyour audio system by a factor of 10. By how many decibels didthis increase the sound level?

81. In any solution, the product of the hydronium ion concentration[H3O+] (moles/L) and the hydroxyl ion concentration IOH ](moles/L) is about 10 ra.

a) Wlat value of [H3O+] minimizes the sum of the concentra-1ie11s, 5 - [H3O+J + IOH ]? (H,rzt.Change notation. Letr = [H:O+].)

b) What is the pH of a solution in which S has this minimumvalue?

c) What ratio of [H3O +] to [OH-] minimizes S?

82. Could log,6 possibly equal 1/log, d? Give reasons for youranswer.

2Z Grapher Explorations83. The equation.x2 :2' has three solut ions:.r :2,.r = 4, and one

other. Estimate the third solution as accurately as you can bygraphmg.

84, Could ,rr'2 possibly be the same as 21n , for r > 0? Graph thetwo functions and explain what you see.

85. The linearization of 2"

a) Find the linearization of f(x) :y at r:0. Then roundirs coefficients to 2 decimal places.

b) Graph the linearization and function together for -3 S r :3 a n d 1 : r < 1 .

86. The linearization of logsx

a) Find the linearization of / (r) : log3 r at, : 3. Then roundits coefncients to 2 decimal places.

b) Graph the linearization and function together in the window0 : r S 8 a n d 2 = x 5 4 .

Calculations with Other Bases@ 87, CALCULATOR Most scientific calculators have keys for log,or

and ln.r. To find logarithms to other bases, we use the equationlog" x = (ln .r)/(ln a).

To find logr,l. f ind ln,r tn ,rand divide by In 2: log, 5 : j-

-- ! 2.32 t9.

To find ln ; given log2 r,multiply by h 2: ln 5 = log, 5 . ll 2 .- 1.6094.

Find the following logarithms ro 5-decimal places.

a) log3 8 b) logr0.5c) logro 17 d) logo5 7e) ln.r, giventhatlogrox : 2.3f) in .x, given that l og2 x :1.4g) ln r, given that 1og2 ,r : -1.5h) ln "r, given that loglo ir : -0.7

88. Conversion factors

a) Show that the equation for conveting base 10 logarithmsto base 2 logarithms is

, l n 1 0 .log2 r =

l;2 logro_r.

b) Show that the equation for converting base a logarithms tobase , logarithms is

, I n a ,r o g r . r = . ; r o g , - r .

Growth and DecayIn this section, we derive the law of exponential change and describe some ofthe applications that account for the importance of logarithmic and exponentialfunctions.

The Law of Exponential ChangeTo set the stage once again, suppose we are interested in a quantity ) (velocity,temperature, electric current, whatever) that increases or decreases at a rate that atany given time / is proportional to the amount present. If we also know the amountpresent at time t : 0, call it y6, we can find ), as a function of t by solving thefollowing initial value problem:

6.5 GroMh and Decay 483

Differential equation:

Initial condition:

dy

a t

y : 1 0( 1 )

when / :0.

If y is positive and increasing, then t is positive, and we use Eq. (1) to say thatthe rate of growth is proportional to what has aheady been accumulated. If y ispositive and decreasing, then t is negative, and we use Eq. (1) to say that the rateof decay is proportional to the amount still left.

We see right away that the constant function y : 0 is a solution of Eq. (1). Tofind the nonzero solutions, we divide Eq. (1) by y:

!.!:ny a r

rntr',t : kr+c i'ii;i:,'^1,;llT;:,ly : ek'*c Exponentiate.

ly) : ec . ekt eu+b - ea , eh

, : * o ( o t ' l t r - r . r h e n \ - = ,

Y : Aek' A ts a more convenleDl name

By allowing A to take on the value 0 in addition to all possible values * ec, wecan include the solution y : 0 in the formula.

We find the right value of ,4 for the initial value problem by solving for Aw h e n y : y 0 a n d r : 0 :

Y o : A e k ' a : A '

The solution of the initial value problem is therefore y : yoek'.

The Law of Exponential Change

y : yr"r'

Growth: /r > 0 Decay: ,t < 0

The number t is the rate constant of the equation.

(2)

The derivation of Eq. (2) shows that the only functions that are their ownderivatives are constant multiples of the exponential function.

Population GrowthShictly speaking, the number of individuals in a population (of people, plants,foxes, or bacteria, for example) is a discontinuous function of time because it takeson discrete values. However, as soon as the number of individuals becomes largeenough, it can safely be described with a continuous or even differentiable function.


If we assume that the propoftion of reproducing individuals remains constant andassume a constant fertility, then at any instant / the birth rate is proportional tothe number jy(t) of individuals present. If, further, we neglect departwes, arrivals,and deaths, the $owth |'aIe dyldt will be the same as the birth rate fy. In otherwords, dy /dt : &y, so that y : yo ekt .As with all kinds of growth, there may belimitations imposed by the surrounding environment, but we wilt not go into thesehere.

EXAMPLE I One model for tire way diseases spread assumes rhat the ratedy / dt at whrch the number of infected people changes is proportional to the numbery. The more infected people there are, the faster the disease will spread. The fewerthere are, the slower it will spread.

Suppose that in the course of any given year the number of cases of a diseaseis reduced by 20Vo.If there are 10,000 cases today, how many years will it take toreduce the number to 1000?

Solution We use the equation y : yoekt. There are tbree things to find:

1. the value of ys,2. the value of t,3. the value of I that makes y : 1000.

Step 1: The value of y0. We are free to count time beginning anywhere we want. Ifwe count from today, then y: 10,000 when l:0, so ye: 10,000. Our equationrs now

) : 10,000 ekr.

Step 2: The value of k. When / : I year, the number of cases will be 807o of itspresent value. or 6uuu. Hence.

8000 = 10.000errr, fl'r,J,,ll,n ,: , ,""

er : 0.8

ln(ek; : 1n 9'3

t : ln 0.8.

At any given time t,

Y : 10,000 e(ho 8)'

Step 3: The value of t that makes y : 1000. We s€t l, equal to 1000 in Eq. (4) andsolve for t:

1000 : 10,000 eono 8)/

, (1n0.8, _ n 1

(3)

(4)

(ln 0.8)/

t

: ln 0.1 Logs of both sidcs

ln 0.1: ln 0-g

- 10.32 years.

It will take a little more than l0 years to reduce the number of cases to 1000, I

involves what is called the indeterminateform 1-. We will see how to evaluate limitsof this type in Section 6.6.

For radon-222 gas, 1 is measur€d in days andk :0.18. For radium-226, which used to bepainted on watch dials to make them glow atnight (a dangerous practice), I is measured inyears and ( . 4.3 . l0 a. The deca) ofradium in the earth's crust is the source of theradon we sometimes find in our basements.

It is conventional to use -t (ft > 0) hererns tead o f I t l . 0 t t o emphas rze t ha l r i sdecreaslng.

6.5 Growth and Decav 485

Continuously Compounded InterestIf you invest an amount .4e of money at a fixed annual interest rate r (expressedas a decimal) and if interest is added to your account t times a year, it turns outthat the amount of money you will have at the end of / yeals is

The interest might be added ("compounded," bankers say) monthly (k : 12), weekly(n : 52), daily (ft : 365), or even morb frequently, say by the hour or by the minute.But there is still a limit to how much vou will eam that wav. and the limit is

l im A,

= A o e " .

The resulting formula for the amount of money in your account after r yearsls

A(t) - Ao e" (6)

Interest paid according to this fomula is said lo be compounded continuously.The number r is called the continuous interest rate.

EXAMPLE 2 Suppose you deposit $621 in a bank account that pays 670 com-pounded continuously. How much money will you have 8 yea$ later?

Sorution We use Eq. (6) with A0 :621, r :0.06, and I :8:

A(8) = 621 e(006) (8) :621eu4u : 1003.58 \ i ' i r f . \ r . c , r r

Had the bank paid interest quarterly (t :4 in Eq. (5)), the amount in youraccount would have been $1000.01. Thus the effect of continuous compounding,as compared with quarterly compounding, has been an addition of $3.57. A bankmight decide it would be worth this additional amount to be able to advertise,"We compound interest every second, night and day-better yet, we compound theinterest continuously."

RadioactivityWhen an atom emits some of itsrc-forms to make an atom of some new element. This process of radiation andchange is called radioactive decay, and an element whose atoms go spontaneouslythrough this process is called radioactiye. Thus, radioactive carbon-1,1 decays intonitrogen; radium, through a number of interyening radioactive steps, decays intolead.

Experiments have shown that at any given time the rate at which a radioactiveelement decays (as measured by the number of nuclei that change per unit time) isapproximately proportional to the number of radioactive nuclei present. Thus, thedecay of a radioactive element is described by the equation dy/dt - ky,k>0.If

)0 is the number of radioactive nuclei present at time zero, the number still presentat anv later time r will be

e,:e,(r+i)r ' (s)

Evaluating

.l im Ao ('* i)"'

_l

mass as rad ia t ion , lhe remainder o f rhe a tom

y : y o e - k t , t > 0 .


EXAMPLE 3 Half-life

The halflife of a radioactive element is the time required for half of the radioactivenuclei present in a sample to decay. It is a remarkable fact that the halflife is aconstant that does not depend on the number of radioactive nuclei initially presentin the sample, but only on the radioactive substance.

To see why, let lyg be the number of radioactive nuclei initially present in thesample. Then the number ) present at any later time / wilt be y : yo e-t,. We seekthe value of / at which the number of, radioactive nuclei present equals half theorisinal number:

I2 ' "12

, 12

- ln2

l n 2' -

k

This value of I is the halflife of the element. It dependsthe number yo does not enter in.

Reciprocal Rule forlogarithms

only on the value of *,.tr

Half-life : ]1?k

(7)

EXAMPLE 4 Polonium-210

The effective radioactive lifetime of polonium-2lo is so short we measure it indays rather than yea$. The number of radioactive atoms remaining after I days ina sample that starts with 1,0 radioactive atoms is

) : l0 e-5xl0r t

Find the element's half-life.

Solution

Half-life : l1?k

l n 25 x 1 0 3

! 139 days

EXAMPLE 5 Carbon-l4

People who do carbon-l4 dating use a figure of 5700 years for its halfJife (moreabout carbon-l4 dating in the exercises). Find the age of a sample in which 107oof the radioactive nuclei originally present have decayed.

Eq (7)

The t fiom polonium's decayequation

B

Carbon-14 dating

The decay of ndioactive elements can

sometimes be used to date events from the

Earth's past. The ages of rocks more than 2

brl l ion years old hare been measured b) the

extent of the radioactive decay of uramum(halfJife 4.5 billion years!). In a living

organism, the ratio of radioactive carbon,

carbon-l4, to ordinary carbon stays lairly

consrant during lhe l i fet ime of the orgrnism.

being approximately equal to the ratio in the

organism's suroundings at the time. After

the organism's death, however, no new

carbon is ingested, and the propofiion of

carbon 14 in the organism's remains

decreases i i ( the carbon-14 decay:.. l t is

possible to estimate the ages ol fairly old

orgcnic remains b1 comparing the proport ion

ol cJrbon- l4 they contain with Lhe propofl ion

assumed to have been in the organism'i

environment at the time it lived.

Archaeologists have dated shells (which

contain CaCOr), seeds, and wooden artifacts

this way. The estimate of 15,500 years tbr

the age of the cave paintings at Lascaux,

France, is based on carbon-14 dating. Alter

generations of controversy, the Shroud of

Turin, long believed by many to be the burial

cloth of Christ, was shown by carbon-l4

dating in 1988 to have been made alter A.D.

1200.

6.5 Growth and Decay 487

so/ution We use the decay equation l' : ys e k' , There are two things to find:

1. the value of k,2. the value of I when yn e t' : 0.9 y0, or e k' - 0.9

step 1: The value oJ k. We use the halt"life equation:

ln2L - _" -

halt'-life

Step 2: The value of t that mdkes e tt :0.9.

e -k ' : 0 .9

e-(ln2/5700)' - 0.9

- t n t , : r nu .o5700

, - - tto,o"';oj \ 8ob )ears

The sample is about 866 years old.

Heat Transfer: Newton's Law of CoolingSoup left in a tin cup cools to the temperature of the sunounding air. A hot silver

ingot immersed in water cools to the temperature of the surrounding water' In

situations like these, the rate at which an obiect's temperature is changing rt rny

given time is roughly proportional to the differcnce between its temperature and

the temperature of the sunounding medium. This obseryation is called Newloz's

I.a,w of cooling, although it applies to warming as well, and there is an equation for

If Z is the temperature of the object at time /, and 75 is the sunounding

temDerature. then

d Tdt

: -k(T Is).

If we substitute I for (? 7s), then

d ! _ J , . _ r ^ , _

dt t lt

d Ti 1

dt

d Tdt

Eq. (8) theretbre reads

and we know that the solution to this diff-erential equation is

Thus, Newton's law of cooling is

T - T s : ( T o T ) e k ' ,

where le is the value of ? at time zero.

ln2(about 1.2 x 10 1)

d T d- - -rr.)d t d t " '

9t) , i L, l t i r . r . r ( l i . rL. t i \ |

I t , ! \ o l l r o t h \ j J . '

5700

rl

(8)

ln tems of ),dtdt

(e)

488 Chaoter 6: Transcendental Fundions

EXAMPLE 5 A hard-boiled egg at 98'C is put in a sink of 18'C water. After5 minutes, the egg's temperature is 38'C. Assuming that the water has not warmedappreciably, how much longer will it take the egg to reach 20'C?

So/utron We find how long it would take the egg to cool from 98'C to 20'C andsubtract the 5 minutes that have already elapsed.

According to Eq. (9), the egg's temperature / minutes after it is put in the sinkis

I : t 8 + ( 9 8 - ' 1 8 ) e { ' : 1 8 - 8 0 " - o ' .

To find ft. we use the information that Z = 38 when l:5:

38 : 18 -|80e 5&

- l n 4

I- 4

I- 5 t : l n ; :+

r : !n+ : o.z rn4

The egg's temperature at time t is 7 : 18 + 80e-(0 zha)'. Now find the time I when

20 : 18 + 80e (o2ln4)r

8oe-(o.2lna\t - 2

-(O'21ft4)t: ln tI

: - ln 40

, : , 'n oo

ry 13 min.O.2lrr4

The egg's temperature will reach 20'C about 13 min after it is put in water to cool.Since it took 5 min to reach 38'C, it will take about 8 min more to reach 20'C.

tr

(about 0.28).

e-tojkan - !

Exercises 6.5

The answe$ to most of the following exercises are in terms of log-arithms and exponentials. A calculator can be helpful, enabling youto express dle answers in decimal form.

l. Human evolution continues. The analysis of tooth shdnkageby C. Loring Brace ard colleagues at the University of Michi-gan's Museum of Anthrcpology indicates that human tooth sizeis continuing to decrease and that the evolutionary process didnot come to a halt some 30,000 yea$ ago as many scientists con-tend. ln northem Europeans, for example, tooth size reductionnow has a late of l% Der 1000 Yea$.

a) If t represents time in years and ) represents tooth size, usethe condition that ), :0.99y0 when / = 1000 to find thevalue of ,t in the equation y : y6 er'. Then use this valueof t to answer the following questrons.

b) In about how many yea$ will human teeth be 9070 of theirpresent size?

c) What will be our descendants' tooth size 20,000 years fromnow (as a percentage of oul present tooth size)?

(So;u\aet LSA Magazine, Spring 1989, Vol. 12, No. 2, p. 19, AnnArbor. MI.)

2, Atmospheric pressure. The earth's atmospheric pressure p is of-ten modeled by assurning that the ftte dp /dh at which p changeswith the altitudel, above sea level is propofiional to p. Supposethat the prcssure at sea level is 1013 millibars (about 14.7 poundsper square inch) and that the pressure at an altitude of 20 km is90 milliba$.

a) Solve the initial value problem

Differential equation: dpldh: kp (t a constant)

Initial condition: p : ps whet h :0

to express p in terms of i. Determine the values of P0 andI ftom the given altitude-pressure data.

b) What is the atmospheric pressure at,:50 km?c) At what altitude does the pressue equal 900 millibars?

3. First order chemical reactions. In some chemical reactions, therate at which the amount of a substance changes with time isproportional to the amount present. For the change of ,-gluconolactone into gluconic acid, for example,

dy: = U.b)A T

when I is measured in hours. If there are 100 grams of d-gluconolactone prcsent when t = 0, how many grams will be left afterthe f,rst hour?

4, The inversion of sugat The processing of raw sugar has a stepcalled "inversion" that changes the sugar's molecular structurc.Once the process has begun, the rate of change of the amount ofraw sugar is proportional to the amount of raw sugar remaining.If 1000 kg of raw sugar rcduces to 800 kg of raw sugar duringthe first 10 h, how much raw sugar will remain after another 14h?

5. Working underwateii The intensity l(r) oflightt feet beneaththe surface of the ocean satisfies the differential equation

dLdx

As a diver, you know from expedence that diving to 18 ft inthe Caribbean Sea cuts the intensity in hall You cannot workwithout artificial light when the intensity falls below. one-tenthof the surface value. About how deep can you expect to workwithout artif, cial light?

6. Voltage in a discharging capacitol: Suppose that electricityis draining from a capacitor at a rate that is proportional to thevoltage y across its terminals and that, ift is measurcd in seconds,

Solve this equation for V, using yo to denote the value of y

when t :0. How long will it take the voltage to drop to l07o of

its original value?

7, Cholera bacteria. Suppose that the bacte a in a colony can grow

unchecked, by fte law of exponential change. The colony starts

with I bacterium and doubles every halfhou. How many bacteria

will the colony contain at the end of 24 h? (Under favorable

laboratorv conditions. the number of cholera bacteria can double

Exercises 6.5 489

every 30 min. In an infected pelson, many bacteda are deshoyed,but this example helps explain why a person who feels well inthe moming may be dangerously ill by evening.)

8. Growth of ba.teria. A colony of bacteria is grown under idealconditions in a laboratory so that the populatiol increases expo-nentially with time. At the end of 3 h there are 10,000 bacteria.At the end of 5 h there are 40,000. How many bacteria werepresent initially?

9. The incidence of a disease (Continuation of Example 7). Srtp-pose that in any given year the number of cases can be reducedby 25Va instead of 20Vo.

a) How long will it take to reduce the number of cases to 1000?b) How long will it take to eradicate the disease, that is, reduce

the number of cases to less than l?

10, The U.S. population. The Museum of Science in Boston dis-plays a mnning total ofthe U.S. population. On May I l, 1993, thetotal was increasing at the rate of I person every 14 sec. The dis-played population figure for 3:45 P.M. that day was 257,313,431.

a) Assuming exponential growth at a constant rate, find themte constant for the population's growth (people per 365-day year).

b) At this rate, what will the U.S. population be at 3:45 P.M.Boston time on May I I, 2001?

ll. oil depletion. Suppose the amount of oil pumped from one ofthe canyon wells in Whittier, Califomia, decreases at the contin-uous rate of lOEa per year When will the well's otltput fall toone-nfth of its present value?

12, Continuous pri.e discounting. To encouage buyers to place1oo-unit orders, your fiIm's sales department applies a continuousdiscount that makes the unit price a function p(t) of the numberof units x ordered. The discount decreases the price at the rateof $0.01 per unit ordered. The price per unit for a 100-unit orderis p(100) : $20.09.

a) Find p(r) by solving the following initial value problem:d o l

Differential equation: i=- rOOO

Initialcondition: p(100):20.09.

b) Find the unit pdce p(10) for a 1o-unit order and the unitprice p(90) for a go-unit order.

c) The sales department has asked you to find out if it is dis-counting so much that the fi1m's revenue, r(x) : x ' p (x),will actually be less for a 100-unit order than, say, lbr a90-unit order. Reassure them by showing that r" has its max-imum value at .r : 100.

ZE a) cneprrEn Graph the revenue function r(x) : "rp({) for0 = x 5 2 0 0 .

13. Continuously compounded interest. You havejust placed A0dollars in a bank account that pays 4qa interest, compoundedcontinuously.

a) How much money will you have in the account in 5 years?b) How long will it take your money to double? to triple?

d v 1 , ,dt 40

490

14.

15.

Chapter 6: Transcendental Functions

John Napier's question. John Napier (1550-1617), the Scottish

laird who invented logarithms, was the first person to answer thequestion What happens if you invest an amount of money at10070 interest, compounded continuously?

a) What does happen?b) How long does it take to triple your money?c) How much can you eam in a year?

Give reasons for youl answers.

Benjamin Franklin's L{,//i. The Franklin Technical Institute ofBoston owes its existence to a prcvision in a codicil to BenjaminFrankl in s wil l . Tn pafl the codici l read:

I wish to be useful even after my Death, if possible, in

. forming and advancing other young men that may be ser-viceable to their Country in both Boston and Philadelphia.To this end I devote Two thousand Pounds Sterling, whichI give, one thousand thereoi to the Inhabitants of the Townol Boston in Massachusetts, and the other thousand to theinhabitants of the City of Philadelphia, in Tmst and forthe Uses, Interests and Purposes hereinafter mentioned anddeclared.

Frannin's plan was to lend money to young apprentices at 57,interest with the provision that each boffower should pay eachyear along

. . . with the yearly Interest, one tenth pafi of the Principal,which sums of Principal and Interest shall be again let tofresh Borowers. . . . If this plan is executed and succeedsas projected without interruption for one hundred Years,the Sum will then be one hundred and thirty-one thousandPounds of which I would have the Managers of the Donationto the Inhabitants of the Town of Boston, then lay out attheir discretion one hundrcd thousand Pounds in PublicWorks. . . . The remaining thirty-one thousand Pounds, Iwould have continued to be let out on Intercst in the mannerabove dirccted for another hundred Years. . . . At the end ofthis second term if no unfoftunate accident has preventedthe operation the sum will be Four Millions and Sixty oneThousand Pounds.

It was not always possible to find as many borrowers asFranklin had planned, but the managers of the trust did the bestthey could. At the end of 100 years from the reception of theFranklin gift, in January 1894, the lund had grown from 1000pounds to almost exactly 90,000 pounds. In 100 years the originalcapital had multiplied about 90 times instead of the 131 timesFranklin had imagined.

What rate of interest, compounded continuously for 100years, would have multiplied Benjamin Franklin's original capitalby 90?

(Continuation of Exercise 15.) In Benjamin Franllin's estimatethat the original 1000 pounds would grow to 131,000 in 100yea$, he was using an annual ftte of 5qo and compoundingonce each year. What rate of interest per year when compounded

continuously for 100 years would multiply the original amountby 131 ?

17. Radon-222. The decay equation for radon-222 gas is known tobe ): )0?

0r8', with / in days. Aboul how long wil l i t takethe radon in a sealed sample of air to fall to 907a of its originalvalue?

18. Polonium-2l0. The hallJife of polonium is 139 days, but yoursample will not be useful to you after 95Vo of the radioactivenuclei present on the day the sample ardves has disintegrated.For about how many days alter the sample arrives will you beable to use the polonium?

L9. The mean life of a radioactive nucleus. Physicists using theradioactivity equation ) : )o e r' call the number llk the meanlile of a radioactive nucleus. The mean life of a rudon nucleus isabout 1/0.18 = 5.6 days. The mean life of a carbon-l4 nucleusis more than 8000 years. Show that 95% of the radioactive nucleioriginally present in a sample will disintegrate within tbree meanlifetimes, i.e., by time t : 3/,t. Thus, the mean life of a mrcleusgives a quick way to estimate how long the radioactivity of asample will last.

20. Californium-2s2. Wlat costs $27 million per gram and can beused to ffeat brain cancer, analyze coal for its sulfur content, anddetect explosives in luggage? The answer is califomium-252, aradioactive isotope so rare that only 8 g of it have been made inthe western world since its discovery by Glenn Seaborg in 1950.The halflife of the isotope is 2.645 years-long enough for auseful service life and short enough to have a high radioactivityper unit mass. One microgram of the isotope releases 170 millionneutrons per second.

a) What is the value of ,t in the decay equation for this isotope?b) What is the isotope's mean life? (See Exercise 19.)c) How long will it take 95Va of a sample's radioactive nuclei

to disintegrate?

2L Cooling soup. Suppose that a cup of soup cooled from 90"C to60'C after l0 minutes in a room whose temperature was 20"C.Use Newton's law of cooling to answer the following questions.

a) How much longer would it take the soup to cool to 35'C?b) Instead of being left to stand in the room, the cup of 90'C

soup is put in a freezer whose temperature is 15"C. Howlong will it take the soup to cool from 90"C to 35"C?

22. A beam of unknown temperature. An aluminum beam wasbrought from the outside cold into a machine shop where thetemperature was held at 65'. After l0 minutes, the beam warmedto 35'F and after another 10 minutes it was 50'F. Use Newton'slaw of cooling to estimate the beam's initial temperature.

23. Surrounding medium of unknown temperature. A pan ofwarm water (46'C) was put in a refrigerator. Ten minutes later,the water's tempemturc was 39"C; 10 minutes after that, it was33'C. Use Newton's law of cooling to estimate how cold thercfiigerator was.

24. Silver cooling in air The temperature of an ingot of silver is60"C above room temperaturc dght now. Twenty minutes ago, it

16.

was 70'C above room tempemture. How far above room tempel-

ature will the silver bea) 15 minutes from now?b) two hours from now?c) When will the silver be 10'C above room temperaturc?

The age of Crater Lake. The charcoal from a tree killed in the

volcanic eruption that formed Crater Lake in Oregon contained

44.5Vo of the carbon-l4 found in living matter About how old

is Cnter Lake?

The sensitivity of carbon 14 dating to meaiurement To see

6.6 L',Hdpital 's Rule 491

the effect of a relatively small error in the estimate of the amount

of carbon-14 in a sample being dated, consider this hypotheticalsituation:

a) A fossilized bone found in cental lllinois in the year A.D2000 contains 1770 of its original carbon-l4 content. Esti-

mate the year the animal died.b) Repeat (a) assuming 187o instead of 177..c) Repeat (a) assuming l67a instead of 1770

Art forgery. A painting athibuted to Vermeer (1632 1675),which should contain no more than 96.270 of its original carbon-14, contains 99.5Va irctead. About how old is the forgery?

27.

lJH6pital's RuleIn the late seventeenth century, John Bernoulli discovered a rule for calculating

limits of fractions whose numeraton and denominators both aPproach zero The

rule is known today as l'H6pital's rule, after Guillaume FranEois Antoine de

l'H0pital (1661-1704), Marquis de St. Mesme, a French nobleman who wrote the

first inftoductory differential calculus text, where the rule first appeared in print'

Indeterminate QuotientsIf functions /(x) and g(r) are both zero at "x : a, then lim*-, /(-r)/g(r) cannot

be found by substituting .r : a. The substitution produces 0/0, a meaningless

expression known as an indeterminate forrn. Our experience so far has been that

limits that lead to indeterminate forms may or may not be hard to lind. It took a

lot of work to find lim,-o (sin -r)/,r in Section 2.4. But we have had remarkable

success with the limit

f ' ( a ) : r im fQ) - J@)) c - a

from which we calculate derivatives and which always produces 0/0 L Hopital's

rule enables us to draw on our success with derivatives to evaluate limits that lead

to indeterminate forms.

Theorem 2lj H6pital's Rule (First Form)Suppose that f(a): g(a):0, that /'(4) and 8i (4) exist' and thar g'(a) +0. Then

_ f'(a)c'(4)

. . f(x),+4 g\x)

( 1 )


Caution

To apply l 'H6pital 's rule to flg, divide thederivative of f by the derivative of 9. Donot fall into the traP of taking thederivative of f/9. The quotient to use isf' lg' , not (flg)' .

A misnamed rule and the firstdifferential calculus text

In 1694 John Bernoulli agreed to accept a

retainer of J00 pound' per yecr from his

former student I'H6pita1 to solve problems

for him and keep him up to date on calculus.

One oi the problems was the so-called 0/0

problem, which Bemoulli solved as agreed.

When I'Hopital published his notes on

calculus in book form in 1696, the 0/0 rule

appeared as a theorem. L H6pital

ackou ledged his debt to Bemoull i cnd. to

avoid claiming authorship of the book's

entire contents, had the book published

anonymously. Bernoulli nevertheless accused

I'H6pital of plagiarism, an accusation

inadvertently supported after I'H6pital's

death in 1704 by the publisher's promotion of

the book as l 'Hdpital 's. By 1721, Bemoull i , a

man so jealous he once threw his son Daniel

out ol the house for accepting a mathematics

prize lrom the French Academy of Sciences,

claimed ro ha\e been the au*lor of lhe enl ire

work. As puzzl ing and 6ckle a' ever. hi ' tory

accepted Bemoulli's claim (until recently),

bul . t i l l named lhe rule af ler lHopilal.

Proof Working backward from ,/'(a) and g'(a), which are themselves limits, we

have

.. "f (.r) f (a)

f ' h ) , , ' - ; x , L7 _ r r rp t d t . . g ( x ) - g ( q )" | l m

x - a

. . / ( x ) - / { a ): l lm ' -

; : ; g ( x ) - g { a }

. . _ f ( x )_o, _a g ( r ) u

'. f (x): llm ;_:.

x+/ g("{ . ,

f (x) - f (a)x - a

g(.,r) - g(a)

x - a

tr

J

EXAMPLE 1

") I's

,n+ i lb) ]'s -,,t - sin .r

")ls r :

z',fr +i | 1= | | : tr t r = o

I c o s x l ^ 03x2 l,:o o

3 - c o s : {Tl.='

1

3 r - s i n x.T

What can we do about the limit in Example 1(c)? A stronger form ofl'Hdpital's

rule says that whenever the rule gives 0/0 we can apply it again, repeating the

process until we get a different result With this shonger rule we get

-{ - sln -r , . l - c o s r s r r | o r r r ,r + 0 J x ' I r l e i r l l j n

0_ r;_ ll l l \,,r, . I

r + 0 O x r r l c ; r r . L i n

cos .r_ t i - _ - 1 , r , t ; t t . .n ,

6 f .sul r . Sl t i t .

Theorem 3UH6pital's Rule (Stronger Form)Suppose that /(a) : 8(a) : 0 and that / and g are differentiable on an open

interval 1 containing a. Suppose also that g'(.r) I 0 on 1 if .r { a' Then

. . I l ^ t , / { ' r )ttm --:-- = lim :-=, (2)' a g l x | r - a g ( x ,

if the limit on the right exists (or is oo or -oo).

You will find a proof of the finitelimit case of Theorem 3 in Appendix 5

6.6 lJH6pital 's Rule 493

EXAMPLE 2

lim. ,n+r -1 - (x /2 ) 0

0

gl2)(1 +x)- t /z -(1/2)r+0 2x

-(1/4)(I + x)-3/2

: l i m sritt 9

When you apply l'Hdpital's rule, lookelse. This is where the limit is revealed.

EXAMPLE 3

Nor l:

l imit is

found f,

from 0/0 to something

2

for a change

limx + x 2

: li-

"il

] : 9 : o \nr l; rimit is rbunr.r - 0 l + Z r 1 u

If we continue to differentiate in an attempt to apply I'Hdpital's rule once more,we get

. . I - cos x . . s in x . . cos. r, ' 0 r l x z , . 0 I + 2 x r - 0 2

which is wrong.

EXAMPLE 4

li- tin" 9

'+0+ x2 0

, . cosr 0= ,uT 2 ,

= - \o , , , , "n

o . ' i \ fo , r l J J

L Hopital's rule also applies to quotients that lead to the indeterminate forma/a.If f (x) and g(x) both approach infinity as,r -> a, then

. . f (x) , . f ' \x ): l rm - ., ' , g ( . r ) ,

" I 'Q)

provided the latter limit exists. The a here may itself be either finite or infinite.

EXAMPLE 5sec t

At llIII ;_7,+ (n /2 ) - \+ tanx

.. sec x tan x ..: l1m ----------: llm Srnrx+(t/z) sec".r x+\n /2)

h x . . l / x . . lD, l lm . . . . . . : : l lm - : l lm ' ' . :

^-* 2Jx ,-* t /Jx ^-* Jx

1, '

tr

- l

- na


Indeterminate Products and DifferencesWe can sometimes handle the indeterminate forms 0 . oo and oo - oo by usingalgebra to get 0/0 or oo/oo instead. Here again, we do not mean to suggest thatthere is a number 0 . oo or oo - oo any more than we mean to suggest that there isa number 0/0 or oo/oo. These forms are not numbers but descriptions of functionbehavior

EXAMPLE 6

lim r cot .t

: l i m

: lim

: lim

[J . co; rewrite i cot j.

1tan -{

t

, _ 0x

tan ,r

I--sec. x

1I tr

1 1sln t x

s m r + r c o s - rsln t

2 c o s x - x s i n x

0

Sti l l -

- o - . '2 t )

Indeterminate PowersLimits that lead to the indeteminate forms 1*. 00. and oo0 can sometimes behandled by taking logarithms first. We use l'H6pital's rule to find the limit of thelogarithm and then exponentiate to find the original function behavior

E0AMPLET Find lin1 (# -;)

Solution II x -+ 0+, then sin; -+ 0+ and1 l

" r n " - i - > o o - o o 'Similarly, if .x -+ 0 , then sinx -+ 0 and

1 l -+ -oo - (-oo) : -oo + oo.SlN X X

Neither form reveals what happens in the limit. To find out, we fint combine thefractions.

-t - Sln t Common denominalor is- .

x s t n r

and then apply l'H6pital's rule to the result:

. . / | l \ . . x - s i n . r .l f m | . - - l = l l m - - - - - - - - - -tr-o \Srn -{ x,/ r-0 I Sln.x

1 - c o s x: lim

: lim

FoIm the Iatio ofthede.ivative of the

numemtor to the dedvativeof the denominator.

Does the limit of the new mtio as-r+4 lead to an indeterminate form?

Stop differentiating. You have founddle limit o. detemined thar the limir

does not exist.

Flowchart 6.1 [Hdpital.s rule

6.6 f Hdpital 's Rule 495

If l im,-, ln ,f (x) : Z, then

l im /(r) : l im e h l \x) - eL .

Here a may be either finite or infinrte.

EXAMPLE 8 Show that t im,-6, (l I x)t/, - ".

Solution The limit leads to the indetenninate form l-. We let f (x) : (l + r) r/,and find l im,-6, ln /(r). Since

ln I (x) = ln (1 + x)r/'

: ] tn ( r + , ) ,

I'H6pital's rule now applies to give

I

l + 'I io+ I

I

Therefore,

l i m { l + r ) ' - l i m / { . \ r = l i m e r ' \ = e - c .J

EXAMPLE 9 Find l im,-- xrl ' .

Solution The limit leads to the indeterminate form oo0. We let /(r) = r,/. andfind l im,-- ln /(-r). Since

I 'H6pital's rule gives

l n f r " r l - - r l r

l n r

.t

t h -

r r m t n / ( , 1 ) - l l m -r+_ r

.. \ /.rI + n I

0

f

Therefore,

l im x ' l ' : l im / ( x ) : l im e r ' r . r ( r ) : e0 : 1 .


Exercises 6.6

Applying I'H6pital's RuleUse i'Hdpital's rule to lind the limits in Exercises 1 42

1. lim- - 2 1 <

r + 5 r + J

t3 l4 . l i m - . .

t - t 4 t ) t I

t - R " 26. Iim #' - € l z r ' + f i t

s in 5 tlt. 11m

r+0 t

s l n , { ,10. lim .

x 2 - 4

t 3 - 4 t + 1 5I t i m

t - 3 l ' - I I l

r-! 7r ' + I

sin t27. t im _

r+0 t

8x2o t i m _

r : 0 C O S i - I

20 -1 t11 . l im -r - n /2 cos \ 2n -U )

3e + l t12 . l im . - -

d- r/ l srn (u + (n/J.))

1 - s i n d1 3 . l i m - -d - ? / 2 l + c o s l t /

x215. lim _

J-0 ln (sec .ir )

r ( l cos t )17. l im -

t - o t - s l n It t l I \

19. l im lx - ; l sec . rr-(r/2) \ t '

) n \20 , l im {= x l t an t

' - t . t t2 l \ 2

3 s i d _ 121. lim -

r2'23. l im ^ .

l n C { + l )25. lim .

. - ! l o g t t

ln (ir2 + 2.x )t ' 7 l i h _

r -0+ ln - r

v ) y + r r - J),

x l14. lim -' .

J+ l ln n - S ln L I

ln (csc r)16. l im -

\ - n /2 l x \ n / t ) ) .

t sin t18. lim -

. + 0 l - C O S t

l t 2 ) , , 1t t I i m

3 , _ l24. bm -

r , o 2 r - l

loe",r26, liol. - --:: -'r-d logr (r + J 1

l n ( c \ l i28. lim #

r+n+ ln I

43. l im rrl(r r)

45. l im (ln x)"*

47. Ilr.]' x t/t"'

49. lim (1 l2x)t/{ztn 4

51. l im r'

/ 3 x + l I \3 4 . l i m l - - - 1

r+0+ \ r srn ir /

/ | 1 \35. lim | . . I

r - l ' \ , r - | l n j t /

36. lim (csc .x - cot i + cos r)

tL, 1 1 I'37. l im / ld t 38 . l im . I ln td t

r+@ J t t rJe r ln ) l : J l

c o s d - l39. l im -

0 - a c o - H - |

e h t 1 + h \$'l,a ^,41. 1in1 1-Jl

t+6 ? ' ' - t

42. l im x2e '

Limits Involving Bases and ExponentsFind the limits in Exercises 43 52.

44. l im rrlG r)

46. lim (ln r)r/(' ')

48, l im rrlr" '

50. l im (e' *.x)1/'

sz. r i . ( r+ l ) 'r -0 - \ x . /

Theory and ApplicationsLH6pital's rule does not help with the limits in Exercises 53-56. Tryit-you just keep on cycling. Find the limits some other way.

v v)a t r

r+@ v,r + I

sec "l55- l im -r+(n /4 tan' x

t;54. lim p

r+u- v stn r|:cot-r

56- l im -

29. lim

30. lim

31. l im

32. lim

, a > 0

57. Which one is conect, and which one is wrong? Give reasons ibr

your answels.

x - 3 . . I Ia , t l m - ̂ ^ - : l l m - : ;

, _ l x . _ J t _ t Z X O

t l 0b ) l i m - - : : - 0

r - 3 a ' - J O

58. Which one is conect, and which one is wrong? Give reasons for

your answe$,

)(ln 2r - ln ("I + l))

(ln n - ln sin t)

33. r. 11- -L)r+0+ \ jr sln "r /

. 2 - r . ) . )a) lim j-- = - lim .r-0 rz - srn.x r-0 2.x - cos.r

2 2_ t i m - _ - I

. - o 2 + s i n r 2 t 0

r: - srnr: r .0 zx - cosr t , - I

Only one of these calculations is conect. Which one? Why arcthe othe$ wrong? Give reasons for your answers.

a ) l i m r l n r : 0 . ( - o o ) : 0

b ) l i m . r l n r : 0 . ( - o o ) : o o

ln .x -ooc) lim .r ln .r : lim -

J-o+ ,-o- (l /.r) oo

l n nd ) l i m t l n r = l i m -

r-0+ '-o+ \I /x)

" (r /'). I i m = [ m ( - . r ) : 0r - 0 t l / . t ' J r ' 0 '

60. Let

f / - \/ lal : 1 [p1 rhat tim l-:] :2.g'(x) '-o g(.{)

contradict 1'H6pital's rule? Give reasons for your

Find a value of c that makes the function( 9r - 3sin3.r

f ( r ) : I s , , ' '+o[ c , r : o

continuous at ,r : 0. Explain why your value of c works.

Find a value of c that makes the functior

I nanefI A + 0g(0): I sin 140, / lt ) '[ c . 0 : O

continuous from the right at 6 : 0. Explain why your value ofc works.

The continuous compaund interest formula. In dedving theformula A(r) : Aoe't in Section 6.5, we claimed that

/ r \ k tl i m A o l l - ; l : A o e " .

This equation will hold if

l i m { l + : I : p " ,t - d \ k l

Exercises 6.6 497

and this, in tum, wiu hold if, r ' t

l i m { l ' ; l : ? ' .

As you can see, the limit leads to the indeterminate form 1-.Verify the limit using l'H6pital's rule.

64. Given that.r > 0, find the maximum value, if any, of

a) at/ 'b) xt/"c\ xth' (n a positive integer)d) show that lim,-6xt/" = | for every positive integer a-

Grapher Explorations65, Determining the value of e.

a) Use l'H6pital's rule to show that

/ I \ 'l i m l l l | - e .

x . /

E u) clrcunroR See how close you can come ro

e:2.'1 1828 1828 45 90 45

by evaluating f (x): (1+ (1/x))' for.x = 10 , 102 , 103,. . . atrd so on. You can expect the approximations to ap-ptoach e at first, but on some calculators they will moveaway again as round-off elrols take thei toll.

c) If you have a gapher, you may prefer to do part (b) bygraphing /(r) = (1 + (l/r))' for large values of "r, usingTRACE to display the coordinates along the $aph. Again,you may expect to find decreasing accuacy as.r increasesand, beyond r : 1010 or so, ermtic behavior.

66. This exercise explores the difference between the limit

/ I \ "l i m l t + ; Ir+@ \ x. . /

and the limit

lim

studied in Exercise 65.

a) Graph

/ r \ ' / r \ '11x ; : { t + - l and g t . r ) : l l + - l

\ x"/ \ x_/

together for .r : 0. How does the behavior of / comparewith that of 8? Estimate the value of lim,-6 /(x).

b) Confirm your estimate of lim,r- /(,r) by calculating itwith l'H6pital's rule.

67. a) Estimate the value of

rrm tr - .r4z +rl

by graphing f(x):x-\E+x over a suitably large in-terval of r-values.

b) lq

atrtta

/(,) : {;,*'' ;13I x ' l I , x # 0

8 r x , : 1 0 , x : 0 .

Show that

lim

Doesn't thisansweN.

61.

( r*1) ' : " ,

63.

498 chapter 6: Transcendental Functions

b) Now connrm your estimate by finding the limit with l'H6pi-

tal's rule. As the first step, multiply /(ir) by the fraction

t , - v ( - , ' / t , ' "4 fJ r r and ' imp l i f y the new nu-

meratol

68. Estimate the value of

x 2 - 4l lnrI J r r 1 5 l

by graphing. Then conflIm your estimate with l'Hopital's rule'

69, Estimate the value ol

lim2x2 - (3x + 1)J, +2

r - I

by graphing. Then confirm your estimate with I'H6pital's rule'

70. a) Estimate the value of

. . ( r - l ) 'j - ' r ln x , r cosr r r

by g raph ing / ( tc ) : (x - l y l@lnr -x cos / t ) near

x : 1 Then confirm your estimate with I'H6pital's rule'

b ) Graph / fo r0 <r . : l l .

71, The continuous extension of (sinx)' to [0 , t']

a) Graph /(r): (sinir)* on the inteNal 0 = t :1t What

value would you assign to / to make it continuous at .{ : 0?

b) Verify your conclusion in (a) by finding lim'-o* /(r) with

I'H6pital's rulec) Retumitg to the graph, estimate the maximum value of /

on [0, ]T]. About where is max / taken on?

d) Sharpen your estimate in (c) by graphing /' in the same

window to see where its graph crosses the t-axis To sim-

plify your work, you might want to delete the exponential

factor from the expression for f' and graph just the tactor

that has a zeroe) Sharpen your estimate of the location of max / further still

by solving the equation /' - 0 numerically

ffi fl cALcULAToR Estimate max / by evaluating / at the lo-

cations you found in (c), (d), and (e) What is your best

value for max f ?

72. The function (sin xfan ". (Continuation of Exercise '7l

)

a) Graph /(;r) : (sinx)t""'on the interval 7 : r 17 How

do you account for the gaps in the graph? How wide are

the gaps?Now graph f on the inteNal D = x :1t. The function is not

defined at ir = n 12, but the graph has no break at this point'

What is going on? What value does the graph appear to give

for f at r : jt 122 (lliat; Use I'H6pita1's rule to find lim f as

t '> (t l2)- afi x > Qt l2)+.)Continuing with the graphs in (b)' find max / and min J as

accurately as yo]r can and estimate the values of t at which they

are taken on.

The place of ln x among the powers of x. The natural loga-

thm/"r 1

l n x _ I d tJ t t

fills the gap in the set of formulas

l t"at : k+0 , ( 3 )

but the formulas themselves do not reveal how well the logarithm

fits in. We can see the nice fit graphically if we select from Eq'

(3) the specific antiderivatives

a r r l - |

I r t - t d r - - L . x > 0 .

J t K

and compare their graphs with the graph of ln x

a) Graph the lunctions f(x) = (tk - 1)/f together with h -r

on t he i n te r va l 0 : j t 550 f o r / < :+1 , +0 '5 ' +01 '

and t 0.05.

b) Show that

1 i , L . : l n r .k-0 k

(Based on "The Place of ln r Among the Powers of -r" by

Henry C. Finlayson, American Mathematical Monthly' YoL

94, No.5, MaY 1987, P.450.)

Confimation of the Iimit in Section 5.7, Exerc6e 42. Estimate

the value of

sind - c| cos dl im --

d+0+ a f -aCOSd

as closely as you can by graphing. Then connrm your estimate

with l'Hopital's rule.

b)

c)

, , * r ,

Relative Rates of GrowthThis section shows how to compare the rates at which functions of 'r grow as r

becomes large and introducas the so-called little-oh and big-oh notation sometimes

used to de;ribe the results of these comparisor$. We restrict oar qttention to

functions whose values eventually become and remain positive as x --+ ce'

6. t4 The graphs o+ e' , 2' , and x2.

6.7 Relative Rates of Growth 499

Relatives Rates of GrowthYou may have noticed that exponential functions l ike 2' and e' seem to grow more

rapidly as -r gets large than the polynomials and rational functions we graphed

in Chapter 3. These exponentials cefainly grcw more rapidly than -r itsell. ardyou can see 2' outgrowing rr as,r increases in Fig. 6.14. In fact, as,t -+ cc, the

functions 2' and e' grow faster than any power ofr, even,t1000000 (Exercise 19).

To get a feeling for how rapidly the values of -t ' = e' grow with increasing

r, think of giaphing the function on a large blackboard, with the axes scaled in

centimeters. Al -r = I cm. the -smph is e I l3 cm above the l-axis. At,r = 6 cm,rhe graph is e" - 403 cm A:4 m high (it is about 1o go through the ceil ing if i thasn't done so already). At ,r = 10 cm, the -snph is ett) x 22.026 cm I 220 nr

high, higher than most buildings. At -r:24 cm. the graph is more than halfway

to the moon, and at,r = 43 cm t'rom the oigin, the grrph is high enough to reachpast the sun's closest stellar neighbor, the red dwarf star Ptoxima Centauri:

e+3 x 4.'73 x l0r8 cm

: ,1 .73 x l0 r r km

- 1.58 r03 lighr-sec'nrrr :ij.ffi;'iJ:JJ **','

- 5.0 l ight-years

The distance to Proxima Centauri is about 4.22 l ight-years. Yet with ,{ - 43 cm

fiom the origin, the graph is sti l l less than 2 feet to the right of the -r'-rxis.In contrast, logarithmic functiols l ike -r': logrr and -r'= ln -r grow morc

slowly as ,r > c! than any positive power of r (Exelcise 2l). With axes scaled in

centimeters, you have to go nearly 5 l ight-years out on the l-axis to find a point

where the graph of t : ln r is even -t, :43 cm high. See Fig. 6.15.These important comparisons ofexponential, polynontial, and logarithnic flnc-

tions can be made plecise by detining whal il means for a flnction I (-r) to grow

faster thall a tunction g(r) as -r + oo.

Definit ion"' Rrt", of Growth as x -+ oo

Let l (.rr) and .g(r) be positive tbr .r sufficiently large.

1, / grows faster than g as ';r -+ co if

'. .l ('r ),lli ,1a

: -

or. equivalently, if

ri- 8 (") = o.,,- l (x)

We also say that I grows slower than ./ as r --+ crl.

2. f and g grow at the same rate as r > cc if

f ( - r )lrrn = f / 0. L finire and nol /ero

6.15 scale drawings of the graphs ofano tn x .


According to these definitions, ) :2x does not grow faster than y :.x. Thetwo functions grow at the same rate because

, . zxu m z : z ,

rrco jr

which is a flnite, nonzero limit. The reason for this apParent disregard of commonsense is that we want "/ grows faster than g" to mean that for large x-values g isnegligible when compared with /.

EXAMPLE I e' grows faster than :r2 as r --) oo because

l i m " : f i m l - : l i m l : o s .t 6 x ' x - " a z x \ - @ z\ - / J \ - / J

e/cb oo/co

EXAMPLE 2a) 3' grows faster than 2i as x --> oo because

, . 3 ' / 1 \ 'r r r r r - - - l i m l l | : o o .

x - 6 z t - @ \ z /

b) As part (a) suggests, exponential functions with different bases nevel grow atthe same rate as .x -+ oo. lf a > b > 0, then a' grows laster than b'. Since(a/b) > r,

- . a ' . . / d \ rl r m - : I t m l - , : c ' o .

x-6 l ) ^ r -6 \ D /

-r2 grows faster than h -r as x -+ oo because

Using I 'H6pi ta1's ru letwice

tr

EXAMPLE 3

EXAMPLE 4

l r m - : l r m - : l l mJ - c o l n x r - @ l / x

2 x 2 : a . L H o p i t c l s r u l e

tr

tr

tr

ln r grows slower than x as ,r: -> oo becauset n Y t l r

Im _::: : Um jl: l Hopi,i,t . -ulc

r+@ r r+oo 1

EXAMPLE 5 In contrast to exponential functions, logarithmic functions with

different bases a and D always grow at the same rate as .r -+ oo:

- . l o g - r . . l n x / l n a l n bItm -:-: llm ;-------;---; = -r . , j . ' l o g h x x ' t u l n x l t n D t n c l

The limiting ratio is always finite and never zero. tr

If / grows at the same rate as I as t -+ m, and g grows at the same rate as

h as x --> oo, then / grows at the same rate as lr as -r -+ oo. The reason is that

rm L : t , and l im 8 - :7 ,,+6 p t+6 n

I

6.7 Relative Rates of crowth 501

together imply

r i r r I = t i ^ L .9 : r , r " .. - - h h

'

It L1 and L2 are finite and nonzero, then so is 11L2.

EXAMPLE 5 Show that v52 + 5 and t2rG 112 grow ar rhe same rate asr \ '€.

Solution We show that the functions grow at the sane rate by showing that theybuth grou at the same rate a5 the luncliun x;

. . v r , ) . . / . )- m \ / t , , - t ._t r i6 v I -

, . r 2 u ( - t t r , . / ' - t ' 7 ' r ' lU n r _ _ | r m |

' v _

' I _ t i n 1 { :

' _ l = + .

.r J-€ \ v/ ir / ,-" \ . / .r , / J

Order and Oh-NotationHere we introduce the "little oh" and "big-oh" notation invented by number theoristsa hundred years ago and now commonplace in mathematical analysis and computersclencg.

DefinitionA function / is of smaller order than I as x + oo if fi^ /1*,) : O. w"" ,-- s(.r)indicate this by writing / = o(g) (".f is little-oh of g").

Notice that saying ./ : o(g) as rthan g asr + oo.

EXAMPLE 7

--+ oo is another way to say that / grows slower

l n r

l im - f :g

ln r : o(x) as r -+ oo because ilX

x2 : ct(xt * 1) as r -+ oo becauseI

DefinitionLet /(x) and g(r) be positive for r sufficiently large. Then .f is of at mostthe order of g as x -+ co if there is a positive integer M for which

I(r\ . A4.g( r ' i

-

for r sumciently large. We indicate this by writing f = O(d ("/ is big-ohof c").

502 chaoter 6: Transcendental Functions

EXAMPLE 8

r + sinr : O(.r) as x -> oo because.x +smx < 2 tor x sulncren y large.

x

EXAMPLE 9

e' + xz : o(e') as x -+ co because "' l" --t 1 as '" > oo,- e x

x : o\e') as t -+ oo becaus" | -+ 0 as x --> oo.e r J

If you look at the definitions again, you will see that -f : o(g) implies J : O(e)

for functions that are positive for .r sufficiently large. Also, if / and g grow at the

same rate, then /: O(g) and g: 917) (Exercise 11).

Sequential vs. Binary SearchComputer scientists sometimes measure the efficiency of an algorithm by counting

the number of steps a computer must take to make the algorithm do something.There can be significant differences in how efficiently algorithms perform, even if

they are designed to accomplish the same task. These differences are often described

in big-oh notation. Here is an example.Webster's Third New Internqtional Dictionary lists about 26,000 words that

begin with the letter a. One way to look up a word, or to leam if it is not there, is

to read through the list one word at a time until you either find the word or detemine

that it is not there. This method, called sequential search, makes no particular use

of the words' alphabetical arrangement. You are sure to get an answer, but it might

take 26,000 steps.Another way to find the word or to learn it is not there is to go shaight to the

rniddle of the list (give or take a few words). If you do not find the word, then go to

the middle of the half that contains it and forget about the half that does not. (You

know which half contains it because you know tbe list is ordered alphabetically.)

This method eliminates roughly 13,000 words in a single step. If you do not find

the word on the second try then jump to the middle of the half that contains it.

Continue this way until you have either found the word or divided the list in half

so many times there are no words left. How many times do you have to divide the

list to find the word or learn that it is not there? At most 15, because

(26.000/2r5) < l .

That cefiainly beats a possible 2fi000 steps.For a list of length n, a sequential search algorithm takes on the order of z

steps to find a word or determine that it is not in the list. A binary search, as the

second algorithm is called, takes on the order of log2 r steps. The reason is that if/"-1 <n:2^, then m - 1<logrn S tZ, and the number of bisections required

to narrow the list to one word will be at most m : flog, rl, the integer ceiling for

log2 n.Big-oh notation provides a compact way to say all this. The number of steps in

a sequential search of an ordered list is O(rz ); the number of steps in a binary searchis O(logrn).In our example, there is a big difference between the two (26,000 vs.

15), and the difference can only increase with n because n grows f-aster than log2 tr

aS n -+ OO.

tr

Exercises 6.7 503

To find an item in a list of length n:

A sequential search takes O(r) steps.

A binary search takes O(log2 n ) steps.

Exercises 6.7

Comparisons with the Exponential e'1. Which ofthe following functions grow faster than ex as r --+ oo?

Which grow at the same rate as e'? Which grow slower?

13 + sin2 r

loglo.r

2. Which ofthe following functions grow faster than e' as,r -+ oo?Which grow at the same late as ?'? Which grow slower?

Ordering Functions by Growth Rates7. Order the following functions from slowest growing to fastest

growrng as r -+ @.

a\ e' b) ,r ' c) ( ln:r) ' d) e' /2

8. Order the following functions from slowest growing to fastestgrowmg as i >oo.

a) 2' b) "t

Big-oh and Little-oh; Order9. True, or false? As x > oo,

a) .r : o(,r)c ) r : O ( r { 5 )e) ex : o(e2')g) ln x: o(ln Lr)

10. True, or false? As r -+ oo,

I I / r \b ) -+ - o l : l

x r ' \ r , /

d ) 2+cos . : r : O(2)

f) ,r ln r : o(.t2)x ) h ) I n U ) : o ( l n ( r 2 l ) )

Show that if positive functions /(r) and g(,r) grow at the samerate as r -+ oo, then / - O(g) a]Jd g: O(f).

When is a polynomial /(r) of smaller order than a polynomialg(.x) as .r + co? Give reasons fot your answer

When is a polynomial /(.x) of at most the order of a polynomialg(-x) as r .+ oo? Give reasons for your anwer.

Simpsons rule and the trapedzoidal rule. The delinitions inthe prcsent section can be made more general by lifting the re-striction that -I'+ oo and considering limits as .r -> d for any

a)c)e)

logr(.t2)t l4xx -2brx

ln (ln r)

b) Iogro l0.rd) | lr2f \

" 'h) ln (2i + 5)

Comparisons with the Power IJ. Which of the following functions grow faster than ,r 2 as ,r . oo?

Which grow at the same rate as r '? Which grou, slowerl

a) 1014 + 30-x + Ic ) /1+J . 'e ) e '8) e-"

a) x2 +4xc ) . , 44+7e) -r ln ;rE) t3e '

a) r ':+ 16c) r2e-'e) x3 t2

c) ( l .1) '

a ) r + 3c) Jie) (3/2)'

e'/2

a) Iog3.{c) ln

"4e) -rc) r /x

b ) x l n x - . rd) (s/2)'f) xe'h ) r ' '

(.r + 3f

b) 10.r,d) togro(r2)f ) (1 /10 fh) "I, + 100-{

c) (ln 2)' d) " '

b)d)f)h )

b)d)f)h )

b ) , t : o ( r i 5 )d) x - O (2x)

f ) r + l n r : O ( x )

D 1F4 : s7 l

4. Which of the following functions grow faster than x/ as x -+ oo?Which grow at the same rate as i2? Which grow slower? "r

- !=o/ l ),r + J \.{,/

1 I / 1 \c ) - - , : o l - lI 1 ' \ j r . /

e ) e ' * x : O ( e ' )g) ln ( ln "r): O(ln

Comparisons with the Logarithm In x5. Which ofthe following functions grow faster than ln.I as r -+ oo?

Which grow at the same rate as ln r? Which grow slower?

b) ln 2.rd) .,trf) 5ln "rh) ?.

6. Which ofthe following functions grow faster than lnr as:r -+ oo?which grow ar the same rare as ln x? Which growilower?

11.


real number.i. Show that the enor ts in the Simpson's ruleapproximation of a definite integral is O(h4) as h + 0 while theerror Er in the trapezoidal rule approximation is O(r'?). Thisgives another way to explain the relative accuracies of the twoapproximation methods.

Other Comparisons15. What do the conclusions we drew in Section 3.5 about the limits

of rational functions tell us about the relative growth of polyno-mials as r -+ oo?

EZ ro. cnapHrn

ally overtake the values of ln r, you have to go way outon the ,-axis before this happens. Find a value of ;r greaterthan 1 for which rll1000,000 > h r. You might start by obsen-ing that when r > I the equation ln x :,rr/r'mo.om is equivalentto the equation ln (ln .r) = (ln x)/1,000,000.

E c) CALCULATOR Even ,rrlro takes a long time to overtakelnir. Expedment with a calculator to find the value of .r atwhich the graphs of.rrllo and ln r cross, or, equivalently, atwhich ln r; l0 ln (ln r). Bracket the crossing point betweenpowers of 10 and then close in by successive halving.

ZZ a) Cnapnen (Continuation of part c.) The value of r at whichln r : 10 ln (ln,r) is too far out for some gmphers and rootfinde$ to identify. Try it on the equipment available to you andsee what happens.

22, The function ln x grows slower than any polynomial. Showthat ln 'I grows slower as r + oo than any nonconstant polyno-mial.

Algorithms and Searches23. a) Suppose you have three different algo.ithms for solving the

same problem and each algorithm takes a number of stepsthat is of the order of one of the functions listed here:

nlogrn, n3/2, n(log.- n)2 .

Which of the algorithms is the most efficient in the longrun? Give reasons for your answer.

E3 n; cnneHfn Graph the functions in parr (a) together to geta sense of how rapidly each one grows.

2l. Repeat Exercise 23 for the functions

n, uElogrn, (logrn)2.

E ZS. CnLCUfATOn Suppose you arc lookrng tbr an item in an ordered list one million items long. How many steps might it raketo find that item with a sequential search? a binary search?

F Ze. CelCUmfOn You are looking lbr an item in an ordered list450,000 items long (the length of Webster's Third New International Dictionary), How many steps might it take to flnd the itemwith a sequential search? a binary search?

a) Investigate

,. ln (-r * l)llm - al'IO

m ' ,

,. ln (r + 999); ; t n r

Then use 1'H6pital's rule to explain what you find.b) Show that the value of

,. ln (,r + a)ln .r

is the same no matter what value you assign to the constanta. What does this say about the relative rates at which thefunctions /(t) : ln (:v * a) and g(t) = ln -x grow?

17. Show that r,/10.t + 1 and .,4 + I grow at the same .ate as r >o<) by showing that they both grow at the same rate as .f as

18. Show that ."/ia + :r and .,6a ,3 grow at the same rute as r ->co by showing that they both grow at the same rate as 12 as

19. Show that e" grows faster as .t + oo than r' for any positiveinteger n, even.rl,000,m0. (I1ir?/. What is the rth dedvative ofr'?)

20, The function e" outgrows any polynomial Show thate'grcwsfaster as r -> ca than any polynomial

a , x n + a n t x n | * . . . + a t x + a o .

21, a) Show that ln r grows slower as -{ -+ at than .trl' for anypositive integer rt, even rll1 000 000.

& bl CALCULAToR Although rhe values of xr'r '000000 eventu-

lnverse Trigonometric FunctionsInverse trigonometric functions arise when we want to calculate angles from sidemeasurements in triangles. They also provide useful antiderivatives and appearfrequendy in the solutions of differential equations. This section shows how thefunctions are defined, graphed, and evaluated.

6,8 InverseTrigonometric Functions 505

Defining the InversesThe six basic trigonometric functions are not one-to-one (their values repeat), but

we can restrict their domains to intervals on which they are one-to-one

Domain Restrictions That Make the Tfigonometric FunctionsOne-to-One

Function Dornain Range

sm'{

cos ir

tan t

cot.t

sec x

csc.r

l - r / z , r /21[0, z]

(-r /2, r /2)(0, ]7)

10. n 12) U (tr 12, rl

l-r 12, 0) | (0, r /21

[ - 1 , 11

[ -1 , l ](-co, oo)

(-oo, oo)

( oo, - l l U [1, oo)

(-oo, - 1l U [1, co)

Since these restdcted functions are now one-to-one, they have inverses, which

we denote by] : s l n ' , r o r ) : a r c s m r

) : cos-lx or l : arc cosiu

) = tan-lx or :t : arc tan 't

) : cot lx or ) : arc cotx

) : sec-tr or l : arc secx

Y : c s c l - r o r ) : a r c c s c r

These equations are read "y equals the arc sine of .r" or "y equals arc sin x" and

so on.

Caution The -1 in the expressions for the inverse means "inverse." It does nol

mean reciprocal. For example, the j 'eciProcal of sin.r is (sinx)-t : 1/sinn: cscx.

The domains of the inverses are chosen to satisfy the following relationships.

sec ' , r : cos ' ( l / , r ) ( 1 )

csc I .{ : sin '(1/x) Q\

c o t t . r : r f 2 - t a n - ' x ( 3 )

We can use these relationships to flnd values of sec-lr, csc r;r, and cot l.n on

calculaton that give only cos-] .r, sin ' ;r, and tan-r.r. As in some of the examples

that follow, we can also find a few of the more common values of sec-r'r, csc rx,

and cot-l r using reference right triangles.

The Arc Sine and Arc CosineThe arc sine of r is an angle whose sine is .r. The arc cosine is an angle whose

cosine is r.

) = c o s r , 0 < . r < 7Domain: [0. ?7]Range: I l . 1]

- 1

506 Chaoter 6: Transcendental Functions

) , . \ '= s in - r . ;< -Y< tDom i\: l-n12. Tl2lRange: l- l , l l

Donain: [- 1. ] jRanqe: L rl2, n/21 Domain:

Range:t r , r l[0, z]

T2

6. t7 The graphs of (a) y = cosx.0 =x =n , and(b) i t s inverse , y : cos 1r .

The graph of cos-1x, obtained byreflection across the l ine y = x, is aportion of the curve x = cos y.

(b)

6.16 The graphs of (a) y : sin x,n12 a x ar12, and (b) its inverse,

y - s tn x . lhe graph o t s rn x , ob tarnedby reflection across the l ine y :;, ;5 gportion of the curve x = sin y.

DefinitionJ = sin-r.r is the number in [-tt /2, z/2] for which sin ] : r.

J = cos-r r is the number in [0, z] for which cos y - .t.

The graph of I : sin rx (Fig. 6.16) is symmetric about the origin (it liesthe graph of "r - sin I). The arc sine is therefore an odd function:

Srn (-x) : - Sln I '

The graph of 1 : qs5 I r (Fig. 6.17) has no such symme11y.

along

(4)

sln ^.r

EXAMPLE 1

,._,; = I

Alt\

0 I

. r " a= f- - u ) .

The angles come from thes i n l r i s I n / 2 , n / 2 ] .

Common values of sint x

^ 5 t t

t 1 21 / )

^5n

1' /3

T l+

n / o

n l4

f_r\. .,5 s i n

[ - ; . ] = ;

ottr

"'f3:rc0)y

lirst and fourth quadraots because the range

EXAMPLE 2

6.8 Inverse Trigonometric Functions 5O7

Common values of cos-1 x

cos '.r

(',I ^^^ r r [ r

12cos -it : cos

P\!o .T :J4 1 2

0 I

The angles come from the first and second quadrants because the range of cos '.t

is [0. z]. tr

ldentit ies Involving Arc Sine and Arc CosineAs we can see from Fig. 6.18, the arc cosine of x satisfies the identity

2

/ i 11

r/21 t 1

-^/ i t '>

- . h n

n l6

5n /6

6 , 7 8 c o s I x + c o s - r ( x ) : z

cos- , r -cos '1 x1 = o .

OI

c o s ' ( - . { ) : z - c o s r .

And we can see from the triangle in Fig. 6.19 that for r > 0,

sin- -r - cos I t = tr 12. (7)

Equation (7) holds for the other values of x in [ 1 , 1] as well, but we cannot

conclude this from the tdangle in Fig. 6.19. It is, however, a consequence of Eqs.(4) and (6) (Exercise 55).

Inverses of tan x, cot x, sec x, and csc xThe arc tangent of x is an angle whose tangent is ,r. The arc cotangent of jr is an

algle whose cotangent is r.

DefinitionJ = tan-l.r is the number in (-t/2,n/2) forwhichtany:-t.

J = cot-r J is the number in (0, z) for which cot y : x.

We use open intervals to avoid values where the tangent and cotangent are undefined.The graph of y : 131-t t is symmetric about the origin because it is a branch

of the graph r : tan ) that is symmetric about the odgin (Fig. 6.20). Algebraically

this means that

tan-r(-t) : tan rx; (8)

the arc tangent is an odd function. The graph of y : cot-1r has no such symmetry(F ig .6 .21) .

(s)

(6)

6.19 In this figure,

sln 1x+cos-1 x : rl2.

Domain: (--, -)Rangei (-T/2, nl2)

6.20 The graph of y = 16n t 1.

]

Domainr (--, -)Range: (0, '')

/" H"-1 l0

*.[3d=-;- .- t_-.-

5.21 The graph of y: cot-'x.


q2

{ y : . . . - ' ,Domain: l.rl > IRange: [0, n/2\ u (r12, Tl

The inverses of the restricted forms of secx and cscr are chosen to be thefunctions graphed rn Figs. 6.22 and 6.23.

Caution There is no general agrcement about how to define sec I x for negativevalues of x. We chose angles in the second quadrant between n f2 and z. Thischoice makes sec r.r : cos-r(1/r). It also makes sec r.r an increasing functionon each interval of its domain. Some tables choose sec I .r to lie in l-r , - n /2)for .r < 0 and some texts choose it to lie in ltt, hr l2) (Frg. 6.24). These choicessirnplify the formula for the derivative (our formula needs absolute value signs) butfail to satisfo the computational equation sec-r x : cos-i( l/.l ).

31r2

, 1 ft - 2

31r- t

6,24 Therc arc several logical choices for the left-hand branch of y = sec-rx. Withchoice A, Eq. (1) holds, but the formula for the derivative of the arc secant iscomplicated by absolute value bars. Choices B and C lead to a simpler derivativeformula, but Eq. (1) no longer holds. Most calculators use Eq. (1), so we chose A.

EXAMPLE 3 Common values of tan-1 x

6.22 The graph of y: sec-1x.

6.23 The graph ol y = g57 t x.

tan-l .r.^-- l r _ .---t- , ,

.F -,

\

A0 15

'y

tan t l - ' v5 l=

/,^(zI

0 V"\, 7

3 6

1

- l

n/4

-n /6

-it /3

The angles come from the first and fourth quadrants because the range of tan-r xis l-r /2. tr /2J. A

Domain: lx > 1Range : 0< )< r ,

\anget [-r12,0\ e (0, Tl2]

6.25 lI d = sin-1(213), then the valuesof the other basic trigonometric functtonsof d can be read from this triangle(Example 4).

The "arc" in arc sine and arc cosine

In case you are wondering about the "arc,',look at the accompanying figure. It gives ageometric intelpretation of y : si1 11 216y : cos-t r for angles in the first quadmnt.For a unit circle, the equation s : r0becomes s : d, so central angles and thearcs they subtend have the same measure. Ifr = sin), then, in addition to being the anglewhose sine is .r. y is also the lengrh of arcon the unit circle that subtends an anglewhose sine is.t. So we call y "the arc whosesine is r." When angles were measured byIntercepted arc len$h!. as they once wrrr.this was a natuml way to speal. Today it cansound a bit strange, but the language hasstayed with us. The arc cosine has a similarinterpretation.

6.8 Inverse Trigonometric Functions 509

EXAMPLE 4 Find coso, tand, seca, csccu, and cotcu if

a : s i n r ] . ( 1 2 )

so/ufion Equation (12) says that srnct:2/3. We picture a as an angle in aright triangle with opposite side 2 and hypotenuse 3 (Fig. 6.25). The length of theremaining side is

We add this information tocompleted triangle:

3)2 - 12Sz - Jn -4: .yB. P\rhasofean theorenr

the figure and then read the values we want from the

"82coscy : - , tancy : -= , Seca:, 45

co tq :

/ / 2 \ \E X A M P L E 5 F i n d c o r { s e c - r { = } l c r c - ' l 2 r ) .. t v 3 t /

So/ution We work from inside out, using reference triangles to exhibit ratios andangles.

Sfep t; Negative values of the secant come from second-quadrant angles:/ , \ / n \

sec r l - - a I : r " . - t l ' =

|\ J3/ \ - . . /3, /

5tr

6

Sfep 2.' Negative values of the cosecant come from fourth-quadrant angles:

csc (-2) : * .- , (+)\ - r , /

Step 3:

cot (sec-r

2'a3, '

3F 'v )

21 l

2

/ - \c s c l , = - 2

"v5

)

' 2 L - 2 - t Arc whose sine is -r

Angle.whose\ /

{ T

cosine is

0

lf

6

/ 2 \ \\ - "g /

* csc ' ( -2 ) ,

/ 5 n z \: c o t l - - - l

\ 6 6 /

/ 2 n \: c o t l - l\ 3 , /

I

a


6.26 Diagram for drift correction(Example 7), with distances rounded tothe nearest mile (drawing not to scare/.

EXAMPLE 6

Solution We let 0 : tan- | (x /3) (to give the angle a name) and pictureright triangle with

tan 0 : opposite/adjaceqt: x/3.

The lengrh of the triangle's hypotenuse is

rina se. (tan-, f).

a

^ \tt + 9sec t/ : ----

^pn/1, / l

I3

sec (tan-' ?, :

,/F+Y: ',FlO.

0 i n a

Thus,

sec 0

^/?Tq hypotenuse

tr

EXAMPLE 7 Drift correction

During an airplane flight from Chicago to St. Louis the navigator detemines that theplane is 12 mi off course, as shown in Fig. 6.26. Find the angle a for a course parallelto the original, conect course, the angle b, and the correction angle c : a ! b,

Solution1 )

a : s in r i ^ O.O0Z radian r 3 .8 'I E U

1 ' )b : sin-r t i - O. tgS radian ry I 1.2'

b2

c : a + b u 1 5 ' .

3

Exercises 6.8

Common Values of InverseTrignonometric FunctionsUse refercnce aiangles like those in Examples l-3 to find the anglesin Exercises l-12.

1 . a ) tan- ' l

2. a)

3. a)

tan-'(-1)

'' '(+)b)

b)

tun-t J5

^ , - , / I \\ ' / 2 /

/ _ l \ran | ---= |

\v3./

sin- ' [ , I

_ / l \br ran-'(-y'3 ) c) ran-' ( J3,)

/ r \ , - ] \a.

" t , i " ' ( ; J b ) s in ' ( - r ,

, _-i\s . a ) * , - ' ( ; ) b r c o s ' ( " ,

/ l \ / l _ \u.

", "or-' | :-) b) cos- 1, /-/

. . / 2 \7, a, sec r(-J2 ) b) sec '

\-JlJ- / -2\

8. a) sec-r \./2 b) cec ' \ Jj/

_ . / - 2 \9. at csc I J2 b) csc-'

\ f i-)- . / 2 \

10 . a ) csc r { -y '2 } b ) csc- ' 1J3 /

I l . a ) co l - r ( l t b ) co t - r rE

12. a) cot-t I b) cot-'(-"/')

Exercises 6.8 51'l

26. sec (cot-l.,/3 + csc 1(-1))

/ 7 r \ \27 . sec r (sec( - i ) ) tThe answer i s nor z /6 )

/ , I \ \28. cot-r (cot (- | ) )

(The ans\ter is rol -n /4.)

Finding Trigonometric ExpressionsEvaluate the expressioN in Exercises 29-40.

c)

c)

c)

c)

c)

c)

c)

c)

s i n ' ; f

, v3 \c o s r ,

. -v3 \cos-' ---

|

sec-11-21

sec /

csc z

csc-r1-2;

cot-r | - I\ J3 . /

' / 1 \c o t ' l - l

\ v3,/

/ , r \29 . sec ( tan

' i J

31. tan (sec r3))

JJ. COS tsrn :X,]

30. sec (tan I 2.r)

:2. t- (*" '])

34. tan (cos-'r)

40. rt" t*-'d=)

15. Given thar d : sec t(-.,5), find sin.r, cosd, tand, cscd, and 45. lim sec-r.r

cotd.

Trigonometric Function Values13. Given that q = sin t(5/13), find cosd, tand, secd, csc4, and

col .'.

14. Giver that a:tan-t(413), find sind, cosc[, seca, csca, and

cot d.

16. Given that q = sec-t(-Jl3l2), f,nd sind, cosq, tano, cscd,

and cot d.

Evaluating Trigonometric and InverseTrigonometric TermsFind the values in Exercises 17-28.

a /a\tz. sin fcos | 5 I 18. sec

\ z /

/ / r \ \to. t - ( ' i " ' ( -2JJ 20. cot

21, csc (sec-r2) + cos (tan-t(-\/3))

22, tari(sec t 1) +sin(csc-'( 2))/ / t i u t r r

23. sin (sin-r ( ;,) *-, '

,,-t|

/ / l \ \

24. cot (sin ' (-i)

- *.-' ,1

25 . sec ( t an t 1+csc ' l )

35, sin (tan-t ̂,F-E\ x z:2

." .," ll-.-, * \-' -' \ '* "41+

l /

r. *' (' i"-' ]) 38.

"o, (sin-' f)

o. 'ln ('""-' f)

LimitsFind the limits in Exercises 41-48. (If in doubt, look at the function'sgraph.)

41, l im sin rx

43. l im tan It

47. l im csc-rx

42. l im cos-r-r

,14. lim tan-r,r

46. lim sec-r jr

4E. l irn csc-t.r

(*-'i)j''tf))

npfil.,ion, and rheory49. You are sitting in a classroom next to the wall looking at the

blackboard at the front of the room The blackboad is 12 ft long

and starts 3 ft from the wall you are sitting next to. Show that

your viewilg angle is) x 1 r

d : c o t _ 1 5 c o r 5

if you are x ft from the front wall.

It 2 '

I3'-ir

l- x --------l

51.


50. The region between the curve ) : sec r.r and the r-axis from.r: I to r:2 (shown here) is rcvolved about the ty-axis togenerate a solid. Find the volume of the solid.

The slant height of the cone shown here is 3 m. How large shouldthe indicated angle be to maximize the cone's volume?

mat angle heregives the laryest

volum€?

52. Find the angle d.

<a Here is an informal proof that tan-r I + tari t2+tan-t3:n.

Explain what is going on.

Twc derivations of the identi ty sec t( x) : :z sec 1x.

a) (Geometric) Hete is a pictorial proof that sec-l (-r) : 7r -sec-l l . See i f you can rel l whal is going on.

b) (Algebruic) Deive the identity sec-r (-r) = / - sec I _r bycombining the following two equations from the text:

c o s l 1 - r ; : 7 t - c o s - l t E q . ( 6 )

S e c I - . c o s { l / x ) I q . / l \

55. The identitysin 1 x + cos 1 x : irl2. Figurc 6.19 establishes theidentity for 0 < r < 1. To establish it for the rest of [-1, 1], ver-ify by direct calculation that it holds for r = 1, 0, and -1. Then,for values of .x in (-1, 0), let.x : -a, a > 0, and apply Eqs. (4)and (6) to the sum sin-'(-a) + cos r(-o).

56. Show that the sum tan I .I + tan-l(l/:r) is constant.

Which of the expressions in Exercises 57-60 are defined, and whichare nor? Give reasons for your answers.

57. a\ tan-12

58. a ) csc- r ]' 2

59. a) sec-r0

ro. "r "or ' /-1' \\ 2 /

ffi Calculator Explorations61. Find the values of

a) sec-r 1.5 b)

62. Find the values of

a) sec '(-3) b)

b)

b )

b)

b)

b)

b)

b)

cos 12

csc -12

s rn v l

cos-r1-51

] : tan (tan-r r)

y = sln lsrn x)

y : cos (cos-l . ;r)

csc-'(-1.5) c) cot ' 2

csc-l 1.7 c) cor-r(-2)

lZ Grapher ExplorationsIn Exercises 63-65, find the domain and range of each compositefunction. Then graph the composites on separate screens. Do thegraphs make sense in each case? Give rcasons for vour answen.Cornnent on any differences you see.

63. a) y : 1a1-t11a1r;

64, a) y: sin-i(sin.r)

65. a) y: cos-t(cos.r)

54.

6raph 1 : sec (sec-rx) :5s6 (cos t(1/j)) Explain what you

see.

Newton's serpentine. Graph Newton's serp entine, ! : 4:t lG2 +

1). Then graph J :2sin (2tan-rr) in the same graphing win-

dow, What do you see? ExPlain.

6.9 Derivatives of lnverse Trigonometric Functions; Integrals 513

68, Graph the rational function ) : (2- x2)1x2. Then graph 1:cos 12 sec I r) in the same graphing window. what do you see?

Explain.

Table 6,5 Derivatives of the inversetri gonometric f unctions

r1(sin-r a) du ldx1 . . - : E - _ . \ u t < l

a x \ / l - u .

d ( cos l a ) du / dxZ . _ , - : _ - = _ _ = . l u t < |

a x \ / t - u '

^ d(tari | ) duldx

d x l + u '

dkot-t u) duldr.4. - . ': -.---:---;

t1x ). + u'

d ( sec La ) du / t l t5 . , -

- - ; - - : . l u l > |ax l u l \ / u . - |

d ( csc - r a ) - du /dx6. . : - - ;---, lul> t

.tx lul,,/u. - |

Derivatives of Inverse Trigonometric Functions;lntegralsInverse trigonometric functions provide antiderivatives for a variety of functions

that arise in mathematics, engineering, an<l physics ln this section we find the

derivatives of the inverse trigonometric functions (Table 6 5) and discuss related

integrals.

EXAMPLE 1

1 1 1x + 2 2 J x + | 2 J x I l ( x + 2 \

) 1 . 1c; f sec- l r - 3xt : -_- -=-- . ' a t - 3 ' r )

d x l - 3 x , V { - J x t 2 - I d x

1

- ^ , t-a

------l

E K A M P L E 2 ' = , o n r r ' ' t " : : l i - 'r l , l a n t r f n / 4 l + r '

1 | = a x : I e ' d u r r ( o ) : o a ( 1 ) = ; r / ' 1J o t - x , J o

: r , l : e " t a - l- l n O

We derive Formulas I and 5 from Table 6.5. The derivation of Formula 3 is similar'

Formulas 2, 4, and 6 can be derived from Formulas 1, 3, and 5 by differentiating

appropriate identities (Exercises 81-83).

The Derivative of Y sin tu

We know that the function r = sin y is differentiable in the inte al -xt /2 < y <

T /2 and that its derivative, the cosine, is positive there Theorem 1 in Section 6'l

therefore assures us that the inverse function y : sin-l x is differentiable throughout

D


.r : sin) \

the interval -1 <.r < 1. We cannot expect it to be differentiable at.x : l orx: -1

because the tangents to the graph are vertical at these points (see Fig. 6.27).We find the derivative of y = si11-r .x as follows:

) : s i n - r ; r < + s i n ) : r

Derivative of both sides with

Chain Rule

We can divide because cos ) > 0t o r - n / 2 < y < r / 2 .

Fig. 6.28

S u l ) : 1s, ( $ n Y ) = r

axdy

c o s y - ; - : rax

d v 1dx cos )

Doma in : l < r< lRw\ge . - d2<J<n /2

6.27 The graph oI y = sin 1 x has verticalt a n g e n t s a t x : - 1 a n d x = 1 .

6.28 ln the reference right triangleabove,

. xS l n Y : - : X .' 1

."1 -Fcosy = - - = \ /1 xz .

6.29 In both quadrants, sec y = x. In thefirst quadrant,

tany=uF- t t ' t=^ /F=.

In the second quadrant,

tany - Jr, - tt{-t|' = -vFl.

The derivative of y: si1-r "

with respect to r is

d 1' ( s i n ' r )

d x ' / l - x '

If & is a differentiable function of .r with lrl < 1, we apply the Chain Ruledy dy dudx du dx

to y : sin-l z to obtain

d . _ , I d ud x ( s r n

t { , : f f i d x , l r l < r .

The Derivativ€ of y= 5s6-1uWe find the derivative of y : sss-r r, lxl > 1, in a similar way.

s e c y : ,

d, (seclJ : r

ax

. d ySec lu ln ) ; - : . lax

1

r/1 - x2

dy

f - = s e c l r + s e c ] = xDerivative of both sideswrth respect to .r

Chair Rule

Since lx > l , ] l ies in(0,ft /2) U Qt /2 , n) andsec l, tan ) I 0.

Fi9.6.29

dx sec / tan y

I

1T-'-

- I " E

T2

6.9 Derivatives of Inverse Trigonometric Functions; lntegrals 515

What do we do about the sign? A glance at Fig. 6.30 shows that for lr > I theslope of the graph of y : sec r,{ is always positive. Therefore,

l l| - l l t > l

d I x J x 2 - 1, (sec x) : | ( l )

d \ | I

I tt l tt - I

With absolute values, we can wdte"Eq. (1) as a single formula:

d 1

/ n r s e c ' r ) =

k ! G r . ' r l ' l '

If z is a differentiable function of r with lal > 1, we can then apply the ChainRule to obtain

d l d u--{sec ,r - ___1-- - . lrrl ' Lax u \ . / uz - l ax

Integration FormulasThe derivative formulas in Table 6.5 yield three useful integration formulas inTable 6.6.

Table 6.6 Integrals evaluated with inverse trigonometric functions

The lollowing formulas hold for any constant a + 0.

1 . [ - : ! : : s i n ' ( 1 ) + cJ J a ) _ u 2 \ a l

(Valid for a' < a'z) (2)

f du l , r . r '2 . 1 - = - t a n ' l - ) - C ( V a l i d f o r a l l a )

J A ' + U ' A ' A l

J . f - L= -1 . . . ' 11 - c Na t id to r u -a ' )J u \ / u ' z a t a t a

(3)

(4)

6,30 The slope of the curve y = sec-1x ispos i t i ve fo r bo thx< I andx> 1 .

The derivative formulas in Table 6.5 have d : l, but in most integrations a + l, andthe fomulas in Table 6.6 are more useful. They are readily verified by differentiatingthe functions on the right-hand sides.

EXAMPLE 3r J l 2 ) " 1 v ' 2

J J 2 2 \ / t - x J r t '

/ -n\ / ,5\='i"'(#)-'i"'(Yl=+-\ - , / \ - , /

t t ) -

I 2 - ,un 1 r , t l - tan ( j ) - ran '10 ; - f -J o l + r , l n 4

I'f dx fJt n n n

J : , , r r , / r 2 - l 1 , , , r 4 6 l 2

tf lf

4 1 2

J'

" - 4b)

c) a


For more about completing the square, seethe end papers of this book.

EXAMPLE 4

" I --!t, - r: I r#-,: 'i" ' (l) + c

h) [ d ' : ! l d '- I t5-4F 2J JA|-A

1 , t u \:7s in ' (_ ) +c

1 , / 2x \: _s i n , I o ]+CZ \ V J , , /

EXAMPLE 5 r.ratuurc [-L.J ,/4x - x2

E q . ( 2 ) w i t h 1 7 : 3 , ! : r

d : v J . r / = a [ .and du/2: dt

Eq. (2)

Solution 'fhe expression I U -P does not match any of the formulas in Table6.6, so we first rewrite 4.x - x2 by completing the square:

4x - x2 : -1xz - 4x) -- -(x2 - 4x + 4) -f 4 : 4 - (x - 2)2.

Then we substitute a = 2, u : x - 2, ard du : dx to set

I d ' : I d*I J4x:v I J4- tx -u,

:t+=J \ , / a ' - u '

: ,in-' /1') + c\ a , l

/ - _ 1 \: s in- ' ( : " - : ) ' C

\ 2 . /

.r = 2. x: r 2. .rnd

Eq. (2)

EXAMPLE 5I d x 1 ' / t = \ * a E q . ( 3 ) w i t h a :

" 1 0 . , , : ,u,

,/ ro + ,, :

.,,aT ,un 'l..rr,0,,

b) t=4=- : t=

[ =ou - " : r t . , : , f 3 t , and t i u / J t=a rJ t +3x . \ / 3 J 4 ' +u t

: i ' ) '^ '( i)* ' Eq (3)

| | - ' / / : ' i: 8'rt ' \7/": -1,* ' (+\ *.-

Ja'- ' \r i /- '

EXAMPLET Evaluare [, = or

J 4 x 2 + $ ( + 2 -

6.9 Derivatives of Inverse Trigonometric Functions; Integrals 5'17

Solution We complete the square on the binomial lf a 4x:

4 x z 1 4 x | 2 : 4 \ x 2 + ^ \ - , = 4 ( l + - - ) ) + z - ]

: +( , +1) ' * , : (2 , r + 1) '?+ 1.\ z /

Then we substitute a : 1, u : Zx + 1, ar'd du/2 : dx to get

f tlx f dx | | du----------------- -; I

J 4 x 2 + 4 ^ 1 2 / ( 2 x + l ) 2 + l 2 J u ' / r c r '

1 l / u \: ' - t a n - ' l - l

2 a \ a , l

1: . t a n ' ( 2 , r * 1 ) t C

2

x : 2 - r + l . a n d

Eq (l)

EXAMPLE 8

Solution

I dxEvaluate , "::.

J x J4x2 - 5

duf dx | 1, - : , - - , 'I x rExr -5 I Y , t , -a ,

2 '

f d u: t -

I u"Qr=dI - ' l r l: s e c _ , 1 _ l _ Ca l d l

f d tEva lua te f - ; : .

J J e 2 ^ - 6

f dx f du /u, : : l :

J J e L _ 6 J J u 2 _ q z

I d ": l -

J uJu2 - a2

I - , t u l- - s e c - ' l l - Ca l a l

:1,.. /4) *.\/5 \ . '/5,/

The 2 s cancel .

Eq. (4)

tr

t t \ : t t u l e ' : d u / u .

"= rc

EXAMPLE 9

Solution

I *. /1) *.J6 \ J6,/ tr

lq . ( lJ

518 Chapter 6: Transcendental Fundions

Exercises 6.9

Finding DerivativesIn Exercises l-22, find the derivative of ), with rcspect to the appro-priate variable.

1 ' 1 : cos 1( r2 )

3 ' y :s in 1 . , '4 t

! . y : s e c ' ( l r + r r

7 . ) = c s c - 1 ( i r ' ? + 1 ) , r > 0

3 ' Y : c s c ' J

I9 . y : s e c - ' : , 0 < t < 1

11. y : co t 114

13. y= ln ( tan- r , r )

15. ;y: q5q tirt.)

17 . y -s " , . /1 -F+cos rs

18. y: ^"[: - 1 .""-t "

19 , y :131 \ .1G71|csc I x ,

. 12 0 . ) : c o t ' - - t a n ' r

2 1 . y - a s i 1 t , : - J 1 x 2- / r \

2 2 . y : l n l x ' / + a ) - x t a n ' ( , )

Evaluating IntegralsEvaluate the integals in Exercises 23-46.

f ) r t i t

B. I -:: 24. I :::J J g _ " 2 J J t 4 x 2 "' I##

,8. [ -: -::J X ! ) X ' - +

zz. [ ---9-I xJ2re2 - 2

P"A/a dst -Jo IT-TF

J,;",

n. l" '"n ,**!!L on' J,4s. [4+

J ' / t - Y '46. [:!J+

J \/l - tanz y

2. :l, : cos t(1/r)

4 . y : s in r (1 - r )

6 ' ) : s e c ' 5 s

?1 0 . y : 5 i 1 r l

1 2 . y : c o t t J t _ 1 ,

14. y = tan-r(ln .{)

16 . ) : sq5 11 . t . )

. r > l

Evaluate the intesrals in Exercises 47-56.

I d,J t r +Ztv \ , +Zf -ZS

f"/2 2cos0 d0

I "/" | + ("t"oY

f d r| - - -J ( tan r ty ) ( l + ) , )

/ '2 sec2 (sec lx)dxt -

Jn ,"F

47. [-::::: 48.I J - x 2 + 4 x 3

4s. [" -:!!: so.J - t J 3 - 2 t - t 2

s. [ . - j !_ s2 .J Y ' 2 Y + )

1 2 a ) -

s:. / -, _-a s4.J ] t ' - L X + Z

ss. [- '1r: s6.I @ + l ) J r z + 2 x

f d x

J J2-x - x2

l t 6 d tl - -

Jv2 J3 +4t - 4t2

f d yt-------------]j-

J v '+6v + to

[4 2dxt -

J t t 2 - 6 t f l 0

t _J r , , -z l , tq t - t t +z

f "[arr C a*

J 1 + * 2

Evaluate the inteerals in Exercises 57-64.

n . ["''csc2 x dx

I + G."tr),

4d t

r0+?;

*"- t "

2tan-t 3x2

ltt

s7. l i= s8.

f { s i n - l i } 2 d t59. I --=_-- 6U.

I J l - x 2

29.

l d r, :--:--;---

[ \ 4ds

f' dt

lo 81 2 t t

f -"4/2

d\I L_

J r y J 4 y 2 - l

| 3 d r

I tT-4i=IFI dtt

Jz*r,-rrId ,

J ex - tt.[2, ttz - q

l-"4/3 d.t34, '

J ztz yJ9t2 - 1

36. [-=:!::l J4- ( r r1 ) '

LimitsFind the limits in Exercises 65-68.

65. lim

Jar,/. llm 't tan -

x

66. lim

68. lim

Integration FormulasVerify the integration formulas in Exercises 69-72.

t t ^ n t r I r t ) _

g n - , . l C6 9 .

J T d r : t n x ; l n ( l + J . r

6r. [ ,---d1-J (S tn ) l v r ) -

f 2 cos (sec - l . r ) / x64. I _

J2 tJ i x \ / x ' - I

39.

38. /

---E-

81.

82,1','

I 14 -5 | .r.o dx

;l xrco' I 5r dx -

'o- cos 15x +;

J ,ffiF

f r r in ' * t 'd" : . r (s in x)2 2x - 2 / l - rz s in- ' x * C

I h lo ' + , ' ) d t : x ln \a2 + xz) - 2x - 2a ran- ' I -c

Exercises 6.9 519

Use the identity, n

c o s ' r : - - s l n - t

z

to dedve the formula for the derivative of cos-La in Table 6.5

ftom the formula for the derivative of sin-r l.

Use the identity

c o l / r : - _ t a n , .

to derive the formula for the dedvative of cot-la in Table 6.5

from the formula for the derivative of tan-r ,.

83. Use the identity

,ec-' ltCSC 1'. : t

S

to derive the formula for the derivative of csc la in Table 6.5

from the formula for the derivative of sec-l lr.

84, Derive the fomula

dy :

dt I +-x2

for the derivative of y : tan-r.r by differentiating both sides ol

the equivalent equation tan ) : it.

85. Use the Derivative Rule in Section 61, Theorem 1, to derive

d , l. . s i n

' x - - - 5 ^ . - l < x < la x \ / t _ x "

86. Use the Derivative Rule in Section 6 1, Theorem 1, to dedve

d r l

d x ' " ^

I 1 1 2 '

87. What is special about the tunctions

Y I

f ( x ) : s i n t " - - 1 , x : 0 . a n d g i x ) : 2 r a o ' t J i i" . x + l

Explain.

88. What is special about the tunctiors

. 1 . 1/ t . l t : , i n - r - - a n d B { r } : t a n

| - ?'

V r * l

Explain.

89. Find the volume of the solid of rcvolution shown here

€3

-'...- - ll

lnitial Value ProblemsSolve the initial value problems in Exercises 73-76.

d ' r l -7 3 . - : _ , y ( u ) : u

d x J l - x ld '

: - l - v { o ) : Id x x 2 * 1

d y l- - = - - - - -== , x> t : Y \ z t= f t4x . r v l ' - I

d t 1 216 .+ : - , - - ^ . Y t0 r -2

a x r + x ' \ / t - x .

Theory and Examples77. (Continuatiotl of Etercise 49, Section 6.8.) You want to position

your chair along the wall to maximize your viewing angle a.

How far from the ftont of the room should you sit?

78, Wlat value of x maximizes the angle 6 shown here? How largeis I at that point? Begin by showing that 6:?I cot-rxcot-'(2 - . ').

Can the integrations in (a) and (b) both be correct? Explain

f d xa ) | - : s r n ' r + L

J . / t - x '

b t [ - i L : = - [ L - - c o s i . t + cJ J l - x z J , / l - x '

Can the integrations in (a) and (b) both be corfect? Explain.

t d x f d x, Jf f i --J-h- 'cos' Irc

80.

f d r t d ub f I - - - : I / - -

J \ / l - x ' J \ / t - \ - u f

[ -du

J \ / r - u '

= cos-r(-x) + C

, ' - jI


90. Find the length of the curve ! : .,4 -7, -ll2 = x = ll2.

Volumes by SlicingFind the volumes of the solids in Exercises 97 and 92.

91. The solid lies between planes perpendicular to the r-axis at r :-1 and x : l. The cross sections perpendicular to the.{-axis are

a) circles whose diameters stretch from the curve @ Sq,y- | lJl +7 to the curve y:l/Nry+}'

b) veftical squares whose base edges run from the curvey: -l1uE+7 to the curve y - U"tr1 7.

92. The solid lies between planes perpendicular to the i-a-\is at AZ gS.x : - \/ 2/ 2 and | - \/ l/2. I he cros\ \ectron\ are

a) circles whose diameterc stetch from the r-axis to the curvey=2/48 7. t lss.

b) squares whose diagonals stretch from the i-axis to the curvej : z / \ / r - x ' '

Calculator and Grapher ExplorationsF gl. cnlcuuroR Use numerical intesration to estimate the

value of

1 0 6 ) .s in 10 .6 : I : : - .

Jo t/7 - x2

For reference, sin r0.6:0.64350 to 5 places.

CALCULATOR Use numerical integration to estimate thevalue of

. f ' In:+ l -:--:----a).

J i I + x '

GRAPHER Graph "f(-r): sin-rr together with its first twoderivatives. Comment on the behavior of / and the shape ofits graph in relation to the signs and values ot f' and, f".GRAPHER Graph f (t):tan-tx together with its f irst twoderivatives. Comrnent on the behavior of ./ and the shape oiits graph in relation to the signs and values of f' and f".

Hyperbolic FunctionsEvery function / that is defined on an interval centered at the odgin can be writtenin a unique way as the sum of one even function and one odd function. Thedecomoosition is

f (x) + f (.-x) f (x) - f (-x)

even pafi odd part

If we write e' this way, we get

#;;The even and odd parts of er, catt"d tfr" nyp"rboti" cosine and hyperbolic sine ofr, respectively, are useful in their own right. They descdbe the motions of wavesin elastic solids, the shapes of hanging electric power lines, and the temperaturedistributions in metal cooling fins. The center line of the Gateway Arch to the Westin St. Louis is a weighted hyperbolic cosine curve.

Definitions and ldentitiesThe hyperbolic cosine and hyperbolic sine functions are defined by the first twoequations in Table 6.7. The table also lists the definitions of the hyperbolic tangent,cotangent, secant, and cosecant. As we will see, the hyperbolic functions bear anumber of similarities to the trigonometric functions after which they are named.(See Exercise 86 as well.)

The notation cosh j is often read "kosh .r;'rh) ming wirh either "go\h r" or "gauche r-"and sinh x is pronounced as if spelled "cinchr" or "shine .x."

Hyperbolic cosine of .ri

Hyperbolic sine of x:

Hyperbolic tangent:

Hyperbolic cotangent:

Hyperbolic secant:

Hyperbolic cosecant:

cosh r :2

sinh x = -

s l n n - r e ' - e "

cosh j r e ' l e '

c o s h j r e ' + e '

s i n h i r e t - e '

l 2

c o s h r e ' l e '

t 2sinh r e' e-t

Table 6.7 The six basic hyperbolic functions (see Fi9. 6.31 for graphs.)

Table 6.9functions

Derivatives of hyperbolic

6.10 Hyperbolic Functions 521

Table 6.8 ldentities for hyperbolicfunctions

l

sinh 2r :2 sinh .x cosh t

cosh 2-r: cosh2r * sinh2.r

sinh2 r -

cosh 2,v * 1

2

cosh 2,r - I

cosh2r - sinh2,r: 1

t a n h 2 r : l - s e c h 2 . r

co th l r : 1 *csch2 . t

ldentit iesHyperbolic functions satisfy the identities in Table 6.8. Except for differences in

sign, these are identities we already know for trigonometric functions.

Derivatives and lntegralsThe six hyperbolic functions, being rational combinations of the differentiable func-

tions e' and e ', have derivatives at every point at which they are defined (Table

6.9). Again, there are similarities with trigonometric functions. The derivative for-

mulas in Table 6.9 lead to the integral formulas in Table 6.10.

EXAMPLE 1

d . . - d u

t (S ln l l l ' l ) : 6osn I

J l

d du-(cosn , l - sinn a

' ;

d . , d u

lJ xll.anfr u) - sech',

t

d , du

1x(coth ,/) - -csch'x

1' l

d du

t ( sech & l : 5gcn u tmn ,

d r r

d d u

t ( cscn t l ) - - cscn , l co tn d

, ;

Table 5.10 Integral formulas fo"hyperbolic functions

I sinh udu: cosh l * C

I cosn utlu: sinh rl + C

Jsecn2udu: tanhr ' r+C

I cscn2rdu: -coth r * C

lsechutanhudu- -sech ri + C

l.scnucothudu: -csch , + c

EXAMPLE 2

t _

f r ( , rn t / t + r ' z ) : sech 'z / t + t : . ; t (J t - t ' z l

. . " u 2 - / t L , 2- ,nTP

f f cosh 5xI coth 5x dx : I -dx =

J J Srnn ):rt l

: i to , l+c = -f f ,

EXAMPLE 3

[ ' , i ,n , ,d , - / ' 'o 'n l ' r , , r , ,Jo J '.t

I f I l s inh 2x l l- i /o

, .o ' t 2x - t \d^ : 21 2 - ,1"

- sinh 2 - ! x o.4o6t2

.l

l5 x +C

I l d ut -

5 J u

ln ]sinh


Evaluating hyperbolic f unctions

Like many standard functions, hyperbolicfunctions and their inverses are easilyevaluated with calculators, which have specialkeys or keystoke sequences for that purpose.

(a.) The hyperbolic sine and itscomponent exponentials.

5,31 The graphs of the six hyperbolfunctions.

EXAMPLE 4f l ' z f l n - r n z

/ 4 c ' s i n h x d x : I a e " - - - ) - d x - I r 2 e 2 ' - 2 t d xJo Jn t Jo

:7e2 ' - 2x1) [2 : (ezt 2 -2In2) ( l -o)

: 4 2 l n 2 I

x t .6 I3J

The Inverse Hyperbolic FunctionsWe use the inverses of the six basic hyperbolic functions in integration. Sinced(sinh x)/dx: cosh ;r > 0, the hyperbolic sine is an increasing function ofx. Wedenote its inverse by

: Y : s i n h r ; r .

For every value of ';r in the interval -crc < r < oo, the value of y : sinh r:r isthe number whose hyperbolic sine is x. The graphs of y : sinh r and ) : sinh I rare shown in Fig. 6.32(a).

tr

(b) The hyperbolic cosine and itscomponent exponentials,

(c) The graphs of] = tanh r and) = coth -! = l/tanh r-

(d) The sraphs of} : cosh -r and

) = sechr - l /coshr.(e) The graphs of ) = sinh jr and) = csch-r = l /s inh.L

6.32 The graphs of the inverse hyperbolicsine, cosine, and secant of x. Notice thesymmetries about the l ine Y = x.

6.33 The graphs of the inverse hyperbolictangent, cotangent, and cosecant of x.

6 .10 Hyperbo l i c Func t ions 523

The function y : cosh r is not one-to-one, as we can see from the graph in

Fig. 6.31 But the restricted function y : cosh r, x - 0, is one-to-one and therefore

has an inverse, denoted by

Y : cosh I 'r '

For every value of l2l, ) = cosh r;r is the number in the inteNal 0 < l ' oo

whose hyperbolic cosine is,r. The graphs of y : qsthr, x : 0, and ) : cosh rr

are shown in Fig. 6.32(b).Like y : cosh x, the function y : sech r : 1/cosh I fails to be one-to-one'

but its restdction to nonnegative values of; does have an inverse, denoted by

J : sech-lx'

For every value of x in the inteNal (0, 11, y : sech-l'r is the nonnegative number

whose hyperbolic secant is r. The graphs of y :5e66,r, .x : 0, and y : sech rr

are shown in Fig. 6.32(c).The hyperbolic tangent, cotangent, and cosecant are one-to-one on their do-

mains and therefore have inverses, denoted by

, : t a n h r x ,

) : c o t h I ' r , ) = c s c h - l - r '

These functions are graphed in Fig. 6.33.

- 6 - 5 - 4 3 - 2 - 1 2 3 4 5 6

i.r : sech l , l > o)

] : s e c h i r , i r > 0

) = coth- l r


Table 6.11 ldentities for inversehyperbolic functions

. , lsecn ,r : cosn -.X

. . _ , Ic s c n r : s r n h -

Tl

cotn r =tann -x

Table 6.12 Derivatives of inversehyperbolic functions

Useful ldentitiesWe use the identities in Table 6.1I to catculate the values of sech-rx, csch-r,r, andcoth-lr on calculators that give only cosh-l.r, sinh-1.r, and tanh-lx.

Derivatives and lntegralsThe chief use of inverse hyperbolic functions lies in integmtions that reverse thederivative formulas in Table 6.12.

The restrictions lul < 1 and lr.rl i 1 on the derivative formulas for tanh-la andcoth ra come from the natural restrictions on the values of these functions. (SeeFigs. 6.33a and b.) The distinction between lal < 1 and lzl > 1 becomes importanrwhen we convert the derivative formulas into integral formulas. If lal < 1, theintegral of 110 - u2) is tanh-l z -l- C. If lal > 1, the integral is coth I u * C.

EXAMPLE 5 Show that if u is a differentiable function of x whose values aregreater than 1, then

d l d ud _ ( c o s h _ , r ) : G = i .

so/ution First we find the derivative of y : cosh-rr for.r > l:

y = cosn -,r

' I : c o s h )

dyl = s t n h ) ,

axd v l

Equivalent equatior

Differentialion with respecr ro .r

S i n c e . r > l . ] > 0 a n d s i n h ) > 0dx sinh y

1

Ji'-

In stron, {tcosh r-t,) : }. the Chain Rule gives the final result:d x J x 2 - I

d. (cosh- 'a ) =

ax

t-:!L:= t--!!:J \ /3 + 4x, J Jaz +u2

: s i * ' (1 )+c

:,r"r,'(ft)+c.

1

",/ u'/ - 1

duE

With appropriate substitutions, the derivative formulasthe integration formulas in Table 6.13.

t t , d xEXAMPLE 6 Er aluate I +:.Jo J3 + 4x2

Solution T\e indefinite integral is

trin Table 6.i2 lead to

u :2x . du : 2dx , a : . . / 5

Formula from Table 6.13

d(sinh-1a)

,1, :

du

.,4 +iF a,d(cosh ' l ) | du , \ ]

dx ,IF=dx' '

dftanh-tu) | du- = _ . , l r i | < l

a x | - u . r l r

d(coth ra) 7 du

d , :

t - r 2 * ' u t t

1(sech-rr) -du /dxdr u,, /7 - u2

/ t c s c h l r t - d u l d r- = ---------: . t+U

dx u ) t / l t u2

Exercises 6.'10 525

Table 5,13 lntegrals leading to inverse hyperbolic functions

Therefore,

1,,2dx 12, \\ J3, /

r3\\ . , /3/

]. :,'*' (fi) -,i"r'-'ror

- 0 - 0.98665.tJ

,,

I o ' : s i n r r ' ( 1 )+c , a>o

I J a 2 + u 2 \ a l

f n, ,

l - ; = : " o t t t ( 1 ) + c ' u > a > oJ \ / u ' - a '

[ ] , u n r ' ' ( ' \ r c i t u z . a ' zI

o , : l o

\ a ' ,

J a z - 4 2 l t . t / u \- l - - * ' ( ; ) . c

i r u t > a 1

t ! ' : -1 . " "h- ' (1 )+c, r - , . , .oJ u J a 2 - P a \ a l

t -L : - r " , "h

' , ! l -c . u7oJ u \ / a ! + u t a t d l

Exercises 6.10

Hyperbolic Function Values and ldentitiesEach of Exercises 1-4 gives a value of sinh ,r or cosh "t. Use thedefinitions and the identity cosh2x - sinh2r : I to find the valuesof the remaining five hyperbolic functions.

to show that

a) sinh 2"r :2 sinh r cosh:v;b) cosh 2.r: cosh2r+ sinh2t.

12. Use the definitions of cosh r and sinh .r to show that

cosh2 r - s inh2x :1 .

DerivativesIn Exercises 13-24, fir]d the derivative of ) with respect to the ap-

Fopriate variable.

13. ) :6 s inl t

tn.

31. sin}l t : - ;

1 73 . c o s h . r :

a r , r t O

42 . s i n h t : t

4. cosh* : f , , ,0

Rewrite the expressions in Exercises 5-10 in tems of exponentialsand simplify the results as much as you can.

5. 2 cosh (ln ,r)

7, cosh 5* f sinh 5.t

9. (sinh .r * cosh -r)a

10. ln (cosh** sinh r) + ln (cosh r-sinh.I)

11. Use the identities

sinh (r + y) : sinh :r cosh ) + cosh i( sinh )

cosh (t + )) : cosh .' cosh ) + sinh .r sinh y

y :2'/7 tanh Jt 16.

) : ln (s inh . , t t .

) : sech 0(1 - ln sech d) 20.l ^

) = l n c o s h D - ; t a n h ' u 2 2 '

6. sinl (2 ln r)

8. cosh 3.r - sinh 3r

I) : s inh (2x + l )

. lj : t ' t ^ f r ,

): ln (cosh z)

y - csch 9(l - ln csch d)I . ,y : l n s l n n u - t c o t n - u

15.

17.

19.

21.


23, y : (x2 + 1) sech (ln r)(iiirtr Beforc differentiating, express in terms of exponentialsand simplify.)

24. y = (4x2 - l) csch (ln2x)

In Exercises 25-36, find the derivarive of y with respect to the ap-prcpriate vadable.

25. y : sinh-1!E

27. y=(1-e) le l t$ - t0

29, y : (l - t) coth l.u/t

31 . y :995- ra-a s " "1 t t t

/ 1 \ o3 3 . v : c s c h r l l l'

\ 2 )

J5. Y= slnn (tan.x)

53, J_t 4

2ee cosh,e de s+. lo"'

+"-u "i* o ae

ss. l"/o o*"n

GaL e) sea2 e de

se. lo"'t

z ,i*(sin d) cos I do

sr. Ir' "o"hi(\ ' dr s8. I,^'"*n o"

ss. l_',^,*"r,(i)a* eo. lo"'o t.'"rr'(|)a,

Evaluating Inverse Hyperbolic Functions andRelated IntegralsWhen hyperbolic function keys are not available on a calculator, it isstill possible to evaluate the invene hyperbolic functions by expressingthem as logarithms as shown in the table below.

s inh- ' r :h ( r+^v ,6r r+ l ) , -oo <r < oo

c o s h ' : r : l n ( : t * J x , - l ) , x > _ t

I t _ L v

r anh r . x : i t " f ; , f = r

s e c h - r : l n I r - i v I - r '

) , 0 . r . t\ r // . t . - - \

c s c h - r x = t n [ 1 * t ' , 1 t '

] , * + o\ x l r | /

1 Y - ! l

c o r n . t : r u 1 , . l a l > t

26 . y=c$I r -12^ lE+ l

28 . y : (02 t20) tanh ' (d + l )

30 . y=(1 - t2 ) co th- r t

32. y:)nx + Jl - 12 sech 1x

34. y: sss\-t2o

36. y : q6s1r - t1 . . . x ) , 0<x<t r /2

Integration FormulasVerify the integration formulas in Exercises 37-40.

ll. a) .l

sech x dr : tan t(sinh ,t) + C

Ib)

J sech x d : t =s in ' ( tanhr ) *C

f , 2 r -3 8 .

; l x s e c h - ' x d x : 1 s e c h ' , -

1 J t . t z - C

| . 2 - 139.

.l x coth-'x dr:

? "o* '* -

.rt - a

f l40. ; f tanh rxd , r -x tanh r . r t

r ln r l -x '? r -C

Indefinite IntegralsEvaluate the integrals in Exercises 4l-50.

ar. I

si* zx dx

a3. /o

cosh (|-nz)a,

es. I tannla,

a7. l sech , ( - ) ) *

4e. lsech'tryJ7-!

Definite Integrals

ez. I sinnla,

u. I

t cosn (3t - tn2) dx

et. | *tn ftae4t.

I cscnz6-iat

-^ fcsch (ln /) corh (ln r)dr

"u' -/ ,

sz. lo'^' o* z, a,

61. sinh-'(-5l12)

63. tanh-'(- 1/2)

65. sech-'(3/5)

Use the formulas in the table here to express the numbers in Exercises6l-66 in terms of natural loearithms.

62. cosh-'(5/3)

64. coth-t(5 /4)

66. csch-' (-1l\43)

Evaluate the inte$als in Exercises 67-74 in terms of (a) inversehyperbolic functions, (b) natural logarirhms.

f ! r . , 1 -

67, IJo \/4 + xt

69. I : "^ -

t3/ t3 ) -7t, I ----:::-

Jys t/ l - I6x2

f t l t 6 , 1 .68. , ---:::-

Jo ./l + 9x2

,0. [ " ' .o ' ,J o t - x -

72. ['-:+J t x \ / 4 + x .

Evaluate the integrals in E).ercises 5l-60.

sr. l'".

com r ax

t"" cos:r r1:r7 l | _" 'J

"/ i+;-

',n L"

I t x \ - ! ( - x t f t x t - f t - ^ l2 ' 2

Applications and Theory75. a) Show that if a function I is delined on an interval symmetric

about the origin (so that / is defined at "r whenever it isdefined at :r). then

Exercises 6.10 527

-r'-axis attached by a rod of unit length to a point P representingthe cab at the origin. As the point P moves up the y-axis, it dragsM along behind it. The curve traced by M, called a lractlt,r lromthe Latin word /,,.1.1!n for "drag," can be shown to be the graphof the function ] = /(.jr) that solves the initial value problem

.l\ I _lDifferential equation: :1 - |' d r x J l - x z J l x 2In i t ia l cond i t ion : )=0 when r :1 .

Solve the initial value problem to find an equation fbr the curve.(You need an inverse hyperbolic function.)

Show that the area of the region in the first quadrant enclosed bythe curve I - (1/d) cosh 1?.r, the coordinate axes, and the line:t - b is the same as the area of a rectangle of height l/d andlength s, where r is the length of the curve tiom r = 0 to r = D.

A region in the first quadrant is bounded above by thc curve

) : cosh ,I, below by the curve I: sinh r! and on the left andright by the _r,-axis and the line r :2, respectively. Find thevolume of the solid generated by revolving the region about thex axis.

dt

rJ l + ( ln r ) '

( 1 )

Then show that (/(r) + /( n))/2 is even and that(f(x) J( -x))/2 is odd.

b) Equation (1) simplifies considerably if / itseli is (i) evenor (ii) odd. What are the new equations? Give reasons foryour answers.

7 6 . l ) e d v e t h e l o r m u l a s i n h r r - l n l r , J - 1 2 , l l - c , o - r < r -\ . /Explain in your derivation why the plus sign is used with thesquare root instead of the minus sign.

77. Skyd iving. If a body ot' mass l|I lalling from rest under the actionof gravity encounters an air resistance proportional to the squareof the velocity, then the body's velocity I seconds into the fallsatisfies the differential equation

dL)m7 -ng ku2 '

a t

where ft is a constant that depends on the body's aerodynamicproperties and the density of the air (We assume that the fall isshort enough so that the vaiation in the air's density will notaflect the outcome.)

a) Show that

Es ,e r \u - . ' r a n h \ / - / l

V f t \ ^ /

satislies the differential equation and the initial conditiont h a t u = 0 w h e n t : 0 .

_ b) Find (he bod)'s / ir i l lr8 uelocinl l im,-o u.H c ) CALCULATOR Fora 160- lbsk)d iver (n8 lo0 , .w i rh r ime

in seconds and distance in feet, a typical value for,t is 0.005.What is the diver's limiting velocity?

A(elerations whose magnitudes are proportional to dis-pla.ement. Suppose that the position of a body moving alonga coordinate line at time / is

a) s : a cos,tt 16 sin t t .b ) s :a cosh t / +b s i nh , t t .

Show in both cases that the acceleration d2s fdt2 is proportionalto s but that in the first case it is directed towad the odgin whilein the second case it is directed away from the odgin.

Tractor trailes and the tractrix. When a tactor trailer turns 81.into a cross street or driveway, its rear wheels follow a curve likethe one shown here. (This is why the rear wheels sometimes rideup over the curb.) we can find an equation for the curve if wepicture the rear wheels as a mass M at the point (1, 0) on the

80.

78,

79,

528

82.

83.

Chapter 6: Transcendental Functions

The region enclosed by the curve y = sech r, the r-axis, andthe lines ir : tln.n6 is revolved about the x-axis to generate asolid. Find the volume of the solid.

a) Find the length ofthe segment ofthe curve )' : (1/2) cosh 2.II I O m - r : U t O . r : I n V l ,

b) Find the length ofthe segrnentof the curve;y - (l/a) cosh a:rf r o m : r : 0 t o : Y : b > 0 .

A minimal surface. Find th€ ar€a of the surface swept out byrevolving about the.r-axis the curve ) : 4 cosh (r/4), - ln 16 :; < I n 8 1 .

. l2

84.

I :4cosh(-r l4)

B(ln 81,6.67)

A ( - l n 16 ,5 )

____------J+.'I

6.34 Since cosh2 u - sinh2u = 1, the point (cosh u, sinh u)lies on the right-hand branch of the hyperbolax' - y' : 1 for every value of u (Exercise 86).

P(cosh , sinh,)is twice the area

l n 81

It can be shown that, ofall continuously differentiable cuNesjoining points A and B in the figure, the curve ) - 4 cosh (r/4)genemtes the surface of least area. Ii you made a rigid wire frameof the end-circles through A and B and dipped them in a soap-Iilm solution, the sudace spanning the circles would be the onegenerated by the curve.

85. a) Find the centoid of the curve l, : cosh .r, - ln 2 = r : ln 2.B l) CafCUfafOR Evaluate the coordinares ro 2 decimal places.

Then sketch the curve and plot the centoid to show its re-lation to the curve.

86. The hyperbolic in hyperbolic functions. In case you arc won-dering where the name hlperbolic comes from, here is the answer:Just as ,r : cos |l and ) : sin z are identified with points (.I, ))on the unit circl€, the functions r : cosh and ]: sini u areidentified with points (,r, )) on the right-hand branch of the unithyperbola, -r2 t'? : I (Fig. 6.34).

Another analogy between hyperbolic and circular functionsis that the variable & in the coordinates (cosh r, sinh &) for thepoints of the right hand branch of the hyperbola,{2 }2 : 1 istwice the area of the sector AOP pictured in Fig. 6.35. To seewhy this is so, carry out the following steps.

a) Show that the area A(,/) of sector AOP is given by thelbrmula

sectorAOP.

6.35 one of the analogies between hyperbolic andcircular functions is revealed by these two diagrams(Exercise 86).

)a) Differentiate both sides of the equation in (a) with respect

to ,/ to show that

b) Solve this last equation for A(4). What is the value of,4(0)?What is the value of the constant of integration C in yoursolution? With C determined, what does your solution sayabout the relationship of u to A(u)'l

Hanging Cables87. Imagine a cable, like a telephone line or TV cable, strung from

one support to another and hanging freely. The cable's weightper unit length is 1,r and the horizontal tension at its lowest pointis a vector of length 11. If we choose a coordinate system forthe plane of the cable in which the r-axis is horizontal, the forceof gravity is straight down, the positive )-axis points straight up.and the lowest point of the cable lies at the point ) : H /u onthe )-axis (Fig. 6.36), then it can be shown that the cable lies

P(cosh ,i, sini r)

P(cos Il, sin ,)

] , c o s h r

A i r r r : I c o s h r . i n h l - I ' C - t a ^ .z J t

6.36' In a coordinate system chosen to match H and u/ inthe manner shown, a hanging cable l ies along thehyperbolic cosine y = (HIw) cosh (wxlH).

along the graph of the hyperbolic cosine

H , u) : - c o s n ; ] :

u n

Such a cufle is sometimes called a chain curve or a catenary,the latter deriving from the Latin catena, meaning "chaillr|'

a) Let P(r, )) denote an arbitrary point on the cable. Figure6.37 displays the tension at P as a vector of length (mag-nitude) T, as well as the tension 11 at the lowest point A.Show that the cable's slope at P is

6.11 Hyperbolic Functions 529

6.37 As discussed in Exercise 87 , T = W in thiscoordinate system.

89. The sag and horizontal tension in a cable. The ends of a cable32 ft long and weighing 2 lb/ft arc fastened at the same level roposts 30 ft apart.

a) Model the cable with the equationI

y - c o s h d r , - 1 5 S x 1 1 5 .a

Use information from Exercise 88 to show that d satisfiesthe equation

16a : sinh 15a. (2)a 1 , .ZI b) GRAPHER Solre Eq. (2) graphically by estimaling lhe co-

ordinates of the points where the gaphs of the equations

_ y:16a and y : sinh 15a intersect in the ay-plane.ff i

") EQUATION SOLVER or ROOT FINDER Solve Eq. (2) ford numerically. Compare your solution with the value youfound in (b).

d) Estimate the horizontal tension in the cable at the cable'slowest point.

tZ e) GRAPIiER craph the catenaryI

) = - c o s n d ' ta

over the interval -15 = r: 15. Estimate the sag in rhecable at its center.

, d y - sinh 9x.H

t8.

b) Using the result from part (a) and the fact that the tensionat P must equal F1 (the cable is not moving), show thatT: rt)). This means that the magnitude of the tension atP(r, )) is exactly equal to the weight of y units of cable.

(Continuation of E ercire 87.) The length of arc AP in Fig. 6.37is s - (1/a) sinh aj, where d : irllI. Show that the coordinatesof P mav be exDressed in terms of s as

I . , _ r. I : - S m n - a s , y - T.J

First Order Differential EquationsIn Section 6.5 we derived the law of exponential change, y: ysekt, as the solutionof the initial value problem dy /dt : fty, y(0) : y6. As we saw, this problem modelspopulation growth, radioactive decay, heat transfer, and a great many other phenom-ena. In the present section, we study initial value problems based on the equation


then

dy ldx : f (x,l,), in which / is a function of both the independent and dependent

variables. The applications of this equation, a generalization of dy /dt : ky (thir]'k

of I as x), are broader still.

First Order Differential EquationsA firsl order dilTerential equalion is a relation

fi": ra.t (1)in which /(-r, y) is a function of two variables defined on a region in the ry-plane.

A solution of Eq. (1) is a differentiable function y : y(-t) defined on an interval

of .r-values (perhaps infinite) such that

d' Y(x) : "f(x' Y('t))

ax

on that interval. The initial condition that y(rO) : y0 amounts to requiring the

solution curve y: y(*) to pass through the point ('I0, )0).

EXAMPLE t The equationd v vdx

- ' x

is a first order clifferential equation in which /('t, y) : 1 - (y/"r). D

EXAMPLE 2 Show that the function1 x

, - r ' 2

is a solution of the initial value problem

d x x

Sorutron The given function satisfies the initial condition because

/ 1 - r \ I 2 3) ( 2 ' : [ _ r - ' : i + . t : j .

' / r = 2

To show that it satisfies the differential equation, we show that the two sides of the

equation agree when we substitute (I/x) + (x12) for y.

Y t z l : t

On the leftd r d / 1 x \ 1 I

n=A(;*Zi:-*,+at l l l x \

O n t h e r i s h r l - 1 : l - - l - F ; lx , r \ x z /

:,_1_i:_i*lz x ' z

The function y : (l/x.) + (x/2) satisfies both the differential equation and

initial condition, which is what we needed to show.

W€ sometimes wdte ),/ - /(r, )) fordy /dx = f(x, y).

6.11 First Order Differential Equations 531

Separable EquationsThe equation y' : J Q, )) is separable if / can be expressed as a product of afunction of;r and a function of y. The differential equation then has the form

4!: s@no).dx

If h(y) * 0, we can separate the variables by dividing both sides by h and mul-tiplying both sides by d.x, obtaining

l ', . . d y : g ( x t d x .n \y )

This groups the ly-terms with dy on the left and the r-tems with dr on the right.We then integrate both sides, obtaining

t t fI - - ay : I g rx tdx .

J N \Y ) J

The integrated equation provides the solutions we seek by expressing y eitherexplicitly or implicitly as a function of ,{, up to an arbitrary constant.

EXAMPLE 3 Solve the differential equation

4!: e+r,)" , .dx

So/ufion Since 1 t y2 is never zero, we can solve the equation by separating thevariables.

d, !_ : r r _ y , t

" ,ax

dy : (r + y2)e" d, $11'd'./fi'X'#Tfil;,,both sides by di.

Divide by (l + )'z).

Integrate both sides.

C represents the combinedconstants of integration,

The equation tan 1y : e'* C gives y as an implicit function of .r. In this case,we can solve for y as an explicit function of ,r by taking the tangent of both sides:

tan (tan-t y) : tan (ei + C)

y : tan (e' t C).

Linear First Order EquationsA first order differential equation that can be written in the form

dyi + P(x)Y : Q6')' (2)ax

where P and Q are functions of x, is a linear first order equation. Equation (2) isthe eouation's standard form.

dy' - . ^ ,1 r

r + y 'r )^, 1| . " ) . : I e ' dx

J I + Y ' J

t a n - l y : s ' a 6

tr


We call u(.r) an integratiog factor for Eq.(3) because its presence makes the equationintegrable.

Put the following equation in standard form

x = = x ' + l ) . x > 0dx

EXAMPLE 4

Solution

x , : x - + 5 yax

. : I + - y u r \ r d e 0 ) r .d x x

dv3d x x '

Notice that P (x) is -3 I x, nor +3 I )c. The standard form is l' * P (x)v : 0(:r), sothe minus sign is part of the formula for P(.r). A

EXAMPLE 5 The equationdyi i : "v

with which we modeled bacterial growth, radioactive decay, and temperature change

in Section 6.5 is a linear first order equation. Its standard form is

* - n , :0 . c1- '1 : k and Q( t ) :0

J

We solve the equation

! * 161r: 91,1ax

by multiplying both sides by a positive function u(r) that ffansforms the left-handside into the derivative of the product u("r) . y. We will show how to find u in amoment, but first we want to show how once found, it provides the solution weseek.

Here is why multiplying by u works:

Standad folm withP( j r ) : 3/ i r and 0(-r ) : i

9+P@)v:oG)d)c

, Al9^ + P (x)u (x) y : u (x) Q@)

(3)

Original equationis in standard fbrm.

Mult ip ly by r( r ) .

u(-t) is chosen to make

r , + P D \ = - l r . \ ld \ J )

litegrate with

!<ufO .y) : t)(x)Q@)dx

r ( . r ) .y : lu@)QQ)dx

, : + [ u r r )Qr * td , so r \e to r 1 (4 )U\X ) J

Equation (4) expresses the solution of Eq. (3) in terms of the functions u(x) and

Q @ ) .Why doesn't the fotmula for P(.r) appear in the solution as well? It does, but

6.'11 First order Differential Equations

inclirectly, in the oonslruction of the positive tunction u(,r). We have

d ,lv- - r r Y ) - u t P u tAI AT

d t J u d y P 'u - r J , - u - , - P , ) ! . . , (

a x a x q x

tll- v , = P u )

ttr

This last equation will hold if

du

dx

J u- = r Q I \ . r . . [ . . t r r , L r ]

[ , l u fr - - I P ' l t '

.l u J

n,: I ca, : , : i , . , . : i l l l l l " : 'I p , , I

( : c r

u - 2l ' c a' (5)

From this, we see that any function u that satisfies Eq (5) will enable us to solve

Eq. (3) with the formula in Eq. (4). We do not need the most general possible u,

only one that will work. Therefore, jt will do no harm to simplily our lives by

choosing the simplest possible antidedYative of P for J P dx.

Theorem 4The solution of the equation

1S

fr*rotr:e1,1

1,= l Iu tx tQtx t ,1x .U('r, J

where

r)(x) : eJ Pa)d'. (8)

In the formula for u, we do not need the most general antiderivative of

P(-rr). Any antiderivative will do.

EXAMPLE 6 Solve the equatlondy

' t i : t ' + : r ' r>o '

533

T h . r e l n \ 1 , r r n ( f L

(6)

(1)


How to Solve a Linear First OrderEquation1. Put it in standard form.2, Find an antidedvative of P(x).

3, Fincl u(r) : "l

teta'.

4. Use Eq. (7) to find ),.

solution We solve the equation in four steps.

Step 1: Put the equqtion in standard form n idcntifu P and Q.dy , ,@) :_1 , e@) :x .i r - - r : x ' x

Step 2: Find an antiderivative of P(x) (any one will do).

I ps ta , : [ -10- : - t [ \a , : -3 rn tx t : -3 rnxI I x J . x

Step 3: Find the integrating fqctor v(x).

ugl : "[

""a' : e-3tn t - etn '-] -

Step 4: Find the solution.1 fy : . I ue \e} ) dx

TIX ) J

I r / l \: ,iA I (.;J,'rr*

: r , . [ ] ^a*J x."

: ' ' (-1* c)\ r /

: _x2 + Cx3

The solution is y: -x2 +Cx3, x >0.

EXAMPLE 7 Solve the equation

x Y ' : x z + 3 Y , x > 0 ,

given the intial condition y(1) : 2.

Solution We first solve the differential equation (Example 6), obtaining

Y : - ' 2 + c x 3 , x > o '

We then use the initial condition to find the right value for C:

Y : - x 2 + c x '

2: - (1)2 + C (1)3 J = 2 when r : I

c : 2 t _ ( l ) 2 : 3 .

The solution of the initial value problem is the function y : -x2 + 3x3.

Resistance Proportional to VelocityIn some cases it makes sense to assume that, other forces being absent, t}Ie resis-tance encountered by a moving object, like a car coasting to a stop, is proportional

to the object's velocity. The slower tlle object moves, the less its forward progress

is resisted by the air through which it passes. We can describe this in mathematicalterms if we picture the object as a mass m moving along a coordinate line with

1--

Example 4

( x>o )

Eq (8)

Eq. (7)

Values from steps 1-3

Don ' t f o rge t t heC . . .

. . . it provides part of the

6.11 First Order Differential Equations 535

position J and velocity u at time t. The resisting force opposing the motion is massx acceleration : m(du/dt), and we can write

dum - : - k u ( k > 0 ) ( 9 )

a t

to say that the force decreases in proportion to velocity. If we rewrite (9) as

d u k; * - u : 0 S t a n d a r d f b r ma t m

(10)

(12)

and let u0 denote the object's velocity at time I : 0, we can apply Theorem 4 toarive at the solution

( 1 1 )

(Exercise 42).What can we leam from Eq. (ll)? For one thing, we can see that ifm is

something large, like the mass of a 20,000-ton ore boat in Lake Erie, it will takea long time for the velocity to approach zero. For anotheq we can integrate theequation to find s as a function of t.

Suppose a body is coasting to a stop and the only force acting on it is aresistance propofiional to its speed. How far will it coast? To find out, we start withEq. (11) and solve the initial value problem

ds- : uoe t k l n ) t

' s (0 ) : 0 'd t

Integrating with respect to I gives

' : -YYn tk/ft)t +C'k

Subst i tu t ing s - 0 when I -0 g ives

U^m0: - ; +c

The body's position at time t is therefore

s{1 = u ! ! , ' t . , 1

un ! :

u : ! !11 - e " ,nu t .

k k k

To find how far the body will coast, we find the limit of s(/) as t -+ oo. Since(k/m) <0, we know that e (k/6r > 0 as / > m, so that

I i m s ( r ) : l i m - ( l - e \ k / n t t )r+N /a

UOm .- ^ . UOnl: , ( l U ) = , .

K K

Distance coasted : 4.k

This is an ideal figure, of course. Only in mathematics can time stretch toinfinity. The number u6 z/k is only an upper bound (albeit a useful one). It is trueto life in one respect, at least-if fi is large, it will take a lot of energy to stop

a n d c : Y n 4 .k

Thus,


Weight vs. massWeight is the force that results from gravitypulling on a mass. The two arc rclated by theequation in Newton's second law'

Weight : mass x acceleration.

To convert mass to weight, multiply by theacceleration of gravity. To convert weight tomass, divide by the acceleration of gravity. Inthe metnc system,

Newtons - kilograms x 9.8

and

Newtons/9.8 : kilograms.

In the English system, where weight ismeasured in pounds. mass is measured inslugs. Thus,

Pounds:s lugsx32

and

Pounds/32 : slugs.

A skater weighing 192 lb has a mass of

192132 : $ 5lug5.

the body. That is why ocean liners have to be docked by tugboats. Any liner ofconventional design entering a slip with enough speed to steer would smash intothe oier before it could stoD.

EX,AMPLE 8 For a 192-1b ice skater, the t in Eq. (l l) is about 1/3 slug/secand m : 192132 = 6 slugs. How long will it take the skater to coast from 11 fVsec(7.5 mph) to 1 ff/sec? How far will the skater coast before coming to a completestop?

Solution We answer the first question by solving Eq.

I l e t / \ 8 : 7

e - ' / t 8 : | / 11

- t / t 8 : I n ( l / 1 1 ) : - l n

t : 18 ln 11 ry 43 sec.

We answer the second question with Eq. (12):

(l l) for r:Eq . (11 ) wnh k : 1 /1 ., ? : 6 , u 0 = l l , u = l

1 1

. v o m t l . 6Dlslance coasteo

k t / 3

: 198 fr.

; : \ - v n r ^ 1 " YR R

tr

RL CircuitsThe diagram in Fig. 6.38 represents an elecnical circuit whose total resistance is aconstant R ohms and whose self-inductance, shown as a coil, is Z henries, also aconstant. There is a switch whose terminals at a ar'd b can be closed to connect aconstant electrical source of V volts.

Ohm's law, V : R1, has to be modified for such a circuit. The modified form

diL ; tR i : v ,

d I

where i is the intensity of the current in amperes and / is the time in seconds. Bysolving this equation, we can predict how the current will flow after the switch isclosed.

EXAMPLE 9 The switch in the RZ circuit in Fig. 6.38 is closed at time , : 0.How will the current flow as a function of time?

So/utron Equation ( 13) is a linear first order differential equation for i as a functionof t. Its standard folm is

d iRv_ - L _ ; - ,d T L L

(14)

and the conesponding solution, from Theorem 4, given that I :0 when t:0, is

(13)

6-38 The Rl circuit in Example 9.(1s )

6.39 The growth of the current in the Rlcircuit in Example 9. / is the currentSsteady state value. The number t: l/R isthe time constant of the circuit. Thecurrent gets to within 5% of its steadystate value in 3 time constants(Exercise 53).

Exerc ises 6.11 537

(Exercise 54). Since R and Z are positive, -(R/Z) is negative and e (RlL)t -> 0as I -) oo. Thus,

l im l : l imv. 0 : -R

(Y-Y",", , , , \ = Y -I\ R R / R R

At any given time, the current is theoretically less thal'V/R, but as time passes thecurrent approaches the steady state value V/R. According to the equation

t !+ni :v .A I

I : V /R rs the current that will flow in the circuit if either Z : 0 (no inductance)or di /dt : 0 (steady cunent, i : constant) (Fig. 6.39).

Equation (15) expresses the solution of Eq. (14) as the sum of two terms: asteady state solution V/R and a transi€nt solution -(V/R)e (R/r)r that tends tozeroas , '> co . u

Exercises 6.11Verifying Solutions

In Exercises I and 2, show that each function ) : /(r) is a solutionof the accompanying differential equation.

l . 2y' - l3y - e- '

a ) Y : e - "b\ ! : e-r + e t3/2h

c ) J :e \ +Ce- (3 /2 ) r

I

In Exercises 5-8, show that each function is a solution of the giveninitial value problem.

Differentialequation

Initialcondition

Solutioncandidate

l : e ' t a l t ( 2 e ' )

y : (x 2)a-,'cos.r

is a solution

, ' 2' ' - t + 4 e 2 ,

- r 1 ' { 1 : 5111 ,, t > 0

, r > l

y ( l n 2 t = !2

v(2) : o/ 7 t \

i ( ; J :o

6.

b ) ) -* * 3 . r + C

In Exercises 3 and 4, show that the function ): /(.r)of the given differential equation.

1 f ' a l

3 . y = : I - - d t .

x 2 J , t r y : e ,x J t t

t f , _ 2 x .4 . v = _ | J t - t a t . y - . ] , y : t

v l + r _ . , r 1 | + . r -

Separable EquationsSolve the differential equations in Exercises 9-14.

! - -ln _r

to . t t+ t t f= ) ( . r - t )


12.

14.

l l .

13.

Linear First Order EquationsSolve the differential equations in Exercises 15-20.

15 . x f t +y :e ' , x>0

1 6 . e ' ; + 2 e ' J = r

r7 . xy ,+3y : $ , , , o

18. y'-1- (tan :r)y : cosz tc, -n /2 < x < n /2) - l

r e . x ; : + 2 y = 1 - 1 . t > 0

2 0 . ( l + r ) y ' + y : J i

First Order EquationsSolve the differential equations in Exercises 21 34.

21. 2y' : e'/2 + J,1,,

22. JV; -eu+J ' . : t .0

23, e2' ), ' + 2e2' y :2x

2 4 . x y ' - y = 2 x l n x

2 5 . s e c r 4 - e l + ' h '

26.

J1

28,

,o

dr cos,r'

d " : * - 2 Y ' x > o

^ d s( r - l ) ' d

t 4 f r - l ) - \ = r + 1 . / > l

t t I l t + - 2 s = J ( r + l ) l L , , ' - 1a . ( a + t ) .

. . _ d,(sec-y 'x) , :vx

A I

n .s inr - t rcos2r r f r :0 . -+ . , . ;

drstn e

de + Gose)r :tan9, 0 < 0 < n /2

drt a n 0

d 0 + r = s i n ' d . 0 < e < n / 2

dvcosh r ; + {s inh -x }y : e - -

d'tsinn .r,; + l(coshr)y : coshr sinhx

Solving Initial Value ProblemsSolve the initial value problems in Exercises 35-40.

Differential Initialcondition

3s. 2 +zv =za ,

x . t ! +zy - t3 , t .oA I

sz . e#+y=s ine , e>o

3 8 . 0 ; - 2 , - 0 r ' e c d r a n g . d > 0

l l . 1 , ' t14) - 2( r r +x)y - - -11. "

, - ld x , . r + 1 .

dyd x + a Y : x

What do you get when you use Theorem 4 to solve the followinginitial value problem for y as a function of r?

dt

dt : ky (k constant), y(0) = yo

42. Use Theorem 4 to solve the following initial value problem foru as a function of t.d u kE

+ ;,

:0 (k atd m positive constanrs), u(0) : uo

Theory and Examples43. Is either of the following equations corlect? Give reasons for

your answers.

f 1a ) x I l d x = x l n l : v l * C

t 1b ) t l : d x = x l n l x l + C x

44. Is either of the following equations corlect? Give reasons foryour answers.

1 la) : I cosx dr: tanx + C

cos r J

1 f cb) ---: I cos r d.r - tan r +cos t J cos.r

45. Blood sugar. If glucose is fed intravenously at a constant rate,the change in the overall concentration c(t) of glucose in theblood with respect to time may be described by the differentialequation

d c G- : _ - k c .dt 100Y

In this equation, G, V, and,t are positive constants, G being the

2 J , y i

= I . x , y > 0

dy

d y :d.x

dy

r -4Y '

2 r 2 + l

) > 0

, t > 0

equation

)(0) = I

)(2) = I

y(n /2) : I

y ( l t / 3 ) : 2

)(0) : 5

Y(0) : -640.

41.

30.

31.

34,

_ a) Solve the initial value problem for r as a function of t.

m b) CALCULATOR About how many years will it take for theamount ln your account to reach $100,000?

47. How lonq will it take a tank to drain? If we drain rhe warerftom a vefiical cylindrical tank by opening a valve at the baseof the tank, the water will flow fast when the rank is full butslow down as the tank drains. It tums out that the rate at whichthe water level drops is proportional to the square root of thewater's depth, ,]l. This means that

dxE : 1000 * 0.10;r, :,.(0):1000.

: -kr1'

mte at which glucose is admitted, in milligrams per minute, andV the volume of blood in the body, in liters (around 5 lite$for an adult). The concenradon c(t) is measured in milligramsper centiliter The term -,tc is included because the glucose isassumed to be changing continually into other molecules at a ratepropotional to its concentration.

a) Solve the equation for c(r), using c0 to denote c(0).b) Find the steady state concenrration, lim,,- c(r).

46. Continuous compounding. You have $1000 with whichto openan account and plan to add $1000 per year. AII funds in rhe ac-count will earn l07o interest per yeat compounded continuously.If the added deposits are also crcdited to your account continu-ously, the number of dollars;r in your account at time / (years)will satisfy the initial value problem

Exerc ises 6.11 539

6,40 Dia am for Exercise 48.

b) Show that if uo : JER, then

RESISTANCE PROPORIIONAL TO VELOCITY

49. For a 1451b cyclist on a l5Jb bicycle on level ground, the,t inEq. (11) is about 1/5 slug/sec and m : 160/32:5 slugs. Thecyclist stafis coasting ar 22 ft/sec (15 mph).a) About how far will the cyclist coast before reaching a com-

plete srop?b) To the nearest second, about how long will it take the cy,

clist's speed to drcp to 1 ftlsec?

50. For a 56,000-ton Iowa class battleship, n = 1,750,000 slugs andthe t in Eq. (11) might be 3000 slugs/sec. Suppose rhe battleshiploses power when it is moving at a speed of 22 ftsec (13.2knots).

a) About how far will the ship coast before it stops?b) About how long will it take the ship's speed to drop ro I

ft/sec?

RT CIRCUITS

51, Current in a closed RL circuit. How many seconds after theswitch in an Rl, circuit is closed will it take the curent i toreach half of its steady state value? Notice that the time dependson R and Z and not on how much voltage is applied.

52. Current in an open RL circuit. If the switch is thrown open afterthe current in an Rl, circuit has built up to its steady state value,

,:n(r+ff,)" 'dtdt

The value of * depends on the acceleration of gravity, the shapeof the hole, the fluid, and the cross-section areas of the tank anddrain hole.

Suppose t is measured in minutes and t = l/10. How longdoes it take the tank to drain if the water is 9 ft deep to star.twith?

8, Escape velocity. "lhe gravitational attraction F exerted by anairless moon on a body of mass n at a distance s from themoon's center is given by the equation F: rng R2s-2, whereg is the acceleration of gravity at the moon's surface and R isthe moon's rudius (Fig. 6.40). The force F is negative becauseit acts in the direction of decreasing .t.

a) If the body is projected vertically upward from the moon,ssurface with an initial velocity uo at time ,:0, use New-ton's second law, F : md, to show that the body's velocityat position "r is given by the equation

. 2p R2r J 2 :

- 6 " + u l ' - Z g n '

Thus, the velocity rematlns poritiu" as long as r.16 1./2gR.The velocity un: /2gE is the moon's escape vetocity. Abody projected upward with this velocity or a greater onewill escape from the moon's gravitational pull.


the decaying curent (graphed here) obeys the equation

diL , + R i - 0 , ( 1 6 )

which is Eq. (13) with Y :0.

a) Solve Eq. (16) to express i as a function of 1.b) How long after the switch is thrown will ir take rhe curent

to fall to half its original value?c) What is the value of the curent when t:4R? (The sig-

nificance of this time is explained in the next exercise.)

B 53. tme constants. Engineers callthenrmber Lf Rthe time constantof the Rl, circuit in Fig. 6.39. The significance of the time con-stant is that the current will reach 95% of its Rnal value within3 time constants of the time the switch is closed (Fig. 6.39).Thus, the time constant gives a built-in measure of how rapidlyan individual circuit will reach equilibrium.

a) Find the value of i in Eq. (15) that coresponds to / - 3a/Rand show that it is about 957o of the steady state valueI - v/R.

b) Approximately what percentage of the steady state cuFentwillbe flowing in the circuit 2 time constants after the switchis closed (i.e., when t = 2L/R)?

54, (De wtion of Eq. (15) in Example 9.)

a) Use Theorem 4 to show that the solution of the equationd i R vd I L L

isv ' . ^ ( R / t ,R

b) Then use the initial condition t(0) = 0 to determine thevalue of C. This will complete the derivation of Eq. (15).

c) Show that i : V /R is a solution of Eq. (14) and that i :

Ce (x/z)/ satisfies the equation

d i R- - L _ i

d t L

MIXTURE PROBLEMS

A chemical in a liquid solution (or dispersed in a gas) runs into acontainer holding the liquid (or the gas) with, possibly, a specified

amount of the chemical dissolved as well. The mixture is kept unifo.mby stirring and flows out of the container at a known rate. In thisprocess it is often important to know the concentration of the chemicalin the container at any given time. The differential equation descdbingthe prccess is based on the fomula

Rate of change / rale ar which \ / rare at $hich \o f a m o u n r = l c h e m i c a t l _ l c h e m i c a l l ( 1 7 )

t r l lIn Contalner \ aITlVes ' \ deparls.

If )(t) is the amount of chemical in the container at time I and y(r) isthe total volume of liquid in the container at time I, then the departurerate of the chemical a1 time t is

Deparlure rate l! . louruoo rare1v l t )

/ concentration in \ -: l l . ( o u l f l o $ r a t e ) .\ container at trme / /

(18)

Accordingly, Eq. (17) becomes

d v v t r )

, ! : (chemical s arr ival rale) -

vtr, ' rortf lo* ralel. (19)

If, say, y is measured in pounds, V in gallons, and t in minutes, theunits in Eq. (19) are

pounds pounds pounds

55. A tank initially contains 100 gal of brine in which 50 lb of saltare dissolved. A brine containing 2 lb/gal of salt runs into thetank at the rate of 5 gaymin. The mixture is kept uniform bystirring and flows out of the tank at the rate of 4 gaymin.

a) At what rate (lb/min) does salt enter the tank at time /?b) What is the volume of brine in the tank at time r?c) At what rate (lb/min) does salt leave the tank at time r?d) Write down and solve the initial value problem describing

the mixing process.e) Find the concentration of salt in the tank 25 min after the

process stans.

56, In an oil refinery a storage tank contains 2000 gal ofgasoline thatinitially has 100 lb of an additive dissolved in it. In prepararionlbr winter weather, gasoline containing 2 lb ofadditive per gallonis pumped into the iank at a rate of 40 gaymin. The well-mixedsolution is pumped out at a rate of 45 gal/min. Find the amountof additive in the tank 20 min after the process starts.

57. A tant contains 100 gal of fresh water. A solution containing Ilb/gal of soluble lawn fenilizer runs into the tank at the rate of1 gaVmin, and the mixture is pumped out of the tank at the rateof 3 gaymin. Find the maximum amount of fertilizer in the tankand the t ime required to reach the maximum.

5E. An executive conference room of a corporation contairs 4500cubic feet of air initially free of carbon monoxide. Starting attime t : 0, cigarette smoke containing 47o carbon monoxide isblown into rhe room ar the rale of 0.3 ft

' /min. A cei l ing fan

keeps the air in the room well circulated and the air leaves theroom at the same rate of 0.3 ft 3/min. Find the time when theconcentration of carbon monoxide in the room reaches 0.0170.

gal

mingalmtn

6.12 Euler's Numerical Method; Slope Fields 541

Euler's Numerical Method; Slope FieldsIf we do not require or cannot immediately find an e.racl solution for an initial valueproblem y' : f(x, y), )(.ro) : yo, we can probably use a computer to generare atable of approximate numerical values ofy for values of ;r in an appropdate interual.Such a table is called a numerical solution of the problem and the method bywhich we generate the table is called a numerical method. Numedcal methodsare generally fast and accurate and are often the methods of choice when exactformulas are unnecessary, unavailable, or overly complicated. In the present section,we study one such method, called Euler's method, upon which all other numericalmethods are based.

Slope FieldsEach time we specify an initial condition 1(_rs) : :!0 for the solution of a differentialequation y' : f(x, y), the solution curve is required to pass through the point(r0, ),0) and to have slope /(x6, ys) there. We can picture these slopes graphicallyby drawing short line segments of slope /(x, y) at selected points (.r, y) in thedomain of /. Each segment has the same slope as the solution curve through (r, l,)and so is tangent to the curve there. We see how the curves behave by followingthese tangents (Fig. 6.41).

Constructing a slope field with pencil and paper can be quite tedious. All ourexamples were generated by a computer. Let us see how a computer might obtailone of the solution curves.

6,41 Slope fields (top row) and selected solution curves (bottom row). In computerrenditions, slope segments are sometimes portrayed with vectors, as they are here.This is not to be taken as an indication that slopes have directions, however, forthey do not.

( b ) ] ' = - (c) )'= (l r)) + i

\ > / / / / / / / / r ' > \ \

\ - 2 / / Z / / / / / 2 - \ \

\ + 2 , / 2 / / , / , ' / / + \ \

\ \ \-/////-\\ \ \

\ \ \ \ \ - - - \ \ \ \ \ i\ \ \ \ \ \ \ \ \ \ \ \ \ \t i l \ \ \ l \ - - i \ \ | il \ \ \ \ i l \ i \ \ l l l\ i \ \ \ 1 . : . \ \ \ \ \ | I\ \ \ \ \ \ \ \ \ \ \ \ \ \

\

\ \ \ \ \ \ \ \ \ \ - / / / / / / , 2 , 2 / /\ \ \ \ \ \ \ \ \ \ - , , 7 / , /\ \ \ \ \ \ \ \ \ \ - - z . r / t / / t / / /\ \ \ \ \ \ \ \ \ \ - . , / / ) / / , t / / ,\ \ \ \ \ \ \ \ \ \ - / / / / / t / / ) /\ \ \ \ \ \ \ \ \ \ - / / / / / / ) / t t

a : ' \ \ I \ \ \ \ \/ , / - \ \ \ \ \ \ \ \/ - 2 - \ \ \ \ \ \ \ a, Z / . * \ \ \ \ \ \ \ \

7 z - \ \ \ \ \ \ \ \

t l t a / . 1 / . 1 -

lilll iiiiii 'i=-i.\\I i i l , r t l t ) | | 1 ' . i t t : - - : , : : : :/ / / / " / / . ' / / /

\ ..:.-s<€;rttiiilii\ I \ \ \ \ \ \ \ \ = + / / / / / 7 / ]

t i I I \ \ t \ \ ' : . ' = t i l i I i i iI I f I l l I \ \ \ \ i : r i l i i , i ,ItIiiIiiil\\-riiiiiII

( a ) y ' = y - x 2

\2/ / ,/1

vtszz€ / r ' + \ \ \: - + \ \ \

\

NII

2t!

I + - r 2

I'\\

\\

It

\\\

I\

,

t,t

tt

tt t

,,t

l t I l1 ' ' , t tJr t t -4t2.


6,42 The equation of the tangent l ine isy = L(x) = yo + f$0, ydk - xd.

6.4j Three steps in the Eulerapproximation to the solution of theinitial value rroblem yt : f(x, y), y = yowhen x : xo. The errors involved usuallyaccumulate as we take more steps.

Using LinearizationsIf we are given a differential eq\atiot dy /dx : f (x, y) and an initial condition

)(x0) : y0, we can approximate the solution curye ): y(x) by its linearization

r t * r : y t ro r - 4 l rx - {o rqx | - _ .

or

Z(,r) : ys t /(.{e, yo) (.r - ro). (1)

The function Z(x) will give a good approximation to the solution y(x) in a shortintewal about xo (Fig. 6.42). The basis of Euler's method is to patch together a

stdng of linearizations to approximate the curve over a longer stretch. Here is how

the method works.We know the point (.n0, ye) lies on the solution curve. Suppose we specity a

new value for the independent variable to be xr : x0 + dx.If the increment d.r is

small, then

yt : L(xt) : yo a f (xx, yd dx

is a good approximation to the exact solution value 1,: y(xl). So from the point(xn, 1,{r), which lies exactly on the solution curve, we have obtained the point(.x1, y1), which lies very close to the point (xr, y(rr)) on the solution curve.

Using the point (,r1, yr) and the slope /(x1, yr), we take a second step. Setting

x2: t + dx, we calculate

y2 : yl + f @r, yt) d:r,

to obtain another approximation (x2, )2) to values along the solution curve y : J(t)(Fig. 6.43). Continuing in this fashion, we take a third step from the point (,r2, y2)

with slope f(xz, y) to obtain the next approximation

h : yz+ f eb y )d .x ,

and so on.

EXAMPLE 1 Find the first three approximations yb y2, h using the Eulerapproximation for the initial value problem

y ' : I * y , : v ( 0 ) : l ,

starting at.r0 :0 and using d.x. :0.1.

Solution

First:

tr

Second:

y 1 : y s l f ( x s , y d d x

: y6a (1 *yq)d. {: 1 + ( 1 + 1 ) ( 0 . 1 ) : 1 . 2

Y 2 : Y l + f @ v , Y 1 ) d x: y 1 1 ( 1 l y ) d x

: 1 . 2 + ( 1 + 1 . 2 ) ( 0 . t ) : 1 . 4 2

) 3 : ) 2 + 0 ' t Y ) d x: 1 . 4 2 + ( 1 + 1 4 2 ) ( 0 . 1 ) : r . 6 6 2

Third:

6.12 Euler's Numerical Method; Slope Fields 543

The Euler MethodTo continue our discussion, the Euler method is a numerical process for generating

a table of approximate values of the function that solves the initial value problem

r' : f(x, y), )(,ro) = yo.

If we use equally spaced values for the independent variable in the table and generaten of them, we first set

\ ! xo+d'x '

x z = x t l d x ,

x n = x a t I t l x .

Then we calculate the solution approximations in tum:

lo + f Oo, Yi dx,

)r + /("rr, lr) dr,

!, : la t ]_ f (x, t, y"_t) dx.

The number n of stePs can be as large as we like, but erors may accumulate if n

is too large.

EX.AMPLE 2 Invcstigate the accumcy of the Euler approximation method for

the initial value problem

y ' : l * l ' ) ' ( o ) = 1

in Example 1 over the interval 0 5:r 5 1, starting at x0 :0 and tahing dr :0 1'

Sorution The exact solution to the initial value problem is y:2g' - 1 (using

either method discussed in Section 6.11). Table 6.14 shows the results of the Euler

approximation method using Eqs. (2) and (3) and compares them to the exact results

(2)

(3)

Tabfe 6.14 Euler so lut ion o i y :1+y 'y(0)- ' l ' increment 517s jv=01

00.10.20.30.40.50.60.'70.80.91.0

J (approxJ

I1 . 2t.42r.662t.92822.22102.54312.89'743.28123.'11594.1875

t (exact)

Ir.21,03r.4428r.699'71.98362.29'142.64423.02'753.451 I3.9t924.4366

Error = J (exact) -J (approx)

00.01030.02280.031'70.05540.0't 640 . 1 0 1 10.13010.16390.20330.249r


rounded to 4 decimal places. By the time we reach r : Iis about 5.67o.

(after 10 steps), the enorJ

EXAMPLE 3 Investigate the accuracy of the Eller method for the initial valueproblem

y ' = I * : y , y ( 0 ) : I

over the interval 0 < ; < l, starting.at r0 :0 and taking dx :0.05.

Solution TabIe 6.15 shows the results and their comparisons with the exact solu-tion. Notice that in doubling the number of steps from t0 to 20 we have reducedthe error. This time when we reach ; : I the error is only abott 2.9i.. J

It might be tempting to reduce the increment size even further to obtain greateraccuracy. However, each additional calculation not only requires additional com-puter time but more importantly adds to the buildup of round-off errors due to theapproximate representations of numbers in the calculations,

The analysis of error and the investigation of methods to reduce it when makingnumerical calculations is important, but appropriate for a more advanced coune.There are numerical methods that are more accurate than Euler,s method. as vouwill see when you study differential equations. In the exercises you will haveiheopportunity to explore the trade-offs involved in trying to reduce error by takingmore but smaller increment steps.

Tabfe 5.15 Euler solution of / = 1 + y, y(O) :1, increment size dx = 9.95

r

00.050.100.150.200.250.300.350.400.450.500.550.600.65o.'100.750.800.850.900.951.00

y (appmx)

Il t1.2051.31531.43101.55261.6802t.81421.95492.102'12.2s'182.42072.591'12.7'7 t32.95993.t5793.36573.58403.81324.05394.3066

J (exact)

11 .10251.21031.3237r.44281.56811.69971.83811.98362.13662.29742.46652.64422.83113.O2'153.23403.451r3.6',7933.91924.1'7144.4366

Error = y (exact) -y (approx)

00.00250.00530.00840.01l80.01550.01950.0239o.02870.03390.03960.04580.05250.05980.06'160.07610.08540.09530.10600 . 1 1 7 50.1300

Exercises 6.12 545

Exercises 6,12

Calculating Euler ApproximationsIn Exercises l-6, use Euler's method to calculate the first three apFox-imations to the given initial value Foblem for the specified incrementsize. Calculate the exact solution and investigate the accuacy of yourapproximations. Round your results to 4 decimal places.

r . y ,= 1 - ! , y ( z ) : - r , dx =osx

2. y' : x(r - y), y(1) :0, dr : 0.2

3. y ' :2 'xy *2y, y(0) :3, dx :0 .2

4, y' : y2(1+bt), y(-1) = 1, dr = 0.5E S. CnlCUmrOn ! ' :2xe" , y(0) :2, h t :o . r

Ee, ca lcu i r roR yt : r+ex -2, y(o) :2, dx:0.5

7. Use the Euler method with d.r :0.2 to estimate t(l) if f' :'yand y(0) : l. what is the exact value of y(l)?

8. Use the Euler method with dr : 0.2 to estimate ](2) if y' :yl:cand jr(l) :2. What is the exact value of y(2)?

E q. ClfCUfefOn Use the Euler merhod with tr = 0.5 to estimatey(5) if f' : f2ltq and y(1) : -1. What is the exact value of)(5) ?

E tO. CnfCUnfOn Use rhe Euler method with dr = 1/3 to estimarey(2) if y' : y - eb and y(0) : 1. What is the exact value ofv(2)1

Slope FieldsIn Exercises 11 14, match the differential equations with the solutioncurves sketched below in the slope fields (a)-(d).

11 . 1 ' : ; y

13 . v ' : y

1 2 . y ' : x a y

14, v ' :a ,

I

iI \

\

\

I

I

I\

\

\(

III

i

N\"+ /

(a)

t l/

I

\\\ \-.','//

I NN:Yill

',lr=\\/2\\\' // , '-\ \\

// /,,-\ \ \\

//.

\

(b)

I\

\\

G)

i lyntl'/,,, ,, hltl I t,l_.,,/ /// /

N\-

i

(d)


In Exercises 15 and 16, copy the slope f,elds and sketch in some ofthe solution curyes.

15. y' : (y { 2)(y - 2)

b) Find the general solution of the differential equation using yourCAS DE solver

c) Graph the solutions for the values of the arbitrary constant C =-2, -1,0, 1,2 superimposed on your slope field plot.

d) Find and gaph the solution that satisfies the specified initialcondition over the intenal [0, ,].

e) Find the Euler numerical apprcximation to the solution of theinitial value problem with 4 subintervals of the r-interval and plotthe Euler approximation superimposed on the graph produced inparr (d).

f) Repeat part (e) for 8, 16, and 32 subintervals. Plot these rhreeEuler approximations superimposed on the graph from part (e).

g) Find the error ), (exact) -), (Euler) at the specified point r =,for each of your four Euler approximations. Discuss the improve-ment in the percentage eror

2 1 . y ' : a 4 y , ) ( 0 ) : - 7 1 1 0 ; 4 3 ) c 3 4 , - 4 = r 3 4 t b : 1

2 2 . y ' : - a 1 y , y ( 0 ) = 2 t - 3 : . r 5 3 , - 3 = y = 3 ; b : 2

23. A log is t i c equat ion . y r=r (2 -y ) , y (0 ) : l /2 ; 0 : ' :4 ,O . y 3 3 i b : 3

24. y ' : G in r ) (s in ) ) , y (0 ) :2 ; -6 .x 56 , -6 : y S 6 ;b :3 t r /2

Exercises 25 and 26 have no explicit solutiol in terms of elementaryfunctions. Use a CAS to explore gaphically each of the differenrialequations, performing as many of the steps (a)-(g) above as possible.

2 5 . y ' : s s s 1 2 r - r ) , y ( 0 ) : 2 ; 0 = , i = 5 , 0 = ) : 5 ; ) ( 2 )26 . A Gomper tz equat ion . y t :y ( l /2 - ln y ) , y (0 )=1/3 ;

0 1 x 5 4 , 0 : y : 3 ; y ( 3 )

27. Use a CAS to lind the solutions of y' + f : f (r) subject to theinitial condition y(0) : 0, if /(,r) n

a) 2x

sin 2t

3ex/z

d) 2e '/2 cos2x,

Graph all four solutions over the interval -2 5 -{ < 6 to comparethe results.

28. a) Use a CAS to plot the slope field of the differential equation

, 3 x 2 + 4 x + 2' 2 ( v - l )

over the region -3 S x =3 and -3 5 y:3.Separate the variables and use a CAS integrator to flnd thegeneral solution in implicit form.Using a CAS implicit function gapher, plot solution curvesfor the a$itlary constant values C : -6, -4, -2,0,2,4 , 6 .Find and graph the solution that satisfles the initial conditionr(0) : -1.

b)

c)

b)

c)

d)

1, L t,t,t ll L,t I L | 1. 1.1,,,, i i , , , t i , , , ,iiiiiiiiiiiiiiirr\t\\t\\l\t\t\} } } I } I I I I ! I I I I Ii t i t i i i i t t t i t i ti l i l i t i i l i l l t i t\\t\\t\\{\\\t\i\ \ \ \ \ \ \ \ \ \ \ \ \ \ \

f,t'Jiliit,',lii'rlT'/,1

tltl,,t,iiii\ \ \ \i lNil{ri\\t

t\? : l, ,

ii\ \

f lI it l\t

16. )' : )(), + 1)() - 1)

n t / t t / / u t /u t / t n t l/ / / / / / / / / / / / / / / / / / / / /,//

"/ . / ,/ ,/ . / / ,/ / / / ,/ ,/ / ,/ ,/ ,/

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ . \ . \ \ ' .\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ "\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \

In Exercises 17-20, sketch part of the slope field. Using the slopefield, sketch the solution curves that pass thrcugh the given points.

r7. y' = y with (a) (0, l), (b) (0, 2), (c) (0, -l)

r8. y' : zly - 41 with (a) (0, 1), (b) (0, 4), (c) (0, 5)

19. y' : y(2 )) with (a) (0, 1/2), (b) (0, 3/2), (c) (0,2),(d) (0, 3)

20. y' : y2 with (a) (0, l), (b) (0,2), (c) (0, -1), (d) (0, O)

S CAS Explorations and ProjectsUse a CAS to explore graphically each of the differential equationsin Exercises 21J4. Pertorm the following steps to help with yourexplorations.

a) Plot a slope field for the differential equation in the given ry-window.

Questions to Guide Your Review 547

CHa.prsn QunsuoNs ro GUrDE YouR Rnvrsw

1. What functions have inverses? How do you know if two functions

/ and I are inverses ol one another? Give examples of lunctionsthat arc (arc not) inverses of one another

2. How are the domains, ranges, and graphs of functions and theirinverses related? Give an example.

3. How can you sometimes express the inverse of a function of .yas a function of iv?

4. Under what circumstances can you be sur€ that the inverse of afunction / is differentiable? How are the derivatives of / and

/-rrelated?

5. What is the natural logarithm function? What are its domain,range, and deivative? What arithmetic properties does it have?Comment on its graph.

6. What is logarithmic differentiation? Give an example.

7. What integrals lead to logadthms? Give examples. What are theintegrals of tan "r and cot r?

8, How is the exponential function e' defined? What are its domain,range, and derivative? What laws of exponents does it obey?Comment on its graph.

9. How are the ftrnctions a' and log. x delined? Are there anyrestrictions on d? How is the graph of log",y related to the graphof In r? What truth is there in the statement that there is reallyonly one exponential function and one logarithmic function?

10. Describe somc of the applications of base 10 logarithms.

11. What is the law of exponential change? How can it be derivedfrom an initial value problem? What are some of the applicationsof the law?

12. How do you compare the growth rates of positive functions asr > co?

13. What roles do the functions e* and ln r play in growth compar-isons?

14, Describe big-oh and little-oh notation. Give examples.

15. Which is more efncient-a sequential search or a binary search?Explain.

16. How are the inverse t gonometric functions defined? How canyou sometimes use right triangles to find values of these func-tions? Give examples.

17. How can you find values of sec l;r, csc l*,andcot-l-r usinga calculator's keys for cos-l.r, sin r;r, and tan lx?

18. What are the derivatives of the inverse t gonometric functions?How do the domains ofthe derivatives compare with the domainsof the functions?

19. What integrals lead to inverse trigonometric functions? How dosubstitution and completing the square broaden the applicationof these integrals?

20. What are the six basic hyperbolic functions? Comment on theirdomains, mnges, and graphs. What are some of the identitiesrelating them?

21. What are the de vatives of the six basic hyperbolic functions?What are the corresponding integral fomulas? What similaritiesdo you see here with the six basic trigonometric functions?

22. How are the inverse hyperbolic functions defined? Comment ontheir domains, nnges, and graphs. How can you nnd valuesof sech-r,r, csch I.t, and coth r.r using a calculator's keys forcosh 1x, sinh-lr, and tanh-lx?

23. What integrals lead naturally to inverse hyperbolic functions?

24. What is a first order differential equation? When is a function asolution of such an equation?

25. How do you solve separable first order differential equations?

26. How do you solve linear first order differential equations?

27. What is the slope lield of a differential equation y': ./(-r, y)?What can we learn from such lields?

28. Describe Euler's method for solving the initial value problemy' - J G, J), y(ro) : ]'0 numerically. Give an example. Com-ment on the method's accumcy. Why might you want to solvean initial value problem numerically?

PRecrrcp ExpRcrsps


DifferentiationIn Exercises l-24, find the dedvative of ) with respect to rhe appro-priate variable.

l ' Y : loe \ /s

3. u: 1r" ' . - -L"..- 4 t 6

l . } = l n ( s l n - u )

7. y : logr(t'/2)

9 ' l = s '

1 1 , ) ] : 5 t 3 6

13. y: (x * 2) '+2

2' v = Ji""6'

6 . ) : l n ( sec '?P )

8 ' y - l o g s ( 3 . r - 7 )

lO . y :Q2 t

12. y = .1ir-"n

14. y : 2(ln x)\/2

- t 5 . ) = S r n V l u ' , U < u < l

16 . r - s i n ' 14 ) . , r r' \ Ju )

1 7 . l : l n c o s ' r

1 8 . 1 : x c o s t z - J t - *

l1 9 , y : 1 1 v 1 ' t -

r l n t

20. y : (1 + t2) cot-t 2t

2 1 , y : x 5 s s - t , n F j , z r t

22. y:21E - l sec 1 Ji

23 . ;y :s5c r (sec 6) , 0 <0 < t /2

24. ! : ( l+x2 \e t * '

Logarithmic DifferentiationIn Exercises 25 30, use logarithmic differentiation to lind the deriva-tive of y with respect to the apptopriate variable.

2s. y -2:!li+t zo. ,: ""13;1

' . / cos 2r \ 2x -4

/ ( r + l ) ( r - l r \ 5" ' ) - \ { ' - 2 i t , + ] ) ) '

' "

28. y :4 -' J u 2 + l

29. y: (.sin 0 )'/e 30. y - (ln ')r/{hn

IntegrationEvaluate the integrals in Exercises 31-50.

zt. | " 'sec'{" '

-Ta,

34. I

er csc (e' + 1) cot (e) + l) d]

s. I

sec2{*)""'ax

37 . I ^ " ^ '

v. 1"" nn \a,

u' 1," fua'$. I

t-l!!d,

4s. I ( tn !3

dx

4t. l! """'rr*rn,ro,

o. |

'2" ax

Evaluate the intesrals in Exercises 51-64.

5r. | -7r

',. 1'(; . *) *ss.

| _-,' e1'*\ a x

57. lo"5

e ' { le ' + t1 3r2ar

36. I

cscl r e""" dx

s . [ "6a*

40. ["" z"or r" d.,Jtrc

42. I -'""L a,J - r l 2 r - S l n l

44. [ :'J U t n u

46. [h (a s)

da. l , )

4s. / ""{1: ''-l a,

so. I

z'^ ' sec2 rax

f32 15 2 . | - d x

s4. [ ' ( : - 1) , "J r \ J a x ' /

sG. fo "" a,

tt./" ee (eo Dt/rde

r ' t , [ " | , ,5 9 .

/ l ( l - 7 l n : , ) - t ' 3 d r u O -

J , , u 4 n , , - . .

6 r . / ' r t n t ' + l ) r ' r ,

J t U + |

$. l,'

tos;e de

62. 12

0 +lr't t)t rn t dt

se. 1,"9! lx4atEvalute rhe integrcls in Erercises 65 78.

3r. J

e' sin (e') dx 32. | "'

cos (3e' - 2\ dtr J / 4 a ) -

6s. I --:J -ya ,/9 - 4xz

f 1 / 5 A n .66. I ---:j]1-

J -tr ./4 - 25x2

Practice Exercises 549

98. Does / grow faster, slower, or at fhe same mte as 8 as j + co?Give reasons fbr your answers.

6s. [ ])-l yJ4y2 - |

f 2 3d t67' J-, a*rp

[ ' t ' dY

l ,e1 t 1 lJ9y t - |

f d x

f ' dt

J u j 3 - l t ' z

f 24dy

J y"F - raa ) / ( r ) : 3 - ' ,b ) / (x ) : l1 la ,

s(.t) : 2-*8(.r) : ln r'?

68.

70.

71,-"G/,8 dt,

2/J5 lyllsy" - z

I " ^J J ;44x - |

f \ 3 d ul -

J - t 4 u ' + 4 u + 4

1 F - 186. lim -

, -s in r _ l88. lim 4

r + 0 e ' - I

4 - 4 e '90- l im _

s inz l zx r92. 1jm _=4

, - 4 e r 4 + J - (

94. l im ?- ' l ) ln )

c) /(r): l1r3 +2r2, g1t]): e'd ) / (x ) : tan t ( l . l r ) , s (x ) : t / ,e) /(x) : sin '111x1, g1t| : t1t2f) /(r) : sech 'r, 8(-r) : e '

99. True, or lake? Give reasons for your answers.74.

76.

r r / t \a ) - * - : O l - lr- \.1' ./

c) .r : o(.x + ln ).)e) tan-r ir : o(1)

I I / 1 \b ) , * " : o l - l: l ' j r - \ r ' , /

d) ln ( ln.r) : o( ln 1)f ) c o s h r : O ( e ' )

f | 2clu1 a l' " ' J_ ,

uz++u+5

T. [ --L-- 78.I ( r + l ) J r t + 2 t - 8

| ,lt

J (3t + l) \ /9t2 + 6t

Solving Equations with Logarithmic orExponential TermsIn Exercises 79-84, solve for y.

7(l ?r' - )t+l

81. 9e2! : x2

8 3 . l n ( y 1 ) : x + l n ) '

Evaluating LimitsFind the limits in Exercises 85-96.

t O r I85. lim .:]]]] j

.sin r

87. lim - -

r + 0 e r - l

5 - 5 c o s n89. lim .

L - t , ? - x I

91. l im

93. lim

t - l n ( 1 + 2 r )P

t - - - t

/ t Y9 5 . l i m l l - - l 9 6 . l i m

r /

Comparing Growth Rates of Functions97. Does / grow faster, slowet, or at the same mte as I as -t -) oo?

Give reasons for your answers.

a) /( t) : log, r, g(t) : log.r.r

If(x): x, C(r|:) :,{ *

;

"f(;) : r/100, sQ): xe "f (x ) : x , 8 ( ,x ) = tan- ' . r

. / (x ) - csc- ' x , gO) :1 /x

/(;v): sinh *, g(-r): e'

100, True, or false? Give reasons for your answers.

I l l l \a ) - : O l - * ; l

x' \ j { ' x- /

| / t 1 \o) 7 : , | ' . . * ; ,J

80. 4-) : 3r+2

8 2 . 3 r : 3 l n . r

84. ln (101n )): ln 5:r

c ) l n r : o ( x l l ) d ) l n 2 x : O ( l n . r )e) sec r-r: O(1) f) sinh.jr : O(e*)

Theory and Applications101. The function f(x): e' + x, being differentiable and one to-

one, has a differentiable inveNe / '(,r). Find the value of

df | /dx at the poin( J (ln 2).

102, Find the inverse ofthe tunction /(r) - I + (1/t), : t 10. Then

show that / t( /( .r)) : /( /- ' (r)) : .x and that

df t I

, 1 , 1 , , , , , I ' t : t t

In Exercis€s 103 and 104, find the absolute maximum and minimum

values of each function on the given interval.

1 0 3 . y : x l n 2 x - x . l ; . ; lL z e z l

104. I : 10-r (2 - ln t), (0, ezl

105. Find the area between the curve y - 2(lrr x)lx and the ir-axis

f r o m r = l t o t : e .

106. a) Show that the area between the curve l: l/r and thej-axis from r : l0 to r :20 is the same as the area

between the curve and the l-axis from r - I to r:2.

b) Show that the area between the cuNe ), - l/t and the 't-

axis from id to fb is the same as the al€a between the curve

and the rc-axis from x :a to r : , (0 < a < b,k > 0).

107, A particle is ffaveling upward and to the right along the cuNey : ln r. Its x-coordinate is increasing at the rate (dx/dt) :

v& n/sec. At what mte is the )-coordinate changing at the pojnt(e2 ,2 )7

108, A girl is sliding down a slide shaped like the curve ):9e */3

Her )-coordinate is changing at the rate dl, /l t : ( l/qJq - fft/sec. At approximately what rate is her x-coordinate changing

when she reaches the bottom of the slide at,r :9 ft? (Take ?'

to be 20 and round your answer to the nearest ft/sec.)

( ' - ' ; )

b)

c)d)e)f)


109, The rectangle shown here has one side on the positive )-axis,one side on the positive,r-axis, and its upper right-hand vefiex

on the curve ) - s ''. What dimensions give the rectangle its

largest area, and what is that area?

110. The rcctangle shown here has one side on the positive )-a\is,one side on the positive.r axis, and its upper right-hand vertex

on the curv€ ) : (ln x)/r'?. What dimensions give the rectangle

its largest area, and what is that area?

111. The functions .l (x) : ln 5r and 8(r) - ln 3:r differ by a con-

stant. What constant? Cive reasons for your answer.

112. a) I f (1n x)lx : ( ln 2)12, I j ] t ] . jst r :21

b) I f (1n -t)/x : 2 1n2.6151;x : l /2?

Give reasons for your answers.

113. The quotient (1og4,r)/(log: -r) has a constant value. What value?

Give rcasons fbr your answer

f,l ttl. tog" (2) vs. tog, (x). How does /(ir) = log,g(r) = log.(x)l Here is one way to find out:

a) Use the equation log,, : ( ln b)l ln a)and 8(,r) in terms of natural logarithms.

b) Craph / and g together Comment on the behavior of fin relation to the signs and values of g.

lZ rrs. cnapHrn Graph the following functions and use what you see

to locate and estimate the extreme values, identify the coordi-

nates of lhe inflection points, and identify the intervals on which

the graphs are concave up and concave down. Then connrm your

estimates bv workins with the functions' derivatives.

a) -r, : (ln x)/"8c ) y = ( l i x ) e '

iZ tte. gnapnfn craph /(i) - :r ln r. Does the function appear tohave an absolute minimum value? Confirm your answer withcalculus.

B ltz, calcunroR what is the age of a sample ofcharcoal in which90% of the carbon-I4 oiginally present has decayed?

E t18. Coot;nq a pie. A deep-dish apple pie, whose intemal temper-ature was 220'F when removed from the oven, was set out on

a breezy 40"F porch to cool. Fifteen minutes later, the pie'sinternal temperature was 180'F How long did it take the pieto cool from there to 70'F?

E 119. tocatlngasolarstation. You are undercontract to build a solarstation at ground level on the east-west line between the twobuildings shown here. How far from the taller building shouldyou place the station to maximize the number of hours it willbe in the sun on a day when the sun passes directly overhead?Begin by observing that

. r , 5 0 - r0 _ , t cor_

60 _ co l ,

l0 .

Then find the value of ,r that ma,rimizes 0.

A round underwater transmission cable consists of a core of

copper wires surounded by nonconducting insulation. If.{ de-

notes the ratio of the radius of the core to the thickness of theinsulation, it is known that the speed of the transmission signalis given by the equation u : x2 ln (l /x). If the radius of thecore is I cm, what insulation thickness ft will allow the greatest

transmission speed?

lni t ial Value ProblemsSolve the initial value problems in Exercises l2l 124.

Differential eouation Initial condition

120.

(2) compare with

to express /().)

t2t.dytlx

d y ) l n )d x 1 * , r 2

r ( 0 ) : 2

r(0) : e':122.

r 2 3 . G + r ) * ] 2 y : x , " , - , ) ( 0 ) : 1

n 4 . x * + 2 y - , 2 + 1 , " ' o

y ( l ) : I

Slope Fields and Euler's MethodIn Exercises 125-128, sketch part of the equation's slope field. Thenadd to your sketch the solution curve that passes through the pointP(1, -1). Use Euler's method with.t6 :7 and dx:0.2 to estimate)(2). Round your answers to 4 decimal places. Find the exact valueof y(2) for comparison.

CrnprBR AnnntoNnr ExsnclsBs-THEoRy. Exarraplss. AppllcetoNs

Additional Exercises-Theory,

125. y' :7

126. y' : l1a

128, y' : l ly

Examples,Appl icat ions 551

LimitsFind the limits in Exercises l-6

lb dxl . l i m r :

b-t Jo Jl 12

2. l im : I tan t td r

3. l im (cos 14;'/ '

4, Iim (a + e')2/'

/ t r r \5 . l im l ' + ' + . . + ' l

a 6 \ a + l a + 2 2 n l

1 / . . . . . . \6. l im 1l et/ , + e2/n + . . . + e(n-t)/n + er/n I, ' - , \

/

7, Let A(t) be the area of the region in the first quadrant enclosedby the coordinate axes, the curve ): e-", and the vertical linex : t, t > 0. Let y(r) be the volume of the solid generated byrevolving the region about the r-axis. Find the following limits.

a ) , l im

A( t ) b ) , l im v { r ) /A{ r ) c )

/ l im v { r } /A{ / )

8. Varying a logarithm's base

a) Find lim log"2 as a -+ 0+, 1-, l+, and oo.a r / . .Zi b) GRAPHER Craph y - log,2 as a function of d over the

in te rva l 0<d<4.

Determining Parameter Values9. Find values of a and , for which

,. sinax + br 4J>o rr 3

10. Find values of a and b for whichd cos i - cos ,.r

Theory and Examples11. Find the areas between the curves y:2(logu -r)/;r and l:

2(lognx)/x and the x-axis from .t: 1 to r - e. What is theratio of the larger area to the smallerl

EZ fZ. CnApUen Graph /(r) = tan I r + ran-r(1/.r) for -5 = r s5. Then use calculus to explain what you see. How would youexpect f to behave beyond the interr'al [-5, 5]? Give reasons foryour answer,

13. For what j > 0doesx(") : (rr)r? Give rcasons foryour answer

n tZ tq , CnapHen Graph / r , r r ' {s in r l " ' o \e r l0 . ln l .E \p la inwharyou see.

15 . F ind / ' (2 ) i f f (x ) :e lo ands( r ) : [ ^ -La t .

J ) \ - f t

16. a) Flnd dfldx rf

f"" 2lr. tf(r) : I - dt.

J t t

b) Find f(0).c) What can you conclude abour the graph of jr? Give reasons

for your answer

17. The figure here shows an informal proof that

- ^ - , I , - - r l ' rt a n t l a n - - -' * '

2 ' " * " 3 - 4 '

,-4-+'/

X / C

B

How does the argument go? (Source: "Beholdl Sums of Arctan,"by Edward M. Harris, College Mathematics "/oanral Vol. 18, No.2, March 1987, p. l4l .)

a) Why does Fig. 6.44 (on the following page) "prove" thatz' < e" ? (Source: "ProofWithout Words," by Fouad Naklil,Mathematics Maga.ine, Vol. 60, No. 3, June 1987, p. 165.)

lim


b) Figure 6.44 assumes that ./(,r) : (ln -r)/i has an absolutemaximum value at,r: ?. How do you know it does?

b)

19. Use the accompanying figure to show that

l " l 2 r f \I s i n r l - r - f s i n - ' r d r .

Jo z Jo

J

I l n , l n d I

b b a a

Explain what is going on in each case.

a)

20. Napier's inequality. Here arle two pictorial proofs that

(Source: Roger B. Nelson, College Mathemdtics JoumaL yol.21.

No. 2, March 1993, p. 165.)

21. Even odd decompositions

a) Suppose that g is an even function of -r and h is an oddtunction of i. Show that if 8(,r) + r(r) :0 for all -r then

8(.r) - 0 for al l x and lz(rc):0 for al l x.b) Use the result in (a) to show that if./(r.) : /F(.r) + l;(.r)

is the sutn of an even function /r(x) and an odd function

/o (i ), then

fFQ) = (f (r) + f ( r))/2 ^nd .fo\): (.f (x) - f (-x))12.

c) What is the signilicance of the result in (b)?

22. Let 8 be a function that is differentiable throughout an open inter,val containing the origin. Suppose 8 has the following properties:

I (.r) + B()) "i ) gf.r * 1) = ." I for al l recl numbers t. t . lndr . g ( . { ) 8 | ] 1

,r + ), in the domain of 8.i i ) hq g (e ) : 0

8 (h )

a) Show that g(0):0.b) Show rhat g'( ir) : 1 + [S(r)] 'c) Find g(x) by solving the differential equation in (b).

Applications23. Find the center of mass of a thin plate of constant density covering

the region in the first and fourth quadrants enclosed by the curves

f : 1 / ( 1 + , r2 ) and ) : 1 / (1 + r . 2 ) and by t he l i nes x : 0 andr - l

24. The region between the curve ) = llQtq) and the r-&\is fromt : | /4 to r - 4 is revolved about the i-axis to generate a solid.

a) Find the volume of the solid.b) Find the centroid of the region.

25. The Rule of 70. If yor use the approximation ln 2 ^, 0.70 (inplace of 0.69314. . .), you can derive a rule of thunb that says,"To estimate how many years it will take an amount of moneyto double when invested at / percent compounded continuously,divide r into 70." For instance, an amount of money invested

6,44 The figure for Exercise 18.

',1

at 59o wrll double in about 70/5 : 14 years. If you want it todouble in 10 years instead, you have to invest it at 70/10 :77o.

Show how the Rule of 70 is derived. (A similar "Rule of 72"uses 72 instead of 70, because 72 has more integer factors.)

Free fall in the fourteenth centu4z In the middle of the four-teenth celltury, Albeft of Saxony (1316-1390) proposed a modelof free fall that assumed that the velocity of a falling body wasproportional to the distance fallen. It seemed reasonable to thinkthat a body that had fallen 20 ft might be moving twice as fast asa body that had fallen l0 ft. And besides, none of the instrumentsin use at the time were accurate enough to prove otherwise. To-day we can seejust how far off Albeft of Saxony's model was bysolving the initial value Foblem implicit in his model. Solve theproblem and compale your solution graphically with the equations: 16t2. You witl see that it describes a motion that starts tooslowly at flrst and then becomes too fast too soon to be realistic.

The best branching angles for blood vesseii andplpes. whena smaller pipe branches off from a larger one in a flow system, wemay want it to run off at an angle that is best from some energy-saving point of view We might require, for instance, that energyloss due to friction be minimized along the section AOB shownin Fig. 6.45. In this diagram, B is a given point to be reachedby the smaller pipe, A is a point in the larger pipe upstreamfrom B, and O is the point where the branching occurs. A lawdue to Poiseuille states that the loss of energy due to friction innonturbulent flow is proportional to the length of the path andinversely proportional to the fourth power of the radius. Thus,the loss along AO is (kd1) / Ra and along OB is (kd) I ra , whereI is a constant, dr is the lcngth ofAO, d2 is the length of OB, Ris the radius of the larger pipe, and r is the radius of the smallerpipe. The angle d is to be chosen to minimize the sum of thesetwo losses:

Addit ional Exercises-Theory, Examples, Applications 553

We can express ihe total loss I as a function 016:

, , l o - b c o t 0 b c s c d \

\ R ' t ' )

a) Show that the critical value of d lbr which dLldg equalszero is

4 : t o t ' 4" R 4

ffi b) ClrCUrnrOR If rhe rario of rhe pipe radii is /./R :5/6,estimate to the nearest degree the optimal branching anglegiven in part (a).

The mathematical analysis described here is also used to explainthe angles at which afiedes branch in an animal's body. (SeeIntroduction to Mathematics for Life Scientists, Second Edition,by E. Batschelet [New York: Springer-Verlag, 1976].)

l2Z ZA, ctoup blood testing. During World War ll it was necessary toadminister blood tests to large numbers ofrecruits. There are twostandard ways to administer ablood test to 1y' people.In method l,each person is tested separately. In method 2, the blood samplesof r people are pooled and tested as one large sample. If thetest is negative, this one test is enough for all r people. If thetest is positive, then each of the I people is tested separately,requiring a total ol r + I tests. Using the second method andsome probability theory it can be shown that, on the average, thetotal number of tests ] will be

With q :0 99 and N : 1000, find the integer value of :t thatminimizes ],. Also find the integer value of .r that maximizes ).(This second result is not impoftant to the.eal-life situation.) Thegroup testing method was used in World War II with a savings of8070 over the individual testing method, but not with the givenvalue of 4.

29. Transport through a cell membrane. Under some conditionsthe result ofthe movement ofa dissolved substance across a cell'smembrane is described by the equation

d v A ,; k . , i ' - - J ) .

In this equation, ) is the concentration ol the substance insidethe cell ard dy /dt is the rate at which t changes over time. Theletters t, A, y, and c stand for constants, k beingthe pemeabilitycoeJrtcient (a property of the membrane), A the surface area ofthe membrane, y the cell's volume, and . the concenhation ofthe substance outside the cell. The equation says that the rate atwhich the concentration changes within the cell is proportionalto the difference between it and the outside concentration.

a) Solve the equation for l(r), using 16 to denote 1(0).b) Find the steady state concentration, lim,-- y(r).

(Based on Some Mathematical Models tr Biolog) by R. M.Thrall, J. A. Mortimer, K. R. Rebman, R. F. Baum, Eds.,Revised Edition, December 1967, PB-202 364, pp. l0l 103;distributed by N.T.l-S., U.S. Department of Commerce.)

i : r ( r - c '+1 )

t L , , _ d zF-" F '

d2sit 0

6.45 Diagram for Exercise 27.

In our model, we assume that AC : a and BC = b are lixed.Thus we have the relations

d t l d z c o s 0 : a d z s i n 0 : b ,

dz - b csc9,

d t - t t - d 2 c o s e - a b c o t e .

chapter 6 thomas 9 ed f 2008

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