Applications of the Derivative - Penn Math

4.6 Applied Optimization 51

A
h1 G1

G2 h2

B

Figure 12

This relation is called Snell’s Law of Refraction.

The time it takes a beam of light to travel from A to B is

ab x 2 + h12 + (L − x )2 + h 2
f (x) = + = 2
. (See diagram below.) Now
v1 v2 v1 v2
x L−x
f (x) = − = 0 yields
v1 x 2 + h 2
1 v2 (L − x )2 + h22

x x 2 + h 2 (L − x) (L − x )2 + h 2 sin θ1 sin θ2 , which is Snell’s Law.
1 2
= or =
v1 v2 v1 v2
h12 2
Since f (x) = + h 2 > 0 for all x, the minimum time is
v1
x2 + h 2 3/2 v2 (L − x )2 + h 2 3/2
1 2

realized when Snell’s Law is satisfied.

A a θ1 L − x
h1 θ1 x

θ2 h2
b θ2

B

52.

Snell’s Law, derived in the previous exercise, explains why it is impossible to see above the
53. TwhateerIniftethrveaalnMgleatotefrvsisioTnhθis2 eisxesurccihsethaantalyzes the solution to Example ?? in a more

hrgeiesgnpheewrcaatliyvseenlteytia.nrLgees. ttStuhpbeperoatshneechtdhiaasnttadvnddcaetnhfderovdmhissattiarhnene(cθtrh2ea)enf>rcsophmevvteo12dP.thltieomhtihtisgehocwnitaythy..eInLdetihrttePreobxaeadmtahnpedlep,hoviigndht=wona2y0t,h(,1e)
vh =(a5)5,Sho=w4thaantdifdth=is9i.nLeqeut axl0itbyehtohledso,ptthimenatlhvearleuiesonfoxanmgilneimθ1izfoinrgwthhiechtimSneeollf’sthLeatwriph.olds.
(a) O(bv)erWwhhaitcdhoclyoosuedthiinntkeryvoaul iws itlhleseoeptiifmyiozuatliooonktathkirnogugphlatchee?water with an angle of vision θ2
(b) FindsaucfhortmhautlaEqf.((x1))fiosrstahtiesfitiemde? of the trip (in terms of vd , vh, , and d) and prove that

x0 = vd / vh2 − vd2.
(c) This formula does not make sense if vh < vd . Why? Where does the minimum of f (x)

in [0, d] occur in this case?

52 Chapter 4 Applications of the Derivative

(d) Observe that the formula for x0 must also be wrong if > vh2 − vd2d/vd since this
would give us x0 > d. Explain why f (x) no longer represents the time of the trip when
x > d. What is x0 in this case?

Ranch City
(s 2+ x2)1/2

s

P d–x
x

Figure 13

(a) The interval of optimization is 0√≤ x ≤ d.
s2 + x2 d − x
(b) The time of the trip is f (x) = + . Solve
vd vh
x1
f (x) = √ − = 0 for x in the stated range to obtain
vd s2 + x 2 vh
s2
x = x0 = svd . Since f (x) = (s2 + x 2)3/2 > 0 for all x, we have that the
minimum vh2 − vd2 vd (x0) where x0 = svd .
vh2 − vd2
time for the trip is given by f

(c) If vh < vd in the formula for x0, then x0 is complex-valued, which is nonsensical. In

this case, f (d) gives the minimum time. In other words, drive in a straight line between

the ranch and the city on the dirt road to reach the destination as fast as possible (since

traveling over the dirt road is faster than traveling on the highway in this case).

d vh2 − vd2 , we then have x0 > d. But f (x0) no longer gives the
(d) In the event that s > vd

minimal time, since x0 > d means that the driver has to backtrack to the city along the

54. highway. In this case, x0 = 0 results in the minimal total time.
Poiseuille’s Law If blood flows into a straight blood vessel of radius r branching off a

55. lTahrgeeprrsotbraleigmhtisbtloopdutveas“sreolooff”roadf isuisdeRsaotnana arencgtlaenθg,lethoef thoetaiglhrteshisatnandcbeaTseobf. tFhienbdltohoed in

tshmeabllreasntclheinnggthvessfsoerl wishgiicvhenthbisyis possible.

a − b cot θ b csc θ
T =C +
R4 s r4

where a, b and C are constants. Show that thehtotaθl resistance is minimized when

cos θ = (r/R)4. b

Figure 14

Place the base of the rectangle along the x-axis in the x y-plane with its center at the origin.
Let a = b/2 be half the length of the base. (This gives a cleaner formulation in what is to
follow.) Let P be the point at the peak of the roof, Q be the point where the “right” side of
the roof (in the first quadrant) intersects the x-axis, and L the full length of the line segment
from P to Q; i.e., the slanted length of the roof on the right side. We’ll minimize the square
of this length L, an equivalent problem with simpler (hand) computations.

The right side of the roof touches the upper right corner of the rectangle at R (a, h), a
point in the first quadrant (i.e., a, h > 0). The line through P and Q is
y = h + m(x − a), where m is its slope. Here m is negative: −∞ < m < 0.

4.6 Applied Optimization 53

Hence P has coordinates (0, h − am) and Q has coordinates a − h , 0 . Via the
m
distance formula, we have that the square of the distance L is

f (m) = a − h 2 + (h − am)2 = (am − h)2 m2 + 1 .
m m2

2 (ma − h) m3a + h h1/3
Solve f (m) = m3 = 0 for m < 0 to obtain m = m0 = − a1/3 . Since
h1/3
f ↑ ∞ as m ↓ −∞ or m ↑ 0, the minimum value of f is f − a1/3 = a2/3 + h2/3 3,

the minimum value of L2, the square of the length L.

Accordingly the minimum value of L is L0 = a2/3 + h2/3 3/2 = (b/2)2/3 + h2/3 3/2.

h1/3 a + a1/3h2/3, 0 . The distance between
For m = m0 = − a1/3 , the coordinates of Q are
Q and R is

s = h2/3 a2/3 + h2/3 or s = h2/3 (b/2)2/3 + h2/3.

56. Let (a, b) be a point in the first quadrant (a, b both nonzero). Find the point (d, 0) on the
57. (p((A1ao9,sd2i−at8ipbv)te,)epdixsp-f.maro1xim8ins7imf–Ao1arP8lw.8roT.h)bhiAlceehmajetnhwisenweslMerurymaddxeoiempsfeiatgnhndaeensdrdoipsnMltaawinnnhscieemtotshafie,nrrRocbmoorg>ptehore√rJato3theharneaoescorponbom,i≤Anptmos√n.(e3M0na,ta.0mtE)h,ax.(dpMael,aooibnnf)t,ghwolyhldy3.5in

the shape of a frustum of a cone of height 1 cm and fixed lower radius r. The upper radius
x can take on any value between 0 and r. Note that x = 0 and x = r correspond to a cone
and cylinder, respectively. As a function of x, the surface area (not including the top and
bottom) is S(x) = π(r + x)s where s is the slant height as indicated in Figure 15. Which
value of x yields the least expensive design (the minimum value of S(x) for 0 ≤ x ≤ r)?

(a) Show that s = 1√+ (r − x)2, and hence S(x) = π(r + x) 1 + (r − x)2.
(b) Show that if r < 2, then S(x) is an increasing function. Conclude that the cone

(x = 0) has minim√al area in this case.
(c) Assume that r > 2. Show that S(x) has two critical points x1 < x2 in (0, r) and that

S(x1) is a local maximum, S(x2) is a local minimum.
(d) Conclude that the minimum value occurs either at x = 0 or at x = x2.
(e) Where does the minimum value occur in the cases r = 1.5 and r = 2?

√
(f) Challenge: le√t c = (5 + 3 3)/4 ≈ 1.5966. Prove that the minimum occurs at x = 0

(cylinder) if 2 < r < c but the minimum occurs at x = x2 if r > c.

x

s
r

Figure 15

(a) Consider a cross-section of the object and notice a triangle can be formed with height
1, hypotenuse s, and base r − x. Then, by the Pythagorean Theorem,

s = 1 + (r − x)2 and the surface area S = π(r + x)s = π(r + x) 1 + (r − x)2.

(b) S (x) = π 1 + (r − x)2 − (r + x)(1 + (r − x)2)−1/2(r − x) =

π 2x2 − 2r x + 1 = 0 yields critical points x = 1 r ± 1 √r2 − 2. If r < √ then there
1 + (r − x)2 2 2 2

are no real critical points. Sign analyses reveal that S(x) is increasing everywhere and

thus the minimum must occur at the left endpoint, x = 0.

54 Chapter 4 Applications of the Derivative

√√
(c) For r < 2, there √are two critical points, x = 1 r ± 1 r2 − 2. Bo√th values are in the
2 2

interval since r > r2 − 2√. Sign analyses reveal that x = 1 r + 1 r2 − 2 is a local
2 2

minimum and x = 1 r − 1 r2 − 2 is a local maximum.
2 2

(d) Depending o√n the value of r, the minimum of S(x) occurs at either x = 0 or
x = 1 r + 1 r 2 − 2.
2 2 √

(e) 1.5 and 2 are both greater than √2. This means that, for both values of r, the local
minimum occurs at x = 1 r + 1 r2 − 2. If r = 1.5, S(x2) = 8.8357 and
2 2
S(0) = 8.4954 , so S(0) = 8.4954 is the local minimum (cylinder). If r = 2,

S(x2) = 12.852 and S(0) = 14.0496, so S(x2) = 12.852 is the minimum.

(f)
58.

For any positive numbers x and y, the quantities on the left and right of the following

inequality are called the geometric mean and the arithmetic mean:
Seismic Prospecting (Adapted from B. Noble, Applications of Undergraduate Mathematics

in Engineering, Mlayaecrmoilflasoni1l 9th6a7t.)liEesxeorncitsoep√so5xf9ya–≤6ro1cxakr+2efocyromnacteironne.dGweoitlhogdiesttesrsmeinndintgwothseound (2)
thickness d of a

pulses from point A to point D separated by a distance s. The first pulse travels directly from A
atoloDngailtosn(sagu)rthfmPaecriosenu,virmeafnaEudcmqet.hov(e2fan)ltuhbueeasicoenkafgrutchpa.ltTcouhDleuss(epacasotfhnodAllpBouwClssD:e)ftoararsveienalcsFhdigfiouwxreend1tpo7o.tshTiethivreoepcnukulsfmoerbtmrearavtSeio,lsnfiw,ntdihthethne

velocity v1 in the soil and v2 in the rock. √xy − x + y
2

for all positive x anAd y satisfyings x + y = S. D

(b) FPrigouvreeit1a6g. aSihnogweothmaetθthric=all√y xbyys,oraeinlfderoribnsgertvoethtθheastehmdiiscinrocltegorefadteiarmtheatenrtxhe+raydiiuns.

(c) Verify the yal)g2,ebanradicusreelatBhtiiosnto4gxriyovce=ka(txhi+rd Cy)2 − (x − y)2. Conclude that 4x y is larger
than (x + proof of Eq. (2).

Figure 17

Figure 16
59. (a) Show that the time required for the first pulse to travel from A to D is t1 = s/v1.

(b) Show that the time required for the second pulse is

2d s − 2d tan θ
t2 = v1 sec θ +
v2

provided that

tan θ ≤ s (3)
2d

(Note: if this inequality is not satisfied, then point B does not lie to the left of C.)
(c) Show that t2 is minimized when sin θ = v1/v2.

Seismic Prospecting.

(a) We have time t1 = distance/velocity = s/v1.

(b) Let p be the length of the base of the right triangle (opposite the angle θ ) and h the

length of the hypotenuse of this right triangle. Then cos θ = d whence h = d sec θ .
d tan θ . Hence h
Moreover, tan θ = p gives p =
d

t2 = tAB + tC D + tBC = h + h + s − 2p = 2d sec θ + s − 2d tan θ
v1 v1 v2 v1 v2

(c) Solve dt2 = 2d sec θ tan θ − 2d sec2 θ = 0 to obtain tan θ = sec θ whence
dθ v1 v2 v1 v2
sin θ/ cos θ = v1 or sin θ = v1 .
1/ cos θ v2 v2

4.7 Newton’s Method 55

60.

61. In this exercise, assume that v2/v1 ≥ 1 + 4(d/s)2.
Cont(ian)ueSwhoitwh tthhaetaisnseuqmuapltiitoyn(o3f) thhoeldpsrefvoirotuhseeaxnegrlceisθe.such that
(a) F(Abin)adnSdthhBoewtahrtiehcakstenptehasesramotefidnthibmeytaoalptdilmiastyeaenfroc, reatsshseu=mse1icn5og0n0tdhfapt.tuvls1e=is.7v2 sin θ = vt21//tv12 . 1.3 when
and that =

(b) The times t1, t2 and the distance s aatrl2ein=meaer2vadf1suu(nr1ec−dtioeknx2po)e1f/r21im/+se.nvsWt2a.lhlya.t The equation in
Exercise 60(c) shows that t2/t1 is might you conclude
if

several experiments were formed for different values of s and the plot of the points
((1c/) sCw, toh2n/erct1el)ukddei=dthnvao1t/tvtl2i2e.=on2ads(t1ra−igkh2t)l1i/n2e+? k.

t1 s

(a) Substituting k = v1/v2 = 0.7, t2/t1 = 1.3, and s = 1500 into (*) gives

1.3 = 2d 1 − (0.7)2 + 0.7. Solving for d yields d ≈ 630.13 ft.
1500

(b) If several experiments for different values of s showed that plots of the points 1 , t2
s t1

did not lie on a straight line, then we would suspect that t2 is not a linear function of 1
t1 s

and that a different model is required.

4.7 Newton’s Method
Preliminary Questions

1. How many iterations of Newton’s Method are required to compute a root if f (x) is a linear
function?

2. What happens if you start Newton’s method with an initial guess that happens to be a zero
of f ?

3. What happens if you start Newton’s method with an initial guess that happens to be a local
minimum or maximum of f ?

4. For which of the following two curves will Newton’s Method converge more rapidly?

Figure 1

5. Is the statement “A root of the equation of the tangent line to f (x) is used as an
approximation to a root of f (x) itself” a reasonable description of Newton’s method?
Explain.

56 Chapter 4 Applications of the Derivative

Exercises

1. Use Newton’s Method to calculate the approximations x1, x2, x3 to the root of the equation
f (x) = x2 − 10. Take x0 = 3.

Let f (x) = x2 − 10. Define g(x) = x − f (x) =x− x2 − 10 = 3, and
f (x) 2x . Take x0

xn+1 = g(xn). We get the following table:

n1 2 3 4

2. xn 3.166666666 3.162280701 3.162277660 3.162277660
Choose an initial guess x0 and use Newton’s Method to calculate the approximations

3. Lx1e,txf2,(x)3 t=o ax3so−lu7tixo+n o3f. tFhiendeqvuaaltuieosnacoasnxd b=sxuc. h that f (a) > 0 and f (b) < 0 by trial and
error. Then use Newton’s Method to find the root of f lying between a and b to three

decimal places.

Let f (x) = x3 − 7x + 3. Define g(x) = x − f (x) =x− x3 − 7x + 3 .

f (x) 3x2 − 7

Taking x0 = −3, we have

n 1 2 3 4 .
xn −2.85 −2.838534619 −2.838469254 −2.838469252

Taking x0 = 0, we have

n 1 2 3 4 .
xn 0.428571429 0.440777577 0.440807711 0.440807712

Taking x0 = 2, we have

n 1 2 3 4 5 .
xn 2.6 2.421084337 2.398036788 2.39766164 2.39766154

4.
Use Newton’s Method to calculate 71/4 to four decimal places.

In Exercises 5–6, calculate x1, x2, x3 using Newton’s Method for the given function and initial
guess. Compare x3 with the value obtained from a calculator.

5. f (x) = x2 − 7, x0 = 2.5

Let f (x) = x2 − 7. Define g(x) = x − f (x) =x− x2 − 7
f (x) 2x . Take x0 = 2.5.

n1 2 3 4

6. xn 2.65 2.645754717 2.645751311 2.645751311
f (x) = x3 − 5, x0 = 1.6
7. Choose an initial guess x0 and calculate x1, x2, x3 using Newton’s Method to approximate a

solution to the equation ex = 3x + 2.

Let f (x) = ex − 3x − 2. Let g(x) = x − f (x) = x − ex − 3x − 2 Let x0 = 0.
f (x) ex −3 .

n1 2 3 4

8. xn -0.5 -.455491111 -.455233364 -.455233355
9.
Find a solution to the equation acsshineincxkE=txhacamot sipt2liexs?ci?no.rtrUheesceti,nNatenerdwvactolon[m0’sp, Maπ2r]e to three decimal places.
LTehtenf (gxu)es=s txh4e −ex6axct2 s+oluxt+ion5, twhoitdh tyooauprproximate the
laaprpgreosxt ipmoastiitoivne. root of f (x) to within an error of at most 10−4. (Refer to Figure ??.)

The largest positive root of f (x) is x = 2.09306435. Let f (x) = x4 − 6x2 + x + 5. Define

f (x) x4 − 6x2 + x + 5 .
g(x) = x − =x−
f (x) 4x3 − 12x + 1

Take x0 = 2.
n1 2 3

xn 2.11111111 2.09356846 2.09306477

4.7 Newton’s Method 57

10.
Use Newton’s Method to calculate 112/3 to four decimal places.

11. Use Figure 2 to choose an initial guess x0 to the unique real root of x3 + 2x + 5 = 0. Then
compute the first 3 iterates of Newton’s Method.

–2 –1 12

Figure 2 Graph of y = x3 + 2x + 5.

Let f (x) = x3 + 2x + 5. Define g(x) = x − f (x) x3 + 2x + 5
f (x) = x − 3x2 + 2 . Take x0 = −1.4,

based on the figure.

n1 2 3

12. xn −1.330964467 −1.328272820 −1.328268856
The equation x3 − 4x + 1 = 0 has three real roots. Use Newton’s Method to approximate

13. tGheUtwoUpsoesiatigvrearpohointsgtocawlciuthlaintoarntoercrhooroosfeaatnminoisttia1l0g−u3.ess for the unique positive root of

x4 + x2 − 2x − 1 = 0. Calculate the first three iterates of Newton’s Method.

Let f (x) = x4 + x 2 − 2Fxig−ur1e. D3efinGeragp(hx )of=f x(x−) f ((xxx3))−=4xx +− x 4 + x2 − 2x − 1 . Take
=f 1. 4x3 + 2x − 2

x0 = 1.

n1 2 3

14. xn 1.25 1.189379699 1.18417128
GU Find the unique root of x3 + 2x√− 14 = 0 using Newton’s Method to within an error
15. Sofhoatwmthoast 1th0e−s3e. qUuseenacegrcaopnhvienrggicnaglctuolatocrotbotcahinoeodsebyoauprpilnyitnigalNgeuwestso.n’s Method is
1 c
defined by xn+1 = 2 (xn + xn ).

Let f (x) = x2 − c. Define g(x) = x − f (x) x2 − c x2 + c 1 x+ c .
=x− = =
f (x) 2x 2x 2 x

1c
16. Then xn+1 = 2 xn + xn .

The first positive solution of sin x = 0 is x = π . Use Newton’s Method to calculate π to
17. fWouhratdehcaipmpaelnsplwacheesn. you apply Newton’s Method to find a zero of f (x) = x1/3? Note that

x = 0 is the only zero.

Let f (x) = x1/3. Define g(x) = x − f (x) =x− x 1/3 = x − 3x = −2x. Take
f (x)
1 x −2/3
3
x0 = 0.5. Then the sequence of iterates is −1, 2, −4, 8, −16, 32, −64, . . . That is, for any
nonzero starting value, the sequence of iterates diverges spectacularly, since xn = (−2)n x0.

18. Thus limn→∞ |xn| = limn→∞ 2n |x0| = ∞.
What happens when you apply Newton’s Method to the equation x3 − 20x = 0 with the

unlucky initial guess x = 2?

Further Insights and Challenges

19. Let c be a positive number and let f (x) = x−1 − c.
(a) Show that x − ( f (x)/ f (x)) = 2x − cx2. Thus Newton’s Method provides a way of
computing reciprocals without performing division.

58 Chapter 4 Applications of the Derivative

(b) Calculate the first three iterates of Newton’s Method with c = 10.324 and the two
initial guesses x0 = .1 and x0 = .5.

(c) Explain graphically why x0 = .5 does not yield a sequence of approximations to the
reciprocal 1/10.324.

(a) Let f (x) = 1 − c. Define g(x) = x − f (x) = x − 1 − c = 2x − cx2.
x f (x) x

−x −2

(b) For c = 10.324, we have f (x) = 1 − 10.324 and thus g(x) = 2x − 10.324x2.
x

Take x0 = 0.1. n 1 2 3
Take x0 = 0.5. xn 0.09676 0.096861575 0.096861682

n 1 2 3
xn −1.581 −28.9674677 −8.72094982

(c) The graph is disconnected. If x0 = .5, (x1, f (x1) is on the other portion of the graph,

which will never converge to any point under Newton’s method.
20.
21. LT−eh3te.5ff8u(3nθ,c).t2=io5n1s3ifn,(θaxn).d=3.133x332.−D4exte+rm1inheatshtehrroeeotzteorowehsiwchhiNchewatroena’pspMroexthimodatceolynverges for

the initial choiθces x0 = 1.85, 1.7, and 1.55. The answer shows that a small change in initial
g(au)esUs sxi0ngcaanghraavpehianbgicgaelcffuelcattoornotrhoetohuetrcmometehodf,Ncehwootosne’asnMineitthiaoldg. uess for a solution to

f (θ ) = .9.

(b) Use Newton’s Method to improve the guess to three decimal accuracy.

Figure 4 Graph of f (x) = 1 x 3 − 4x + 1.
3
Let f (x) = sin x and let 0 < α < 1.

x

(a) Let h(x) = sin x − α. Define
x
sin x − α
g(x) = x − h(x) =x− x = x (x cos x − 2 sin x + αx) . The
h (x)
x cos x − sin x · 1 x cos x − sin x

x2
sequence of iterates is given by xn+1 = g (xn).

(b) For x0 = 0.8, we have f (x0) − α = sin 0.8 − 0.9 ≈ −0.0033. So let’s use x0 = 0.8 for
0.8
our initial guess.

n 1 2 3
(c) Then xn 0.786779688 0.786683077 0.786683072

In the next two exercises (adapted from Animating Calculus, Packel and Wagon, p. 79), we
consider a metal rod of length L inches that is fastened at both ends. If you cut the rod and weld
on an additional m inches of rod leaving the ends fixed, the rod will bow up into a circular arc
of radius R (unknown) as indicated in Figure 5.

h

L
Rθ

Figure 5 The bold circular arc has length L + m.

4.8 Antiderivatives 59

22.

Our goal is to determine the maximum vertical displacement of the rod h when the

23. Lqueat nLtit=ies1Lanadndmm=a3re. given.

(a) U(as)e SNheowwtotnh’ast ML e=tho2dRtsoinfiθndanthdecaonngclleudθestahtiastfying Eq. (1) to within an error of 10−3.

Use a graphing calculator to make an initial guess.
L(1 − cos θ )
(b) Estimate the value of h. h = 2 sin θ

Let LFE(brx=o)emrSc1hiEsaoexnwed2r1tcmhiwsae=titt?hh3?eα.(ba=n),gwl41eepθdroesdadtuuicscfieesetshthateht efsoienθlqsloθiuθnwa−θtiino=gnLN+LLewm+Ltom=n sin θ −1 = 0. Applying
(a) 0 or 4 converge to θ .

θ which (1)
iterates,

n0 1 2 3 4 . That is,
θn Hin2t: sh2o.4w70fi0r2st7t0h0a8t L +2.4m74=5724R6θ2.2 2.474576787 2.474576787

θ ≈ 2.474567687.

(b) Accordingly, h = L (1 − cos θ ) ≈ 1.443213471.
2 sin θ
24.

Quadratic Convergence to Square Roots
Let f (x) = x2 − c.

(a) Show that if xn = 0, then the formula for xn+1 in Newton’s Method is

4.8 Antiderivatives 1c
Preliminary Questions xn+1 = 2 xn + xn

Exercises (b) Let en = √ = en2/2xn.
xn − c √be the error in√xn. Prove that en+1
(c) Show that if x0 > c√then xn > c for all n. Explain this result graphically.
(d) aAnstisduemriivnagtitvheaot fx0th>e ≤ en2 /2c.
1. Find an c, show (thxa)t=en+01.
function f

2. What is the difference, if any, between finding the general antiderivative of a function f (x)
and evaluating f (x) dx?

3. Joe happens to know that two functions f (x) and g(x) have the same derivative, and he
would like to know if f (x) = g(x). Does Joe have sufficient information to answer his
question?

4. Is y = x2 a solution to the differential equation with initial condition:

dy = 2x, y(0) = 1
dx

5. Write any two antiderivatives of cos x. Which initial conditions do they satisfy at x = 0?

6. Suppose that F (x) = f (x) and G (x) = g(x). Are the following statements true or false?
Explain.

(a) If F and G differ by a constant, then f = g.
(b) If F = G, then f = g.
(c) If F and G differ by a constant, then f and g differ by a constant.
(d) If f = g, then F = G.

1. Which of (A) or (B) is the graph of an antiderivative of f (x) in Figure 1?

60 Chapter 4 Applications of the Derivative

f (x) (A) (B)

Figure 1

The graph of the antiderivative F(x) of f (x) is (A); because the graph of f (x) = F (x) is

2. that of the slopes of F(x). (C) is not the graph of the antiderivative of f (x) in Figure 2?
Which of the graphs (A), (B),

Find the general antiderivative of the following functions in Exercises 3–6.

3. 12x Figure 2

12x dx = 12 x dx = 12 · 1 x 2 + C = 6x 2 + C.
2
4.
x2
5. x2 + 3x + 2

x2 + 3x + 2 dx = x2dx + 3 x1 dx + 2 x0 dx = 1 x 3 + 3 · 1 x 2 + 2· 1 x 1 + C =
3 2 1
1 3
3 x 3 + 2 x 2 + 2x + C.

6. 8x −4

7. Evaluate (x2 + 1) dx. Check your answer by differentiation.

x2 + 1dx = 1 x 3 + x +C since
3
d 1 d 1 d d
3 x3 + x + C = dx 3 x 3 + dx (x) + dx (C ) = x2 + 1 + 0 = x2 + 1.
8. dx
Evaluate (cos x + 3 sin x) dx. Check your answer by differentiation.
9. Find the solution of the differential equation d y/dx = x3 with initial condition y(0) = 2.

Since dy = x3, we have y = x3 dx = 1 x 4 + C. Thus 2 = y(0) = 1 (0)4 + C, whence
dx 4 4
dy
C = 2. Therefore, y = 1 x 4 + 2. As a check, note that dx = d 1 x 4 + 2 = x 3 and
4 dx 4
(0)4
10. y(0) = 1 + 2 = 2, as required.
4

Find the solution of the differential equation d y/dx = cos 2x with initial condition

11. Ay(0p)ar=tic3le. located at the origin at t = 0 begins moving along the x-axis with velocity

v(t ) = 1 t 2 − t ft/s. Let s (t ) be the particle’s position at time t. State the differential
2
equation with initial condition satisfied by s(t) and find s(t).

Given that v(t ) = ds and that the particle starts at the origin, the differential equation is
dt
ds 1
dt = 2 t 2 − t and the initial condition is s(0) = 0. Accordingly, we have

s(t) = 1 t 2 − t dt = 1 t2 dt − t dt = 1 · 1 t 3 − 1 t 2 + C = 1 t 3 − 1 t 2 + C. Thus
2 2 2 3 2 6 2
1 (0)3 1 (0)2 1 1
0 = s(0) = 6 − 2 + C, whence C = 0. Therefore, the position is s(t) = 6 t 3 − 2 t 2 .

12.

A particle begins moving along the x-axis with velocity v(t) = 25t − t2 ft/s at t = 0. Let
13. sA(tp)abrteictlheelpoacrattiecdlea’st tphoesiotrioigninatatimt =e t0. begins moving in a straight line with acceleration

a(t)(=a)4S−tat21et tfhte/sd2.ifLfeertevn(ttia)lbeeqtuhaetivoenlowciitthy iannitdiasl(ct)onbdeittihoendsiasttiasnficeedtrbayvesl(etd).at time t.
(a) S(bta)teFainndd ss(otl)veastshuemdiinffgerthenetpiaalrteiqculeatiisolnocaassteudmaitngx t=he5paatrttic=le 0is. at rest at t = 0.
(b) F(cin)dFsi(ntd).s(t) assuming the particle is located at x = 5 at t = 2.

(a) The differential equation is a(t) = dv =4− 1 t and its initial condition is v(0) = v0
dt 2
where v0 is the (unspecified) initial velocity of the particle. We have

v(t) = 4 − 1 t dt = 4 dt − 1 t dt = 4t − 1 t 2 + C. Thus
2 2 4
1 (0)2
v0 = v(0) = 4 (0) − 4 + C, whence C = v0. Therefore, the velocity is

v(t ) = 4t − 1 t 2 + v0.
4

4.8 Antiderivatives 61

(b) In (a), we saw that v(t) = ds = 4t − 1 t 2 + v0. Accordingly,
dt 4

s(t) = 4t − 1t2 + v0 dt = 4· 1t2 − 1 · 1t3 + v0t + K = 2t 2 − 1 t3 + v0t + K
4 2 4 3 12

Thus 8 = s(0) = 2 (0)2 − 1 (0)2 + v0 (0) + K, whence K = 8. Therefore, the position
12
1
is s (t ) = 2t 2 − 12 t 3 + v0t + 8.

14.

Match each of the functions

In Exerci(sae)s 1si5n–x30, evaluate the indefinite integral.
(b) x sin(x2)

15. (x (+c)1)sidnx(1 − x)

16. x +(d)1 adxnxstii=dnexr21ivxa2ti+vex: + C.
with its
17. ((t95−(+i)53xct)o+sd(x21)−dtx)
(ii) − cos x
t 5(i+ii)3t−+21 2codst(x=2)61 t 6 + 3 2 + +
2 t 2t C.
18.
8t(−iv4)dtsin x − x cos x
19. t −9/5 dt

t −9/5 dt = − 5 t −4/5 + C.
4
20.
(4 sin t − 3 cos t) dt
21. (5x 3 − x −2 − x 3/5) d x

5x 3 − x −2 − x 3/5 d x = 5 x 4 + x −1 − 5 x 8/5 + C.
4 8
22.
2 dx

23. √1 dx
x

24. √ x −1/2 dx = 2x 1/2 + C = √ + C.
1/ x dx = 2x

(5x − 9) dx

25. (4 sin x − 3 cos x) dx

4 sin x − 3 cos x dx = −4 cos x − 3 sin x + C.
26.

sin 4x dx
27. (x 3 + 4x −2) d x

x3 + 4x −2 d x = 1 x 4 − 4x −1 + C.
4
28.
√x −2 d x
29. x dx

√ x 1/2 dx = 2 x 3/2 + C.
x dx = 3
30.
(x + 3)−2 d x

31. Use the formulas for the derivatives of tan x and sec x to evaluate the integrals

(a) sec2(3x) d x

(b) sec(x + 3) tan(x + 3) dx

Recall that d (tan x) = sec2 x and d (sec x) = sec x tan x.
dx dx

(a) Accordingly, we have sec2(3x) d x = 1 tan(3x) + C.
3

32. (b) Moreover, we see that sec(x + 3) tan(x + 3) dx = sec(x + 3) + C.

Use the formulas for the derivatives of cot x and csc x to evaluate the integrals
(a) csc2 x d x
(b) csc x cot x dx

62 Chapter 4 Applications of the Derivative

33. A 900 kg rocket is fired from a spacecraft. As it burns fuel, its mass decreases and its
velocity increases. Let v be the velocity as a function of mass m. Find the velocity when
m = 500 if

dv/dm = −m−1/2

Assume that v(900) = 0 (the velocity when first fired is 0).

Since dv = −m −1√/2 , we have v(m) = −m−1/2 dm = −2m1/2 + C . Thus √
dm

0 = v(900) = −2 900 + C =√−60 + C, when√ce C = 60. Therefore, v(m) = 60 − 2 m.
Accordingly, v(500) = 60 − 2 500 = 60 − 20 5 ≈ 15.28.
34. Find constants c1 and c2 such that F(x) = c1x sin x + c2 cos x is an antiderivative of

35. (fa()x)Sh=owx cthoastxt.he functions f (x) = tan2 x and g(x) = sec2 x have the same derivative.

(Hbi)ntdW:icrheaaclcttlucyla.anteyoFu(cxo)nacnluddseolavbeofuotrthce1,rce2l.ation between f and g? Verify this conclusion

Let f (x) = tan2 x and g(x) = sec2 x.

(a) Then f (x) = 2 tan x sec2 x and g (x) = 2 sec x · sec x tan x = 2 tan x sec2 x, whence
f (x) = g (x).

(b) Accordingly, the antiderivatives f (x) and g(x) must differ by a constant; i.e.,
f (x) − g(x) = tan2 x − sec2 x = C for some constant C. To see that this is true
directly, divide the identity sin2 x + cos2 = 1 by cos2 x. This yields
tan2 x + 1 = sec2 x, whence tan2 x − sec2 x = −1.

In Exercises 36–45, solve the differential equation with initial condition.
36. d y
= x; y(0) = 5
37. dd xy = 0; y(3) = 5

dt

Since dy = 0, we have y = 0 dt = C. Thus 5 = y(3) = C, whence C = 5. Therefore,
dt
y ≡ 5.
38.
dy
39. ddyt = 5 − 2t2; y(1) = 2
= 8x3 + 3x2 − 3; y(1) = 1
dx

Since dy = 8x3 + 3x2 − 3, we have y = 8x3 + 3x2 − 3 d x = 2x4 + x3 − 3x + C. Thus
dx
1 = y(1) = 0 + C, whence C = 1. Therefore, y = 2x4 + x3 − 3x + 1.
40.
dy
= 4√t + 9; y(0) = 1
41. ddyt = t; y(1) = 1

dt √
t
Since dy = = t 1/2, we have y = t 1/2 dt = 2 t 3/2 + C. Thus 1 = y(1) = 2 + C,
dt 3 3
1 2 t 3/2 1
42. whence C = 3 . Therefore, y = 3 + 3 .
43.
dy = sin x; y π =1
dd xy = sin 2z; y 2π =4

dz 4

Since dy = sin 2z, we have y = sin 2z dz = − 1 cos 2z + C. Thus 4 = y π = 0 + C,
dz 2 4

44. whence C = 4. Therefore, y = 4− 1 cos 2z.
2

dy = cos 5x; y(π )π= 3
= sec2 3x; y =
45. dd xy 2

dx 4

Since dy = sec2 3x, from Exercise 31(a) we have y = sec2 3x dx = 1 tan 3x + C. Thus
dx 3
π 1 7 7 1
2=y 4 = − 3 + C, whence C = 3 . Therefore, y = 3 + 3 tan 3x.
46.
Find the general antiderivative of sin 3t.

Hint: try to guess the answer, based on the formula d/dt(cos 3t) = −3 sin 3t.

4.8 Antiderivatives 63

47. Find the general antiderivative of cos 9t.

48. cos 9t dt = 1 sin 9t + C.
9

Find the solution of the differential equation d y/dt = cos 2t with initial condition

49. Fyi(n0d) t=he3g. eneral antiderivative of (2x + 9)10.

Hint: try to guess the answer, based on the formula d (2x + 9)11 = 22(2x + 9)10.
dx

By the Chain Rule, we have d/d x (2x + 9)11 = 11 (2x + 9)10 · 2 = 22 (2x + 9)10. Hence

d 1 (2x + 9)11 = (2x + 9)10. Thus (2x + 9)10 dx = 1 (2x + 9)11 + C.
22 22
50. dx
Find the general antiderivative of cos(2x + 9).

Further Insights and Challenges

51. Suppose that F (x) = f (x) and G (x) = g(x). Is it true that F(x)G(x) an antiderivative of
f (x)g(x)? Confirm or provide a counterexample.

Let f (x) = x2 and g(x) = x3. Then F(x) = 1 x3 and G(x) = 1 x4 are antiderivatives for
3 4
f (x) and g(x), respectively. Let h(x) = f (x)g(x) = x5, the general antiderivative of

which is H (x ) = 1 x 6 + C. There is no value of the constant C for which
6
1
F (x )G (x ) = 12 x 7 equals H (x ). Accordingly, F (x )G (x ) is not an antiderivative of

h(x) = f (x)g(x). Therefore, it is not true that P (x) = p(x) and Q (x) = q(x) imply that

P(x)Q(x) is an antiderivative of p(x)q(x).
52.

Suppose that F (x) = f (x).

53. Find(aan) aSnhtoidwertihvaattiv12 eFf(o2rx)f (isxa)n=an|xti|d. erivative of f (2x).

Let (b) )F=ind|xth| e=gen−xerxal afnotridxer≥iva0tive of f (kx) for any constant k. f (x)
f (x . Then the general antiderivative of is
for x < 0

x dx for x ≥ 0 = 1 x 2 + C for x ≥ 0
−x dx for x < 0 2 for x < 0
F(x) = f (x) dx =
− 1 x2 + C
2

54. The Power Rule for antiderivatives does not give us a formula for an antiderivative F(x) of
f (x) = x−1. However, we can see what the graph of an antiderivative would have to look

like. Which of the graphs in Figure 3 could plausibly represent F(x)?

Figure 3