Chapter 8 Dynamics of Change: Differential...

Copyright © Houghton Mifflin Company. All rights reserved. 475

Chapter 8 Dynamics of Change: Differential Equations and Proportionality

Section 8.1 Differential Equations and Slope Fields 1. c = kg

2. dc

p kdp

⋅ = so dc k

dp p=

3. dp

daka=

4. dc

kdw

=

5. a.

One possible answer: The displayed particular solutions go through the points (0, 0.5), (2, 2) and (4, 5).

b. All particular solutions are horizontal lines, each passing through the chosen initial condition.

c. y C= , where C is a constant

6. a.


b. All particular solutions are lines with slope 1, each passing through the chosen initial condition.

c. y x C= + , where C is a constant

7. a.


b. All particular solutions are parallel lines with slope –1 and differing vertical shifts.

c. y x C= − + , where C is a constant

476 Chapter 8: Dynamics of Change: Differential Equations Calculus Concepts

Copyright © Houghton Mifflin Company. All rights reserved.

8. a.


b. All particular solutions are lines

with slope 12 , each passing through

the chosen initial condition.

c. 12y x C= + , where C is a constant

9. a.

One possible answer: The displayed particular solutions go through the points (0, 1), (1, 3) and (2, 6.5).

b. All particular solutions are concave-up parabolas with minimum points on the vertical axis.

c. 20.25y x C= + , where C is a

constant

10. a.

One possible answer: The displayed particular solutions go through the points (0, 1), (1, −3) and (2, 6.5).

b. All particular solutions are parabolas, each passing through the chosen initial condition.

c. 2y x C= + , where C is a constant

11. a.

One possible answer: The displayed particular solutions go through the points (0, 1), (2, 3) and (2, 6.5).

b. All particular solutions are concave-down parabolas with maximum points on the vertical axis.

c. 20.5y x C= − + , where C is a

constant 12. The particular solutions with initial

condition (0, 0) are all lines that pass through the point (0,0). The slopes of the lines differ.

13. The particular solutions with initial

condition (0, 0) all pass through the point (0,0). Each particular solution has

Calculus Concepts Section 8.1: Differential Equations and Slope Fields 477


a constant term zero, but the nonconstant terms all differ.

14. a. df

dxk= b. f(x) = kx + C

c. Taking the derivative of the function in part b, we get d

dxkx C k( )+ = . Thus we have the

identity k = k, and our solution is verified.

15. a. Let p be the energy production in

the United States in quadrillion Btu t years after 1975. dp

dt= 0 98. quadrillion Btu/year

t years after 1975

b. p(t) = 0.98t + C quadrillion Btu t years after 1975

c. p(5) = 0.98(5) + C = 64.8, so C = 59.9 p(t) = 0.98t + 59.9 quadrillion Btu

t years after 1975

d. p(0) = 59.9 and dp

dt= 0 98. when

t = 0. In 1975 the production was 59.9 quadrillion Btu and was increasing at a rate of 0.98 quadrillion Btu per year.

e. Particular solution with p(0)=59.9

16. a. Let c be the energy consumption in the United States in quadrillions of Btu t years after 1975. dc

dt= 108. quadrillion Btu per year

t years after 1975

b. c(t) = 1.08t + C quadrillion Btu t years after 1975

c. c(5) = 1.08(5) + C = 76.0, so C = 70.6 c(t) = 1.08t + 70.6 quadrillion Btu t years after 1975

d. c(0) = 70.6 and dc

dt= 108. for t = 0.

In 1975 the consumption was 70.6 quadrillion Btu and was increasing at a rate of 1.08 quadrillion Btu per year.

e. Particular solution through (5, 76).

The graph crosses the vertical axis

around 70.5. Thus we estimate that the energy consumption in 1975 was 70.5 quadrillion Btu.

17. a. Let c be the amount of arable and

permanent cropland in millions of square kilometers t years after 1970. dc

dt= 0 0342. million square

kilometers per year t years after 1970

b. c(t) = 0.0342t + C million square kilometers t years after 1970



c. c(10) = 0.0342(10) + C = 14.17, so C = 13.828 c(t) = 0.0342t + 13.828 million square kilometers t years after 1970

d. c(0) = 13.828, c(20) = 14.512 and dc

dt= 0 0342. when t = 0 and t = 20.

Cropland was increasing at a rate of 0.0342 million square kilometers per year in both 1970 and 1990. In 1970 there were 13.828 million square kilometers of cropland, and in 1990, there were 14.512 million square kilometers of cropland.

18. a. dp

daa= ⋅ −( . )37 10 5 inches of

mercury per foot at an altitude of a feet

b. 5 2( ) (1.85 10 )p a a C−= ⋅ + inches of mercury at an altitude of a feet

c. p C( ) ( . )0 185 10 0 305 2= ⋅ + =− , so

C = 30

p a a( ) ( . )= ⋅ +−185 10 305 2 inches of mercury at an altitude of a feet

d. Particular solution with p(0) = 30

19. a. v(t) = −32t feet per second t seconds

after the object is dropped

b. Let s be distance. ds

dtt= −32 feet per second t seconds

after the object is dropped

c. s t t C( ) = − +16 2 feet t seconds after the object is dropped

d. s C( ) ( )0 16 0 352= − + = , so C = 35

s t t( ) = − +16 352

When s(t) = 0, t ≈ ±1.479, where only the positive answer makes sense in this context. It takes approximately 1.5 seconds after the object is dropped for the object to hit the ground. v(1.479) ≈ −47.3

The object has a terminal velocity of approximately −47.3 feet per second. (The negative sign on the velocity indicates downward motion.)

20. a,b.

c. The graphs show the position of the object t seconds after it is thrown. The point at which the graph reaches the horizontal axis corre-sponds to the time when the object hits the ground.

21. a. df

dxkx=

b. f xk

x C( ) = +2

2

c. Taking the derivative of f, we get

2 (2 ) 02 2

d k kx C x kx

dx + = + =

Thus we have the identity kx = kx , and our solution is verified.



22. a. dw

dt

k

t= pounds per month

b. w t k t C( ) ln= + pounds after t

months

w k C( ) ln1 1 6= + = , so C = 6.

w k( ) ln9 9 6 80= + = , so k = 74

9ln.

Thus w t t( )ln

ln= +74

96 pounds

after t months.

c. 74

(3) ln3 6 43ln9

w = + ≈ pounds

74

(6) ln6 6 66ln9

w = + ≈ pounds

d. The rate of increase of the weight of the dog will slow until it eventually becomes zero. This differential equation indicates that the rate increases infinitely.

23. a. dh

dt

k

t= feet per year after t years

b. h t k t C( ) ln= + feet in t years

h(2) = k ln 2 + C = 4 and

h(7) = k ln 7 + C = 30 Solving this system of equations, we get k ≈ 20.75 and C ≈ −10.39. h(t) = 20.75 ln t − 10.39 feet in t years

c. h(15) ≈ 45.8 feet Over time, the tree will continue to grow, but the rate of increase will be smaller and smaller.

24. a.

b. The output values on the graph corresponding to t = 3 and t = 6 are approximately 42 pounds and 65 pounds, respectively.

25. a.

b. The output value on the graph corresponding to an input of t = 15 is (15) 46h ≈ feet.

26. a. df kxdx

= b. f x k x C( ) ln= +

c. Taking the derivative in part b, we

get ( )lnd kxdx

k x C+ = . Thus we

have the identityk kx x= , and the

solution is verified.



27. a.

i. One possible answer: The particular solutions shown go through (1, 1.5), (−2, 1), and

(1, 0).

ii. When x > 0, the graph of a par-ticular solution rises as x gets larger. When x < 0, the solution graph rises as x gets smaller. The particular solution graphs are concave down.

iii. The family of solutions appears to increase rapidly as x moves away from the origin (in both direc-tions), and then the increase slows down. The line x = 0 (lying on the y-axis) appears to be a vertical asymptote for the family.

b.

i. One possible answer: The

particular solutions shown go through (10, 0), (−10, 0), and

(5, 5).

ii. When x > 0, the graph of a par-ticular solution rises as x gets larger. When x < 0, the solution graph rises as x gets smaller. The

particular solution graphs are concave down.

iii. The family of solutions appears to behave the same as that in part a, but the slope at each point on a particular solution graph is 10 times the slope at the corre-sponding point on a particular solution graph in part a. Again, the line x = 0 appears to be a ver-tical asymptote for the family.

c.

i. One possible answer: The particular solutions shown go through (1, 1.5), (−2, 1), and

(1, 0).

ii. When x > 0, the graph of a par-ticular solution falls as x gets larger. When x < 0, the solution graph falls as x gets smaller. The particular solution graphs are concave up.

iii. The slope at each point on a par-ticular solution graph is the nega-tive of the slope at a correspond-ing point on a particular solution graph in part a. The family of so-lutions appears to decrease rap-idly as x moves away from the origin (in both directions), and then the decrease levels off. The line x = 0 appears to be a vertical asymptote for the family.



d.

i. One possible answer: The particular solutions shown go through (0.1, 0), (−0.1, 0), and (0.05, 0.05).

ii. When x > 0, the graph of a par-ticular solution rises as x gets larger. When x < 0, the solution graph rises as x gets smaller. The particular solution graphs are concave down.

iii. The family of solutions appears to behave the same as that in part a, but the slope at each point on a

particular solution graph is 110

times the slope at the corre-sponding point on a particular solution graph in part a. Again, the line x = 0 appears to be a ver-tical asymptote for the family.

28.



29.

30.

31.

Section 8.2 Separable Differential Equations

1. dT k

dt T=

This is a separable differential equation.

211 22

TdT kdt

TdT kdt

T c kt c

=

=

+ = +

∫ ∫

2 2

2

T kt C

T kt C

= +

= ± +

Because thickness can’t be negative,

Calculus Concepts Section 8.2: Separable Differential Equations 483


T t kt C( ) = +2 .

2. dP

dtkP C P= −( )

This differential equation has a logistic function of the form P tC

Ae Ckt( ) =

+ −1as its general

solution.

3. dA

dtkA=

This a separable differential equation.

( )( )

1 2

1

1

ln

ln

note that

where

kt C kt C kt C

kt C

dA kdtA

dA kdtA

A c kt c

A kt C

A e e e e

A ae a e

+ +

=

=

+ = += +

= = ⋅

= =

∫ ∫

(Note that A is positive, so we omitted the absolute value signs from the natural logarithms,)

A t aekt( ) =

4. dh

da

k

h=

This is a separable differential equation. hdh kda=

hdh kda=∫ ∫

211 22

h c ka c+ = +

2 2h ka C= +

2h ka C= ± +

Because height can’t be negative, h a ka C( ) = +2 .

5. dx

dtkx N x= −( )

This differential equation of the form dP

dtkP C P= −( ) has a logistic function of the form

P tC

Ae Ckt( ) =

+ −1as its general solution.

Thusx tN

Ae Nkt( ) =

+ −1



6. Accumulation rate = kt where k is a constant Packing-down rate = hS where h is a constant dS

dtkt hS= −

7. Flow-in rate = where is a constantk

kD

Flow-out rate = where is a constanthD h dD

dt

k

DhD= −

8.

One possible answer: The particular solutions that are sketched go through (1,−2), (−1, 0), and (1, 2).

9.

One possible answer: The particular solutions shown go through (−2, 2), (−2, −1), and (1, 1).

10.



One possible answer: The particular solutions shown go through (−1, 2), (0, −1), and (1, 1). 11.

One possible answer: The particular solutions shown go through (0, −2), (−2, −4), and (−1, 6).

12.

One possible answer: The particular solutions shown go through (−2, 6), (1, −4), and (0, 0).

13.

One possible answer: The particular solutions shown go through (−2, 1), (2, 1), and (1, −1).



14.

One possible answer: The particular solutions shown go through (10, 2), (12, 1), and (20, 3). 15.

One possible answer: The particular solutions shown go through (0, 0), (−1, −4), and (1, 3).

16.

One possible answer: The particular solutions shown go through (6, 2), (2, 2.5), and (7, 0.5). 17.

One possible answer: The particular solutions shown go through (0, 3), (1, 0), and (2, 1).



18. Solve by antidifferentiation.

dy kx dx=∫ ∫

2( )2

ky x x C= +

19. Solve by separation of variables.

1 2

1

1

ln

ln

( )

kx C

kx

dy kdxy

dy kdxy

y c kx c

y kx C

y e

y x ae

+

=

=

+ = +

= +

= ±

= ±

∫ ∫


211 22

2 2

( ) 2

ydy kdx

ydy kdx

y c kx c

y kx C

y x kx C

=

=

+ = +

= +

= ± +

∫ ∫

21. Solve by antidifferentiation.

k

dy dxx

=∫ ∫

y x k x C( ) ln= +




2 211 22 2

2 2

2( )

k

ydy kxdx

ydy kxdx

y c x c

y kx C

y x kx C

=

=

+ = +

= +

= ± +

∫ ∫


1 2

ln

1

1

ln ln

ln ln

( )

kxC

k

k

kdy dx

y x

kdy dx

y x

y c k x c

y k x C

y e e

y a x

y x ax

=

=

+ = +

= +

=

=

= ±

∫ ∫


21 22

1

1

ln k

dy kxdxy

dy kxdxy

y c x c

=

=

+ = +

∫ ∫

2

22

22

/ 2

ln

( )kx

k

kx C

y x C

y e

y x ae

+

= +

= ±

= ±

25. a. dq

dtkq= milligrams per hour

b. Solve by separation of variables.

1

1

dq kdt

dq kdt

q

q

=

=∫ ∫

ln ln

Note that ln can be rewritten as ln .

Now change the equation from logarithmic form to

exponential form. . Replace

with and remove the absolute value signs.

k

k x C k x C

C

k x C x C

y e e e

e a

+

+ +

= = ⋅

2 2/ 2 / 2Recall that . Replace

with and remove the absolute value signs.

kx C kx C

C

e e e

e a

+ = ⋅



1 2ln

ln

( )

kt C

kt

q c kt c

q kt C

q e

q t ae

+

+ = +

= +

=

=

When t = 0, q = 200, and when t = 2, q = 100. Thus, we have

q a( )0 200= = and (2)q = 2

2

2 12

12

1 12 2

100

200 100

2 ln

ln

k

k

k

ae

e

e

k

k

=

=

=

=

=

Thus we have a = 200 and 1 12 2ln 0.346574k = ≈ − .

0.346574( ) 200 tq t e−= milligrams after t hours

c. After 4 hours, q(4) ≈ 50 milligrams will remain. After 8 hours, q(8) ≈ 12.5 milligrams will remain.

26. a. dr

dtkr= milligrams per year


1 2

1

1

ln

ln

kt C

kt

dr kdtr

dr kdtr

r c kt c

r kt C

r e

r ae

+

=

=

+ = +

= +

= ±

= ±

∫ ∫

When t = 0, r = 0.1, and when

t = 210,000, r = 0.05. Thus, we have r a( ) .0 01= = and r ae k( ) .,2 0 05210 000= = .

Solving this system of equations, we get a = 0.1 and k = ≈ − ⋅ −1210 000

12

6330 10,

ln .

6(3.3010 )( ) 0.1 tr t e−− ⋅= milligrams after t years

27. a. da

dtka= units per day




1 2

1

1

ln

ln

( )

kt C

kt

da kdta

da kdta

a c kt c

a kt C

a e

a t ce

+

=

=

+ = +

= +

=

=

∫ ∫

a(0) = c = the initial amount. When t = 3.824,

a = half of the initial amount = c2

. Thus, we have 3.824

3.824

21212

1 13.824 2

ln 3.824

ln

0.181262

k

k

c ce

e

k

k

=

=

=

=

≈ −

a t ce t( ) .= −0181262 units after t days

c. Here, c = 1 gram. After 12 hours = 0.5 days, a(0.5) ≈ 0.91 gram will remain. After 4 days, a(4) ≈ 0.48 gram will remain. After 9 days, a(9) ≈ 0.20 gram will remain. After 30 days, a(30) ≈ 0.004 grams will remain.

28. a. df

dxkf=


1 2

1

1

ln

ln

( )

kx C

kx

f

f

df kdx

df kdx

f c kx c

f kx C

f e

f x ae

+

=

=

+ = +

= +

= ±

= ±

∫ ∫

c. Taking the derivative of the function f in part b, we get ( ) ( )kx kxdae ake k f x

dx± = ± = .

Substituting this derivative into the differential equation in part a gives the identity kf = kf, and the solution is verified.



29. a. Let N be number of countries that issued stamps. dN

dtN N= −0 0049 37. ( ) countries per year

b. This differential equation has a logistic function as its solution.

0.0049(37)

0.1813

37( )

137

countries1

t

t

N tAe

Ae

−

−

=+

=+

t years after 1800

c. Because N(55) = 16, we have the equation 0.1813(55)

37(55) 16

1N

Ae−= =+

. Solving this

equation for A, we get

0.1813(55)

9.9715

9.9715

3716

1

37 16 16

21

1628,097.439

Ae

Ae

Ae

−

−

−

=+

= +

=

≈

0.1813

37( )

1 28,097.439 tN t

e−=+

countries t years after 1800

d. N(40) ≈ 2 and N(60) ≈ 24; There were 2 countries in 1840 and 24 countries in 1860.

e. The upper asymptote is N(t)=37, and the lower asymptote is N(t) = 0.

f.

30. a. Let N be number of patents obtained.

dN

dtN N= ⋅ −−( . ) ( )7 52 10 27005 patents per year


5 0.20304(7.5210 )2700

2700 2700( ) patents

11tt

N tAeAe

− −− ⋅= =

++ t years after 1865



c. Because N(18) = 980, we have the equation

NAe

( ).

182700

1980

3 65472=

+=

−.

Solving this equation for A, we get

Ae

= ≈−

1720

98067 8464

3 65472.. .

The particular solution is N te t

( ). .

=+ −

2700

1 678464 0 20304 patents t years after 1865.

d. N(35) ≈ 2558; We estimate that there were 2558 patents issued in 1900.

e. From the graph we estimate that the asymptotes are 0 and 2700 patents.

f.

31. a. df

dxkf L f= −( )

b. This differential equation has a logistic function as its solution: f xL

Ae Lkx( ) =

+ −1

c. Taking the derivative in part b, we get

2

2

2 2

(1 ) ( )

( )(1 ) (1 )

Lkx Lkx

LkxLkx

Lkx Lkx

dfL Ae Ae Lk

dx

L L kAeLkAe

Ae Ae

− − −

−−

− −

= − + −

−= − =+ +

Substituting f xL

Ae Lkx( ) =

+ −1 into kf(L − f) and simplifying gives



( )2

2

( ) =1 1

(1 )

1 1 1

1 1 1

Lkx Lkx

Lkx

Lkx Lkx Lkx

Lkx Lkx

Lkx Lkx Lkx

L Lkf L f k L

Ae Ae

L L Ae Lk

Ae Ae Ae

L L LAe L kL Aek

Ae Ae Ae

− −

−

− − −

− −

− − −

− − + +

+ = − + + +

+ − = = + + +

Thus we have the identity

( ) ( )kL Ae

Ae

kL Ae

Ae

Lkx

Lkx

Lkx

Lkx

2

2

2

21 1

−

−

−

−+=

+, and our solution is verified.



Section 8.3 Numerically Estimating by Using Differential Equations: Euler’s Method

1. a. dy

dx= 1

2, initial condition (0, 0) , step size

4 0

22

− = steps

units per step

x Estimate of y(x) Slope at x 0 0 0.5

2 0 + 2(0.5) = 1 0.5

4 1 + 2(0.5) = 2

Thus we estimate y(4) ≈ 2.

b. dy

dxx= 2 , initial condition (1, 4) , step size

7 1

23

− = steps

units per step

x Estimate of y(x) Slope at x 1 4 2(1) = 2

4 4 + 3(2) = 10 2(4) = 8

7 10 + 3(8) = 34

Thus we estimate y(7) ≈ 34.

2. We can easily find the exact solution to the differential equation in part a of Activity 1. If

dy

dx= 1

2, then y x C= +1

2. If the initial condition is (0, 0), then 0 01

2= +( ) C , and C = 0. Thus

the value of y when x is 4 is found as y = =12

4 2( ) , and we see that the estimate in part a of

Activity 1 is exact. Euler’s Method used for a differential equation of the form dy

dxk= will

always produce an exact solution. We can also find the exact solution to the differential equation in Activity 1 part b. If

dy

dxx= 2 , then y x C= +2 . Substituting the initial condition (1, 4) gives 4 12= + C , so C = 3.

The value of y when x is 7 is found as y = + =7 3 522 . The estimate in part b of Activity 1 is

not exact (and not very close to the actual value).

3. a. dy

dx y= 5

, initial condition (1, 1), step size is 5 1

22

− = units per step

x Estimate of y(x) Slope at x

1 1 5

51

=

3 1 + 2(5) = 11 5

0.454511

≈

5 11 + 2 5

11

≈ 11.91

We estimate that y(5) ≈ 11.91.

Calculus Concepts Section 8.3: Euler’s Method 495


b. dy

dx x= 5

, initial condition (2, 2), step size is 8 2

23

− = units per step

x Estimate of y(x) Slope at x

2 2 5

2.52

=

5 2 + 3 5

2 = 9.5 5

15

=

8 9.5 + 3(1) = 12.5

We estimate that y(8) ≈ 12.5. 4. For Activity 3 part a:

a. dy

dx y= 5

can be solved by separating the variables:

211 22

212

5

5

5

5

ydy dx

ydy dx

y c x c

y x C

=

=

+ = +

= +

∫ ∫

Substituting (1, 1) into the general solution, we have 12

21 5 1( ) ( )= + C , and C = −4.5.

The particular solution is 12

2 5 4 5y x= − . or y x= ± −10 9 .

b.

c.

d. The steepness of the curve at x = 1, followed by a leveling off, causes the Euler estimates to be significantly greater than the actual solution. The Euler estimate was approximately 11.91. The actual solution is approximately 6.4.

For Activity 3 part b:

a. dy

dx x= 5

can be solved by separating the variables:

5

5ln

y dxx

y x C

=

= +

∫

Substituting (2, 2) into the general solution, we have 2 5 2= +ln C and C = 2 −5 2ln ≈ −1.4657.

The particular solution is y x= + −5 2 5 2ln ln or y x≈ −5 14657ln .



b.

c.

d. The steepness of the curve at x = 2, followed by less steepness, causes the Euler estimate of approximately 12.5 to be significantly greater than the actual solution of approximately 8.93.

5. a. dw

dt t= 3367885.

pounds per month after t months

b. Initial condition (1, 6), step size 0.25 month

t (months)

Estimate of w(t)

(pounds)

Slope at t (pounds per

month)

t (months)

Estimate of w(t)

(pounds)

Slope at t (pounds per month)

1 6.000 33.679 3.25 48.768 10.363

1.25 14.420 26.943 3.5 51.359 9.623

1.5 21.155 22.453 3.75 53.764 8.981

1.75 26.769 19.245 4 56.010 8.420

2 31.580 16.839 4.25 58.115 7.924

2.25 35.790 14.968 4.5 60.100 7.484

2.5 39.532 13.472 4.75 61.967 7.090

2.75 42.900 12.247 5 63.739 6.736

3 45.961 11.226 5.5 67.027 6.123

3.25 48.768 10.363 5.75 68.558 5.857

3.5 51.359 9.623 6 70.022

The Euler estimates using a step size of 0.25 month are w(3) ≈ 46.0 lbs; w(6) ≈ 70.0 lbs

c. Initial condition (1, 6), step size 1 month w(3) ≈ 56.52 pounds; w(6) ≈ 82.90 pounds

t (months)

Estimate of w(t)

(pounds)


month)

t (months)

Estimate of w(t)

(pounds)


month) 1 6 33.679 4 67.745 8.420

2 39.679 16.839 5 76.164 6.736

3 56.518 11.226 6 82.90

The Euler estimates using a step size of 1 month are w(3) ≈ 56.5 lbs; w(6) ≈ 82.9 lbs

d. The answer to part b should be more accurate because it uses a smaller step size.



6. a. dC

dtC C= −0 0049 37. ( ) countries per year t years after 1800

b. Initial condition (55, 16), step size = 40 55

53

− = years per step

t (years)

Estimate of C(t) (countries)

Slope at t (countries per year)

55 16 0.0049(16)(21) = 1.6464

52 16 − 3(1.6464) = 11.0608 0.0049(11.0608)(25.9392) = 1.4059

49 11.0608 − 3(1.4059) = 6.8432 0.0049(6.8432)(30.1568) = 1.0112

46 6.8432 − 3(1.0112) = 3.8096 0.0049(3.8096)(33.1904) = 0.6196

43 3.8096 − 3(0.6196) = 1.9509 0.0049(1.9509)(35.0491) = 0.3350

40 1.9509 − 3(0.3350) = 0.9458

We estimate that in 1940, only 1 country had issued postage stamps.

c. Initial condition (55, 16), number of steps = 40 55

53

− = years

steps

t (years)


Slope at t (countries per year)

55 16 0.0049(16)(21) = 1.6464

50 16 − 5(1.6464) = 7.7680 0.0049(7.768)(29.232) = 1.1127

45 7.768 − 5(1.1127) = 2.2045 0.0049(2.2045)(34.7955) = 0.3759

40 2.2045 − 5(0.3759) = 0.3252

We estimate that in 1940, no countries had issued postage stamps.

d. We expect the answer to part b to be more accurate because of the smaller step size.

7. a. 3.55 1.351353.935 tdpt e

dt−= thousand barrels per year t years after production begins, where

p(t) is the total amount of oil produced after t years

b. Initial condition: (0, 0); Step size: 0.5 year

t Estimate of p(t)

p′′′′(t) t Estimate of p(t)

p′′′′(t)

0 0 0 3 4.9680 3.3734

0.5 0 0.1709 3.5 6.6547 2.9668

1 0.0855 1.0187 4 8.1381 2.4251

1.5 0.5948 2.1865 4.5 9.3507 1.8744

2 1.6881 3.0891 5 10.2879

2.5 3.2326 3.4707



After 5 years of production, the well has produced approximately 10.3 thousand barrels.

c.

The graph of the differential equation is the slope graph for the graph of the Euler estimates. Similarly, the graph of the Euler estimates is an approximation to the accumulation graph of the differential equation graph.

8. a. 0.705989

0.705989 2

6,608,830

(1 925.466 )

x

x

df e

dx e

−

−=+ worker hours per week, where f is the total number of

worker hours used by the end of the xth week

b.

The graph is always positive so the total number of worker hours is increasing. There are no critical points. Since the graph is decreasing, the rate that the number of worker hours is increasing is getting smaller.

c. Using technology (a calculator and the Euler program available for download at the Calculus Concepts website), the total number of worker hours used by the twentieth week is approximately 10,098.



d. Estimate of f(x)

0

2000

4000

6000

8000

10000

12000

0 5 10 15 20

weeks

wo

rker

ho

urs

Accuracy can be improved by using more intervals.

9. a. ( )dT

k T Adt

= − °F per minute after t minutes.

b. Solve the equation k(98 − 70) = −1.8 to get k ≈ −0.064.

c. t Estimate of p(t)

p′′′′(t) t Estimate of p(t)

p′′′′(t)

0 98 −1.8 8 86.4552 −1.05784

1 96.2 −1.68429 9 85.3974 −0.989833

2 94.5157 −1.57601 10 84.4076 −0.926201

3 92.9397 −1.47470 11 83.4814 −0.866659

4 91.465 −1.37989 12 82.6147 −0.810945

5 90.0851 −1.29119 13 81.8038 −0.758813

6 88.7939 −1.20818 14 81.0449 −0.710032

7 87.5857 −1.13051 15 80.3349

After 15 minutes, the temperature of the object is approximately 80.3°F.

10. a. dp

dtp= −100 percentage points per hour

b. t Estimate of p(t) ( )p t′

0 0 100

0.25 25 75

0.5 43.75 56.25

0.75 57.81 42.19

1 68.36 31.64

1.25 76.27 23.73



1.5 82.20 17.80

1.75 86.65 13.35

2 89.99

The percentage of the task that is learned is approximately 90%.

c.

The rate of learning appears to be slowing as time increases.

11. Euler’s method uses tangent-line approximations. Tangent lines generally lie close to a curve near the point of tangency and deviate more and more as you move farther and farther away from that point. Thus, smaller steps generally result in better approximations. Compare the graphs in Activity 4b.

12. a-b. With 40 steps, the step size is 0.375.

The Euler estimate for C(t) is 1.671916734. That is, one country was issuing postage stamps in 1840. With 80 steps, the step size is 0.1875. The Euler estimate for C(t) is 1.720710909. That is, one country was issuing postage stamps in 1840.


0

2

4

6

8

10

12

14

16

18

40 45 50 55

years since 1800

cou

ntr

ies


02468

1012141618

40 45 50 55

years since 1800

cou

ntr

ies

Calculus Concepts Section 8.4: Second-Order Differential Equations 501


c. One possible answer: The second estimate using smaller steps is likely to be more accurate, but the answers are very close and the interpretations of the two answers equivalent. 13.a-b. With monthly intervals, there are 60

intervals with step size 1/12. The Euler estimate for p(5) is approximately 10.594 thousand barrels.

Estimate of p(t)

0

2

4

6

8

10

12

0 1 2 3 4 5

years since 1800

cou

ntr

ies



With weekly intervals, there are 260 intervals with step size 1/52. The Euler estimate for p(5) is approximately 10.639 thousand barrels. With daily intervals, there are 1825 intervals with step size 1/365. The Euler estimate for p(5) is approximately 10.650 thousand barrels.

0

2

4

6

8

10

12

0 1 2 3 4 5

year

tho

usa

nd

bar

rels

0

2

4

6

8

10

12

0 1 2 3 4 5

year

tho

usa

nd

bar

rels

c. One possible answer: The estimates using smaller steps are close to the first estimate. Using smaller steps does not greatly change the estimate. 14. a-b. With 140 intervals, the step size is

20/140. The Euler estimate for f(20) is approximately 10,097 worker hours.

Estimate of f(x)

0

2000

4000

6000

8000

10000

12000

0 5 10 15 20

week

wo

rker

ho

urs

c. One possible answer: Using 140 steps does not greatly improve the estimate.



15. a-b. There are 790 one-second intervals

with step size 1/60. The Euler estimate for T(15) is approximately 80.6696349 or 80.7 °F.

Estimate of T(t)

70

75

80

85

90

95

100

0 3 6 9 12 15

minutes

tem

per

atu

re

c. The estimate using 790 intervals instead is 0.4 degrees higher than the estimate using 15

intervals. The estimate using 790 intervals is probably more accurate than the estimate using 15 intervals since the function has no really steep areas of descent and no change of curvature.



Section 8.4 Second-Order Differential Equations

1. 2

2 2

d S k

dt S=

2.2

2

d DkD

dt=

3. 2

2

d Pk

dy= ; Taking the antiderivative, we get

dPky C

dy= + .

Taking the antiderivative of dP

dy, we get 2( )

2

kP y y Cy D= + + .

4. 2

2

d h k

ydy=

Taking the antiderivative, we get lndh

k y Cdy

= + .

We have not studied a method for finding the antiderivative of this function, so we cannot find a general solution.

5. a. 2

26.14

d R

dt= jobs per month per month in the tth month of the year

b. Taking the antiderivative of 2

2

d R

dt, we get 6.14

dRt C

dt= + .

When t = 1, 0.87.dR

dt= − Solving for C, we get 0.87 6.14(1)

7.01

C

C

− = += −

6.14 7.01dR

tdt

= − jobs per month in the tth month of the year

Taking the antiderivative of dR

dt, we get 2( ) 3.07 7.01R t t t C= − + .

When t = 2, R = 14. Solving for C, we get 214 3.07(2 ) 7.01(2)

1 475.

C

C

= − +=

The particular solution is 2( ) 3.07 7.01 15.74R t t t= − + jobs in the tth month of the year

c. R(8) ≈ 156 and R(11) ≈ 310 We estimate the number of jobs in August to be approximately 156 and the number in November to be 310.

6. a. 2

20.007

d a

dt= year of age per year2

t years after 1990



b. Taking the antiderivative, we have 0.007da

t Cdt

= + .

When t = 1, 0.412da

dt= . Solving for C, we get C = 0.405, so 0.007 0.412

dat

dt= + years of

age per year t years after 1990.

Taking the antiderivative, we get a(t) = 0.0035t2 − 0.412t + C . When t = 0, a = 32. Solving for C, we get C = 32. The particular solution is

2( ) 0.0035 0.412 32a t t t= − + years of age t years after 1990.

c. In 2007, the median age is predicted to be 26; in 2008, the median age is predicted to be 25.7; in 2008, the median age is predicted to be 25.4.

7. a. 2

22009

d A

dt= − cases per year per year, where t is the number of years since 1988

b. Taking the antiderivative of 2

2

d A

dt, we get 2099

dAt C

dt= − + . When t = 0, 5988.7

dA

dt= , so

C = 5988.7. The particular solution is 2099 5988.7dA

tdt

= − + cases per year, where t is the

number of years since 1988.

Taking the antiderivative of dA

dt, we get 2( ) 1049.5 5988.7A t t t C= − + + . When t = 0,

A = 33,590, so C = 33,590. The particular solution is 2( ) 1049.5 5988.7 33,590A t t t= − + + cases, where t is the number of years since 1988

c. When t = 3, 308.3dA

dt= − and A(3) ≈ 42,111.

We estimate that in 1991 there were 42,111 AIDS cases and the number of cases was decreasing at rate of 308.3 cases per year.

8. a. 2

20.022

d p

dt= cent per year squared t years after 1919

b. Taking the antiderivative, we have 0.022dp

t Cdt

= + . When t = 39, 0.393dp

dt= . Solving for

C, we get C = −0.465, so 0.022 0.465dp

tdt

= − cents per year t years after 1919.

Taking the antiderivative again, we have 2( ) 0.011 0.465p t t t C= − + When t = 0, p = 2. Solving for C, we get C = 2. The particular solution is

2( ) 0.011 0.465 2p t t t= − + cents t years after 1919.

c. In 2007, the first-class postage for a 1-ounce letter is predicted to be 39 cents and the rate of change in 2007 is predicted to be 1.295 cents per year; in 2008, the first-class postage for a 1-ounce letter is predicted to be 40 cents and the rate of change in 2008 is predicted to be 1.315 cents per year; in 2009, the first-class postage for a 1-ounce letter is predicted to be 41 cents and the rate of change in 2009 is predicted to be 1.335 cents per year.



9. a. 2

2

d fk

dx=

b. Taking the antiderivative, we have df

kx Cdx

= + .

Taking the antiderivative of the result, we get 2( )2

kf x x Cx D= + + .

c. Taking the derivative in part b, we get 2 (2 ) 02 2

d k kx Cx D x C kx C

dx + + = + + = +

Taking the derivative of this result, we get ( )dkx C k

dx+ = so we have the identity k = k,

and our solution is verified.

10. a. 2

2

d x kx

mdt= − , where

30 15

32 16m = = , so

1516

1516

k

m= = and

2

216

d xx

dt= − .

b. ( ) sin(4 )x t a x c= + and ( ) 4 cos(4 )x t a x c′ = + Because x(0) = 2 and x′(0) = 0, we have 2 sina c= and 0 4 cosa c= . The second equation

is true if cos 0c = . This occurs at π

2c = . Substituting this in to the first equation, we have

π

22 sin (1)a a a= = = . Thus a = 2 and 2

cπ= . The particular solution is

( ) 2sin 42

x t xπ = +

feet beyond its equilibrium point after t seconds.

c.

The graph shows that over time the spring oscillates from 2 feet beyond its equilibrium point and back again.

d. x(t) ==== 0 when π

24 πt + = . Solving for t yields π

8t =

( ) π π

2 2π

8 8cos( ) 8( 1) 8 feet per secondx′ = + = − = −

The mass is moving at a speed of 8 ft/sec when it passes its equilibrium point.

11. a. 2

20.212531

d EE

dt= − mm per day per month per month, where E(t) is the amount of

radiation in mm per day and t is measured in months



b. The general solution to the equation is of the form ( ) sin( )E t a kt c= + with

0.212531 0.461011k = ≈ . In June, the amount of radiation is 4.5 mm above the expected value, and in December, the amount of radiation is 4.7 mm below the expected value. Thus we have 4.5 = a sin(0.461011(6) + c) and −4.7 = a sin(0.461011(12) + c). We solve for a in the

first equation 4.5

sin(0.461011(6) )

ac

=+

and substitute this into the second equation:

−4.7 = 4.5

sin(0.461011(6) ) c+

sin(0.461011(12) + c).

Using technology, we find c ≈ 2.24801 and a ≈ −4.71284.

( ) 4.71284sin(0.461011 2.24801)E t t= − + mm per day, where t is the month of the year.

c. ( ) 4.71284sin(0.461011 2.24801)E t t= − + + 12.5 mm/day where t is the month of the year

d. In March the amount is f (3) ≈ 14.7 mm per day, and in September the amount is f(6) ≈ 12.0 mm per day. The model over-predicts by 2.7 mm per day in March and slightly underpredicts by 0.5 mm per day in September.

12. a. 2

2

d fkf

dx= − (where k > 0)

b. The general solution to the equation is of the form ( ) sin( )f x a k x c= + .

c. Taking the derivative in part b, we get ( sin( )) ( )cos( )d

a k x c a k k x cdx

+ = + .

Taking the derivative of this result, we get

( )( )cos( ) ( )( )sin( )

sin( )

dfa k k x c a k k k x c

dx

ka k x c kf

+ = − +

= − + = −

so we get the identity −kf = −kf, and our solution is verified.



Chapter 8 Concept Review 1. a. The relative risk of having a car accident is changing with respect to blood alcohol level at

a rate that is proportional to the risk of having a car accident at certain blood alcohol level.


1 2

1

1

ln (Because the risk is always positive, we omit the absolute value.)

ln

( )

kb C

kb

kb

R

R

dR kdb

dR kdb

R c kb c

R kb C

R e

R ae

R b ae

+

=

=

+ = += +

=

=

=

∫ ∫

c. (0) 1R a= = and 0.14(0.14) 20kR ae= = . Solving these equations, we get a = 1 and

ln 2021.398

0.14k = ≈ .

21.398( ) bR b e= percent, where b is the proportion of alcohol in the blood stream

d. A certain occurrence corresponds to a relative risk of 100%, so R = 100. We solve 21.398 100be = for b: 21.398 100

21.398 ln100

ln100

21.3980.215

be

b

b

==

=

≈

Thus according to the model, a crash is certain to occur when the blood alcohol level is approximately 21.5%.

e.

2. a. 0.001175 (16.396 )dP

P Pdx

= − million people per year x years after 1800

Calculus Concepts Chapter 8: Concept Review 509



0.001175(16.396) 0.019265

16.396 16.396( )

1 1x xP x

Ae Ae− −= =+ + million people is the population of Ireland

x years after 1800

c. We use the fact that P(−20) = 4.0, to solve for A:

0.385306

0.385306

0.385306

0.385306

16.3964

1

16.396 4(1 )

12.396 4

12.3962.108

4

Ae

Ae

Ae

Ae

=+

= +

=

= ≈

Thus 0.019265

16.396( )

1 2.108 xP x

e−=+

million people is the population of Ireland x years after

1800.

d. P(40) ≈ 8.3 and P(50) ≈ 9.1 We estimate that there were 8.3 million people in 1840 and 9.1 million people in 1850.

3. a. x

(years after 1800) Estimate of P(x) (million people)

Slope at x (million people per year)

−20 4 0.058

−10 4.583 0.064

0 5.219 0.069 10 5.904 0.073 20 6.632 0.076 30 7.393 0.078 40 8.175 0.079 50 8.965

We estimate that P(40) ≈ 8.18 million people; P(50) ≈ 8.97 million people.

b. x

Estimate of P(x)

Slope at x

x

Estimate of P(x)

Slope at x

−20 4 0.058 20 6.684 0.076

−15 4.291 0.061 25 7.066 0.077

−10 4.596 0.064 30 7.453 0.078

−5 4.915 0.066 35 7.844 0.079

0 5.247 0.069 40 8.239 0.079 5 5.590 0.071 45 8.633 0.079 10 5.945 0.073 50 9.027 15 6.310 0.075

We estimate that P(40) ≈ 8.24 million people and P(50) ≈ 9.03 million people.



c. The 1940 estimates in this question are 0.12 and 0.06 million people less than population found in the Activity 2. The 1950 estimates are 0.13 and 0.07 million people less than the Activity 2 answer.

4. a. 0.008307 (7.154 )dQ

Q Qdx

= − − million people per year, where the population is

P(x) = Q(x) + 4.4 million people and x is the number of years since 1800


0.008307(7.154) 0.059428

7.154 7.154( )

1 1x xQ x

Ae Ae= =

+ + million people is the population of Ireland x

years after 1800

Because Q(100) = 4.5, we have the equation0.59428

7.1540.1

1 Ae=

+. Solving this equation for A

(as illustrated in Activity 2 part c), we get5.9428278

7.0540.185

0.1A

e= ≈ .

0.059428

7.154( )

1 0.185 xQ x

e=

+ million people, where x is the number of years since 1800

c.

0.019265

0.059428

16.396 million people when 40

1 2.108( )

7.1544.4 million people when 50

1 0.185

x

x

xe

P x

xe

− ≤ += + ≥ +

is the population of Ireland where x is the number of years since 1800

d. Using the model in part c, P(50) ≈ 6.0 We estimate that there were 6.0 million people in 1850. This answer is significantly smaller than the one found in part d of Activity 2.

Chapter 8 Dynamics of Change: Differential...

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