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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.
Calculus Concepts Section 8.1: Differential Equations and Slope Fields 479
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.
480 Chapter 8: Dynamics of Change: Differential Equations Calculus Concepts
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.
Calculus Concepts Section 8.1: Differential Equations and Slope Fields 481
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.
482 Chapter 8: Dynamics of Change: Differential Equations Calculus Concepts
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=
b. Solve by separation of variables.
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.
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 .
496 Chapter 8: Dynamics of Change: Differential Equations Calculus Concepts
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
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.
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.
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
502 Chapter 8: Dynamics of Change: Differential Equations Calculus Concepts
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.
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.
504 Chapter 8: Dynamics of Change: Differential Equations Calculus Concepts
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.
506 Chapter 8: Dynamics of Change: Differential Equations Calculus Concepts
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.
508 Chapter 8: Dynamics of Change: Differential Equations Calculus Concepts
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
b. This differential equation has a logistic function as its solution.
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.