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Calculus An Applied Approach 9th Edition Larson Test Bank

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Calculus An Applied Approach 9th Edition Larson Test Bank

ISBN-13: 978-1133109280

ISBN-10: 1133109284

 

Description

Calculus An Applied Approach 9th Edition Larson Test Bank

ISBN-13: 978-1133109280

ISBN-10: 1133109284

 

 

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Chapter 11: Differential Equations
1.
Determine whether
3
2
y x
is a solution of the differential equation
3
0y y
x
  .
A)
3
 
2 2
y x
6 and 6
3
x
2
x
0
  

x
B)
3
 
2 2
y x
6 and 2
3
x
2
x
0
  

x
C)
4 4
3
x x
3
 
and
2
0
4 4
y x
x
  

D)
2 2
3
 
2 2
y x
and
3
x
2
x
0
  

3 3
x
E)
 
2 2
32 3
and 2
y x
x
2
x
0
  

3
x
Ans: A

2. Determine whether
ln
4
y x xCx  is a solution of the differential equation
 
1 40
x y
y
    .
A)
 
 


y xC
ln 1 and
 
xy
1

y
4
xxC
ln 1
1
xxCx
ln
4 4 0
 




B)

 
 


y xxC
ln
 
and
xy
1

y
4
xxC
ln 1
1
xxCx
ln
4 4 0
 




C)
 
 


y xC
ln 1
 
and
xy
1

y
4
xxxC
ln
1
xxCx
ln
4 4 0
 




D)
 
 


y xC
ln 1
 
and
xy
1

y
4
xxC
ln 1
1
xxC
ln
4 4 0
 



E)

 


  
y xxC
2 2
 
ln
and
xy
1
y
4
xxxC
ln
1
xxCx
ln
4 4
 




Ans: A

3. Determine whether 1 2
y C xC
sin
cos
x 
is a solution of the differential equation
0y y
  .

A) 1 2
y C x C

sin
cos and
x
y y C x C x C
sin
cos
sin
xC
cos 0
x


 




1
2
1
2

B) 1 2 1 2
y C x C

sin
cos and
x
y y C x C x C xC
sin
cos
sin
cos 0
x
 





1
2

C) 1 2
y C x C

sin
cos and
x
y y C x C
sin
cos
xC x C
sin
cos 0
x


 




1 2 1 2
D)

y C x C

sin
cos and
x
y y C x C x C x C
sin
cos
sin
cos 0
x


 




1
2
1 2 1 2

E) 1 2 1 2
y C x C

sin
cos and
x
y y C x C
sin
cos
xC xC
sin
cos 0
x
 





1
2
Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 433

4. Determine whether
4 x
y Ce
is a solution of the differential equation
y
4
y
 
.
A)
4
x
y Ce
4 4
y
  

B)
4
x
y Ce
4
y
  

C)
3
x
y Ce
4 4
y
  

D)
3
x
y Ce
3 4
y
  

E)
4
x
Ce
y
4
y
  

4
Ans: A

5. Which of the following is a solution of the differential equation 3 3 0y y
  ?
A) 1 2
– sin– cos
C xC
x
B)
–4 x
x
e e

C)
x x
 
– sin–
Ce
x C e
cos
x

1 2
D)
2 2
x
4
x e x

E)
sin ln sec
tan
x x x 

Ans: A

6. Which of the following is a solution of the differential equation
(4)
1296 0y y ?
A)
2
x
(6
)
y x e 

B)
y
5ln
x

C)
–6 x
y e

D)
y x x
–6 cos

E)
–6
x
3 sin
y e xx 

Ans: C

7. Find the particular solution of the differential equation 15 4
0x yy
  that satisfies the
initial condition y = 8 when x = 3, where
2 2
15 4
x y C
 
is the general solution.
A)
2 2
15 4 996
x y 

B)
2 2
15 4 391
x y 

C)
2 2
15 4 199
x y 

D)
2 2
15 4 167
x y 

E)
2 2
15 4
265
x y 

Ans: B

Larson, Calculus: An Applied Approach (+Brief), 9e Page 434

8. Use integration to find a general solution of the differential equation.

3
5 – 3
dy
x
x

dx
A)
4 2
5 – 3
x x C

B)
2
15 – 3
x
C

C) 4 25 3

4 2
x x C

D)
2
5 – 3
x
C

E) 4 25 3

4 2
x x xC 

Ans: C

9.
Use integration to find a general solution of the differential equation
dy
11
x
dx
2
15
x


.
A)
 
22
2
ln 15 x C
 

2
x
B)
2
11
x
C

 
2
ln 15
x

C)
11
C

 
x x
2
ln 15

D)
 
2
11
ln 15
2
x C 

E)
 
2
11
ln 26
2
x C
x
 

Ans: D

10. Determine whether the function
2 x

y e

is a solution of the differential equation
 4
16 0y y .
A) Solution
B) Not a solution
Ans: A

11.
Determine whether the function
4
y

is a solution of the differential equation
x
 4
16 0y y .
A) Solution
B) Not a solution
Ans: B

Larson, Calculus: An Applied Approach (+Brief), 9e Page 435

12.
Determine whether the function
2
2
9
x

y
xe

is a solution of the differential equation

y yy
3 20
 .
A) Solution
B) Not a solution
Ans: B

13. Determine whether the function
x
y xe
is a solution of the differential equation

y yy
3 20
 .
A) Solution
B) Not a solution
Ans: B

14. Find the general solution satisfies the differential equation. Then find the particular
solution that satisfies the initial condition.
General Solution:
2 x

y Ce

Differential equation: 2 0y y
  
Initial condition: 3 when 0y x 
A)
y
2 2
x
2 , so
Ce
y y
20;3
x
 
 
y e

  

B)
2 2
x
y Ce
2 , so
y y
20;3
x


y e
  

C)
y
2 2
x
2 , so
Ce
y y
20;
x
 
 
y e

  

D)
2 2
x
y Ce
2 , so
y y
20;3
x


y e
  

E)
y
2 2
x
2 , so
Ce
y y
20;
x
 
 
y e

  

Ans: A

15. Find the general solution satisfies the differential equation. Then find the particular
solution that satisfies the initial condition.
General Solution: 1 2
ln
y C C x
 

Differential equation:

xy y
0
 
Initial condition: 5 and
y y x
0.5 when 1
 

A)
 
 
1
y C x
1 and
y C x
2
 


1
, so
xy y
0; 5
y
ln
x
 


2 2
2
B)
 
 
1
y C x
1 and
y Cx
2
 


1
, so
xyy
0; 5
y
xx
ln
 


2 2
2
C)
 
 
1
y C x
1 and
y Cx
2
 


1
, so
xyy
0;
y
5
ln
x
 


2 2
2
D)
 
 
1
y C x
1 and
y C x
2
 


1
, so
xy y
0;
y
5
xx
ln
 


2 2
2
E)
 
 
5 1
y C x
1 and
y Cx
2
 


1
, so
xyy
0;
y
ln
x
 


2 2
x
2
Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 436

16.
Use the integration to find the general solution of the differential equation
2
3
dy
x

.
dx
A)
3
y x C 

6y xC 
B)

C)
3

y x C
 

D)
3
6
y x C 

E) 6
y x C
 

Ans: A

17.
Use the integration to find the general solution of the differential equation
1
1
dy
dx
x


.
A)
ln 1y xC 

B)
1

y
C
 
1
x
 

C)
1

y
C
 
2
1
x
 

D)
ln
y x xC 

E)
 
y x C
1
  

Ans: A

18.
Use the integration to find the general solution of the differential equation
cos 4
dy
x
dx

.
A)
 
1
sin 4
4
y
x

B)
 
y
4sin 4
x

C)
 
1
sin 3
4
y
x

D)
 
1
sin 4
4
y
x


E)
 
y
4sin 4
x


Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 437

19. Some of the curves corresponding to different values of C in general solution of the
differential equation
2 3
y Cx xy y
, 2
3 0
 are shown in the figure given below. Find
the particular solution that passes through the points plotted on the graph.

A)
2
5 1617
y x 

B)
2
5 1617
y x 

C)
2
16 17
y x 

D)
2
16 17
y x 

E)
2
5 16y x 

Ans: A

20. Some of the curves corresponding to different values of C in general solution of the
differential equation
x
y Cey y
,
0
 are shown in the figure given below. Find the
particular solution that passes through the points plotted on the graph.

A)
x
3
y e

B)
x
y
3
e

C)
x

y e
3

D)
3x
y e

E)
3x

y e

Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 438

21. Write the the differential equation by separating the variables.
dy
x
dx y
3

A) ( 3)
y dy xdy 

B) ( 3)y dxxdy 

C) ( 3)
y dy xdx 

D) ( 3)
x dy ydx 

x dy xdy 
E) ( 3)

Ans: C

22. Write the the differential equation by separating the variables.
1
1
dy
dx x
 
A)
 
1
1
dy
dx
  

y
 
B)
 
1
1
dx
dy
x
  
 

C)
xdy
1
dx 

D)
 
1
1
dy
dx
x
  
 

E)
 
1
1
dy
dx
x
  
 

Ans: D

23. Write the the differential equation by separating the variables.
dy
x y
dx
 
A)
 
1
1
dy
dx
y
  
 

B)
 
1
1
dx
dy
  

 
x
C) 1
xdy
dx 

D) No, the variables cannot be separated
E)
 
1
1
dy
dx
x
  
 

Ans: D

Larson, Calculus: An Applied Approach (+Brief), 9e Page 439

24. Use separation of variables to find the general solution of the differential equation.
2
dy
x

dx
A)
3
y x C 

B)
2

y x C
 

C)
2
y x C 

D)
3

y x C
 

E) y xC 

Ans: C

25. Use separation of variables to find the general solution of the differential equation.
1
3
dy x
dx y

A)
4 2
2 4
y x xC 

B)
4 2
y x xC 

C)
2
2 4
y x xC 

D)
2

y x xC
2 4
 

E)
2
2 4
y x xC 

Ans: A

26. Use separation of variables to find the general solution of the differential equation.
dy
2
3 1
y
dx

A) y x C 

B)
3
dy
x C
dx
 

C)
3
y x C 

D)
3 2
y x C 

E)
3
( )
dy
x C
dx
 

Ans: C

Larson, Calculus: An Applied Approach (+Brief), 9e Page 440

27. Use separation of variables to find the general solution of the differential equation.
2
4 0
dy
x y
dx
 
A)
31
6
y xC 

B)
1
6
dy
x C
dx
 

C) 2 31
6
y x C
 

D)
31
6
dy
x C
dx
 

E)
2 1
3
y x C 

Ans: C

28. Use separation of variables to find the general solution of the differential equation.
‘ 0y xy 
A)
2
x
y Ce

B)
2
x
/2
dy
Ce
dx

C)
2
x
/2
y Ce

D)
2
x
dy
Ce
dx

E)
2
x
/2
dy e

Ans: C

29. Use separation of variables to find the general solution of the differential equation.
2
y dy
3 1
e t
dt
 
A)
2
y Int
tC 

B)
3
y In t
t C 

C)
3
y In t
t C 

D)
2
y Int
tC 

E)
2
y In t
t C 

Ans: B

Larson, Calculus: An Applied Approach (+Brief), 9e Page 441

30. Use separation of variables to find the general solution of the differential equation.
1
dy
y
 

dx
A)
1
2
x
 
y C
   
 

B)
1
2
x
 
y C
   
 

C)
 
y C x
1
  

D)
 
y C x
1
  

E)
1
2
x
 
y C
   
 

Ans: E

31. Use separation of variables to find the general solution of the differential equation.
(2
) ‘ 2
x y y 

A) (2
)
y C x 

B)
2
(2
)
y C x 

C)
2
(2
)
y C x 

D)
y C
(2
)
x 

E)
2
(4
)
y C x 

Ans: C

32. Use separation of variables to find the general solution of the differential equation.
dy
y
sin
x

dx
A)
2sin
y x C 

B)
cos
y x C 

C)
2
y
2cos
x C


D)
2
2cos
y x C 

E)
2
cos
y x C 

Ans: C

Larson, Calculus: An Applied Approach (+Brief), 9e Page 442

33. Find the general solution of the differential equation
dy
x
dx

A)
1
2
y xC 

B)
21
2
y x C 

C)
2
y x C 

D)
21
2
y C x 

E)
2
y C x 

Ans: B

34. The isotope
14
C has a half-life of 5,715 years. Given an initial amount of 11 grams of
the isotope, how many grams will remain after 1,000 years? After 10,000 years? Round
your answers to four decimal places.
A) 6.8205 gm, 2.2896 gm
B) 3.8974 gm, 1.3083 gm
C) 9.7436 gm, 3.2708 gm
D) 11.6923 gm, 3.9250 gm
E) 5.8462 gm, 1.9625 gm
Ans: C

35. Use the initial condition to find the particular solution of the differential equation.
‘ 0
x
yy e  , 4
y  when 0
x 
A)
2
x
12
y e 

B)
2
x
2 14
y e 

C)
2
x
2 14
y e 

D)
2 2
x
2 14
y e 

E)
2
x
2 14
y ye 

Ans: C

36. Use the initial condition to find the particular solution of the differential equation.
 4 ‘0
x y
y  ,
5
y  when 0
x 
A)
2
y e
x
/2

4
 

B)
2
x
y e
/2
4
 

C)
2
y e
x
/2

4
 

D)
x
y e
/2

4
 

E)
2
x
4
y e 

Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 443

37. Use the initial condition to find the particular solution of the differential equation.
2
16 ‘ 5
x y x  ,
2
y  when 5
x 
A)
2
5 1716
y x 

B)
5 1716
y x 

C)
2
5 1617
y x 

D)
5 1617
y x 

E)
2
2 1617
y x 

Ans: C

38. Use the initial condition to find the particular solution of the differential equation.
cos
dy
y x
dx

, 1
y  when 0
x 
A)
2cosx

y e

B)
sin x
y e

C)
2sin x

y e

D)
cos x
y e

E)
sin x

y e

Ans: B

Larson, Calculus: An Applied Approach (+Brief), 9e Page 444

39. Find an equation of the graph that passes through the point and has the specified slope. Then
graph the equation.
Point:  
1, 1
,
9
x
Slope:
y
y


16

A) 2 2
16 9 25y x 

y
2.5
2
1.5
1
0.5
x
-2.5 -2 -1.5 -1 -0.5
0.5 1 1.5 2 2.5
-0.5
-1
-1.5
-2
-2.5
B) 2 2
9 16 25y x 

Larson, Calculus: An Applied Approach (+Brief), 9e Page 445

y
2.5
2
1.5
1
0.5
x
-2.5 -2 -1.5 -1 -0.5
0.5 1 1.5 2 2.5
-0.5
-1
-1.5
-2
-2.5
C) 2 2
16 9 25y x 

y
2.5
2
1.5
1
0.5
x
-2.5 -2 -1.5 -1 -0.5
0.5 1 1.5 2 2.5
-0.5
-1
-1.5
-2
-2.5
D) 2 2
9 16 25y x 

Larson, Calculus: An Applied Approach (+Brief), 9e Page 446

y
2.5
2
1.5
1
0.5
x
-2.5 -2 -1.5 -1 -0.5
0.5 1 1.5 2 2.5
-0.5
-1
-1.5
-2
-2.5
E) None of the Above
Ans: A

40. Solve the differential equation to find velocity v as a function of time t if 0
v  when
0
t  The differential equation models the motion of two people on a toboggan after
consideration of the forces of gravity, friction, and air resistance.
12.5
43.2 1.25
dv
v
dt
 

A)
 
v e
34.56 1
t
 

B)
 
24.56 1v e
0.1
 

C)
 
v e
34.25 1
0.1
t
 

D)
 
v e
24.56 1
0.1
t
 

E)
 
v e
34.56 1
0.1
t
 

Ans: E

Larson, Calculus: An Applied Approach (+Brief), 9e Page 447

41. Write the first-order linear differential equation in standard form.

3 2
x xyy
2 30
 
A)
3
x

  
2
2 2
y y
x

B)
3
x

  
2 2
y y
x

C)
3

  
2
y y x
x
D)
3
2 2
x
y y
x
  
E)
3
2
2
y y x
x
  
Ans: A

42. Write the first-order linear differential equation in standard form.

xy yxe
x
  

A)
1
2
y ye
x
  

B)
y xy e
x

  

C)
1
x
y y e
x
  

D)
1
2
y
x
xy e
  

E)
1
2
2
y ye
x

  

Ans: C

43. Write the first-order linear differential equation in standard form.

1 ( 1)y xy
  

A) 1 2
1 1
y y
x
x
  
 

B) 1 1
1 1
y y
x
x
  
 

C) 1 2
1 1
y y
x
x
  
 

D) 1 1
1 1
y y
x
x
  
 

E) 1 1
1 1
y y
x
  

x
 
Ans: E

Larson, Calculus: An Applied Approach (+Brief), 9e Page 448

44. Find the general solution of the first-order linear differential equation.

3 6
dy
y
dx
 
A)
y Ce
3
x

1
 

B)
y
3
x

2
Ce
 

C)
y
3
x
2
Ce 

D)
3
x
1
y Ce 

E)
3
y
x
3
Ce 

Ans: B

45. Find the general solution of the first-order linear differential equation.

4 x
dy
y e
dx
 

A)
3
1
( )
3
x

y e C
 

B)
31
( )
3
x
y e C 

C)
3
1
( )
3
y eC
x


D)
3
1
( )
3
x x
y e e C 

E)
31
( )
3
x
y e C


Ans: D

46. Find the general solution of the first-order linear differential equation.

2
dy x
dx
3

x
A) 1 1
ln
2 3
y x
xC
 

B) 1 1
ln
2 3
y x xC
 

C)
1
3ln
2
y x xC
 

D)
21
3ln
2
y x
xC
 

E)
21
y x xC
3ln
 

2
Ans: D

Larson, Calculus: An Applied Approach (+Brief), 9e Page 449

47. Find the general solution of the first-order linear differential equation.

2 10
y xy
x
  

A)
2 2
x x

5 (
e e C
)

B)
2 2
x x
5 (
)
e e C

C)
2 2
x x

10 (
e e C
)

D)
2 2
x x

10 (
e e C
)
 

E)
2 2
x x
 
5 (
e eC
)
 

Ans: A

48. Find the general solution of the first-order linear differential equation.

( 1)
x yyx
2
1
 
A)
2
x x C
2
 
y
( 1)
x

B)
2
x x C
2
 
y
x
3( 1)

C)
3
x x C
3
 
y
x
3( 1)

D)
2
x x C
3
 
y
x
3( 1)

E)
3
x x C
3
 
y
x
3( 1)

Ans: C

49. Find the general solution of the first-order linear differential equation.

1
2
3
2
x
x y ye
  

A)
1
2
1
 
x
2
2
y e C
x
  
 

B)
1
2
1
 
x
2
2
y e C
x
  
 

C)
1
2
1

 
x
2
2
y e C
x
  
 

D)
1
2
1
 
x
3
  
2
y e C
x

 
E)
1
2
1
 
x
3
2
y e C
x
  
 

Ans: B

Larson, Calculus: An Applied Approach (+Brief), 9e Page 450

50.
A calf that weighs 80 pounds at birth gains weight at the rate
(1100
),
dw
k w
dt
  where
w is weight in pounds and t is time in years. Use a computer algebra system to solve the
differential equation for
1.2
k 
.
A)
1100 1020
1.2
t t

w e
e
 

B)
1.2
w e
t

80

C)
w e
1020 1.2
t
 

D)
w e
1100 1020
1.2
t
 

E)
w e
1.2
t

1020

Ans: D

51. Find the particular solution of the differential equation
y x
(2 – 4) 0
y
 
 that satisfies
the boundary condition  4 2
y  .
A)
2
4 –
x x
2
y e

B)
2
–4 + 2
x x
2
y e

C)
2
4 + 2
x x
2
y e

D)
2
–4 –
x x
4
y e

E)
2
4 –
x x
4
y e

Ans: A

52.
Find the particular solution of the differential equation
3 3
5
dy
x y x
dx

 passing through
13
the point
0,
2
 
 
 
.
A)
4
–1.25
1 63
5 10
x
y e 

B)
4
–1.251 3
5 10
x
y e 

C)
4
1 67
–1.25
x
y e 

5 10
D)
4
1 63
–1.25
x
y e 

10 10
E)
4
1 67
–1.25
x
y e 

10 10
Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 451

53. Solve for y in two ways.

4y y
  
A)
x

4
y Ce
 
B)
x
4
y Ce 
C)
x
4
y Ce 
D)
x

y Ce
4
 
E)
x
x

4
y Ce
e
 

Ans: A

54. Solve for y in two ways.

y xy x
2 2
  

A)
2
x
2
y Ce 
B)
2
x

y Ce
1
 
C)
2
x
1
y Ce 
D)
2
x

y Ce
2
 
E)
2
x

y Ce
2
 
Ans: C

55. Find the solution of the differential equation 2 0y x
   , without solving it.
A)
2
y x C
2
 

B)
2
2
y x C 

C) 2
y x C 

D)
2
y x C 

E)
y x C
2
 

Ans: D

56. Find the solution of the differential equation 2 0y xy
   , without solving it.
A)
2
x
2
x Ce 

B)
2
x
2
x Ce

C)
x
Ce
D)
2
2
x
x Ce

E)
2
x
Ce
Ans: E

Larson, Calculus: An Applied Approach (+Brief), 9e Page 452

57. Find the particular solution that satisfies the initial condition.

y y e
x
6
  
;

Initial Condition: 3
y  when 0
x 
A)
x
y e
B)
x
2
y e

C)
x

y e
2

D)
x
3
y e

E)
x

3
y e

Ans: D

58. Find the particular solution that satisfies the initial condition.

xy y
0
  ;

Initial Condition: 2
y  when 2
x 
A)
2
x

y
B) 4
xy 
C)
2
xy 
D)
2
xy 
E)
4
x

y
Ans: B

59. Find the particular solution that satisfies the initial condition.

2 2
y x y x
3 3
  
;

Initial Condition: 6
y  when 0
x 
A)
2
x
3 3
y e 

B)
3
x

y e
1 5
 

C)
3
y e
x

1 3
 

D)
2
x

y e
1 3
 

E)
2
x

y e
3 3
 

Ans: B

Larson, Calculus: An Applied Approach (+Brief), 9e Page 453

60. Find the particular solution that satisfies the initial condition.

xy yx
2
2 3 5
x
   ;

Initial Condition: 3
y  when
1
x  
A)
3 5
5
2
2
y x
4 412
x
x
 

B)
3 5
1
2
2
y x
4 412
x
x
 

C)
3 5
7
2
2
y x
4 412
x
x
 

D)
3 5
7
2
2
y x
4 312
x
x
 

E)
23 5
1
y x
4 412
x
x
 

Ans: D

Larson, Calculus: An Applied Approach (+Brief), 9e Page 454

61. Use the chemical reaction model described to find the amount y (in grams) as a function
of time t (in hours). Then use a graphing utility to graph the function.
45
y  grams when 0
t  ; 4
y  grams when 2
t 
A)
360
8 4
y
t

B)
350
8 4
y
t

C)
360
7 3
y
t

Larson, Calculus: An Applied Approach (+Brief), 9e Page 455

D)
360
8 2
y
t

E)
350
8 5
y
t

Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 456

62. Use the advertising awareness model described to find the number of people y (in
millions) aware of the product as a function of time t (in years).
0
y  when 0
t  ; 0.75
y 
when 1
t 

A)
1.386

1y e
 

B)
1.386
y
1
e 

C)
1.386
y e
t

1
 

D)
1.386
y e
t
1
 

E)
1.386
y e
t

1
 

Ans: E

Larson, Calculus: An Applied Approach (+Brief), 9e Page 457

63. Use the Gompertz growth model described to find the population y as a function of
time t (in years).
100
y  when 0
t  ; 150
y  when 2
t 

A)
0.4397
t
0.6931
y e
e
200

B)
0.4397
0.6931
200y e

C)
0.4397
t

0.6931
y e
e

200

D)
0.4397

0.6931
y e
e
20

E)
0.4397
t

y e
0.6931
2000
e

Ans: C

Larson, Calculus: An Applied Approach (+Brief), 9e Page 458

64. Use the hybrid selection model described to find the percent y (in decimal form) of the
population that has the indicated characteristics as a function of time t (in generations).
0.1
y  when 0
t  ; 0.4
y  when 4
t 
A)
(2
) 19
0.50634
t
2
(1
) 81
y y
e
y


B)
(1
) 18
0.50634
t
2
(1
) 81
y y
e
y


C)
(2
) 19
0.1
t
2
(1
) 81
y y
e
y


D)
(1
) 19
0.50634
t
2
(2
) 81
y y
e
y


E)
(2
) 81
0.50634
t
2
(1
) 19
y y
e
y


Ans: A

65. Assume that the rate of change in y is proportional to y. Solve the resulting differential
equation /dy dx ky and find the particular solution that passes through the points
(0,1),(3, 2)
A)
( ln 2) /8
0.2310
x
y x
x
e 

B)
( ln 2) /10
0.2310
x
y e
y
e 

C)
( ln10) / 3
0.2310y x
y e
e 

D)
( ln 2) / 3
0.2310
x
y e
x
e 

E)
( ln2)
0.2310
x
y e
e 

Ans: D

Larson, Calculus: An Applied Approach (+Brief), 9e Page 459

66. Assume that the rate of change in y is proportional to y. Solve the resulting differential
equation /dy dx ky and find the particular solution that passes through the points
(0, 4),(4,1)
A)
( ln 4) /10
4 4
0.3466
x
y e
x
 
 

B)
( ln 4) / 4
0.3466
x
y e
x
 
4
e
 

C)
( ln 4) / 4
4 4
0.3466
x
y e
x
 
e
 

D)
( ln 4) / 4
4 4
0.3466
y x

y e
e
 

E)
( ln 4) / 4
2 8
0.3466
x
y e
x
 
e
 

Ans: C

67. Assume that the rate of change in y is proportional to y. Solve the resulting differential
equation /dy dx ky and find the particular solution that passes through the points
(2,2),(3,4)
A)
(ln 2)
0.69311 1
3 3
x
y e
x
e 

B)
(ln 2)
0.69311 1
2 2
x
y e
x
e 

C)
(ln 2)
0.69311 1
4 4
x
y e
x
e 

D)
(ln10)
0.69311 1
4 4
y e
x
x
e 

E)
(ln10)
0.69311 1
2 2
y e
x
x
e 

Ans: B

68. During a chemical reaction, a compound changes into another compound at a rate
proportional to the unchanged amount y . Write the differential equation for the
chemical reaction model. Find the particular solution when the initial amount of the
original compound is 20 grams and the amount remaining after 1 hour is 16 grams.
A)
0.2231
, 20
t
dy
y y
dt
 

B)
0.2231
, 20
t
dx
kx y

e
 

dt
C)
0.2231
, 2
t
dy
k y e
dt
 

D)
0.2231
, 20
t
dy
ky y

e
 

dt
E)
, 20
dy
ky y
e
 

dt
Ans: D

Larson, Calculus: An Applied Approach (+Brief), 9e Page 460

69. The rate of change of the population of a city is proportional to the population P at any
time (in years). In 2000, the population was 200,000, and the constant of proportionality
was 0.015. Estimate the population of the city in the year 2020.
A) 260,972 people
B) 268,972 people
C) 263,972 people
D) 266,972 people
E) 269,972 people
Ans: E

70. A wet towel hung from a clothesline to dry loses moisture through evaporation at a rate
proportional to its moisture content. After 1 hour, the towel has lost 40% of its original
moisture content. How long will it take the towel to lose 80% of its original moisture
content?
A)  2.15 h
B)  4.95 h
C)  3.95 h
D)  3.15 h
E)  4.15 h
Ans: D

71. The rate of change in sales S (in thousands of units) of a new product is proportional
to the difference between L and S at any time t (in years), where L is the maximum
number of units of the new product available. When 0, 0t S  Write and solve the
differential equation for this sales model.
A)
S L e
kt

(1
)
 

B)
S L e
k

(1
)
 

C)
kt

S e
(1
)
 

D)
L S e
kt

(1
)
 

E)
L S e
k

(1
)
 

Ans: A

72. A population of eight beavers has been introduced into a new wetlands area. Biologists
estimate that the maximum population the wetlands can sustain is 60 beavers. After 3
years, the population is 15 beavers. The population follows a Gompertz growth model.
How many beavers will there be in the wetlands after 10 years?
A) 34 beavers
B) 30 beavers
C) 33 beavers
D) 31 beavers
E) 39 beavers
Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 461

73. At any time (in years), the rate of growth of the population N of deer in a state park is
proportional to the product of N and L N , where 500
L 
is the maximum number
of deer the park can maintain.
(a) Use a symbolic integration utility to find the general solution.
(b) Find the particular solution given the conditions
100
N  when 0
t  and
200
N  when 4
t 
(c) Find N when 1
t 
(d) Find t when 350
N 

A)
50
a.
kt
1
N
Ce


500
b.
0.2452
1 4
t
N
e


c. 121 deer

d. 9.1yr

B)
500
a.
N

kt

1
Ce

500
b.
0.2452
1 4
t
N
e


c. 121 deer
d. 9.1yr

C)
500
a.
kt
1
N
Ce


50
b.
0.2452
1 4
t
N
e


c. 121 deer
d. 9.1yr

D)
500
a.
kt
1
N
Ce


500
b.
0.2452
1 4
t
N
e


c. 131 deer

d. 9.1yr

E)
500
a.
kt
1
N
Ce


500
b.
0.2452
1 4
t
N
e


c. 121 deer
d. 10.1yr

Ans: B

Larson, Calculus: An Applied Approach (+Brief), 9e Page 462

74. A 100-gallon tank is full of a solution containing 25 pounds of a concentrate. Starting at
time 0
t  distilled water is admitted to the tank at the rate of 5 gallons per minute, and
the well-stirred solution is withdrawn at the same rate.
(a) Find the amount Q of the concentrate in the solution as a function of by solving the
Q
 
differential equation

5
100
Q
  
 

(b) Find the time required for the amount of concentrate in the tank to reach 15 pounds.
A) a.
0.05
Q e
t

25

b. 1.22 min

B) a.
0.05
Q e
t

15

b. 10.22 min

C) a.
0.05
Q e
t

25

b. 8.22 min

D) a.
0.05
t

Q e
5

b. 10.22 min

E) a.
0.05
Q e
t

25

b. 10.22 min

Ans: E

75. When predicting population growth, demographers must consider birth and death rates
as well as the net change caused by the difference between the rates of immigration and
emigration. Let P be the population at time t and let N be the net increase per unit
time due to the difference between immigration and emigration. So, the rate of growth
of the population is given by
,
dP
kP N N
dt
 
is constant. Solve the differential equation
to find P as a function of t.
A)
kt N
 
p Ce
k

B)
kt N

p Ce
k

C)
kt N
 
p e
k

D)
kt N
 
p Ce
k

E)
N
 
p Ce
k

Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 463

76. A large corporation starts at time 0
t  to invest part of its profit at a rate of P dollars
per year in a fund for future expansion. Assume that the fund earns r percent interest
per year compounded continuously. The rate of growth of the amount A in the fund is
given by
dA
rA P
dt
  where 0A  when 0
t  and r is in decimal form. Solve this
differential equation for A as a function of t .
A)
( 1)
P
rt
A e
r
  

B)
( 1)
P
rt
A e
r
  
C)
( 1)
rtP
A e
r
 

D)
( 1)
rtP
A e
r
 

E)
P
( 1)
A e
 

r
Ans: D

77.
Use this equation ( 1)
rtP
A e
r
  and Find A for each situation.
(a) $100, 000 , 0.12
5
.
P
r
and t years
 

(b) $250, 000 , 0.15
10
.
P
r
and t
years
 

A) a. 685099.00
b. 5802815.12
B) a. 585099.00
b. 5802815.12
C) a. 685099.00
b. 6802815.12
D) a. 585099.00
b. 4802815.12
E) a. 485099.00
b. 5802815.12
Ans: A

78.
Use this equation ( 1)
rtP
A e
r
  and Find P if the corporation needs $120,000,000 in
8 years and the fund earns 8% interest compounded continuously.
A) $110,708,538.49
B) $10,500,538.49
C) $10,708,538.49
D) $100,708,538.49
E) $1,708,538.49
Ans: C

Larson, Calculus: An Applied Approach (+Brief), 9e Page 464

79. A medical researcher wants to determine the concentration C (in moles per liter) of a
tracer drug injected into a moving fluid with flow R (in liters per minute). Solve this
problem by considering a single-compartment dilution model (see figure). Assume that
the fluid is continuously mixed and that the volume V (in liters) of fluid in the
compartment is constant.

If the tracer is injected instantaneously at time 0
t  , then the concentration of the fluid
in the compartment begins diluting according to the differential equation
,
dC R
C C C
 


when 0
t 
dt
0
V
 
 
(a) Solve this differential equation to find the concentration as a function of time.
(b) Find the limit of C as t 
A) a.
Rt V
/
C Ce

0
b. 1
B) a.
Rt V
C Ce
/

0
b. 0
C) a.
/Rt V

C e

b. 2
D) a.
Rt

C Ce

0
b. 0
E) a.
Rt V
/

C C

0
b. 1
Ans: B

80. A 300-gallon tank is full of a solution containing 35 pounds of concentrate. Starting at
time 0,
t 
distilled water is added to the tank at a rate of 30 gallons per minute, and the
well-stirred solution is withdrawn at the same rate. Find the amount of concentrate Q in
the solution as a function of t.
A)


1/10
t

35
Q e

B)


1/10
t

30
Q e

C)


Q e
1/30 t

D)


1/35 t

Q e

E)


1/10
t

300
Q e

Ans: A

Larson, Calculus: An Applied Approach (+Brief), 9e Page 465