BC Calc III
DE Quiz
Name:
key
Calculator not allowed.
You must show enough work so that I can recreate your results.
#1.
a. Write a differential equation (Initial Value Problem) describing the temperature
as a function of time of a can of soda taken out of a 40  F refrigerator and left in a
65  F room.
y  k ( y  65); y (0)  40 , where y represents the temperature of the soda at time t.
b. Make a sketch of the solution to the above Initial Value problem. Mark the
scale on the temperature axis clearly. You do not need a scale on the time axis.
Temp
65
40
time
#2.
Solve the following differential equation using separation of variables.
y   y 2 x( x 2  2)4
dy
  y 2 x( x 2  2) 4
dx
dy
 2   x( x 2  2) 4 dx
y
dy
1
  2    2 x( x 2  2) 4 dx
y
2
y   y 2 x( x 2  2) 4 
1
1 1

   ( x 2  2)5   C
y
2 5

10
y 2
( x  2)5  C

BC CALC III
#3
A national park in Tanzania can sustain a population of water buck of about 800.
After several years of drought, the population was only 100, but now is beginning
to grow logistically again. (Let this be time t = 0.)
a. Express the differential equation for this situation (in terms of k).
P  kP(800  P); P(0)  100 where P represents the population of water buck at
time t.
The solution to this differential equation is given by: P 
C
de
 kCt
1
, where
P = number of water buck at time t (in years).
b. Find the constant of integration d.
P(0)  100  100 
C
de
 kC 0
1

800
 100d  700  d  7
d 1
c. Four years later the population had grown to 300. Find the constant k. [Your
answer should involve a natural log].
 e3200 k
800
 2100e3200 k  500
7e
1
5
1
 5
 k 
ln  
21
3200  21 
P(4)  300  300 
 k 8004
d. What will be the population of water buck when the population is growing
most rapidly?
The population will be ½ the carrying capacity = ½ (800) = 400
e. What would happen to the water buck if their population was 900 at some
point? Explain your answer making explicit reference to the Differential
equation.
Since k > 0, if P = 900 then P  kP(800  P)  k 900  (100)  0 . Since P  0 the
population will decrease. As P gets closer to 800, P  0 , so again we will have
a horizontal asymptote at P = 800.
BC CALC III
Match each slope field with the correct differential equation.
(a)
(b)
(1)
y  xy
(h)
(2)
(c)
y  x  y
(a)
(d)
3)
y  x  (2  y )
(e)
(4)
(e)
 y 
y  sin 

 3 
(b)
(f)
(5)
y 
1
( x  2)( y  3)
2
(d)
(6)
y  x 2
(c)
(g)
(h)
(7)
y  y 2
(g)
  y2 

(8) y  sin 

 2 
(f)
BC CALC III