(9/1/08)
Math 10B. Lecture Examples.
Section 11.5. Growth and Decay
†
Example 1
Match problems
(I)
through
(IV)
below to differential equations (a) through (d) and to
the slope fields in Figures 1 through 4.
(I)
The thickness of the ice on a lake grows at a rate that is proportional to the reciprocal of its
thickness. Find the thickness
y
=
y
(
t
) as a function of the time
t
.
(II)
A population grows at a rate proportional to its size. Find the population
y
=
y
(
t
) as a
function of the time
t
.
(III)
A hot potato is taken out of the oven at time
t
= 0 into a kitchen that is at 20
◦
Celsius.
The rate of change of the potato’s temperature is proportional to the difference between its temperature
and that of the kitchen. Find the temperature
y
=
y
(
t
) of the potato as a function of
t
.
(IV)
Find a function
y
=
y
(
t
) whose rate of change with respect to
t
is

2
t
.
(a)
dy
dt
= 0
.
2
y
(b)
dy
dt
=
20
y
(c)
dy
dt
=

2
t
(d)
dy
dt
=

2(
y

20)
t
5
10
y
100
200
300
400
t
500
1000
y
50
100
150
200
FIGURE 1
FIGURE 2
t
1
2
3
4
5
6
y
10
20
30
40
50
t
1
2
3
y
10
20
30
40
50
60
FIGURE 3
FIGURE 4
Answer:
Problem I goes with equation (b) and the slope field in Figure 2.
•
Problem II goes with equation (a) and
Figure 1.
•
Problem III goes with equation (d) and Figure 4.
•
Problem IV goes with equation (c) and Figure 3.
†
Lecture notes to accompany Section 11.5 of
Calculus
by HughesHallett et al.
1
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Math 10B. Lecture Examples. (9/1/08)
Section 11.5, p. 2
Example 2
(a)
Solve the differential equation
(a)
dy
dt
= 0
.
2
y
from Example 1 (a population) with
the initial condition
y
(0) = 50 and draw its graph with the corresponding slope field.
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 Summer '10
 reed
 Derivative, Harshad number, dy, 49 meters per second, 98 meters per second

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