SUTCLIFFE’S NOTES: CALCULUS 2 SWOKOWSKI
.
7.6 Laws of Growth and Decay
Suppose y  y t  , i.e. a quantity y is a function of time and suppose the rate of change of
dy / dt is directly proportional to the magnitude of the quantity at time t, i.e.
dy / dt  cy .
Goal: To find an expression for y.
dy
dy
dy
 cy 
 cdt  
 cdt  ln y  ct  d (assume y  0), for some constant d
dt
y
y 
 y  ect d  ed ect
If at t  0, y  y0 (the initial value) , then y0  ed e0  ed  y  y0ect .
Thus, we have proved:
Theorem Let y be a differentiable function of t such that y 0 for every t , and let y 0 be the
value of y at t 0 . If
dy / dt  cy
for some constant c , then
 
y  y0ect .
Note: The theorem says that if the rate of change of y  y t  with respect to t is directly
proportional to y, then y may be expressed as an exponential function.
If c 0 then   is a law of growth. (ex, bacterial population)
If c 0 , then   is a law of decay.
(ex, radioactive decay)
Ex 8 The rate at which salt dissolves in water is directly proportional to the amount that remains
undissolved. If 10 pounds of salt are placed in a container of water and 4 pounds dissolve in 20
minutes, how long will it take for two more pounds to dissolve?
Solution:
Let y = the amount of undissolved salt
Use y  y0ect with y 0  10 .
Then y  10ect . Find c. Use the fact that y  6 when t  20 minutes .
6 10e20c  0.6  e20c  20c  ln0.6  c 
t
ln 0.6 
t
ln 0.6
ln0.6
20
y  10e 20
Substitute y  4 and solve for t.
t
ln 0.6
t
20ln0.4
 35.87 min
ln0.6
Thus, it will take 15.87 minutes for 2 more pounds to dissolve.
4  10e 20
 0.4  e 20
  0.6 20  t 
1
Note: Instead of solving for c, one can also solve for ec . In the above example, we could have
done the following:
1
6 10e20c  0.6  e20c  ec   0.6  20
t
So, the equation we will get is y  10  0.6  20 . Convince yourself that this is equivalent to the
equation we obtained above.
Ex 4 The population of a city is increasing at the rate of 5% per year. If the present population is
500,000 and the rate of increase is proportional to the number of people, what will the population
be in 10 years?
Solution:
y  y0ect with y0  500,000 and c  0.05
y  500,000e0.05t
y 10  500,000e.05(10)  824,361
Ex 2 The polonium isotope 210 Po has a half-life of approximately 140 days. If a sample weighs
20 mg initially, how much remains after t days? Approximately how much will be left after two
weeks?
Solution:
Use y  20ect
When t  140, y  10 . Use this to find c.
1
ln2
 e140c  140c   ln2  c  
2
140
So the amount of the isotope left after t days is
10  20e140c 

ln2

ln2
(14)
140
t
y  20e 140 .
After 2 weeks or 14 days, the amount left is
y  20e
 20e

ln2
10
 18.66 mg
Example 3 Newton’s Law of Cooling
The rate at which an object cools is directly proportional to the difference in temperature between
the object and the surrounding medium.
dT / dt  c(T  S) ,
where T is the temperature of the object at any time t and
S is the temperature of the surrounding medium (ex, air)
Ex 6 A metal plate that has been heated cools from 1800 F to 1500F in 20 minutes when
surrounded by air at a temperature of 600 F . Use Newton’s law of cooling to approximate its
temperature at the end of one hour of cooling. When will the temperature be 1000 F ?
2
Solution:
dT
 c  dt
T  60
 ln T  60  ct  d  T  60  ect d  T  60  ect d
dT / dt  c(T  60)  
When t  0, T  180 : 180  60  ed  ed  120. Thus, T  60  120ect
When t  20, T  150 :
3
3
1 3
150  60  120e20c  90  120e20c   e20c  20c  ln  c 
ln
4
4
20 4
1
Thus, the equation is T  60  120e 20
Find t when T  100 :
100  60  120e
1 3
ln t
20 4
 40  120e
1 3
ln t
20 4
3
ln t
4
ln t
1
1 1 3
20ln3
20 4
e
 ln 
ln  t  t 
 76 min
3
3 20 4
ln3  ln4
1

3
Ex 16 In the study of lung physiology, the equation used to describe the transport of a substance
across a capillary wall is
dh
V h 
 
,
dt
Q  k  h 
where h is the hormone concentration in the bloodstream, t is the time, V is the maximum
transport, Q is the volume of the capillary, and k is a constant that measures the affinity between
the hormones and enzymes that assist the transport process. Find the general solution of the
differential equation.
Solution:
dh
V h 
 
dt
Q  k  h 
Separate the variables h and t:
k h
V
dh   dt
h
Q
V
 k
 1 h  dh   Q  dt
V
h  k ln h   t  C
Q
3