Section 3.8 Exponential Growth and Decay
Ruipeng Shen
Feb 27
1
General Theory
dy
= ky is sometimes called the law of natural growth (if k > 0) or the law of
dt
natural decy (if k < 0). We can solve this separable equation:
The equation
dy
= kdt
y
=⇒ |y| = eC ekt
=⇒ ln |y| = kt + C
=⇒ y = ±eC ekt
Thus we have a general solution y = Aekt . In fact, a basic calculation shows
dy
d −kt −kt
e y =e
− ky = 0,
dx
dt
thus any solution to this equation satisfies e−kt y = A, where A is a constant. This proves the
following theorem
Theorem 1. The only solutions of the differential equation dy/dt = ky are the exponential
function y(t) = Aekt . Here the constant A can be determined by the initial condition A = y(0).
2
Examples
Example 2 (Population Growth). Use the fact that the world population was 2560 million in
1950 and 3040 million in 1960 to model the population of the world in the second half of the
20th century. What is the growth rate? Predict the population in 1993.
Solution Let us measure the time in years and let t = 0 in the year 1950. By the theorem
above, we know the population is given by the formula
y(t) = 2560ekt
Plugging y(10) = 3040, we have
3040 = 2560e10k
=⇒
10k = ln
3040
2560
=⇒
k=
1
3040
ln
≈ 0.017185.
10 2560
Thus we have
y(43) ≈ 2560e0.017185×43 ≈ 5360.
1
Example 3 (Radioactive Decay). The half life of radium-226 is 1590 years.
(a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of the sample
that remains after t years.
(b) Find the mass after 1000 years correct to the nearest milligram.
(c) When will be the mass reduced to 30 mg?
Solution
(a) We know the formula is m(t) = 100ekt . Since the half-life is 1590 years, we have
50 = 100e1590k
=⇒
1590k = ln(1/2)
Thus we have
=⇒
k=
− ln 2
.
1590
t ln 2
m(t) = 100e− 1590 = 100 × 2−t/1590
(b) The mass after 1000 years is
m(1000) = 100 × 2−1000/1590 ≈ 65 mg.
(c) We solve the equation
30 = 100 × 2−t/1590
3
=⇒
2t/1590 =
10
3
=⇒
t = 1590 ·
ln(10/3)
≈ 2762 years.
ln 2
Newton’s Law of Cooling
Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not
too large. This can be formulated by
dT
= k(T − Ts )
dt
Here t is the time, T is the temperature of the object, Ts is the temperature of surroundings,
and k is a negative constant.
Example 4. A bottle of soda pop at room temperature (72◦ F) is placed in a refrigerator where
the temperature is 44◦ F. After half an hour the soda pop has cooled to 61◦ F. How long does
it take for the soda pop to cool to 50◦ F?
Solution
First of all, the equation above is equivalent to
d(T − Ts )
= k(T − Ts ).
dt
Thus we have T − Ts = Aekt . Here Ts = 44. Plugging in t = 0 and T = 72, we have A = 28.
Therefore
T − 44 = 28ekt .
Plugging in t = 30 and T = 61, we have
17
1
17
=⇒ k =
ln
≈ −0.01663
28
30 28
Therefore we can find the time to cool to 50◦ F by solving
17 = 28e30k
=⇒
30k = ln
50 − 44 = 28e−0.01663t
2
→
t ≈ 92.6 min.