Exponential Models Exponential Models
Introduction
Exponential growth is modelled by differential equations of the form:
\frac{dy}{dt} = ky
Where
k
is a constant.
In other words:
The rate of change of a quantity is proportional to the size of the quantity. We can solve this differential equation to find the function
y(t)
that models exponential growth.
\begin{align*}
\frac{dy}{dt} &= ky \\ \text{} \\
\frac{dy}{ky} &= dt \\ \text{} \\
\int \frac{dy}{y} &= \int k \, dt \\ \text{} \\
\ln|y| &= kt + C_1 \\ \text{} \\
y &= e^{kt + C_1} \\ \text{} \\
y &= e^{kt}e^{C_1} \\ \text{} \\
y &= C_2e^{kt}
\end{align*}
e^{C_1}
is just some real number. Instead of writing that constant out, we replace it with
C_2
.
y(0)
is the size of the quantity at the beginning, when
t=0
. Since
y(0)=C_2e^{k(0)}=C_2
, the constant
C_2
is the initial size of the quantity.
Example
In the first lesson of this module, we used the following differential equation to model a population increasing at a rate of 2% per year:
\frac{dP}{dt} = 0.02P
The equation which represents a population undergoing this growth is:
P = C_2e^{0.02t}
If the population starts with
42,000
people, the equation becomes:
P = 42000e^{0.02t}
After
10
years, the population would be:
P = 42000e^{0.02(10)} = 42000e^{0.2} = 51299
Exponential Decay
Exponential decay is the same as exponential growth but with a negative rate of change.
For example, here is a model for a car's value decreasing by
0.5\%
per
1000
miles driven:
V = C_2e^{-0.005m}
If the car's starting price was
\$20,000
, the equation becomes:
V = 20000e^{-0.005m}
After being driven
100,000
miles, the car's value would be:
V = 20000e^{-0.005(100)} = 20000e^{-0.5} = \$12,130.61