# Exponential Growth and Decay

Exponential growth is where a function’s growth rate is proportional to the function’s current value. Exponential growth can be seen in things like human population or virus spreading. Exponential decay occurs when the decay rate is proportional to the function’s current value, i.e. the function is decreasing in a similar fashion instead of increasing.

Geometric growth or decay is the same as exponential growth or decay except the function is only evaluated at discrete values. Example, recall a geometric series that has a start value of 5 and a common ratio of 2, i.e. $5+10+20+40+80+...$ is an example of a series that exhibits geometric growth. The series $10+1+0.1+0.01+0.001+...$ is an example of a series that exhibits geometric decay.

Example – A rare species of dandelion is being observed. The population, p, after a given number of years, T, is given by

$p(t)=\frac{500e^{0.3t}}{1+4e^{0.3t}}$

a) Find the population of dandelion at the beginning of the study.
b) Find the rate at which the dandelion population changes after 5 years. Give your answer to two decimal places.
c) Explain why the population of dandelion does not exceed 125.

Solutions:
a) The initial population of dandelion can be found by finding p(0), i.e. by substituting t=0 into p. The initial population is

$p(0)=\frac{500}{1+4}=100.$

b) The rate at which the population changes after 5 years is found by substituting t=5 into the expression for $\frac{dp}{dt}$. The simplest way to find $\frac{dp}{dt}$ is to write p as

$p(t)=\frac{500e^{0.3t}}{1+4e^{0.3t}}=\frac{500}{e^{-0.3t}+4}=500\left(e^{-0.3t}+4\right)^{-1}$

by dividing top and bottom by $e^{0.3t}$. We can then use the chain rule to differentiate p:

$\frac{dp}{dt}=500\times -1\times\left(e^{-0.3t}+4\right)^{-2}\times -0.3e^{-0.3t}=\frac{150e^{-0.3t}}{\left(e^{-0.3t}+4\right)^{2}}.$

See more on differentiating e to the x. Substituting t=5 into this expression gives

$\frac{dp(5)}{dt}=\frac{150e^{-0.3\times 5}}{\left(e^{-0.3\times 5}+4\right)^{2}}=\frac{150e^{-1.5}}{\left(e^{-1.5}+4\right)^{2}}=1.88.$

c) From part b), we have that p can be written as

$p(t)=\frac{500}{e^{-0.3t}+4}$

Recall that $e^{-x}$ is a number between 0 and 1 for any positive x value. It follows that $e^{-0.5t}$ is between 0 and 1 (not including) for any positive t value. Hence, $e^{-0.5t}+4$ is a number between 4 and 5 and so we are dividing 500 by a number that is between 4 and 5. The result is a number between 100 and 125 and so the population does not exceed 125.