# Math 22 Exponential Growth and Decay

## Exponential Growth and Decay

 If ${\displaystyle y}$ is a positive quantity whose rate of change with respect to time
is proportional to the quantity present at any time ${\displaystyle t}$, then  is of the form
${\displaystyle y=Ce^{kt}}$
where ${\displaystyle C}$ is the initial value and ${\displaystyle k}$ is the constant of proportionality.
Exponential growth is indicated by ${\displaystyle k>0}$ and exponential decay by ${\displaystyle k<0}$.


## Guidelines for Modeling Exponential Growth and Decay

 1. Use the given information to write two sets of conditions involving ${\displaystyle y}$ and ${\displaystyle t}$
2. Substitute the given conditions into the model ${\displaystyle y=Ce^{kt}}$ and
use the results to solve for the constants ${\displaystyle C}$ and ${\displaystyle k}$.
3. Use the model ${\displaystyle y=Ce^{kt}}$ to answer the question.


Exercises

1) The number of a certain type of bacteria increases continuously at a rate proportional to the number present. There are 150 bacteria at a given time and 450 bacteria 5 hours later. How many bacteria will there be 10 hours after the initial time?

Solution:
At the initial time: ${\displaystyle y_{0}=Ce^{kt_{0}}}$, so ${\displaystyle 150=Ce^{kt_{0}}}$
After 5 hours: ${\displaystyle y_{1}=Ce^{k(t_{0}+5)}}$, so ${\displaystyle 450=Ce^{k(t_{0}+5)}=Ce^{kt_{0}+5k}=Ce^{kt_{0}}e^{5k}}$
Take the second equation and divided it by the first equation, we get:
$\displaystyle \frac{450}{150}=\frac{Ce^{kt_0}e^{5k}}{Ce^{k(t_0+5)}}$
${\displaystyle 3=e^{5k}}$
Now, consider the equation after 10 hours: ${\displaystyle y_{2}=Ce^{k(t_{0}+10)}=Ce^{k[(t_{0}+5)+5]}=Ce^{k(t_{0}+5)}e^{kt}=450.3=1350}$

2) A sports utility vehicle that costs $33000 new has a book value of$20856 after 3 years. Assume the value of the vehicle exponential decay over time. Find this exponential model.

Solution:
Consider: ${\displaystyle y=Ce^{kt}}$
So, ${\displaystyle 20856=33000e^{k3}}$, then ${\displaystyle {\frac {20856}{33000}}={\frac {79}{125}}=e^{3k}=(e^{k})^{3}}$
Then, ${\displaystyle {\sqrt[{3}]{\frac {79}{125}}}=e^{k}}$
Then, ${\displaystyle \ln e^{k}=\ln {\sqrt[{3}]{\frac {79}{125}}}}$
Therefore, ${\displaystyle e=\ln {\sqrt[{3}]{\frac {79}{125}}}}$