# Math 22 Exponential Growth and Decay

## Exponential Growth and Decay

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


## Guidelines for Modeling Exponential Growth and Decay

 1. Use the given information to write two sets of conditions involving $y$ and $t$ 2. Substitute the given conditions into the model $y=Ce^{kt}$ and
use the results to solve for the constants $C$ and $k$ .
3. Use the model $y=Ce^{kt}$ 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: $y_{0}=Ce^{kt_{0}}$ , so $150=Ce^{kt_{0}}$ After 5 hours: $y_{1}=Ce^{k(t_{0}+5)}$ , so $450=Ce^{k(t_{0}+5)}=Ce^{kt_{0}+5k}=Ce^{kt_{0}}e^{5k}$ Take the second equation and divided it by the first equation, we get:
$\displaystyle \frac{450}{150}=\frac{Ce^{kt_0}e^{5k}}{Ce^{k(t_0+5)}}$
$3=e^{5k}$ Now, consider the equation after 10 hours: $y_{2}=Ce^{k(t_{0}+10)}=Ce^{k[(t_{0}+5)+5]}=Ce^{k(t_{0}+5)}e^{kt}=450.3=1350$ 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: $y=Ce^{kt}$ So, $20856=33000e^{k3}$ , then ${\frac {20856}{33000}}={\frac {79}{125}}=e^{3k}=(e^{k})^{3}$ Then, ${\sqrt[{3}]{\frac {79}{125}}}=e^{k}$ Then, $\ln e^{k}=\ln {\sqrt[{3}]{\frac {79}{125}}}$ Therefore, $e=\ln {\sqrt[{3}]{\frac {79}{125}}}$ 