Math 22 Exponential Growth and Decay

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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:
${\frac {450}{150}}={\frac {Ce^{kt_{0}}e^{5k}}{Ce^{k(t_{0}+5)}}}$
$3=e^{5k}$
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, $e=\ln {\sqrt[{3}]{\frac {79}{125}}}$

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