To model this situation, we are going to use the decay formula:
![A=Pe^{rt}](https://tex.z-dn.net/?f=A%3DPe%5E%7Brt%7D)
where
![A](https://tex.z-dn.net/?f=A)
is the final pupolation
![P](https://tex.z-dn.net/?f=P)
is the initial population
![e](https://tex.z-dn.net/?f=e)
is the Euler's constant
![r](https://tex.z-dn.net/?f=r)
is the decay rate
![t](https://tex.z-dn.net/?f=t)
is the time in years
A. We know for our problem that the initial population is 1,250, so
![P=1250](https://tex.z-dn.net/?f=P%3D1250)
; we also know that after a year the population is 1000, so
![A=1000](https://tex.z-dn.net/?f=A%3D1000)
and
![t=1](https://tex.z-dn.net/?f=t%3D1)
. Lets replace those values in our formula to find
![r](https://tex.z-dn.net/?f=r)
:
![A=Pe^{rt}](https://tex.z-dn.net/?f=A%3DPe%5E%7Brt%7D)
![1000=1250e^{r}](https://tex.z-dn.net/?f=1000%3D1250e%5E%7Br%7D)
![e^{r}= \frac{1000}{1250}](https://tex.z-dn.net/?f=e%5E%7Br%7D%3D%20%5Cfrac%7B1000%7D%7B1250%7D%20)
![e^{r}= \frac{4}{5}](https://tex.z-dn.net/?f=e%5E%7Br%7D%3D%20%5Cfrac%7B4%7D%7B5%7D%20)
![ln(e^{r})=ln( \frac{4}{5} )](https://tex.z-dn.net/?f=ln%28e%5E%7Br%7D%29%3Dln%28%20%5Cfrac%7B4%7D%7B5%7D%20%29)
![r=ln( \frac{4}{5} )](https://tex.z-dn.net/?f=r%3Dln%28%20%5Cfrac%7B4%7D%7B5%7D%20%29)
![r=-02231](https://tex.z-dn.net/?f=r%3D-02231)
Now that we have
![r](https://tex.z-dn.net/?f=r)
, we can write a function to model this scenario:
![A(t)=1250e^{-0.2231t}](https://tex.z-dn.net/?f=A%28t%29%3D1250e%5E%7B-0.2231t%7D)
.
B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.
C.
- The function is decreasing
- The function doe snot have a x-intercept
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0:
![A(0)=1250e^{(-0.2231)(0)](https://tex.z-dn.net/?f=A%280%29%3D1250e%5E%7B%28-0.2231%29%280%29)
![A_{0}=1250e^{0}](https://tex.z-dn.net/?f=A_%7B0%7D%3D1250e%5E%7B0%7D)
![A_{0}=1250](https://tex.z-dn.net/?f=A_%7B0%7D%3D1250)
- Over the interval [0,10], the function will have a minimum at t=10:
![A(10)=1250e^{(-0.2231)(10)](https://tex.z-dn.net/?f=A%2810%29%3D1250e%5E%7B%28-0.2231%29%2810%29)
D. To find the rate of change of the function over the interval [0,10], we are going to use the formula:
![m= \frac{A(0)-A(10)}{10-0}](https://tex.z-dn.net/?f=m%3D%20%5Cfrac%7BA%280%29-A%2810%29%7D%7B10-0%7D%20)
where
![m](https://tex.z-dn.net/?f=m)
is the rate of change
![A(10)](https://tex.z-dn.net/?f=A%2810%29)
is the function evaluated at 10
![A(0)](https://tex.z-dn.net/?f=A%280%29)
is the function evaluated at 0
We know from previous calculations that
![A(10)=134.28](https://tex.z-dn.net/?f=A%2810%29%3D134.28)
and
![A(0)=1250](https://tex.z-dn.net/?f=A%280%29%3D1250)
, so lets replace those values in our formula to find
![m](https://tex.z-dn.net/?f=m)
:
![m= \frac{134.28-1250}{10-0}](https://tex.z-dn.net/?f=m%3D%20%5Cfrac%7B134.28-1250%7D%7B10-0%7D%20)
![m= \frac{-1115.72}{10}](https://tex.z-dn.net/?f=m%3D%20%5Cfrac%7B-1115.72%7D%7B10%7D%20)
![m=-111.572](https://tex.z-dn.net/?f=m%3D-111.572)
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.