Question

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated that the population will be 1,000. After 3 years, it is estimated that the population will be 640. a. Write a function to model this scenario. b. Create a graph to show the bee population over the next 10 years. c. Identify the key features of the function. Identify the x- and y-intercepts. Determine the maximum, the minimum, whether the function is increasing or decreasing, the rate of change of the function over the interval [0, 10], and any asymptotes.

Answer 1

To model this situation, we are going to use the decay formula: $A=Pe^{rt}$
where 
$A$ is the final pupolation
$P$ is the initial population 
$e$ is the Euler's constant
$r$ is the decay rate 
$t$ is the time in years

A. We know for our problem that the initial population is 1,250, so $P=1250$; we also know that after a year the population is 1000, so $A=1000$ and $t=1$. Lets replace those values in our formula to find $r$:
$A=Pe^{rt}$
$1000=1250e^{r}$
$e^{r}= \frac{1000}{1250}$
$e^{r}= \frac{4}{5}$
$ln(e^{r})=ln( \frac{4}{5} )$
$r=ln( \frac{4}{5} )$
$r=-02231$

Now that we have $r$, we can write a function to model this scenario:
$A(t)=1250e^{-0.2231t}$.

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: 
$A(0)=1250e^{(-0.2231)(0)$
$A_{0}=1250e^{0}$
$A_{0}=1250$
- Over the interval [0,10], the function will have a minimum at t=10:
$A(10)=1250e^{(-0.2231)(10)$
$A_{10}=134.28$

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: $m= \frac{A(0)-A(10)}{10-0}$
where 
$m$ is the rate of change 
$A(10)$ is the function evaluated at 10
$A(0)$ is the function evaluated at 0
We know from previous calculations that $A(10)=134.28$ and $A(0)=1250$, so lets replace those values in our formula to find $m$:
$m= \frac{134.28-1250}{10-0}$
$m= \frac{-1115.72}{10}$
$m=-111.572$
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.