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m_a_m_a [10]
3 years ago
9

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha

t 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.

Mathematics
1 answer:
8_murik_8 [283]3 years ago
7 0
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.

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