A lawn sprinkler sprays water in a circular area of grass. The diameter of the circular area of grass is 15 ft. What is the a
1 answer:
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To model this situation, we are going to use the decay formula: 
where
is the final pupolation
is the initial population
is the Euler's constant
is the decay rate
is the time in years
A. We know for our problem that the initial population is 1,250, so
; we also know that after a year the population is 1000, so 
and 
. Lets replace those values in our formula to find :
Now that we have, we can write a function to model this scenario:
.
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:
- Over the interval [0,10], the function will have a minimum at t=10:
D. To find the rate of change of the function over the interval [0,10], we are going to use the formula:
where
is the rate of change
is the function evaluated at 10
is the function evaluated at 0
We know from previous calculations that
and 
, so lets replace those values in our formula to find :
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.
where
A. We know for our problem that the initial population is 1,250, so
Now that we have
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:
- Over the interval [0,10], the function will have a minimum at t=10:
D. To find the rate of change of the function over the interval [0,10], we are going to use the formula:
where
We know from previous calculations that
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.