Let’s model this with a function:
V=4/3• π•(2x)^3, where x=elapsed time and r=2ft./min multiplied by time (x).
We are given tho find volume when the radius is 3ft.
So, we need to find the volume when r=2 and r=3 to compare the volume and find the increase:
When r=2:
V=4/3•π•(2)^3
Simplify:
V=4/3•π•8
V=4(π•8)/3
V=33.51ft.^3
When r=3:
V=4/3•π•(3)^3
Simplify:
V=4/3•π•9
V=4(π•9)/3
V=113.1ft.^3
So, let’s find the increase in volume by taking the difference of the two volumes calculated above:
when r=3 - when r=2:
113.1 - 33.51 = 79.59
So, the volume increased 79.59ft^3.
Now, we must find the time interval to produce the rate of increase based on time:
Recall that when r=2x, x=minutes elapsed.
So, let’s set 2x=3 to find x (time in minutes)
Solve for x:
x=3/2
x=1.5
So, 1.5 minutes have elapsed when for the radius to equal 3.
This means we can create a rate of increase with the volume and time:
79.59/1.5
This ratio means that the volume is increasing at 79.59in^3 per every 1.5 minutes. We can simplify and find the unit rate by dividing the numerator by the denominator:
53.06in^3/min
So, the volume is increasing at 53.06in^3 per minute.
