1. The formula for the volume of a tank is V= (pi)r^3 where r is the radius of the tank?

Question

If water is flowing in at the rate of 15 cubic feet per minute,
find the rate at which the radius is changing when the radius is 3 feet?

Answer 1

V=(pi)r^3
therefore dV/dt=3(pi)r^2 dr/dt
or dr/dt=dV/dt*1/3(pi)r^2

for water flowing at the rate 15cubic feet/min at r= 3 feet we have
dr/dt=15/3*3.14*3^2

or dr/dt=35/198 or 0.17

Answer 2

I wonder if the information in your question is correct. According to the log book the formula for the volume of a tank is:
V = π r² h. Do you mean what rate h is changing? If so:

V = π r² h But the water enters the tank at 15ft^3 per min.
15 = π(3)² h
h = 15/π(3)²
h = 0∙530516477 ft/min.

If the height of the cylinder is the same as the radius, then you could have the volume = π r^3. However, there are two different R's value. One will change with the water flow, the other won't.

Answer 3

Formula For Volume Of Tank

Answer 4

The obvious way I can see the Volume being equal to pi r^3 is if the tank is cylindrical with a height that happens to match its radius. In that case, the radius won't change.

However, the height will change by dV/(pi r^2), or:

15/(pi*9) =5/(3*pi)

Alternatively, the tank might be a cone whose height is 3 times greater than the radius. In that case, the rate of change in the volume will be 3 * pi r^2 dr. You can rearrange to make:

dV/(3*pi*r^2) = dr
15/(3*pi*9) = 5/(9 pi)

You really need the shape of the object.

Answer 5

V= pi r^3
dV/dt =pi 3r^2 dr/dt
15 = pi 3 * 3^2 dr/dt
dr/dt = (15*7) / (22*27)

so get the answer

Answer 6

How the fixed tank radius can be changed? You asked a worng question.

Answer 7

if r=3, V=pi x 3^3=27pi cubic feet.
then the tank wil be full after 27pi/15 minutes (5'39")