Water Tank Volume Rate of Change at Specific Depth

[jackscholar]

Homework Statement

Water runs into a cylindrical tank at the rate of 0.6m^3/s. The tank has a heigh of 2.5m and a base radius of 1.2m. How fast is the water level rising when the water level is 1.6m deep.

Homework Equations

V=∏ x r^2 x h

The Attempt at a Solution

I know that 0.6m^3/s is dV/dt which should mean that the equation looks something like this: dV/dt=(dV/dh)(dh/dt). When differentiating volume with respect to height though, the h is eliminated and is makes no sense if the question is giving the h variable.

 

[Doc Al]

Yes, the rate of water level rise is independent of height. (Trying to trick you, no doubt.)

 

[jackscholar]

So then that means 0.6=(∏ x 1.2^2)(dh/dt)
0.6/(∏ x 1.2^2)=dh/dt
0.132m^3/s=dh/dt

 

[Doc Al]

So then that means 0.6=(∏ x 1.2^2)(dh/dt)
0.6/(∏ x 1.2^2)=dh/dt
0.132m^3/s=dh/dt

Looks good. Except for the units!

 

[jackscholar]

oh yes, sorry haha. Height isn't a volume :S