Related Rates-Conical Reservoir

[amme636]

Homework Statement

A conical reservoir has a depth of 24 ft and a circular top of radius 12 ft. It is being filled so that the depth of the water is increasing at a constant rate of 4ft/hour. Determine the rate of which the water is entering the reservoir when the depth is 5 ft.

Homework Equations

V=(1/3)pi r^2 h

The Attempt at a Solution

I'm not really sure where to start even, I drew a picture but that's it.

 

[Dick]

The rate of water coming in is dV/dt, right? Your first job is probably to express V in terms of a single variable, say h. Given the proportions of your cone, r and h are proportional. Can you write an expression relating them?

 

[amme636]

The rate of water coming in is dV/dt, right? Your first job is probably to express V in terms of a single variable, say h. Given the proportions of your cone, r and h are proportional. Can you write an expression relating them?

So, is it just r=h/2 and V=(1/3)pi (h/2)^2 h, which is (pi h^3)/12?
Than V'(dh/dt) would be dV/dt right?
or am I missing something still?

 

[Dick]

Sure, r=h/2 and V(t)=pi*h(t)^3/12. Now find dV/dt in terms of h(t) and h'(t) and solve for dV(t)/dt given h(t)=5 ft.