Thread: Related rates and how fast the height of a pile's increasing?

The question:

Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 19 feet high? Recall that the volume of a right circular cone with height h and base radius r is given by V=(1/3)(pi)(r^2)(h)
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I've attempted this problem four times and I'm really just confusing myself at this point. Can anyone tell me what I'm doing wrong?

Calculations so far:

diameter = height
radius = height/2

V = (1/3)pi(h/2)^2(h)
V = (1/3)pi((h^2)/4)(h)
V = (1/3)pi((h^3)/4)
V = (pi*h^3)/12

Implicit differentiation:

dV/dt = (3pi/12)h^2(dh/dt) = (pi/4)h²(dh/dt)
dh/dt = (dV/dt)/(pi/4)h^2

dV/dt = 40 cubic feet per minute
h = 19 feet

dV/dt = (40)/(pi/4)*(19^2)
dV/dt = 40/(361*pi)/4
dV/dt = 40/22.5625

but that answer's apparently incorrect.

What do I need to do?

where at

Thanks for the prompt reply. I don't really understand where that's coming from or how I use it to get the answer, though. What do I need to do, exactly..?

The first line of my response is the chain rule in terms of the variables given in the problem.

Do you know this rule?

I know what the chain rule's formula is, but I'm kind of sticky on when to apply it practically and what the use of it is.

Well you have 3 variables in your problem, height, time and volume, thats how I constructed the chain rule for this one.

Ohh, I think I see what it's saying. So:

dh/dt =
dh/dt =
dh/dt =
dh/dt =

And the rate of change for height with respect to time is 10/361

?