1. ## related rates

I worked this all out, but the answers not right. I don't know if I even went about it the right way. Here's the problem:

Gravel is being dumped from a conveyor belt at a rate of 10 ft^3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 13 ft high? Give your answer correct to three decimal places.

So I know the formula for the volume of a cone is V=1/3(pi)(r^2)h, dv/dt=10ft^3/min, and h=13 ft and I have to find dh/dt. Right?

And since they say that h=diameter, I figured that I can replace the radius in the formula to h, so r^2 in the equation would be h^2/4.

Is this ok so far?

2. Originally Posted by kwivo
I worked this all out, but the answers not right. I don't know if I even went about it the right way. Here's the problem:

Gravel is being dumped from a conveyor belt at a rate of 10 ft^3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 13 ft high? Give your answer correct to three decimal places.

So I know the formula for the volume of a cone is V=1/3(pi)(r^2)h, dv/dt=10ft^3/min, and h=13 ft and I have to find dh/dt. Right?

And since they say that h=diameter, I figured that I can replace the radius in the formula to h, so r^2 in the equation would be h^2/4.

Is this ok so far?
yes, that is fine so far

3. Thanks. Nevermind the rest. I made a simple mistake and I got it now.

4. Originally Posted by kwivo
Thanks. Nevermind the rest. I made a simple mistake and I got it now.
are you using "Calculus" by James Stewart?

5. Here's a quick way to solve one of these.

Since the diameter and height are the always the same, then when it is 13 feet high the radius must be 13/2 feet.

That gives the area of the base as ${\pi}(13/2)^{2}=\frac{169\pi}{4}$

Now, divide that into the dV/dt:

$\frac{10}{\frac{169\pi}{4}}=\frac{40}{169\pi}$

Which is the same answer as if you subbed r=h/2 into the volume of a cone formula, differentiated, subbed in your knowns and solved for dh/dt. This way is just a little quicker.

This works because the rate of change of the volume is equal to the area of the surface times the change in height, $\frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}$. If we solve for dh/dt, we get:
$\frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}$

Which is what we used above.

6. Originally Posted by Jhevon
are you using "Calculus" by James Stewart?
Yes I am.