# Homework Help: Related Rates Problem that I am getting wrong.

1. Jun 24, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

Sand is pouring from a pipe at the rate of 16 cubic feet per second. If the falling sand forms a conical pile on the ground whose altitude is always 1/4 the diameter of the base, how fast is the altitude increasing when the pile is 4 feet high?

2. Relevant equations

$V = \frac{1}{3}\pi r^{2}h$

3. The attempt at a solution

Firstly, since I am given that the altitude is 1/4 of the diameter, I rearrange the volume formula.

1/4(2)r = h
1/2r = h
r = 2h

So:

$V = \frac{1}{3}\pi 2h^{2}h$
or

$V = \frac{1}{3}\pi 2h^{3}$

Second, I'm given the rate of volume increase as 16, so $\frac{dV}{dt}=16$

Differentiation of both sides of the volume equation gives me:

$16 = 2\pi 2h^{2}\frac{dh}{dt}$

So it follows that

$\frac{dh}{dt}= \frac{16}{2\pi h^{2}}$

Solving this for h = 4 gives me ~= .159155

Which is incorrect.

Can someone show me where I'm messing this up?

Thanks!

2. Jun 24, 2011

### eumyang

Here's the problem, I think. You need parentheses around the 2h, like this:
$$V = \frac{1}{3}\pi (2h)^{2}h = \frac{1}{3}\pi 4h^{3} = \frac{4}{3}\pi h^{3}$$

Ignoring the mistake from earlier, I don't know where that extra 2 came from.

Now that extra 2 disappeared.

You really need to take care in typing in the equations.

Last edited: Jun 24, 2011
3. Jun 24, 2011

### Skins

$$V \, = \, \frac{1}{3} \pi (2) h^{3}$$

I think you made a small mistake in there. All your work seems ok to that point.

(P.S. sorry for basically a repeat of what the poster above pointed out. At the time I clicked on "reply" there were no responses posted)

Last edited: Jun 24, 2011
4. Jun 26, 2011

### 1MileCrash

My problem was indeed the volume formula. I needed (2h)(2h).