# Related rates prob.

1. Oct 19, 2008

### elitespart

Related rates prob. [solved]

A hemispherical bowl of radius 8 in. is being filled with water at a constant rate. If the water level is rising at the rate of 1/3 in./s at the instant when the water is 6 in. deep, find how fast the water is flowing in by using the fact that if V is the volume of the water at time t, then dV/dt = (pi)(r)^2(dh/dt)

I just need someone to check my work. From my picture, I have:

dV/dt = (pi)(64-(8-h)^2) dh/dt.

just plug in 6 for h and 1/3 for dh/dt and solve right? Thanks.

Last edited: Oct 19, 2008
2. Oct 19, 2008

### Staff: Mentor

The first formula you show looks fine, but can you tell us how you arrived at your second formula, namely dV/dt = pi (64 - (8 - h)^2) dh/dt?
You mentioned that it comes from your picture, but since I can't see your picture, it's not clear to me how h is related to the radius of the disk you are using for your incremental volume. I'm not saying your formula is wrong, but you haven't provided enough information so that I can tell.

3. Oct 19, 2008

### elitespart

http://img363.imageshack.us/img363/3018/problemab0.jpg [Broken] http://g.imageshack.us/img363/problemab0.jpg/1/ [Broken]

I suck at drawing on paint but this is kind of how my pic looks. y + h is radius of the bowl which is 8. 8 - h = y and then i just used c^2 - a^2 = b^2. And thanks for your help.

Last edited by a moderator: May 3, 2017
4. Oct 19, 2008

### Staff: Mentor

OK, thanks for including the drawing. Now I know how h and y are related. I don't see anything wrong in your second formula.

Last edited by a moderator: May 3, 2017
5. Oct 19, 2008

### elitespart

Thanks again.