# Connected rate of change & Calculus

1. Jul 26, 2009

### Chewy0087

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

3. The attempt at a solution

Managed part a) just fine, but the trouble comes with part b. Erm, the thing is i've got the answer just fine using the chain rule;

i worked out dh/dt = 1/(dV/dh) * dV/dt

Giving me 18 * $$\frac{1}{\frac{pi}{3}(18h-3h^2)}$$

Now, I thought, "Okay, it says something about 3 so i'll insert that" and i did, giving me the correct answer, but when I was explaining to my friend about it and I thought i could try to explain it to him geometrically by drawing out the semi-circle and the line x = 3 - h I got stuck myself. I just can't justify it to myself geometrically because;

you have the semi-circle passing through (-3 , 0) (0, 3) & (3 , 0) and the vertical line x = 3 - h, now when it's 'full' surely h will equal 6? So the line goes back to (-3,0)...

I think i'm generally confused.

I know it's a long problem but i'd really appreciate any help

2. Jul 26, 2009

### HallsofIvy

This is a hemisphere of radius 3. When it is full, the height of the water equals the radius, 3.

3. Jul 26, 2009

### Chewy0087

Yeah, which is the same reasoning that I used when answering it, however when i sketch the actual graph of y = (9 - x²)^0.5 & x = 3 - h. You'll see what I mean, sureley only when h is 6 will the whole volume of the hemisphere be given when you integrate the volume...

4. Jul 26, 2009

### Chewy0087

never minddddd :PP i got it!

Is it that when you integrate you would only need the cross-section for 1/4 of the hemisphere? (Because it rotates it around)

Thank you!! :P

Last edited: Jul 26, 2009