# Rate of Change Problem

• Jul 30th 2010, 09:13 PM
Etherlite
Rate of Change Problem
Hi all, first time here. Not sure whether this is the right category...

I am working with water levels and a truncated cone.
I have been told that V = (pi/3)(((x+2)^3)/4)-2000).
I am then told to find dV/dx, which = (pi(x^2+40x+400)/4).

This is where I am stuck, because it then asks "Hence find the rate of change of height (x) in terms of x.
Would I use Chain Rule, or is it simply dV/dx inverse?
I'm not sure at all.

Thanks for the help.

• Jul 30th 2010, 09:19 PM
Math Major
Sorry, but I disagree with your derivative. If

$V = \frac{\pi}{3}(\frac{(x+2)^3}{4} - 2000)$, then

$\frac{dV}{dx} = \pi \frac{x^2 + 4x + 4}{4}$/
• Jul 30th 2010, 09:38 PM
Etherlite
Whoops, meant to write (x+20)^3, not (x+2)^3.
• Jul 31st 2010, 06:43 AM
skeeter
Quote:

Originally Posted by Etherlite
Hi all, first time here. Not sure whether this is the right category...

I am working with water levels and a truncated cone.
I have been told that V = (pi/3)(((x+20)^3)/4)-2000).
I am then told to find dV/dx, which = (pi(x^2+40x+400)/4).

This is where I am stuck, because it then asks "Hence find the rate of change of height (x) in terms of x.
Would I use Chain Rule, or is it simply dV/dx inverse?
I'm not sure at all.

any information given about how the volume is changing w/r to time?
• Jul 31st 2010, 04:39 PM
Etherlite
Quote:

Originally Posted by skeeter
any information given about how the volume is changing w/r to time?

Don't worry all, I realised on a previous page that they gave me the rate at which water is dripping into the bucket.
Apply chain rule, and the answer's there.

Thanks all.