# Thread: Rate of Change Problem

1. ## 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.

2. Sorry, but I disagree with your derivative. If

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

$\frac{dV}{dx} = \pi \frac{x^2 + 4x + 4}{4}$/

3. Whoops, meant to write (x+20)^3, not (x+2)^3.

4. 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?

5. 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.