# Quick Derivative Question

1. Oct 28, 2007

### momogiri

I just have a quick question on derivatives
If I was to take $$\frac{d}{dt}$$ of V = pi*h*r^2, the pi would be 0, right? Since pi is a constant, and derivatives of constants are 0, right?
In need of some confirmation, thanks :D

Last edited: Oct 28, 2007
2. Oct 28, 2007

### rocomath

i don't want to say yes b/c it seems as if there is more to your problem that isn't showing in the latex

if it's just $$\frac{d}{dt}\pi=0$$ then yes, but idk about your problem

3. Oct 28, 2007

### Gib Z

It seems you wanted $$\frac{dV}{dt}$$, where $$V= \pi hr^2$$.

Well, there is no t in the expression for V, so really dV/dt is 0, since the other things are all constants with respect to t.

4. Oct 28, 2007

### momogiri

Sorry, I've fixed it though XD

It's actually a "Related Rates" problem I'm working on, and my prof said to take d/dt of both sides of V = (pi)hr^2
So I'm just wondering if the pi portion would automatically be 0

Last edited: Oct 28, 2007
5. Oct 28, 2007

### Gib Z

Give us the actual problem!! $$\frac{d}{dx} \pi = 0$$ but $$\frac{d}{dx} \pi x = \pi$$

6. Oct 28, 2007

### rocomath

lol ... :D

7. Oct 28, 2007

### momogiri

Ok, so the actual problem is:
An upright cylindrical tank with radius 6 m is being filled with water at a rate of 4 m3/min. How fast is the height of the water increasing?

So I wanted to take the d/dt of the equation V = (pi)hr^2, since I figured that's what the prof said, and now.. I'm just trying to figure out what the d/dt is :/