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Quick Derivative Question

  1. Oct 28, 2007 #1
    I just have a quick question on derivatives
    If I was to take [tex]\frac{d}{dt}[/tex] 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. jcsd
  3. Oct 28, 2007 #2
    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 [tex]\frac{d}{dt}\pi=0[/tex] then yes, but idk about your problem
  4. Oct 28, 2007 #3

    Gib Z

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    It seems you wanted [tex]\frac{dV}{dt}[/tex], where [tex]V= \pi hr^2[/tex].

    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.
  5. Oct 28, 2007 #4
    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
  6. Oct 28, 2007 #5

    Gib Z

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    Give us the actual problem!! [tex]\frac{d}{dx} \pi = 0[/tex] but [tex]\frac{d}{dx} \pi x = \pi[/tex]
  7. Oct 28, 2007 #6
    lol ... :D
  8. Oct 28, 2007 #7
    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 :/
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