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Differentiate both sides with respect to t

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-01-28at83636AM.png





    3. The attempt at a solution

    I really need some help here. I've been able to get through about 50 pages of calculus without really understanding what it means to differentiate both sides with respect to t, now I've run out of space to hide and I can't succeed any more without understanding this concept. I don't see why here

    33.png

    differentiating both sides with respect to t forces me to take the inverse of (pi(h^2))/4. any help would be really appreciated.

    If you want me to differentiate (pi(h^2))/4, then to me is

    simply h, because pi as a number turns to zero, 4 turns to zero and h squared turns to h.
     
    Last edited: Jan 28, 2012
  2. jcsd
  3. Jan 28, 2012 #2
    This is what you have:

    [tex]
    V=\frac{\pi}{12} \cdot h^{3}
    [/tex]

    The constant-multiple rule says that if you have a constant multiplying a variable, then the derivative is the same constant times the derivative of the variable. This is:

    [tex]
    \frac{d}{dx} c\cdot x = c\cdot \frac{d}{dx}x

    [/tex]

    Your constant here is simply [tex] \frac{\pi}{12} [/tex]

    Can you try it now? You should really review the properties of derivatives.
     
  4. Jan 28, 2012 #3
    I still can't figure it out because I don't know what d/dx refers to, nor what x refers to.

    I understand how to get from

    (pi/12)h^3 to (pi/4)h^2

    I do not understand how to get from

    (pi/4)h^2 to 4/((pi)h^2)
     
  5. Jan 28, 2012 #4

    Curious3141

    User Avatar
    Homework Helper

    They're skimming over steps.

    They start with this:

    [tex]V = \frac{\pi}{12}h^3[/tex]

    Then they differentiate both sides wrt t. But the RHS is in terms of h, not t. So you have to use Chain Rule to differentiate it.

    [tex]\frac{dV}{dt} = (\frac{dV}{dh})(\frac{dh}{dt})[/tex]

    You should already know how to differentiate a simple polynomial expression, so:

    [tex]\frac{dV}{dh} = \frac{\pi}{12}(3h^2) = \frac{\pi}{4}h^2[/tex]

    Hence,

    [tex]\frac{dV}{dt} = \frac{\pi}{4}h^2{\frac{dh}{dt}}[/tex]

    And the next part is only an algebraic rearrangement.

    [tex]\frac{dh}{dt} = \frac{4}{{\pi}h^2}\frac{dV}{dt}[/tex]

    Hope it's clear now. :smile:
     
    Last edited: Jan 28, 2012
  6. Jan 28, 2012 #5
    I still think you need to freshen up your properties of derivatives. If you don't understand what d/dx means then you're in trouble. But I'll take your word that you understand how to take the derivative of [tex] \frac{\pi}{12}h^{3} [/tex]
    For the second line, we have:
    [tex]
    \frac{dV}{dt} = \frac{\pi}{4}h^{2}\frac{dh}{dt}
    [/tex]
    What they did is just solve for [tex] \frac{dh}{dt} [/tex]
    So we divide both sides by [tex] \frac{\pi}{4}h^{2} [/tex] to get:

    [tex]
    \frac{dh}{dt} = \frac{dV}{dt}\frac{4}{\pi}h^{2}
    [/tex]

    Does this make sense to you? All that really happened is that to divide both sides by [tex] \frac{\pi}{4}h^{2} [/tex]
    we just multiply by the reciprocal:
    [tex] \frac{4}{\pi * h^{2}} [/tex]
     
  7. Jan 28, 2012 #6
    ok, it looks like in order to get dh/dt by itself I just multiply both sides of the equation by the inverse. I didn't know that that's what differentiating both sides with respect to t meant, but now I do, thanks.
     
  8. Jan 28, 2012 #7

    Curious3141

    User Avatar
    Homework Helper

    NO!!!

    That's just algebraic rearrangement. Differentiation is something else entirely. Did you see and understand my post?
     
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