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Differentials and Rates of Change; Related Rates

  1. Oct 23, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    http://i4.minus.com/jboxzSadIJVVoi.jpg [Broken]

    2. Relevant equations

    Product rule; implicit differentiation.

    Volume of cylinder, V = pi(r^2)(h)

    3. The attempt at a solution

    dV/dt = 0 = pi[2r(dr/dt)(h) + (dh/dt)(r^2)]

    Solve the equation after plugging in r = 5; h = 8, and dh/dt = -2/5. Solve for dr/dt.

    0 = 80pi(dr/dt) - 10pi

    1 = 8(dr/dt)

    dr/dt = 1/8
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 23, 2013 #2
    You are not trying to find [itex] \frac{dr}{dt} [/itex]. Each part of the question wants you to find [itex] \frac{dV}{dt} [/itex] for a given rate at which the radius is changing (think about what represents the rate of change of the radius).
     
  4. Oct 23, 2013 #3
    It says to find the rate of change of the radius in the first part of the question.

    I think you just have to have it like this: [itex]\frac{dV}{dt}=2\pi r\frac{dr}{dt}\frac{dh}{dt}[/itex], then just plug in values.

    Edit: Oh forgot to add, the V is constant, so what does that mean the value of dV/dt is?
     
    Last edited: Oct 23, 2013
  5. Oct 23, 2013 #4

    haruspex

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    It's multiple choice. V is given as constant, dh/dt is a given value, so the obvious approach is to calculate dr/dt and see which choice matches. Of course, you could run it the other way: for each choice compute dV/dt and see which one gives 0, but that probably takes longer on average.
    You mean V is constant, so what value is dV/dt, right?
     
  6. Oct 23, 2013 #5
    Yes, my mistake.
     
  7. Oct 27, 2013 #6

    Qube

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    Is 1/8 the correct answer? I've redone the work again below:
     
  8. Oct 27, 2013 #7

    Qube

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  9. Oct 27, 2013 #8

    haruspex

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    Yes.
     
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