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Change in x (Differentiation) - Calc III needed?

  1. Oct 8, 2013 #1

    Qube

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

    http://i.minus.com/jDtSwpCGlrMhP.jpg [Broken]

    2. Relevant equations

    Solving for dx/dy (change in x with respect to y) I get a solution that isn't one of the answer choices.

    3. The attempt at a solution

    3y^2 = 8x' + x'/x

    Coordinate:

    x = 1 (given)
    y = 2 (solved for in original equation)

    3(4) = 9x'
    12/9 = dx/dy
    4/3 = dx/dy (change of x with respect to y).
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 8, 2013 #2

    Dick

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    You want to think about dy/dt and dx/dt not dy/dx. For example, d/dt(y^3)=3y^2*dy/dt. Try that again. There is a correct answer in there.
     
    Last edited by a moderator: May 6, 2017
  4. Oct 8, 2013 #3
    No matter what derivatives you take, please make sure you apply the chain rule properly. I don't think you did that in your attempt.
     
  5. Oct 9, 2013 #4

    Mark44

    Staff: Mentor

    The tipoff that they're looking for time rates of change (dx/dt and dy/dt) is the units in the answers - units/sec.
     
  6. Oct 9, 2013 #5

    Qube

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    I'm not seeing the issue when I take the derivative of the equation with respect to y.
     
  7. Oct 9, 2013 #6

    Qube

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    Thank you. That makes more sense. I found dx/dt and plugged in dy/dt which is -1/2.

    3y^2(dy/dt) = 8(dx/dt) + dx/dt*(1/x)

    y=2, as always.

    3(4)(-1/2) = 9(dx/dt)

    -6 = 9(dx/dt)
    -2/3 = dx/dt

    It appears the answer is D.
     
  8. Oct 9, 2013 #7

    Dick

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    Yes, it is.
     
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