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Homework Help: Need help with chain rule for relating ds/dt to dx/dt and dy/dt

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution


    The problem with that is its only d/dx if y is a set number. I don't know how to differentiate the entire thing properly. I have been hacking at this for 8 hours. I feel like mental jello.
  2. jcsd
  3. Jun 3, 2012 #2
    You are taking the derivative with respect to t.

    So d/dt of 3x2 = 6x * dx/dt, not 6x.

    Maybe this helps figure out the whole derivative?
  4. Jun 3, 2012 #3


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    Well the relevant equation under 2. Relevant equations would be an expression of the chain rule.

    d/dt(f(g(t)) = f'(g(t))*g'(t)


    Let g(t) = g(x(t),y(t)) and f = √

    One could also write the original equations as s2 = 3x2 + 6y2, and differentiate each term with respect to t.
  5. Jun 3, 2012 #4
    That doesn't really seem like it would get me anywhere. I know I am wrong, but why would that work?

    Thank you for the links.
    Last edited: Jun 3, 2012
  6. Jun 3, 2012 #5
    I have narrowed down my question, specifically to the area I have highlighted on this picture (bear in mind, I can't post pics under 10 posts):

    http:// i.imgur .com /62erw.png

    Where did all the dx/dt and dy/dt come from on the right side? I don't understand that step. I know how to do this when thinking about it in function form, but it confuses me to think about it in fraction form, which is whats required to answer.
  7. Jun 3, 2012 #6


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    If s is a function of two variables, x and y, which are themselves functions of t. The "chain rule" says
    [tex]\frac{ds}{dt}= \frac{\partial s}{\partial x}\frac{dx}{dt}+ \frac{\partial s}{\partial y}\frac{dy}{dt}[/tex]

    Here, [itex]s(x,y)= \sqrt{3x^2+ 6y^2}= (3x^2+ 6y^2)^{1/2}[/itex]
    What are [itex]\partial s/\partial x[/itex] and [itex]\partial s/\partial y[/itex]?
    Last edited by a moderator: Jun 3, 2012
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