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Differentiation Question.

  1. May 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Given [tex]f(x,y)=ycosx,x(t)=t^2,y(t)=sint[/tex]

    Calculate [tex]\frac{df((x(t),y(t))}{dt}[/tex]
    [tex]t∈ℝ[/tex]

    2. Relevant equations
    For parametric equations I know
    [tex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

    3. The attempt at a solution
    So far I have found
    [tex]\frac{∂f}{∂x}=-ysinx[/tex]
    [tex]\frac{∂f}{∂y}=cosx[/tex]
    [tex]\frac{dx}{dt}=2t[/tex]
    [tex]\frac{dy}{dt}=sint[/tex]

    but I'm not too sure where to go afterwards.
     
  2. jcsd
  3. May 17, 2015 #2
    Why do you just write f(x(t),y(t)) just change x but x(t) and y(t) and you'll get f as a WHOLE function of time, all you need there is takeout the derivative :)
     
  4. May 17, 2015 #3

    pasmith

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    Apply the chain rule.
     
  5. May 17, 2015 #4
    oh so just [tex]f(x,y)=sintcost^2[/tex]
     
  6. May 17, 2015 #5
    Yes, But write f(t) = sin(t)cos(t2) and pull out the derivative :3
     
  7. May 17, 2015 #6

    HallsofIvy

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    If [itex]f(x,y)= y cos(x)[/itex] with [itex]x= t^2[/itex] and [itex]y= sin(t)[/itex], then, yes, [itex]f(x)= sin(t)cos(t^2)[/itex] and you can differentiate that.

    But I think it is more likely that you expected to use the chain rule:
    [tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}[/tex]
     
  8. May 17, 2015 #7

    vela

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    It's curious why you would calculate these derivatives if you didn't have some idea already in mind about how to solve the problem. Perhaps you just weren't clear on the details. When you reached this point, a good idea would have been to consult your notes or textbook for an example like this problem to see how to put it all together.
     
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