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2nd year Calculus: partial derivatives

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data
    See attatched image.

    2. Relevant equations
    I just don't know where to start...


    3. The attempt at a solution
    Any help would be appreciated! :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 21, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    images don't clear for a while, try writing the problem for a quicker reply

    there's tips on using tex floating around or just click on other etex to see how its written
     
  4. Oct 21, 2009 #3
    If u=f(x,y), where x=escos(t) and y=essin(t), show that:
    second derivative of u wrt x + second derivative of u wrt y = e-2s(second derivative of u wrt s + second derivative of u wrt t)
     
  5. Oct 22, 2009 #4

    Mark44

    Staff: Mentor

    Is this the question?
    [tex]\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2}[/tex]
    [tex]=~e^{-2s}(\frac{\partial^2u}{\partial s^2} + \frac{\partial^2u}{\partial t^2})[/tex]

    Do you know the form of the chain rule for multivariable functions? If so, start by taking the partials of u w. r. t. x and y, and then take the partials of the first with respect to x and of the second with respect to y, then add them together.
     
  6. Oct 22, 2009 #5
    Ok, well I got :
    [tex]\frac{\partial^2u}{\partial x^2}[/tex] =-2essin(t) and

    [tex] \frac{\partial^2u}{\partial y^2}[/tex] = 2escos(t)
     
  7. Oct 22, 2009 #6

    Mark44

    Staff: Mentor

    So what do you get for
    [tex]\frac{\partial^2u}{\partial s^2}[/tex]
    and
    [tex]\frac{\partial^2u}{\partial t^2}[/tex]
     
  8. Oct 22, 2009 #7
    well, from the formula,
    [tex](\frac{\partial^2u}{\partial s^2} + \frac{\partial^2u}{\partial t^2})[/tex] should = 2e3s(cos(t) - sin(t))
    But I still don't understant how to get to the partails wrt s and t
    I have u=f(x,y) so u=f(escost,essin t)
     
  9. Oct 22, 2009 #8

    Mark44

    Staff: Mentor

    So u is a function of x and y, and x and y are each functions of s and t. Here's the form of the chain rule for one of the partials you need.

    [tex]\frac{\partial u}{\partial s}~=~\frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial s} [/tex]

    The other partial derivative is very similar, but any partial with respect to s should be with respect to t.
     
  10. Oct 22, 2009 #9
    Ok, well when I tried to do that I got answerd without sin and cos in them.. is what i did so far right?
    and is there a formula for finding the second partials wrt s and t?
     
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