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Partial derivative using differentials

  1. Aug 1, 2014 #1
    1. The problem statement, all variables and given/known data
    If [itex] xs^2 + yt^2 = 1 [/itex] and [itex] x^2s + y^2t = xy - 4 [/itex], find [itex] \frac{\partial x}{\partial s}, \frac{\partial x}{\partial t}, \frac{\partial y}{\partial s}, \frac{\partial y}{\partial t} [/itex], at (x, y, s, t) = (1, -3, 2, -1)


    2. Relevant equations




    3. The attempt at a solution
    I took the differential of both equations and substituted in the values and I got:

    [tex] 4ds + 4dx + 6dt + dy = 0 [/tex] and
    [tex] ds + 7dx + 9dt + 5dy = 0 [/tex]

    but now I don't know how do find the individual partial derivatives that are being asked for. For example, for [itex] \frac{\partial x}{\partial s} [/itex], I assumed t and y are constant and made dt and dy = 0 and tried to solve for ds and dx but that doesn't work.

    Any ideas on how to proceed in finding these partials?
    Thanks!
     
  2. jcsd
  3. Aug 1, 2014 #2
    ##s## and ##t## don't happen to be independent, so that ## \frac{\partial s}{\partial t} = \frac{\partial t}{\partial s} = 0##? That would make things much easier.
     
  4. Aug 1, 2014 #3

    HallsofIvy

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    In order that this problem make any sense, s and t have to be independent so Quesadilla's comment is crucial. You should NOT assume y is constant since it is one of the dependent variables. But, taking the derivatives with respect to s you should assume t is constant (I would have said "treat t as a constant").

    So from [itex]4ds+ 4dx+ 6dt+ dy= 0[/itex] and [itex]ds+ 7dx+ 9dt+ 5dy= 0[/itex] you can get
    [tex]4\frac{\partial s}{\partial s}+ 4\frac{\partial x}{\partial s}+ 6\frac{\partial t}{\partial s}+ \frac{\partial y}{\partial s}= 4+ 4\frac{\partial x}{\partial s}+ \frac{\partial y}{\partial s}= 0[/tex]
    and
    [tex]\frac{\partial s}{\partial s}+ 7\frac{\partial x}{\partial s}+ 9\frac{\partial t}{\partial s}+ 5\frac{\partial y}{\partial s}= 1+ 7\frac{\partial x}{\partial s}+ 5\frac{\partial y}{\partial s}= 0[/tex]
    Giving you two equations to solve for [itex]\partial x/\partial s[/itex] and [itex]\partial y/\partial[/itex].

    Do the same thing to find the derivatives with respect to t.

    Personally, I wouldn't have use "differentials". Just differentiating [itex]xs^2+ yt^2= 1[/itex] with respect to s (and treating t as a constant) gives
    [tex]s^2\frac{\partial x}{\partial s}+ 2xs+ t^2\frac{\partial y}{\partial s}= 0[/tex]
    and differentiating [itex]x^2s+ y^2t= xy- 4[/itex] with respect to s gives
    [tex]2sx\frac{\partial x}{\partial s}+ x^2+ 2ty\frac{\partial y}{\partial s}= x\frac{\partial y}{\partial s}+ y\frac{\partial x}{\partial s}[/tex]
    again giving two equations to solve for [itex]\partial x/\partial s[/itex] and [itex]\partial y/\partial s[/itex].
     
    Last edited: Aug 1, 2014
  5. Aug 1, 2014 #4

    Ray Vickson

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    Why do you claim are s and t not independent? If we arbitrarily specify values of s and t we can then use the equations to find x and y. In fact, I would say we can take s and t as independent, but with x and y being functions of them: x = x(s,t) and y = y(s,t).
     
    Last edited: Aug 1, 2014
  6. Aug 1, 2014 #5
    If you mean that we should look to apply the implicit function theorem that sounds like an excellent idea!
     
  7. Aug 1, 2014 #6
    Great, thanks everyone!
     
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