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Partial derivatives of implicitly defined functions

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data
    If the equations

    [tex]x^2 - 2(y^2)(s^2)t - 2st^2 = 1[/tex]
    [tex]x^2 + 2(y^2)(s^2)t + 5st^2 = 1[/tex]

    define s and t as functions of x and y, find [tex]\partial^2 t / \partial y^2[/tex]

    3. The attempt at a solution

    Equating the two, we get 4y^2*s^2*t = -7s*t^2. My main problem is, as simple as this sounds, how do I implicitly differentiate a single term with three variables? (i.e. 4y^2*s^2*t) Because if I can implicitly differentiate it in terms of y, I can probably find [tex]\partial t / \partial y[/tex] in terms of the other partial derivatives.
     
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  3. May 30, 2009 #2

    djeitnstine

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    Partial differentiation is not implicit differentiation if that's what you mean. You simply treat all other variables as constant. In this case you get 8y*s^2 etc...
     
  4. May 30, 2009 #3
    But how would I get the partial derivative of t wrt y from there? I mean, aren't we assuming t is a function of x and y (and with s as well?) ?


    So let me try this out:

    8y*s^2*t + 4y^2([tex]2 \partial s / \partial y[/tex])t + 4y^2*s^2*([tex] \partial t / \partial y[/tex]) = -7([tex] \partial s / \partial y[/tex])*t^2 - 7s([tex]2 \partial t / \partial y[/tex])
     
  5. May 30, 2009 #4

    gabbagabbahey

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    Don't combine your two original equations; instead differentiate each one with respect to [itex]y[/itex]--- you will end up with two equations, each involving [itex]\frac{\partial s}{\partial y}[/itex] and [itex]\frac{\partial t}{\partial y}[/itex], which you can then solve for [itex]\frac{\partial t}{\partial y}[/itex].
     
  6. May 30, 2009 #5
    Sounds like a plan. So let me try this again...


    [-4y[itex]s^2[/itex]t - 2[itex]y^2[/itex]([itex]2s\frac{\partial s}{\partial y})t[/itex] - 2[itex]y^2 s^2[/itex]([itex]\frac{\partial t}{\partial y}[/itex])] + [-2([itex]\frac{\partial s}{\partial y})t^2[/itex] - 2s([itex]2 \frac{\partial t}{\partial y})[/itex]] = 1

    [4y[itex]s^2[/itex]t + 2[itex]y^2[/itex]([itex]2s\frac{\partial s}{\partial y})t[/itex] + 2[itex]y^2 s^2[/itex]([itex]\frac{\partial t}{\partial y}[/itex])] + [5([itex]\frac{\partial s}{\partial y})t^2[/itex] + 5s([itex]2 \frac{\partial t}{\partial y})[/itex]] = 1


    [3([itex]\frac{\partial s}{\partial y})t^2[/itex] + 3s([itex]2 \frac{\partial t}{\partial y})[/itex]] = 2

    Is there a way to eliminate the [itex]\frac{\partial s}{\partial y}[/itex] term?
     
  7. May 30, 2009 #6

    gabbagabbahey

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    Careful, when you differentiate an equation, you need to differentiate both sides of it.

    [tex]\frac{\partial}{\partial y} (1)\neq 1[/tex]

    Sure....you have 2 equations and 2 unkowns; think back to your high school algebra :wink:
     
  8. May 31, 2009 #7
    Ok, so now I have

    [tex]3t^2 \frac {\partial s}{\partial y} = -6s \frac{\partial t}{\partial y}[/tex]
    [tex]\frac {\partial s}{\partial y} = \frac{-2s}{t^2} \frac{\partial t}{\partial y}[/tex]

    I sub this back into the first equations to get [itex]\frac{\partial t}{\partial y}[/itex]. However, I get a different value for [itex]\frac{\partial t}{\partial y}[/itex] when I plug it into the second equation. If I'm going to write the answer, would I write them both down? Because I feel there really should only be one answer...
     
    Last edited by a moderator: May 31, 2009
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