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Partial differentiation and partial derivatives

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data

    If ##xs^2 + yt^2 = 1## (1) and ##x^2s + y^2t = xy - 4,## (2) find ##\frac{\partial x}{\partial s}, \frac{\partial x}{\partial t}, \frac{\partial y}{\partial s}, \frac{\partial y}{\partial t}## at ##(x,y,s,t) = (1,-3,2,-1)##.



    2. Relevant equations

    Pretty much those just listed above.



    3. The attempt at a solution

    Alright so I spent quite a while on this one. First of all, I took the differentials of both the equations, (1) and (2), and came up with $$s^2\,dx + 2sx\,ds + t^2\,dy + 2ty\,dt = 0 $$ for (1) and $$ 2xs\,dx + x^2\,ds + 2yt\,dy + y^2\,dt = y\,dx + x\,dy$$ for (2).

    Then, substituting in ##(x,y,s,t) = (1,-3,2,-1)## for the respective variables I came up with $$4\,dx + 4\,ds + \,dy + 6\,dt = 0 $$ for (1) and $$4\,dx + \,ds + 6\,dy + 9\,dt = -3\,dx + \,dy$$ for (2).

    Simplifying and moving things around a bit, I got $$ 4\,dx + 4\,ds = -\,dy - 6\,dt $$ for (1) and $$ 7\,dx + \,ds = -5\,dy - 9\,dt $$ for (2).

    Then I used Cramer's Rule to solve these two equations for ##\,dx## and ##\,ds.##. For ##\,dx## I got $$ \,dx = \frac{
    \begin{vmatrix}
    -\,dy - 6\,dt & 4 \\
    -5\,dy - 9\,dt & 1
    \end{vmatrix}
    }{
    \begin{vmatrix}
    4 & 4 \\
    7 & 1
    \end{vmatrix}
    }

    = \frac{-\,dy - 6\,dt + 4(5\,dy + 9\,dt)}{-24} = \frac{-\,dy - 6\,dt + 20\,dy + 36\,dt}{-24} = \frac{19\,dy + 30\,dt}{-24} = \frac{-19\,dy - 30\,dt}{24}$$

    $$ \,ds = \frac{
    \begin{vmatrix}
    4 & -\,dy - 6\,dt \\
    7 & -5\,dy - 9\,dt
    \end{vmatrix}
    }{
    \begin{vmatrix}
    4 & 4 \\
    7 & 1
    \end{vmatrix}
    } = \frac{4(-5\,dy - 9\,dt) + 7(\,dy + 6\,dt)}{-24} = \frac{-20\,dy - 36\,dt + 7\,dy + 42\,dt}{-24} = \frac{-13\,dy + 6\,dt}{-24} = \frac{13\,dy - 6\,dt}{24}.
    $$

    Then I changed the equations around to $$\,dy + 6\,dt = -4\,dx - 4\,ds$$ for (1) and $$5\,dy + 9\,dt = -7\,dx - \,ds$$ for (2). Using Cramer's Rule again to solve for ##\,dy## and ##\,dt## I got
    $$\,dy = \frac{-6\,dx + 30\,ds}{21}$$
    $$\,dt = \frac{-13\,dx - 19\,ds}{21}.$$

    Then I wanted to find ##\frac{\partial x}{\partial s}##, so I took what I got for ##\,dy## and ##\,dt## plugged it into (1), ##4\,dx + 4\,ds + \,dy + 6\,dt = 0,## and got $$4\,dx + 4\,ds + \frac{-6\,dx + 30\,ds}{21} + 6(\frac{-13\,dx - 19\,ds}{21}) = 0.$$ Multiplying this all out, everything just cancels to zero and I can't find ##\frac{\partial x}{\partial s}##. I've checked the math for the applications of Cramer's Rule and can't find anything wrong there, so... what am I doing wrong?
     
    Last edited: Feb 24, 2014
  2. jcsd
  3. Feb 24, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I cannot understand what you are trying to do, or why you want to do it. The easiest way is just to differentiate directly. Letting ##x_s = \partial x/\partial s##, etc, we have:
    [tex] (1) \Longrightarrow s^2 x_s + 2 s x + t^2 y_s = 0 \\
    (2) \Longrightarrow x^2 + 2 s x x_s + 2 t y y_s = x y_s + y x_s
    [/tex]
    which you can solve for ##x_s,y_s##. It is much easier if you first substitute in the numerical values of ##x,y,s,t##.
     
  4. Feb 24, 2014 #3
    As in solve for ##y_s## in (1) and then plug that value for ##y_s## into (2)?

    Edit: Also, I know that the answer for ##\frac{\partial x}{\partial s}## is ##\frac{-19}{13}##
     
  5. Feb 24, 2014 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You have two equations in the two unknowns ##x_s, y_s##, and these equations are linear in these unknowns. They happen to be nonlinear in ##s,t,x,y##, but that does not matter, since we are already given numerical values for them. The things we don't know are ##x_s, y_s## and ##x_t,y_t##.
     
  6. Feb 24, 2014 #5
    Yeah, I just did it that way and got the right answer. Thanks!
     
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