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Implicit differentiation

  1. Dec 6, 2005 #1
    [tex]x^2+y^2+r^2-2s=13=0[/tex]
    [tex]x^3-y^3-r^3+3s+59=0[/tex]

    How do I find the partial derivatives of x(r,s) or y(r,s) implicitly? I tried implicit differentiation and I got 2 different answers for either. Can someone show me any of the 4 derivatives step-by-step?
     
  2. jcsd
  3. Dec 6, 2005 #2
    Please, show us what you have done

    marlon
     
  4. Dec 6, 2005 #3
    [tex]2x\frac{\partial x}{\partial r}+2r=0[/tex]
    [tex]3x^2\frac{\partial x}{\partial r}-3r^2=0[/tex]

    [tex]2x\frac{\partial x}{\partial r}+2r=3x^2\frac{\partial x}{\partial r}-3r^2[/tex]

    [tex]\frac{2r+3r^2}{3x^2-2x}=\frac{\partial x}{\partial r}[/tex]

    This is what I get if I equate them. If I solve for dx/dr individually and then add the equations, I get something else.
     
  5. Dec 6, 2005 #4
    An answer before 2PM tomorrow will be appreciated! (Final exam)
     
  6. Dec 6, 2005 #5

    NateTG

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    Hmm, I'm not positive about this, but isn't it something like
    [tex]F(x,y,s,r)=x^2+y^2+r^2-2s+13=0[/itex]
    So
    [tex]\frac{\partial x}{\partial s}=-\frac{\frac{\partial F}{\partial s}}{\frac{
    \partial F}{\partial x}}=-\frac{-2}{2x}=\frac{1}{x}[/tex]
    so
    [tex]\frac{\partial}{\partial r}\frac{\partial x}{\partial s}=-\frac{1}{x^2} \times \frac{\partial x}{\partial r}[/tex]
    And
    [tex]\frac{\partial x}{\partial r}=-\frac{2r}{2x}=-\frac{r}{x}[/tex]
    so we have
    [tex]\frac{\partial}{\partial r \partial s} =\frac{r}{x^3}[/tex]
     
  7. Dec 6, 2005 #6

    HallsofIvy

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    y is also a function of r. You are differentiating both sides of the equation with respect to r. The derivative of x2+ y2+ r2- 2s+ 13= 0 with respect to r is
    [tex]2x\frac{\partial x}{\partial r}+ 2y\frac{\partial y}{\partial x}+ 2r= 0[/tex]
     
  8. Dec 6, 2005 #7
    I think it should be dy/dr. It gives me the right answer when it is. Thanks everyone!
     
    Last edited: Dec 6, 2005
  9. Dec 7, 2005 #8

    NateTG

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    Potentially stupid question, but that doesn't make sense to me. I would expect that to be:
    [tex]2x\frac{\partial x}{\partial r}+ 2y\frac{\partial y}{\partial r} + 2r - 2 \frac{\partial s}{\partial r}=0[/tex]

    Since [itex]s[/itex] is also a function of [itex]r[/itex]. Can you clarify why there is no [itex]\frac{\partial s}{\partial x}[/itex] term?
     
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