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Help with Partial Differentialtion

  1. Jun 5, 2009 #1
    1. The problem statement, all variables and given/known data
    I am given [tex]x^2 + y^2 = r^2[/tex]


    Need to find [tex]\frac{\partial y}{\partial x}=-\frac{1}{r}sin(\frac{v}{r}t)[/tex]

    2. Revelant equations
    x parametriztion of circle is:
    [tex]x=rcos(\frac{v}{r}t)[/tex]


    Trigonometric Identity
    [tex]sin^2t + cos^2t=1[/tex]
    [tex]r^2sin^2t=r^2-r^2cos^2t[/tex]
    [tex]1 + \frac{cos^2t}{sin^2t} = \frac{1}{sin^2t}[/tex]

    3. The attempt at a solution
    so [tex]y = (r^2 - x^2)^\frac{1}{2}[/tex]


    [tex]\frac{\partial y}{\partial x}=\frac{1}{2}(r^2 - x^2)^\frac{-1}{2}(-2x)[/tex]
    simplify: [tex]\frac{\partial y}{\partial x}=-x(r^2 - x^2)^\frac{-1}{2}[/tex]


    [tex]\frac{\partial ^2y}{\partial x^2}=-(r^2 - x^2)^\frac{-1}{2}+\frac{-1}{2}(r^2 - x^2)^\frac{-3}{2}(-2x)(x)[/tex]
    Simplify: [tex]\frac{\partial ^2y}{\partial x^2} = -\frac{1}{(r^2 - x^2)^\frac{1}{2}}-\frac{x^2}{(r^2 - x^2)^\frac{3}{2}}[/tex]


    Substitute parametrization to second partial derivative:
    [tex]\frac{\partial ^2y}{\partial x^2} = -\frac{1}{(r^2 - r^2cos^2(\frac{v}{r}t))^\frac{1}{2}}-\frac{r^2cos^2(\frac{v}{r}t)}{(r^2 - r^2cos^2(\frac{v}{r}t))^\frac{3}{2}}[/tex]
    [tex]\frac{\partial ^2y}{\partial x^2} = -\frac{1}{rsin(\frac{v}{r}t)}-\frac{r^2cos^2(\frac{v}{r}t)}{r^3sin^3(\frac{v}{r}t)}[/tex]
    [tex]\frac{\partial ^2y}{\partial x^2} = -\frac{1}{rsin(\frac{v}{r}t)}(1 + \frac{r^2cos^2(\frac{v}{r}t)}{r^2sin^2(\frac{v}{r}t)})[/tex]
    [tex]\frac{\partial ^2y}{\partial x^2} = -\frac{1}{rsin(\frac{v}{r}t)}\frac{1}{sin^2(\frac{v}{r}t)}[/tex]

    For some reason i ended up with [tex]\frac{1}{sin^2(\frac{v}{r}t)}[/tex] instead of [tex]sin^2(\frac{v}{r}t)[/tex]. Did i do any intermediate steps wrong?
     
  2. jcsd
  3. Jun 5, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I don't think you are doing anything wrong. I get y''=(-r^2)/y^3 which seems to agree with you final answer using implicit differentiation.
     
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