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Chain rule, transforming PDE

  1. Nov 14, 2009 #1
    If [tex]z = f(x,y)[/tex] and [tex]x = r \cos{v}[/tex], [tex] y = r\sin{v}[/tex] the object is to show that [tex] d = \partial [/tex] since it's easier to do on computer

    Show that:
    [tex] \frac{d^2 z}{dr^2} + \frac{1}{r} \frac{dz}{dr} + \frac{1}{r^2} \frac{d^2 z}{dv^2} = \frac{d^2 z}{dx^2} + \frac{d^2 z}{dy^2} [/tex]

    It's from Adams calculus, will show where I get lost.

    We have from the chain rule
    [tex] \frac{dz}{dr} = \frac{dz}{dx} \frac{dx}{dr} + \frac{dz}{dy} \frac{dy}{dr} [/tex]

    But [tex] \frac{dx}{dr} = \cos{v} [/tex] and [tex] \frac{dy}{dr} = \sin{v} [/tex], so gives:

    [tex] \frac{dz}{dr} = \frac{dz}{dx} \cos{v} + \frac{dz}{dy} \sin{v} [/tex]

    Now we need [tex] \frac{d^2 z}{dr^2} [/tex], where I end up in trouble. We have

    [tex] \frac{d^2 z}{dr^2} = \frac{d}{dr} \cos{v} \frac{dz}{dx} + \frac{d}{dr} \sin{v} \frac{dz}{dy} [/tex] which can be written as

    [tex] \frac{d^2 z}{dr^2} = \cos{v} \frac{d}{dr} \frac{dz}{dx} + \sin{v} \frac{d}{dr} \frac{dz}{dy} [/tex]

    Now my books says that

    [tex] \frac{d}{dr} \frac{dz}{dx} = \cos{v} \frac{d^2 z}{dx^2} + \sin{v} \frac{d^2 z}{dy dx} [/tex] if I understand correct. But I don't see how you get to this ...
  2. jcsd
  3. Nov 14, 2009 #2
    Since z is a function of (x,y), the derivative of z in respect to x (or y), is another function of (x,y). Therefore the operator d/dr acts on the derivative of z the same way it acts on z.

    Look at this:

    Substitude z by dz/dx and you get the final result which you have written.
  4. Nov 14, 2009 #3
    Thank you very much, the equation for d^2 z/dv^2 got even messier. But I think I understand, I solved it and I'm going to try other examples on this.
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