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(adsbygoogle = window.adsbygoogle || []).push({});

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 ...

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# Chain rule, transforming PDE

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