- #1
mrcleanhands
Homework Statement
Use [itex]\frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}[/itex]
and [itex]\frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y}[/itex] to show that
[itex] \frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{\partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial ^ {2}z}{\partial x \partial y}+\cos^{2}\theta\frac{\partial ^ {2}z}{\partial y^{2}} [/itex]
Homework Equations
The Attempt at a Solution
[itex]
\frac{\partial z}{\partial\theta}\frac{\partial z}{\partial\theta}=\frac{\partial z}{\partial\theta}(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})
[/itex]
[itex]
\frac{\partial^{2}z}{\partial\theta^{2}}=(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})[/itex]
[itex]
\frac{\partial^{2}z}{\partial\theta^{2}}=r^{2}\sin^{2}\theta\frac{ \partial ^ {2}z}{\partial x^{2}}-2r^{2}\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+r^{2}\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}[/itex]
[itex]
\frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}[/itex]
but [itex] \frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}[/itex]
and if I add that I get:
[itex] \frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{ \partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}+\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}[/itex] so somehow [itex] \frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}[/itex] is supposed to be 0?
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