Question

Show that (partial of z/partial of x)^2 + (partial of z/partial of y)^2 = (partial of z/partial of r)^2 + [(1/r^2)(partial of z/partial of theta)^2]. I know it has to done with polar coordinates and such I just have no idea where to start.

Answer 1

this might help.

Answer 2

wait for me, I am finishing off my scone.
Ok, I am free now.
Let
\[ x=r\sin\theta \\
y=r\cos\theta \]
We have
\[ \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r} \\
\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta} \]
Therefore,
\[ \frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\sin\theta+\frac{\partial z}{\partial y}\cos\theta \]
and
\[ \frac{1}{r} \frac{\partial z}{\partial \theta}=\frac{1}{r}\left(\frac{\partial z}{\partial x}r\cos\theta+\frac{\partial z}{\partial y}r(-\sin\theta) \right)= \frac{\partial z}{\partial x}\cos\theta-\frac{\partial z}{\partial y}\sin\theta\]

Now, your job is easy, squaring both sides, we have the desired equality.

Answer 3

Thank you so much!