Question

suppose that z=f(x,y) has continuous partial derivatives. let x=(e^r)cos(theta) and y=(e^r)sin(theta). show that (z partial x)^2 + (z partial y)^2=(1/e^2r)[(z partial r)^2 + (z partial theta)^2]

Answer 1

Zx - z partial x and so on ..
using the chain rule we get :
Zr = Zx * (e^r)cos(t) + Zy * (e^r)sin(t)
Zt = Zx * (e^r)(-sin(t)) + Zy * (e^r)cos(t)

square both equations:
(Zr)^2 = (Zx)^2 * (e^2r)cos^2(t) + 2ZxZye^(2r)cos(t)sin(t) + (Zy)^2 * e^(2r)sin^2(t)
(Zt)^2 = (Zx)^2 * (e^2r)sin^2(t) - 2ZxZye^(2r)cos(t)sin(t) + (Zy)^2 * e^(2r)cos^2(t)

add both equations :
(Zr)^2 + (Zt)^2 = (Zx)^2 * (e^2r) + (Zy)^2 * e^(2r)
divide both sides by e^(2r) and there you have it :)