Question: Calculus

1. Oct 8, 2006

Mathman23

Hi

Given a function

z = f(x,y), where x = r * cos(\phi) and y = r * sin (\phi)

First I show that

$$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} cos(\phi) + \frac{\partial z}{\partial y} sin (\phi)$$

and

$$\frac{\partial z}{\partial \phi} = - \frac{\partial z}{\partial x} r \cdot sin(\phi) + \frac{\partial z}{\partial y} r \cdot sin(\phi)$$

Finally I need to show that

$$(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2} (\frac{\partial z}{\partial \phi}) ^2$$

How do I approach this part of the problem?

Sincerley

Fred

Last edited: Oct 8, 2006
2. Oct 8, 2006

arildno

Hmm..square both previous equations and add them together, mayhap?

3. Oct 8, 2006

Mathman23

This is my solution for (b) please look at them at see if I made a mistake:

solving (b)

$$(\frac{\partial z}{\partial x}^2) ^2 + \frac{1}{r^2} (\frac{\partial z}{\partial \phi})^2 = (\frac{\partial z}{\partial x})^2 \cdot cos ^2 (\phi) + (\frac{\partial z}{\partial y})^2 \cdot sin(\phi) + 2 \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y} sin(\phi) \cdot cos(\phi) + (\frac{\partial z}{\partial x}) ^2 \cdot \frac{r^2 \cdot sin^2 (\phi)}{r^2} + (\frac{\partial z}{\partial y})^2 \cdot \frac{r^2 \cdot cos^2 (\phi)}{r^2} - 2 \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y} \cdot \frac{r^2 sin(\phi) \cdot cos(\phi)}{r^2} =$$

$$= (\frac{\partial z}{\partial x})^2 \cdot (cos^2(\phi) + sin^2 (\phi) \cdot (\frac{\partial z}{\partial y})^2 sin^2 (\phi) = (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2$$

Sincerely Yours
Fred

Last edited: Oct 8, 2006