# Partial Derivative

Whoops got it now, didn't carry out my substitutions far enough.

## Homework Statement

$$z = x^2 + 2y^2$$
$$x = rcos(\theta)$$
$$y = rsin(\theta)$$

## The Attempt at a Solution

Find $$(\partial z/\partial x)$$ (theta is constant)

dz = 2xdx + 4ydy
dx = cos$$(\theta)$$dr - rsin$$(\theta)$$d$$\theta$$
dy = sin$$(\theta)$$dr + rcos$$(\theta)$$d$$\theta$$

Unfortunately I'm not really quite sure where to go from here, I know that
$$(\frac{ \partial z } { \partial x} )$$ is 2x when y is constant. But how to factor in theta being constant?
I suppose I could reduce
dx to dx = cos$$(\theta)$$dr
and dy = sin$$(\theta)$$dr

Last edited:

HallsofIvy
Homework Helper
Whoops got it now, didn't carry out my substitutions far enough.

## Homework Statement

$$z = x^2 + 2y^2$$
$$x = rcos(\theta)$$
$$y = rsin(\theta)$$

## The Attempt at a Solution

Find $$(\partial z/\partial x)$$ (theta is constant)

dz = 2xdx + 4ydy
dx = cos$$(\theta)$$dr - rsin$$(\theta)$$d$$\theta$$
dy = sin$$(\theta)$$dr + rcos$$(\theta)$$d$$\theta$$

Unfortunately I'm not really quite sure where to go from here, I know that
$$(\frac{ \partial z } { \partial x} )$$ is 2x when y is constant. But how to factor in theta being constant?
If $\theta$ is a constant, then $d\theta= 0$

I suppose I could reduce
dx to dx = cos$$(\theta)$$dr
and dy = sin$$(\theta)$$dr
Yes, that is exactly correct. Then dz= dx+ dy= ?