Partial Derivative of Polar Coordinates Proof

**ghow90** · Apr 7th 2010, 07:16 AM

The question is

Let $z=f(x,y)$ be a differential function of x and y and let x=rcosθ and y=rsinθ

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

**Jester** · Apr 7th 2010, 08:02 AM

Originally Posted by ghow90

The question is

Let $z=f(x,y)$ be a differential function of x and y and let x=rcosθ and y=rsinθ

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

Use Jacobians to determine how the derivatives change.

$<br /> z_x = \frac{\partial(z,y) }{\partial(x,y)} = \frac{\partial(z,y)}{\partial(r,\theta)}/ \frac{\partial(x,y)}{\partial(r,\theta)} = \frac{z_r y_{\theta} - z_{\theta} y_r}{x_r y_{\theta} - x_{\theta} y_r} = \cos \theta z_r - \frac{\sin \theta}{r} z_{\theta}.<br />$

Similar for $z_y$ . Then square both, add together and simplify.

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