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Math Help - Partial Derivative of Polar Coordinates Proof

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    Partial Derivative of Polar Coordinates Proof

    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
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    Quote Originally Posted by ghow90 View Post
    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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