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I was studying about partial derivatives and I got confused by this problem. I'm asked to prove that $$(\frac{\partial \omega}{\partial \theta})^2+ \frac{1}{r^2}(\frac{\partial \omega}{\partial r})^2=(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2$$ when I'm given that $\omega=f(x,y)$, $x=rcos\theta $ and $y=rsin\theta$.

I know how to find $\frac{\partial \omega}{\partial \theta}$ and $\frac{\partial \omega}{\partial r}$ but I don't know how to find out $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. Some help is greatly appreciated.

SlayerDiAngelo
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1 Answers1

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Let’s abbreviate $c=\cos\theta$, $s=\sin\theta$ and $v=v(r,\theta)=(rc,rs)$. First calculate the gradient of $\omega(r,\theta)=(f\circ v)(r,\theta)$ via chain rule: $$\nabla\omega=\begin{pmatrix}c&s\\ -rs &rc\end{pmatrix}\begin{pmatrix}f_x(v)\\f_y(v)\end{pmatrix}=\begin{pmatrix}cf_x(v)+sf_y(v)\\ -rsf_x(v)+rcf_y(v)\end{pmatrix}.$$ Now compute $$\begin{align} &(\color{navy}{\omega_{\theta}^2+\frac{1}{r^2}\omega_r^2})(r,\theta)=\\ &\quad=\bigl(\omega_{\theta}(r,\theta)\bigr)^2+ \frac{1}{r^2}\bigl(\omega_{r}(r,\theta)\bigr)^2\\ &\quad=(cf_x(v)+sf_y(v))^2 +\frac{1}{r^2}\left(-rsf_x(v)+rcf_y(v)\right)^2\\ &\quad=c^2f_x^2(v)+2csf_(v)f_y(v)+s^2f_y^2(v)+\\ &\qquad{}+s^2f_x^2(v)-2csf_x(v)f_y(v)+c^2f_y^2(v)\\ &\quad=f_x^2(v)+f_y^2(v)\\ &\quad=\bigl(\color{navy}{(f_x\circ v)^2+(f_y\circ v)^2}\bigr)(r,\theta) \end{align} $$ what was to be shown.

Michael Hoppe
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