Chapter 13.5, Problem 57ES

Suppose that the equation $z=f\left(x,y\right)$ is expressed in the polar form $z=g\left(r,\theta \right)$ by making the substitution $x=r\cos \theta \text{ and }y=r\sin \theta$ .

(a) View r and $\theta$ as functions of x and y and use implicit differentiation to show that $\frac{\partial r}{\partial x}=\cos \theta \text{and}\frac{\partial \theta }{\partial x}=-\frac{\sin \theta }{r}$

(b) View r and $\theta$ as functions of x and y and use implicit differentiation to show that $\frac{\partial r}{\partial y}=\sin \theta \text{and}\frac{\partial \theta }{\partial y}=\frac{\cos \theta }{r}$

(c) Use the results in parts (a) and (b) to show that

$\begin{array}{l}\frac{\partial z}{\partial x}=\frac{\partial z}{\partial r}\cos \theta -\frac{1}{r}\frac{\partial z}{\partial \theta }=\sin \theta \\ \frac{\partial z}{\partial y}=\frac{\partial z}{\partial r}\sin \theta +\frac{1}{r}\frac{\partial z}{\partial \theta }=\cos \theta \end{array}$

(d) Use the result in part (c) to show that ${\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2={\left(\frac{\partial z}{\partial r}\right)}^2+\frac{1}{r^2}{\left(\frac{\partial z}{\partial \theta }\right)}^2$

(e) Use the result in part (c) to show that if $z=f\left(x,y\right)$ satisfies Laplace’s equation $\frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial y^2}=0$ then $z=g\left(r,\theta \right)$ satisfies the equation $\frac{{\partial }^{2}z}{\partial r^2}+\frac{1}{r^2}\frac{{\partial }^{2}z}{\partial {\theta }^2}+\frac{1}{r}\frac{\partial z}{\partial r}=0$ and conversely. The latter equation is called the polar form of Laplace’s equation.