202104162114 Homework 1 (Q3)

Suppose $z=f(x,y)$ is differentiable. Find the expression

$\displaystyle{\bigg( \frac{\partial z}{\partial x} \bigg)^2 + \bigg( \frac{\partial z}{\partial y} \bigg)^2}$

in terms of polar coordinates $r$ and $\theta$ .

Solution.

Setup.

$\begin{aligned} & \begin{cases} \displaystyle{\frac{\partial z}{\partial r}} & = \displaystyle{\frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}} \\ \displaystyle{\frac{\partial z}{\partial \theta}} & = \displaystyle{\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}} \\ \end{cases} \\ \Rightarrow & \begin{cases} \displaystyle{\frac{\partial z}{\partial r}} & = \displaystyle{\cos\theta\frac{\partial z}{\partial x} + \sin\theta\frac{\partial z}{\partial y}} \\ \displaystyle{\frac{\partial z}{\partial \theta}} & = \displaystyle{-r\sin\theta\frac{\partial z}{\partial x} + r\cos\theta\frac{\partial z}{\partial y}} \\ \end{cases} \\ \Rightarrow & \begin{bmatrix} \displaystyle{\frac{\partial z}{\partial r}} \\ \displaystyle{\frac{\partial z}{\partial \theta}}\end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{bmatrix} \begin{bmatrix} \displaystyle{\frac{\partial z}{\partial x}} \\ \displaystyle{\frac{\partial z}{\partial y}} \end{bmatrix} \\ \end{aligned}$

(to be continued)

Lemma.

_{Wikipedia on Cramer’s rule}

(continue)

$\begin{aligned} \frac{\partial z}{\partial x} & = \frac{\begin{vmatrix}\displaystyle{\frac{\partial z}{\partial r}} & \sin\theta \\ \displaystyle{\frac{\partial z}{\partial \theta}} & r\cos\theta \end{vmatrix}}{\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix}} \\ \\ & = \frac{r\cos\theta\displaystyle{\frac{\partial z}{\partial r}}-\sin\theta\displaystyle{\frac{\partial z}{\partial \theta}}}{r\cos^2\theta + r\sin^2\theta} \\\\ & = \cos\theta\displaystyle{\frac{\partial z}{\partial r}} - \frac{1}{r}\sin\theta\displaystyle{\frac{\partial z}{\partial \theta}} \\\\ \frac{\partial z}{\partial y} & = \frac{\begin{vmatrix}\cos\theta & \displaystyle{\frac{\partial z}{\partial r}} \\ -r\sin\theta & \displaystyle{\frac{\partial z}{\partial \theta}} \end{vmatrix}}{\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix}} \\\\ & = \frac{1}{r}\cos\theta\frac{\partial z}{\partial \theta} + \sin\theta\frac{\partial z}{\partial r} \end{aligned}$

Thus,

$\begin{aligned} & \quad\bigg( \frac{\partial z}{\partial x}\bigg)^2 + \bigg( \frac{\partial z}{\partial y}\bigg)^2 \\ \\ & = \bigg[ \cos\theta\frac{\partial z}{\partial r} - \frac{1}{r}\sin\theta\frac{\partial z}{\partial \theta} \bigg]^2 + \bigg[ \frac{1}{r}\cos\theta\frac{\partial z}{\partial \theta} + \sin\theta\frac{\partial z}{\partial r} \bigg]^2 \\ \\ & = \cos^2\theta \bigg( \frac{\partial z}{\partial r}\bigg)^2 - \frac{\sin 2\theta}{r}\bigg(\frac{\partial z}{\partial r}\bigg)\bigg(\frac{\partial z}{\partial \theta}\bigg) + \frac{1}{r^2}\sin^2\theta \bigg(\frac{\partial z}{\partial \theta}\bigg)^2 \\ & \qquad\qquad + \frac{1}{r^2}\cos^2\theta \bigg(\frac{\partial z}{\partial \theta}\bigg)^2 + \frac{\sin 2\theta}{r}\bigg(\frac{\partial z}{\partial r}\bigg)\bigg(\frac{\partial z}{\partial \theta}\bigg) + \sin^2\theta \bigg( \frac{\partial z}{\partial r}\bigg)^2 \\ \\ & = \bigg( \frac{\partial z}{\partial r} \bigg)^2 + \frac{1}{r^2} \bigg( \frac{\partial z}{\partial \theta} \bigg)^2 \end{aligned}$

M	T	W	T	F	S	S
			1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
19	20	21	22	23	24	25
26	27	28	29	30

Leave a Reply Cancel reply