91Ó°ÊÓ

Begin by obtaining the partial derivatives with respect to $\mathrm{θ}$keeping x and y constant. All that is required is to rewrite the equation in terms of the relevant variables, after which we can directly determine the partial derivative in each case separately.

$$\begin{gathered} y=r\sin \mathrm{θ},x=r\cos \mathrm{θ} \\ {\left(\frac{∂y}{∂\mathrm{θ}}\right)}_r=r\cos \mathrm{θ}=r×\frac{x}{r}=x \end{gathered}$$

$$\begin{gathered} y=r\sin \mathrm{θ},x=r\cos \mathrm{θ} \\ y=\left(\frac{x}{\cos \mathrm{θ}}\right)\sin \mathrm{θ}=x\tan \mathrm{θ} \\ {\left(\frac{∂y}{∂\mathrm{θ}}\right)}_x=r{\sec }^2\mathrm{θ}=x×\frac{r^2}{x^2}=\frac{r^2}{x} \end{gathered}$$

Thus, the required answer is ${\left(\frac{∂y}{∂\mathrm{θ}}\right)}_r=x,{\left(\frac{∂y}{∂\mathrm{θ}}\right)}_x=\frac{r^2}{x}$.

$$\begin{gathered} y=r\sin \mathrm{θ},x=r\cos \mathrm{θ} \\ \tan \mathrm{θ}=\frac{y}{x}⇒\mathrm{θ}={\tan }^{-1}\left(\frac{y}{x}\right) \end{gathered}$$

$$\begin{gathered} {\left(\frac{∂\mathrm{θ}}{∂y}\right)}_x=\frac{1}{x\left(1+\frac{y^2}{x^2}\right)} \\ =\frac{x}{r^2} \end{gathered}$$

Wherein utilizing the derivative formula for the inverse tangent function to get the final expression.

Use the method of differentials as follows:

$$\begin{gathered} y=r\sin \mathrm{θ},x=r\cos \mathrm{θ} \\ dx=\cos \mathrm{θ}dr-r\sin \mathrm{θ}d\mathrm{θ} \\ dy=r\cos \mathrm{θ}d\mathrm{θ}+\sin \mathrm{θ}dr \end{gathered}$$

$$\begin{gathered} dx=0 \\ \cos \mathrm{θ}dr-r\sin \mathrm{θ}d\mathrm{θ}=0 \\ dr=r\tan \mathrm{θ}d\mathrm{θ} \end{gathered}$$

Put for $dr$in $dy$the equation as follows:

$$\begin{gathered} dy=r\cos \mathrm{θ}d\mathrm{θ}+\sin \mathrm{θ}\left(r\tan \mathrm{θ}d\mathrm{θ}\right) \\ d\mathrm{θ}+{\tan }^2\mathrm{θ}d\mathrm{θ}=\frac{dy}{r\cos \mathrm{θ}} \\ {\left(\frac{∂\mathrm{θ}}{∂y}\right)}_x\left(1+{\tan }^2\mathrm{θ}\right)=\frac{1}{r\cos \mathrm{θ}} \end{gathered}$$

$$\begin{gathered} {\left(\frac{∂\mathrm{θ}}{∂y}\right)}_x=\frac{1}{r\cos \mathrm{θ}\left(1+{\tan }^2\mathrm{θ}\right)} \\ =\frac{1}{r×\frac{r}{x}\left(1+\frac{y^2}{x^2}\right)} \\ =\frac{x}{r^2} \end{gathered}$$

Thus, the required answer is ${\left(\frac{∂\mathrm{θ}}{∂y}\right)}_x=\frac{x}{r^2}$.

The general rule would be that both $\frac{dy}{d\mathrm{θ}}and\frac{d\mathrm{θ}}{dy}$aren’t reciprocal unless they each share the same independent variable that is satisfied in the case. Throughout the evaluation of the two expressions in this problem, maintained the same variable (X) constant.

Thus, they are reciprocal when they both share the same independent variables.