Question

If$\mathit{z}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathit{x}\mathbf{=}\mathit{r}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{,}\mathbf{}\mathit{y}\mathbf{=}\mathit{r}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{,}$find the following partial derivatives.

role="math" localid="1659095485411" ${\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}}_{\mathit{\theta }}$.

Short answer

The value of provided equation is ${\left(\frac{{\partial }_Z}{\partial r}\right)}_{\theta }=2r.{\sin }^2\theta$.

Step-by-step solution

Given data

The given data are listed below

$$\begin{gathered} z=x^2+2y^2...\left(1\right) \\ x=r\cos \theta ...\left(2\right) \\ y=r\sin \theta ...\left(3\right) \end{gathered}$$

Partial differentiation

The procedure of calculating the partial derivative of a function is called partial differentiation. This method is used to get the partial derivative of a function with respect to one variable while keeping the other constant.

Calculation

From equation 1, 2 and 3,

$\begin{array}{l}z={(r\cos \theta )}^2+2\left(r\sin \theta 2\right)\\ z=r^2({\cos }^2\theta +2{\sin }^2\theta )\end{array}$

Take the partial derivative of z with respect to 'r',

role="math" localid="1659095903768" $$\begin{gathered} {\left(\frac{\partial z}{\partial r}\right)}_{\theta }=\frac{\partial }{\partial r}\left[r^2\left(1+{\sin }^2\theta \right)\right] \\ {\left(\frac{\partial z}{\partial r}\right)}_{\theta }=2r\left(1+{\sin }^2\theta \right) \end{gathered}$$

Hence,

${\left(\frac{\partial z}{\partial r}\right)}_{\theta }=2r\left(1+{\sin }^2\theta \right)$