Question

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

${\left(\frac{\partial z}{\partial \theta }\right)}_{\mathbf{y}}$.

Short answer

The value of provide equation is ${\left(\frac{\partial z}{\partial \theta }\right)}_y=-2r^2\sin 2\theta$.

Step-by-step solution

Given data

The given data are listed below

$z=x^2+2y^2$ ...(1)

$x=r\cos \theta$ ...(2)

$y=r\sin \theta$ ...(3)

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 and 2,

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

Take the partial derivative of 'z' with respect to '$\theta$' in above equation;

role="math" localid="1659092547223" $$\begin{gathered} {\left(\frac{\partial z}{\partial \theta }\right)}_y=\frac{\partial }{\partial \theta }\left[r^2{\cos }^2\theta +2y^2\right] \\ =2r^2\left(\sin \theta ·\cos \theta \right) \\ {\left(\frac{\partial z}{\partial \theta }\right)}_y=2r^2\sin 2\theta \end{gathered}$$

Hence,

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