Question

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

${\left(\frac{\partial z}{\partial y}\right)}_{\mathbf{r}}\mathbf{=}\mathbf{2}\mathit{y}$.

Short answer

Hence the value of provide equation is${\left(\frac{\partial z}{\partial y}\right)}_r=2y$

Step-by-step solution

Explanation of solutionStep 1: Given data

The given data are listed below

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

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

$y=rsin\mathrm{\theta }$ ...(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

Square the equation 2 and 3, and then add.

$x^2+y^2=r^2$ ...(4)

From equation (1),

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

Take the partial derivative of z with respect to y,

$$\begin{gathered} {\left(\frac{\partial z}{\partial y}\right)}_r=\frac{\partial }{\partial y}(r^2+y^2) \\ =2y \end{gathered}$$

Hence,

${\left(\frac{\partial z}{\partial y}\right)}_r=2y$