Question

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

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

Short answer

The value of provided equation is ${\left(\frac{\partial z}{\partial r}\right)}_y=2r$.

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

Square the equation 2 and 3, then add.

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

From equation (1),

${}_{z=r^2+y^2}^{z=\left[x^2+y^2\right]+y^2}z=x^2+2y^2$

Take the partial derivative of z with respect to ,

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

Hence,${\left(\frac{\partial z}{\partial r}\right)}_y=2r$