Question

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

$\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }\mathbf{r}\mathbf{\partial }\mathbf{x}}$.

Short answer

The value of provided equation is $\frac{{\partial }^{2}z}{\partial r\partial x}=0$.

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),

$$\begin{gathered} z=\left(x^2+y^2\right)+y^2 \\ z=r^2+x^2{\tan }^2\theta \end{gathered}$$

Take the partial derivative of z with respect to ,

$$\begin{gathered} \frac{\partial z}{\partial r}=\frac{\partial }{\partial r}\left(r^2+x^2{\tan }^2\theta \right) \\ \frac{\partial z}{\partial r}=2r \end{gathered}$$

Take the partial derivative of $\frac{\partial z}{\partial r}$with respect to x,

role="math" localid="1659090151351" $$\begin{gathered} \frac{{\partial }^{2}z}{\partial r\partial x}=\frac{{\partial }^2}{\partial r\partial x}2r \\ \frac{{\partial }^{2}z}{\partial r\partial x}=0 \end{gathered}$$

Hence,

$\frac{{\partial }^{2}z}{\partial r\partial x}=0$