Question

If $\mathit{z}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}$, find${\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}}_{\mathbf{\theta }}\mathbf{,}{\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{\theta }}\mathbf{\right)}}_{\mathbf{r}}\mathbf{,}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }\mathbf{r}\mathbf{\partial }\mathbf{\theta }}\mathbf{,}{\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{x}}\mathbf{\right)}}_{\mathbf{y}}$

Short answer

The value of${\left(\frac{\partial z}{\partial r}\right)}_{\theta }=2r{\sin }^2\theta ,{\left(\frac{\partial z}{\partial \theta }\right)}_r=r^2\sin 2\theta ,\frac{{\partial }^{2}z}{\partial r.\partial \theta }=2r\sin 2\theta \text{and}{\left(\frac{\partial z}{\partial x}\right)}_y=0$

${\left(\frac{\partial z}{\partial r}\right)}_{\theta }=2r{\sin }^2\theta ,{\left(\frac{\partial z}{\partial \theta }\right)}_r=r^2\sin 2\theta ,\frac{{\partial }^{2}z}{\partial r.\partial \theta }=2r\sin 2\theta \text{and}{\left(\frac{\partial z}{\partial x}\right)}_y=0$${\left(\frac{\partial z}{\partial r}\right)}_{\theta }=2r{\sin }^2\theta ,{\left(\frac{\partial z}{\partial \theta }\right)}_r=r^2\sin 2\theta ,\frac{{\partial }^{2}z}{\partial r.\partial \theta }=2r\sin 2\theta \text{and}{\left(\frac{\partial z}{\partial x}\right)}_y=0$.

Step-by-step solution

Given Information

Given that,$z=r^2-x^2$.

Formula Used

The value of ${\mathit{r}}^{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}$ ...(1)

$$\begin{gathered} \mathit{x}\mathbf{=}\mathit{r}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta } \\ \mathit{y}\mathbf{=}\mathit{r}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta } \end{gathered}$$

Solving the given equation

The given equation

$z=r^2-x^2$ ...(3)

From equation 1 and 2,

$$\begin{gathered} z=x^2+y^2-x^2 \\ z=y^2 \end{gathered}$$

Substitute the value of y in the above equation.

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

Differentiate with respect to r,

$$\begin{gathered} \frac{\partial z}{\partial r}=2r{\sin }^2\theta \\ {\left(\frac{\partial z}{\partial r}\right)}_{\theta }=2r{\sin }^2\theta \end{gathered}$$

We know that$z=r^2{\sin }^2\theta$

Differentiate with respect to$\theta$

$$\begin{gathered} \frac{\partial z}{\partial \theta }=2r^2\sin \theta .\cos \theta \\ \frac{\partial z}{\partial \theta }=r^2\sin 2\theta \\ {\left(\frac{\partial z}{\partial \theta }\right)}_r=r^2\sin 2\theta \end{gathered}$$

Solving further

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial r.\partial \theta }=\frac{\partial }{\partial r}\left(\frac{\partial z}{\partial \theta }\right) \\ \frac{{\partial }^{2}z}{\partial r.\partial \theta }=\frac{\partial }{\partial r}\left(r^2\sin 2\theta \right) \\ \frac{{\partial }^{2}z}{\partial r.\partial \theta }=2r\sin 2\theta \end{gathered}$$

We know that$z=y^2$

$$\begin{gathered} \frac{\partial z}{\partial x}=0 \\ {\left(\frac{\partial z}{\partial x}\right)}_y=0 \end{gathered}$$

Hence${\left(\frac{\partial z}{\partial r}\right)}_{\theta }=2r{\sin }^2\theta ,{\left(\frac{\partial z}{\partial \theta }\right)}_r=r^2\sin 2\theta ,\frac{{\partial }^{2}z}{\partial r.\partial \theta }=2r\sin 2\theta \text{and}{\left(\frac{\partial z}{\partial x}\right)}_y=0$ .