Question

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

Short answer

The value of${\left(\frac{\partial \mathrm{z}}{\partial \mathrm{x}}\right)}_{\mathrm{r}}=-6\mathrm{x},{\left(\frac{\partial \mathrm{z}}{\partial \mathrm{\theta }}\right)}_{\mathrm{x}}=2{\mathrm{x}}^2\mathrm{tan\theta }.{\sec }^2\mathrm{\theta }\frac{{\partial }^2\mathrm{z}}{\partial \mathrm{x}\partial \mathrm{\theta }}=4\mathrm{x}\mathrm{tan\theta }.{\sec }^2\mathrm{\theta }$ .

Step-by-step solution

Given Information

Given that$z=y^2-2x^2$ and the polar coordinates are$x=r\cos \theta ,y=r\sin \theta$

Formula Used

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

Finding the values of (∂z∂x)r,(∂z∂θ)x,∂2z∂x∂θ

Using the formula,

$$\begin{gathered} \frac{\partial \mathrm{z}}{\partial x}=-6x \\ {\left(\frac{\partial \mathrm{z}}{\partial \mathrm{x}}\right)}_r=-6x \end{gathered}$$

We know that$\tan \theta =\frac{y}{x}\Rightarrow y=x\tan \theta$

Then

$$\begin{gathered} z=x^2{\tan }^2\theta -2x^2 \\ \frac{\partial z}{\partial \theta }=2x^2{\tan }^2\theta -2x^2 \\ {\left(\frac{\partial z}{\partial \theta }\right)}_x=2x^2\tan \theta .se{c}^2\theta \end{gathered}$$

Solving further

$$\begin{gathered} \frac{{\partial }^{2}z}{\partial x\partial \theta }=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial \theta }\right) \\ \frac{{\partial }^{2}z}{\partial x\partial \theta }=\frac{\partial }{\partial x}\left(2x^2\tan \theta .se{c}^2\theta \right) \\ \frac{{\partial }^{2}z}{\partial x\partial \theta }=4x\tan \theta .se{c}^2\theta \end{gathered}$$

Hence ${\left(\frac{\partial \mathrm{z}}{\partial \mathrm{x}}\right)}_{\mathrm{r}}=-6\mathrm{x},{\left(\frac{\partial \mathrm{z}}{\partial \mathrm{\theta }}\right)}_{\mathrm{x}}=2{\mathrm{x}}^2\mathrm{tan\theta }.{\sec }^2\mathrm{\theta }\frac{{\partial }^2\mathrm{z}}{\partial \mathrm{x}\partial \mathrm{\theta }}=4\mathrm{x}\mathrm{tan\theta }.{\sec }^2\mathrm{\theta }$.