Question

Find $\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{y}}{\mathbf{\partial }\mathbf{x}\mathbf{\partial }\mathbf{\theta }}$.

Short answer

The value of $\frac{{\partial }^{2}y}{\partial x\partial \theta }$ is 1.

Step-by-step solution

Given Information

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

Concept

To find out the value of $\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{y}}{\mathbf{\partial }\mathbf{x}\mathbf{\partial }\mathbf{\theta }}$differentiate$\frac{\mathbf{\partial }\mathbf{y}}{\mathbf{\partial }\mathbf{\theta }}\mathbf{=}\mathit{r}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$.

Find the value of ∂2y∂x∂θ

Differentiate the given polar coordinates

$$\begin{gathered} \frac{\partial y}{\partial \theta }=r\cos \theta \\ \frac{\partial y}{\partial \theta }=x \\ \frac{{\partial }^{2}y}{\partial x\partial \theta }=\frac{\partial }{\partial x}\left(\frac{\partial y}{\partial \theta }\right) \\ \frac{{\partial }^{2}y}{\partial x\partial \theta }=\frac{\partial }{\partial x}\left(x\right) \\ \frac{{\partial }^{2}y}{\partial x\partial \theta }=1 \end{gathered}$$

Therefore the value of $\frac{{\partial }^{2}y}{\partial x\partial \theta }=1$.