Question

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

Short answer

The value of $\frac{{\partial }^{2}r}{\partial \theta \partial y}\text{is}\cos \theta$

Step-by-step solution

Given Information

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

find the value of ∂2r∂θ∂y

The value of r is expressed as,

.$r=\sqrt{x^2+y^2}$

The above equation differentiates with respect to y.

$$\begin{gathered} \frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}} \\ \frac{\partial r}{\partial y}=\frac{r\sin \theta }{\sqrt{r^2}} \\ \frac{\partial r}{\partial y}=\sin \theta \\ \backslash \end{gathered}$$

Now,

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

Hence the value of $\frac{{\partial }^{2}r}{\partial \theta \partial y}=\cos \theta$.