Question

Solve equations (11.11) to get equations (11.12).

Short answer

Hence, the required derivatives are:

$\frac{\partial F}{\partial x}=\cos \theta \frac{\partial F}{\partial r}-\frac{1}{r}\sin \theta \frac{\partial F}{\partial \theta }$and $\frac{\partial F}{\partial y}=\sin \theta \frac{\partial F}{\partial r}+\frac{1}{r}\cos \theta \frac{\partial F}{\partial \theta }$

Step-by-step solution

Partial derivatives

If any function $f\left(x,y\right)$is having two variables, then its derivative with respect to x nd y will be considered as partial derivatives of $f\left(x,y\right)$.

Find Partial derivative

The given equations are:

$$\begin{gathered} \frac{\partial F}{\partial r}=\cos \theta \frac{\partial F}{\partial x}+\sin \theta \frac{\partial F}{\partial y}...\left(1\right) \\ \frac{\partial F}{\partial \theta }=-r\sin \theta \frac{\partial F}{\partial x}+r\cos \theta \frac{\partial F}{\partial y}...\left(2\right) \end{gathered}$$

Now, multiply eq. (1) by $r\sin \theta$and eq. (2) by $\cos \theta$. Then, we have:

$$\begin{gathered} r\sin \theta \frac{\partial F}{\partial r}=r\sin \theta \cos \theta \frac{\partial F}{\partial x}+r{\sin }^2\theta \frac{\partial F}{\partial y}...\left(3\right) \\ \cos \theta \frac{\partial F}{\partial \theta }=-r\sin \theta \cos \theta \frac{\partial F}{\partial x}+r{\cos }^2\theta \frac{\partial F}{\partial y}...\left(4\right) \end{gathered}$$

Adding eq. (3) and (4), we get:

$$\begin{gathered} r\frac{\partial F}{\partial y}=r\sin \theta \frac{\partial F}{\partial r}+\cos \theta \frac{\partial F}{\partial \theta } \\ \frac{\partial F}{\partial y}=\sin \theta \frac{\partial F}{\partial r}+\frac{1}{r}\cos \theta \frac{\partial F}{\partial \theta } \end{gathered}$$

Simplify further

Again, multiply eq. (1) by $-r\cos \theta$ and eq. (2) by $\sin \theta$. Then, we have:

$$\begin{gathered} -r\cos \theta \frac{\partial F}{\partial r}=r{\cos }^2\theta \frac{\partial F}{\partial x}-r\sin \theta \cos \theta \frac{\partial F}{\partial y}...\left(5\right) \\ \sin \theta \frac{\partial F}{\partial \theta }=-r{\sin }^2\theta \frac{\partial F}{\partial x}+r\sin \theta \cos \theta \frac{\partial F}{\partial y}...\left(6\right) \end{gathered}$$

Adding eq. (5) and (6), we get:

$\frac{\partial F}{\partial x}=\cos \theta \frac{\partial F}{\partial r}-\frac{1}{r}\sin \theta \frac{\partial F}{\partial \theta }$

Hence, the required derivatives are:

$\frac{\partial F}{\partial x}=\cos \theta \frac{\partial F}{\partial r}-\frac{1}{r}\sin \theta \frac{\partial F}{\partial \theta }$and$\frac{\partial F}{\partial y}=\sin \theta \frac{\partial F}{\partial r}+\frac{1}{r}\cos \theta \frac{\partial F}{\partial \theta }$