Question

6. If $\mathit{w}\mathbf{=}\mathit{f}\left(x,y\right)$and $\mathit{x}\mathbf{=}\mathit{r}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{,}\mathit{y}\mathbf{=}\mathit{r}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$, find formulas for $\frac{\mathbf{\partial }\mathbf{w}}{\mathbf{\partial }\mathbf{r}}\mathbf{,}\frac{\mathbf{\partial }\mathbf{w}}{\mathbf{\partial }\mathbf{\theta }}\mathbf{,}$and $\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{w}}{\mathbf{\partial }{\mathbf{r}}^{\mathbf{2}}}$.

Short answer

The required answer is $\frac{\partial w}{\partial r}=f_x\cos \left(\theta \right)+f_y\sin \left(\theta \right),\frac{\partial w}{\partial \theta }=r\left(f_y\cos \theta -f_x\sin \theta \right),$ and $\frac{{\partial }^{2}w}{\partial r^2}=f_{xx}{\cos }^2\theta +2f_{xy}\cos \theta \sin \theta +f_{yy}{\sin }^2\theta$.

Step-by-step solution

Determine the value of ∂w∂r

Begin by using the differentials of the equation $w=f\left(x,y\right)$then substitute from one equation into the other to determine $\frac{\partial w}{\partial r}$as follows:

localid="1664331253335" $$\begin{gathered} w=f\left(x,y\right) \\ dw=\frac{\partial f\left(x,y\right)}{\partial x}dx+\frac{\partial f\left(x,y\right)}{\partial y}dy \\ \frac{\partial w}{\partial r}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial r}...\left(1\right) \end{gathered}$$

Differentiate the functions x and y partially as follows:

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

Substitute equation (2) into equation (1) as follows:

$$\begin{gathered} \frac{\partial w}{\partial r}=\frac{\partial f\left(x,y\right)}{\partial x}\cos \theta +\frac{\partial f\left(x,y\right)}{\partial y}\sin \theta \\ \frac{\partial w}{\partial r}=f_x\cos \left(\theta \right)+f_y\sin \left(\theta \right) \end{gathered}$$

Thus, the required answer is $\frac{\partial w}{\partial r}=f_x\cos \left(\theta \right)+f_y\sin \left(\theta \right)$.

Determine the value of ∂2w∂r2

Using the exact same method as in the preceding part, derive a formula for $\frac{\partial w}{\partial \theta }$as follows:

localid="1664331279875" $$\begin{gathered} w=f\left(x,y\right) \\ \frac{\partial w}{\partial \theta }=\frac{\partial w}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial w}{\partial y}\frac{\partial y}{\partial \theta }...\left(1\right) \end{gathered}$$

Differentiate the functions x and y partially as follows:

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

Substitute equation (2) into equation (1) as follows:

$$\begin{gathered} \frac{\partial w}{\partial \theta }=-r\frac{\partial f\left(x,y\right)}{\partial x}\sin \theta +r\frac{\partial f\left(x,y\right)}{\partial y}\cos \theta \\ \frac{\partial w}{\partial \theta }=r\left(f_y\cos \theta -f_x\sin \theta \right) \end{gathered}$$

Thus, the required answer is $\frac{\partial w}{\partial \theta }=r\left(f_y\cos \theta -f_x\sin \theta \right)$.

Determine the value of ∂w∂θ

The same method can be used to obtain an expression for the second derivative with respect to r we have the first derivative of the expression with respect to r, denoted as$G(x,y)$.

$$\begin{gathered} \frac{\partial w}{\partial r}=f_x\cos \theta +f_y\sin \theta =G\left(x,y\right) \\ dG\left(x,y\right)=\frac{\partial G\left(x,y\right)}{\partial x}dx+\frac{\partial G\left(x,y\right)}{\partial y}dy \\ \frac{\partial G}{\partial r}=\frac{\partial G}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial G}{\partial y}\frac{\partial y}{\partial r}...\left(1\right) \end{gathered}$$

Differentiate the functions G with respect to x and y partially as follows:

$$\begin{gathered} \frac{\partial G}{\partial x}=\frac{\partial }{\partial x}\left(f_x\cos \theta +f_y\sin \theta \right)=f_{xx}\cos \theta +f_{yx}\sin \theta \\ \frac{\partial G}{\partial y}=\frac{\partial }{\partial y}\left(f_x\cos \theta +f_y\sin \theta \right)=f_{xy}\cos \theta +f_{yy}\sin \theta \end{gathered}$$

Differentiate the function x and y as follows:

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

Substitute equation (2) into equation (1) as follows:

$$\begin{gathered} \frac{\partial G}{\partial r}=\left(f_{xx}\cos \theta +f_{yx}\sin \theta \right)\cos \theta +\left(f_{xy}\cos \theta +f_{yy}\sin \theta \right)\sin \theta \\ \frac{\partial G}{\partial r}=\frac{{\partial }^{2}w}{\partial r^2}=f_{xx}{\cos }^2\theta +f_{yx}\sin \theta \cos \theta +f_{xy}\cos \theta \sin \theta +f_{yy}{\sin }^2\theta \\ \frac{{\partial }^{2}w}{\partial r^2}=f_{xx}{\cos }^2\theta +2f_{xy}\cos \theta \sin \theta +f_{yy}{\sin }^2\theta \end{gathered}$$

Where we utilized the relation $f_{xy}=f_{yx}$to get the final expression.

Thus, the required answer is $\frac{{\partial }^{2}w}{\partial r^2}=f_{xx}{\cos }^2\theta +2f_{xy}\cos \theta \sin \theta +f_{yy}{\sin }^2\theta$.