91Ó°ÊÓ

In the given expression, replace the Cartesian coordinates (x, y) with their corresponding polar coordinate expressions (r*cos(φ), r*sin(φ)). So we have: $$ \begin{aligned} x &= r \cos \varphi \\ y &= r \sin \varphi \end{aligned} $$ Now we need to replace these expressions into the given expression of W.

Operating the given transformation \(u(r, \varphi) = U(r \cos \varphi, r \sin \varphi)\), we need to compute the partial derivatives \(\frac{\partial U}{\partial x}\) and \(\frac{\partial U}{\partial y}\) in terms of the partial derivatives of u with respect to r and φ: $$ \begin{aligned} \frac{\partial U}{\partial x} &= \frac{\partial u}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial u}{\partial \varphi} \frac{\partial \varphi}{\partial x} \\ \frac{\partial U}{\partial y} &= \frac{\partial u}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial u}{\partial \varphi} \frac{\partial \varphi}{\partial y} \end{aligned} $$ Now let's find the expressions for \(\frac{\partial r}{\partial x}\), \(\frac{\partial r}{\partial y}\), \(\frac{\partial \varphi}{\partial x}\), and \(\frac{\partial \varphi}{\partial y}\) using the relationships of x and y in Step 1: $$ \begin{aligned} \frac{\partial r}{\partial x} &= \frac{\partial}{\partial x}\left(r \cos \varphi\right) = \frac{\cos \varphi}{\sqrt{\cos^2 \varphi + \sin^2 \varphi}} = \cos \varphi \\ \frac{\partial r}{\partial y} &= \frac{\partial}{\partial y}\left(r \sin \varphi\right) = \frac{\sin \varphi}{\sqrt{\cos^2 \varphi + \sin^2 \varphi}} = \sin \varphi \\ \frac{\partial \varphi}{\partial x} &= -\frac{\sin \varphi}{r} \\ \frac{\partial \varphi}{\partial y} &= \frac{\cos \varphi}{r} \end{aligned} $$ Now, substitute them into the expressions for the partial derivatives of U with respect to x and y: $$ \begin{aligned} \frac{\partial U}{\partial x} &= \cos \varphi \frac{\partial u}{\partial r} - \frac{\sin \varphi}{r} \frac{\partial u}{\partial \varphi} \\ \frac{\partial U}{\partial y} &= \sin \varphi \frac{\partial u}{\partial r} + \frac{\cos \varphi}{r} \frac{\partial u}{\partial \varphi} \end{aligned} $$

Finally, substitute the expressions for the partial derivatives of U with respect to x and y into the expression of W: $$ \begin{aligned} W &= \frac{1}{\sqrt{x^{2}+y^{2}}}\left(x \frac{\partial U}{\partial x} + y \frac{\partial U}{\partial y}\right) \\ &= \frac{1}{\sqrt{r^2}} \left( r\cos \varphi\left(\cos \varphi \frac{\partial u}{\partial r} - \frac{\sin \varphi}{r} \frac{\partial u}{\partial \varphi}\right) + r\sin \varphi\left(\sin \varphi \frac{\partial u}{\partial r} + \frac{\cos \varphi}{r} \frac{\partial u}{\partial \varphi}\right)\right) \\ &= \frac{1}{r}\left( r\cos^2 \varphi \frac{\partial u}{\partial r} - r\sin^2 \varphi \frac{\partial u}{\partial r} + r\sin^2 \varphi \frac{\partial u}{\partial r} + r\cos^2 \varphi \frac{\partial u}{\partial r} \right) \\ &= \frac{1}{r}\left( r(\cos^2 \varphi + \sin^2 \varphi) \frac{\partial u}{\partial r}\right) \\ &= \frac{\partial u}{\partial r} \end{aligned} $$ So, the expression W transformed to polar coordinates is: $$ W = \frac{\partial u}{\partial r} $$