Derive the velocity potential for a doublet; that is$,$ derive Equation $(3\mathrm{.}88).$

Hint: The easiest method is to start with Equation $(3\mathrm{.}87)$ for the stream function and extract the velocity potential. Continuing with the problem, show that the equipotential lines are circles with centers on the x axis. To prove this fact, convert from polar to cartesian

coordinates. or

$=-\frac{}{2}\frac{sin}{r}$

Equation $(3\mathrm{.}87)$ is the stream function for a doublet. In a similar fashion, the

velocity potential for a doublet is given by $($see Problem $3\mathrm{.}14)$

$=\frac{}{2}\frac{cos}{r}$

The streamlines of a doublet flow are obtained from Equation $(3\mathrm{.}87)$: