Discussion
For a potential flow, the fluid velocity is given by \(\vec{V}(x, y) = u\hat{i} + v\hat{j}\). The slope of the potential line at \((x, y)\) is
Answer: C
The slope is \(-\frac{u}{v}\).
Along a potential (equipotential) line \(\phi\) is constant, so
\[ \begin{aligned} d\phi &= \frac{\partial\phi}{\partial x}dx + \frac{\partial\phi}{\partial y}dy \\ &= u\,dx + v\,dy = 0 \end{aligned} \]
giving \(\frac{dy}{dx} = -\frac{u}{v}\).
\(\frac{v}{u}\) is the slope of a streamline, which is tangent to the velocity. The two slopes multiply to \(-1\), the familiar result that streamlines and equipotentials cross at right angles. \(\frac{u}{v}\) and \(-\frac{v}{u}\) have the right pieces in the wrong arrangement.
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