I have this homework question and I get a different answer to the solutions.
In Cylindrical polar coordinates $(r,\theta,z)$, the velocity potential of a flow is given by:
$$\phi = -\frac{Ua^2r}{b^2-a^2}(1+\frac{b^2}{r^2})cos\theta$$
Find the velocity.
I get the velocity as:
$$v = (-\frac{Ua^2}{b^2-a^2}(1+\frac{b^2}{r^2})cos\theta + \frac{2Ua^2b^2}{(b^2-a^2)r^2}cos\theta)e_r + (\frac{Ua^2}{b^2-a^2}(1+\frac{b^2}{r^2})sin\theta) e_{\theta}$$
The answer misses out the second term in the $r$ direction, but I can't see where I've gone wrong. Any help appreciated.
Your answer is correct!
Maybe they gathered the term in the solution as below ( and maybe made an error of sign).
$$v = (-\frac{Ua^2}{b^2-a^2}(1-\frac{b^2}{r^2})cos\theta) e_{r} + (\frac{Ua^2}{b^2-a^2}(1+\frac{b^2}{r^2})sin\theta) e_{\theta}$$