I don't understand how to get from .54 to .55, any help?
Edit: Wolfram alpha gives the answer as this http://www.wolframalpha.com/input/?i=grad+Ucos%28theta%29%28r%2Ba%5E2%2Fr%29
I can't seem to convert their answer to the answer needed even though it is similar.
Hint: Remember that $x = r\cos(\theta)$ in cylindrical coordinates. You have $$ \phi = U \, r\cos(\theta) + U \, \frac{a^2}{r} \cos(\theta) = U\,x+U\,\frac{a^2}{r}\cos(\theta) $$ Take this gradient in cylindrical coordinates (with $x$ the rectangular coordinate) $$ \nabla \phi = \frac{\partial \phi}{\partial x}\mathbf{\hat{e}_x} + \frac{1}{r} \frac{\partial \phi}{\partial \theta} \mathbf{\hat{e}_{\theta}} + \frac{\partial \phi}{\partial r} \mathbf{\hat{e}_{r}} \\ = U\mathbf{\hat{e}_{x}} + U \frac{1}{r}\left( - \frac{a^2}{r} \sin(\theta) \right)\mathbf{\hat{e}_{\theta}} + U \left( - \frac{a^2}{r^2} \cos(\theta) \right) \mathbf{\hat{e}_{r}}$$ Now, after minor algebraic manipulation you will see that the equation you are curious about is exactly $- \nabla \phi$.
