Why is dU^b/dτ = U^a ∇_a U^b?

[MathematicalPhysicist]

I have next question, we have in the next printcreen the equation:
$$\frac{dU^b}{d\tau}=U^a \nabla_a U^b$$
why is this right?
(the printscreen is from the book of Woodhouse in GR, page 34)

Thanks.

 

[MathematicalPhysicist]

Ok, I think I can see why this is right, we pick a frame where $$U^a=(1,0,0,0)$$.

Clumsy me.
:-)