I have the following vector function:
$ r(t)=(t-\sinh t\cosh t)\,\partial_x+2\cosh t\,\partial_y. $
I computed its velocity to be as such:
$ r'(t)=-2\sinh^2t\,\partial_x+2\sinh t\,\partial_y. $
Therefore, its speed is as follows:
$ v(t)=2\sinh t\cosh t. $
However, when I computed this using Maple, the program (after simplifying) gave me this:
$ v(t)=2\sqrt{\sinh^2t\cosh^2t}. $
Why did it not get rid of the radical?
Edit: Here is the Maple code: