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: