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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:

                                             enter image description here

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Let's look in the Maple code. –  Berci Oct 3 '12 at 21:04
    
Does Maple assume that $t$ is complex by default? It may avoid reducing the square root in case $t$ has an imaginary part. –  Alexander Gruber Oct 3 '12 at 22:16

1 Answer 1

up vote 0 down vote accepted

Probably because the speed is $$ 2 \; \cosh t \; \; | \sinh t | $$

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