I have been trying to figure out if this problem could be simplified more.
$\displaystyle sqrt(cos^2(t) cos(2 sin(t))-sin^2(t))$
Thanks
AHHH, so when I square I take the negative with it, right?
for $\displaystyle r'(t)=cost \cdot cos(sint) i - cost \cdot sin(sint) j- sint k$
for $\displaystyle |r'(t)|=\sqrt{(cost \cdot cos(sint))^2 +(-cost \cdot sin(sint))^2 +(-sint)^2}$
$\displaystyle |r'(t)|=\sqrt{cos^2t \cdot (cos^2(sint) + sin^2(sint)) +sin^2t}$
$\displaystyle |r'(t)|=\sqrt{cos^2t \cdot (1) +sin^2t}$
$\displaystyle |r'(t)|=1$