r(t) = <8$sint^2$,8$cost^2$> Solve for T and N.
I know that T is equal to v/|v|. and I believe that v is equal to
<$16sin(t)cos(t)$,$-16cos(t)sin(t)$>
and |v| is equal to $sqrt((16sin(t)cos(t))^2+(-16cos(t)sin(t))^2)$
so that means T is equal to $<8sin(t)^2$/$sqrt((16sin(t)cos(t))^2+(-16cos(t)sin(t))^2)$,
$8cos(t)^2/sqrt((16sin(t)cos(t))^2+(-16cos(t)sin(t))^2))$
MyMathLab does not accept this answer and says it is wrong, maybe because I didn't simplify it? I know that sin(t)cos(t) is equal to sin(2x) but even with that I'm not sure how to simplify it. MyMathLab says that the correct answer for T is $< cos(t)^2, -sin(t)^2>$
I know that to solve for N you do (dT/dt)/(|dT/dt) but first I need to know how to solve for T. I could do this problem if r was equal to something like $<5t,4t^2$ but I don't know how to do it with sin and cos and sorry for the formatting I don't know how to do that either.
I think that you are complicating things right from the start. Try to simplify $x(t)$ and $y(t)$ first:
$$x(t)=8\sin^2t=4(1-\cos2t)$$
$$y(t)=8\cos^2t=4(1+\cos2t)$$
$$\dot{x}=8\sin2t$$
$$\dot{y}=-8\sin2t$$
$$v=\sqrt{\dot{x}^2+\dot{y}^2}=8\sqrt2 |\sin2t|$$
So the components of unit vector $\vec{T}$ are:
$$\vec{T}=\frac{\sqrt2}{2}\frac{\sin2t}{|\sin2t|}\vec i-\frac{\sqrt2}{2}\frac{\sin2t}{|\sin2t|}\vec j$$
In other words, for $\sin2t>0$, the unit vector is:
$$\vec{T}=\frac{\sqrt2}{2}\vec i-\frac{\sqrt2}{2}\vec j\tag{1}$$
It looks strange because $\vec{T}$ is not a function of $t$ but it makes perfect sense. From your starting equations it is obvious that $x+y=8$ so the point moves along a straight line and the unit vector $\vec{T}$ depends only on the direction of movement.
Hope this helps.