Unit Tangent Vector and Unit Normal Vector I was trying to compute the unit normal vector of $\alpha(t)=(cos^3(t),sin^3(t))$ and was finding that it gets particularly unpleasant after the curve is reparameterized by unit length.  Are there any substitutions or tricks I could use to avoid making the unit normal vector so unpleasant?
 A: You don't need to find a unit speed parametrisation in order to compute the normal vector.  Just find the tangent vector of the curve, normalize it to get a unit vector, and then rotate the resulting vector by the matrix $\pmatrix{0 & -1 \\ 1 & 0}$ (namely, switch the coordinates and change the sign of one of them).
A: According to my calculations, the computations you need are not in fact so messy, provided a little trick I'll mention below is used.  The tangent vector of $\alpha(t)$ is easily seen to be
$\dot {\alpha}(t) = 3(-\cos^2 t \sin t, \sin^2t \cos t), \tag{1}$
from which it follows that
$\langle \dot {\alpha}(t), \dot {\alpha}(t) \rangle = 9(\cos^4 t \sin^2 t + \sin^4 t \cos^2 t), \tag{2}$
whence
$\Vert \dot{\alpha} (t) \Vert = 3\sqrt{\cos^4 t \sin^2 t + \sin^4 t \cos^2 t} = 3\sqrt{(\cos^2 t \sin^2 t)(\cos^2 t + \sin^2 t)}$
$ = 3\cos t \sin t = \frac{3}{2}\sin 2t, \tag{3}$
provided $t$ is taken to be in a subset $I \subset \Bbb R$ such that $\sin t, \cos t > 0$, which I shall assume in what follows.  This being the case, we have the unit tangent vector $\mathbf u(t)$ to $\alpha(t)$:
$\mathbf u(t) = \frac{\dot{\alpha}(t)}{\Vert \dot{\alpha}(t) \Vert} = (-\cos t, \sin t). \tag{4}$
We now invoke the little trick I mentioned, which is explained in my answer to this question, viz., that 
$\frac{ds}{dt} = \Vert \dot{\alpha}(t) \Vert = 3\cos t \sin t, \tag{5}$
where $s$ is the arc-length along $\alpha(t)$; whence
$\frac{dt}{ds} = \frac{1}{\Vert \dot{\alpha}(t) \Vert}, \tag{6}$
and if we, in agreement with the common practice, let $\kappa$ denote the curvature and
$\mathbf n$ denote the (standard) unit normal to $\alpha(t)$, then we obtain
$\kappa \mathbf n(t) = \frac{d\mathbf u}{ds} = \frac{d\mathbf u}{dt}\frac{dt}{ds} = \frac{1}{3\cos t \sin t}(\sin t, \cos t); \tag{7}$
apparently 
$\mathbf n(t) = (\sin t, \cos t)  \tag{8}$
and
$\kappa = (3\sin t \cos t)^{-1}. \tag{9}$
The "trick" allows us to avoid re-parametrizing $\alpha(t)$ by $s$.
We see that our $\mathbf n$ is agreement with the answer of user72694 up to a sign which is explained by the fact that $\mathbf n(t)$ actually points in the direction of bending of $\alpha(t)$.  
I think the above covers the basics; I leave the cases in which $\sin t$ and/or $\cos t$ are non-positive to my readers, as well as the treatment of the possibility that there exist $t$ such that $\dot{\alpha}(t) = 0$.
Cheers, and as always
Fiat Lux!
