Q:

Find Formula for Curvature and Max and Min Values

If $C$ is a smooth curve in 3-space parametrized by arc length, then then curvature $k(t)$ is defined as $$k(t)=\frac{||r'(t)\times r''(t)||}{||r'(t)||^3}$$ Let $C$ be the curve parametrized by $x=\cos t, y=\sin t, z=\cos t, 0\le t\le 2\pi$.
How do I find the formula for $k(t)$, the curvature. And, how do I find the max and min values of the radius of curvature for $C$?

A:

Your expression $$\kappa(t) = \frac{\|{\bf r}'(t)\times{\bf r}''(t)\|}{\|{\bf r}'(t)\|^3}$$ is actually good for curves with any parametrization, not only for unit speed curves (i.e., curves that are parametrized by arc-length). When I read the question earlier, in a bit of a hurry, I thought that you were asking about a deduction of said formula - as I mentioned in the comments, John Oprea's Differential Geometry and Its Applications is a good reference. That being said, you have: $$\begin{align} {\bf r}(t) &= (\cos t, \sin t, \cos t) \\ {\bf r}'(t) &= (-\sin t, \cos t, -\sin t) \\ {\bf r}''(t) &= (-\cos t, -\sin t, -\cos t) = -{\bf r}(t)\end{align}$$
Don't be afraid, plug everything in the expression for $\kappa(t)$ above. You don't have to think much, it is mechanical. Now, remember that if $\kappa(t)$ is the curvature, then $R(t) = 1/\kappa(t)$ is the radius of curvature. Once you find $\kappa(t)$, you automatically get $R(t)$.
And for finding max/min for $R(t)$, note that although we've been dealing with vectors so far, $R$ is a function that receives a real number $t$ and spits a real number $R(t)$. So you'll use what you have learned in single variable calculus - meaning: solve $R'(t_0) = 0$, then see the sign of $R''(t_0)$ (max.:  $<0$, min.: $>0$, saddle point:$=0$). 
Just take it slowly and you can do it.

