Why does the curvature approach $\infty$ at cusps? I found the curvature of the astroid $(\cos^3 t, \sin ^3 t)$ to be:
$$\kappa(t) = \frac1{3|\sin t \cos t|}$$
The astroid has $\gamma(\pm \pi/2) = (0, \pm 1)$ and $\gamma(0)$ (resp. $\gamma(\pi)$) $= (\pm 1, 0)$ as cusps. 
When $t$ approaches any of these values, $\kappa \to \infty$. Why is this so? What does it mean (geometrically)?
 A: Curvature measures the rate at which the tangent rotates as you move along the curve. A large curvature value means that the tangent is turning very quickly. Or, saying it another way, curvature measures the change in the tangent direction per unit step along the curve. At a cusp, the tangent rotation is infinitely fast because the tangent direction jumps from one value to another in a zero length step.
Another explanation: curvature is the reciprocal of radius of curvature. At a cusp, radius of curvature is zero, so curvature is infinite.
A: A cusp is just the opposite of a straight line as it forms when a point rotates about itself instantaneously. Examples : When a circle rolls without slipping on a plane or another circle momentarily it has zero radius of curvature or an infinite curvature ( cycloids, epi-, trochoids etc.) . The velocity direction changes at this juncture like when you drive an automobile, at the cusp point you change from reverse gear to the forward after a momentary stop. The ground trace is always a sharp spike.
3D space curves (e.g., helix) also produce cusps when aspected between progressive and regressive projections. 
Also in 3D surfaces we call such a line as a cuspidal edge or edge of regression.
