Conditions for a smooth parametric curve A curve defined by $x=f(t), y=g(t)$ is smooth if $f'(x)$ and $g'(x)$ are continuous and not simultaneously zero.
Why do we have the second condition(simultaneously zero)?
 A: That second condition is so that the parametrisation is regular. Informally speaking, you want the curve to be "drawn" by the parametrisation in a nice way. In particular, you don't want the parametrisation to turn back on itself. If you have a regular parametrisation then you know that the only way that $f'(t) = g'(t) = 0$ is when the curve itself is singular.
Consider two examples. Take the cusp $\gamma(t) = (t^2,t^3)$; this has a regular parametrisation. If you take the derivative $\gamma'(t) = (2t,3t^2)$ and so the singular points come from $2t = 3t^2 = 0$, i.e. $t=0$. So the singular point of the curve is $\gamma(0) = (0,0)$, i.e. the cusp point.
Now consider the smooth curve $y=x^2$ with the non-regular parametrisation $\alpha(t) =  (t^3,t^6)$. We have $\alpha'(t) = (3t^2,6t^5)$. It's tempting to think that $t=0$ gives a singular point, but it doesn't. The curve is smooth. It's the parametrisation that causes the problem. Informally, the pen stops for an instant at the origin.
A: Intuitively, we want the curve to have no corners. There are two ways it could have a corner: 
* 
* Its parameterization could fail to be differentiable at a point  
* The parameterization could slow to a stop, and then start up again in a completely different direction.
It might seem counter-intuitive that you could have the latter without the former, but there are functions that go from being constantly zero to being nonzero, and yet stay completely smooth. (They are called "bump functions").
A: I think the reason behind that to preserve the orientation as a well defined orientation. If the derivative is zero then we have singularity, which means we have more than one orientation on that curve.
