'smooth' curves with cusps in 3d

[inkliing]

While reviewing basic calculus, I noticed that the curve (1+t^2,t^2,t^3), which clearly has a cusp at (1,0,0), has a derivative curve (2t,2t,3t^2) which is clearly smooth. This struck me as odd since differentiation usually seems to turn cusps into discontinuities, whereas integration smoothes out a curve, especially a curve described by polynomials. In fact, in general I have always taken a curve to be smooth iff it has a continuous derivative, which this curve has, and yet a cusp cannot be smooth in any sensible sense. I suspect the explanation is relatively simple - just something I'm missing.

 

Thx in advance.

[43de]

Very interesting question – it took me a minute to figure this one out. Let's look at a simpler case of 2 variables. Let x=t³ and y=t². Notice that the derivatives of both functions (with respect to t, don't forget the principles you are actually applying) are indeed smooth curves as you pointed out.

 

However, to find the derivative with respect to x we can use the chain rule, or alternatively solve for y in terms of x and then differentiate w.r.t. x. The first way is easier and yields:

 

dy/dx=2/(3t), which is indeed discontinuous at t=0.