Smooth curves with cusps in 3d

[inkliing]

"smooth" curves with cusps in 3d

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.

 

[tiny-tim]

hi inkliing!

… differentiation usually seems to turn cusps into discontinuities …

no, that's only for cusps that have a non-zero angle

a cusp with a zero angle is often an illusion

consider a point on the wheel of a steadily moving car …

in the frame of reference of the car, it's going in a uniform circle (you can't get any smoother than that!), with https://www.physicsforums.com/library.php?do=view_item&itemid=27" of constant magnitude v²/r

but in the frame of reference of the ground, it follows a cycloid (see http://en.wikipedia.org/wiki/Cycloid" for a neat .gif), with a cusp whenever that point contacts the ground …

it moves vertically down just before contact, and vertically up just after …

but it still obviously has acceleration of constant magnitude v²/r

(can you find a frame of reference in which your curve has no cusp? )

 

[HallsofIvy]

Note also that your "cusp" is at t= 0 where the derivative is (0, 0, 0) so that is NOT a proper parameterization of the curve.

 

[inkliing]

Thx tiny tim for the very straightforward frame-of-reference refrence. I understand it much better now :)

 

[LCKurtz]

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.

The condition you need to avoid such "smooth" cusps is that R'(t) ≠ 0. If think of an object moving, if you allow it to smoothly come to a stop then smoothly take off in a different direction, you can get sharp corners. But if you have a continuous non-zero derivative for R(t), that can't happen, and that is the definition of a smooth parametric curve.