Find the curvature and the torsionSolutionSolution Suppose

Solution

Solution :

Suppose, without loss of generality, that is a unit speed curve.

Suppose is part of a circle. Then since every point on is a fixed distance rr away from some point p, we can write:

((t)p)((t)p)=r^2

Differentiate this expression several times with respect to t. See if you can\'t use the expressions you obtain to show that 0, i.e., >0. Then see if you can\'t get an expression involving dds and use this expression to conclude dds is 0, i.e., is constant.

Now suppose has positive constant curvature. See if you can\'t show that the curve (t)=(t)+1N(t) is actually constant, i.e., a point p. From here, see if you can\'t use this to show that is a fixed distance away from from p using what you know about N.

the curvature of a circle of radius r is 1/r. So in example #1. curvature is 1/1 = 1