Because of our choice of normal vector, we can assign a sign to the curving of a curve.

If the velocity X^'(t) is non-zero, then there is a well-defined line in the direction of that vector through the point X(t) called the tangent line to X at X(t). For fixed t, this tangent line can be parametrized by u to give Y(u) = X(t) + uX^'(t). The direction of the velocity vector gives an orientation to the tangent line, so there is a well-defined left-hand half-plane determined by the tangent line.

In general we expect that the velocity vector and the acceleration vector will be linearly independent, so that they are not scalar multiples of each other. If the acceleration vector at a point is in the left half-plane of the tangent line at that point, we say that the curve is curving positively and if the acceleration vector is in the right half-plane, then the curve is said to be curving negatively.

Demonstration 8: Signed Curvature

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The vector U^' is the unit normal vector that points into the positive half space determined by the oriented tangent line. It follows that the curve is curving positively at a point if X^''(t)U(t)>0 and curving negatively if X^''(t)U(t)<0.

Using the curve t->[t², t³ + t/2] where t goes from -1 to 1, find the set of points for which the acceleration vector is a multiple of the velocity vector.

For the curves discussed earlier, indicate for which t the curve is curving positively and when it is curving negatively. Do the same for the family of functions X(t) =(t^m,tⁿ) for pairs of positive integers (m,n). What is the effect of reparametrization on these examples? In particular, what can we say about Y(t) =X(t³+t), or, more generally, Y(t) =X(u(t))?

Show that if a parametrized curve has constant speed, then the acceleration vector at each point is perpendicular to the velocity vector. Show that if X(t) is a curve of constant speed k, then Y(t)=X(tk) is a curve of constant speed 1.