1.3: Acceleration and Normal Vectors

In the same way that our position vector traces out a curve as t changes, so does our tangent vector. This suggests that we could discover much from differentiating the tangent vector to obtain the acceleration vector, X''(t). If the curve is X(u(t)), the first derivative is X'(u(t))u'(t) and so by the chain rule and the product rule, the acceleration vector is

X''(u(t))(u'(t))2+ X'(u(t))(u''(t)).

The acceleration vector on the other hand can change both length and direction underreparametrization. Nonetheless there are some properties that are independent of reparametrization: if the acceleration vector is in the left-hand half-plane determined by the velocity vector in one parametrization, then it remains so in any (orientation preserving) reparametrization. It turns out that there is another very geometric quantity associated with the acceleration that is independent of reparametrization, namely, the curvature, a function that measures deviation from straightness at each point of the curve. In order to isolate this curvature function, we want to relate the acceleration to a coordinate system given by a specially constructed pair of mutually orthogonal unit vectors that move along the curve, a "moving frame" particularly well adapted to the curve. To acheive this we do the following: at each point along a curve where the velocity vector is not the zero vector, we divide the velocity vector by its length to define the unit tangent vector,

T(t) = X'(t)/|X'(t)| = X'(t)/s'(t).

A vector perpendicular to the velocity vector at a point is said to be normal to the curve at that point. At each point of a smooth curve, there are two unit normal vectors, pointing in opposite directions. We choose one by defining the unit normal vector U(t) to be the unit tangent vector rotated by 90 degrees in the counter-clockwise direction, so that if T(t)=(a(t),b(t)) , then U(t) = (-b(t),a(t)).


This demo allows you to examine the effects of reparametrization of the domain of a curve X(t)=X(u(t)). The tapedeck controls the value of t0. The first window shows the curve in green and the tangent vector to the curve in cyan. There is a second window that shows the unit circle and the unit tangent vector (in orange) beginning at the origin (so that it always lies on the circle). The remaining portion of the tangent vector is shown in cyan.



1. Consider the circle X(t) = R(cos(t),sin(t)), with X'(t) = R(-sin(t),cos(t)). The velocity vector has constant length R and it is always perpendicular to the position vector. However if we consider Y(t) = R(cos(t2),sin(t2)), then Y'(t) = 2tR2(-sin(t2),cos(t2)) so the length of the velocity vector equals 2abs(t)R, a changing quantity; however Y'(t) is still perpendicular to Y(t) for all t.

2. Investigate the functions Y(t) = X(t3+t) for the circle, the exponential spiral, the arithmetic spiral, and (t3,t4). When you enter a reparametrization, you want the graph to have the same endpoints as it did before. To achieve this you will need to divide the suggested t by some constants and adjust the domain accordingly.

3. More generally, investigate Y(t) = X(u(t)) for a reparametrization u(t) such that u'(t) > 0 for all t.


This demo allows you to examine the unit tangent and normal vectors of a curve. Running the tapedeck in the control panel makes the parameter t run through its domain.



1. Change the slider c to 1 on the default curve, to make the curve a unit circle. What is the relationship between normal vectors and position vectors? What is the relationship between velocity vectors and unit tangent vectors?

2. Choose a curve with an inflection point, like X(t) = (t,t3). What happens to the normal vectors and to the tangent vectors that emanate from the neighborhood of the inflection point?

3. What happens if the curve has a cusp; for example, X(t) =(t2,t3)? What happens at the point t = 0 for the curve (cos(5t),t3)?
Top of Page