# 13.4: Motion Along a Curve

We have already seen that if \(t\) is time and an object's location is given by \({\bf r}(t)\), then the derivative \({\bf r}'(t)\) is the velocity vector\index{velocity vector} \({\bf v}(t)\). Just as \({\bf v}(t)\) is a vector describing how \({\bf r}(t)\) changes, so is \({\bf v}'(t)\) a vector describing how \({\bf v}(t)\) changes, namely, \({\bf a}(t)={\bf v}'(t)={\bf r}''(t)\) is the **acceleration vector**.

Example 13.4.1

Suppose \({\bf r}(t)=\langle \cos t,\sin t,1\rangle\). Then \({\bf v}(t)=\langle -\sin t,\cos t,0\rangle\) and \({\bf a}(t)=\langle -\cos t,-\sin t,0\rangle\). This describes the motion of an object traveling on a circle of radius 1, with constant \(z\) coordinate 1. The velocity vector is of course tangent to the curve; note that \({\bf a}\cdot{\bf v}=0\), so \({\bf v}\) and \({\bf a}\) are perpendicular. In fact, it is not hard to see that \({\bf a}\) points from the location of the object to the center of the circular path at \((0,0,1)\).

Recall that the unit tangent vector is given by \({\bf T}(t)= {\bf v}(t)/|{\bf v}(t)|\), so \({\bf v}=|{\bf v}|{\bf T}\). If we take the derivative of both sides of this equation we get

$${\bf a}=|{\bf v}|'{\bf T}+|{\bf v}|{\bf T}'.$$

Also recall the definition of the curvature, \(\kappa=|{\bf T}'|/|{\bf v}|\), or \(|{\bf T}'|=\kappa|{\bf v}|\). Finally, recall that we defined the unit normal vector as \({\bf N}={\bf T}'/|{\bf T}'|\), so \({\bf T}'=|{\bf T}'|{\bf N}= \kappa|{\bf v}|{\bf N}\). Substituting into equation 13.4.1 we get

$${\bf a}=|{\bf v}|'{\bf T}+\kappa|{\bf v}|^2{\bf N}.$$

The quantity \(|{\bf v}(t)|\) is the speed of the object, often written as \(v(t)\); \(|{\bf v}(t)|'\) is the rate at which the speed is changing, or the scalar acceleration of the object, \(a(t)\). Rewriting equation 13.4.2 with these gives us

$${\bf a}=a{\bf T}+\kappa v^2{\bf N}=a_{T}{\bf T}+a_{N}{\bf N};$$

\(a_T\) is the **tangential component of acceleration** and \(a_N\) is the **normal component of acceleration**.

We have already seen that \(a_T\) measures how the speed is changing; if you are riding in a vehicle with large \(a_T\) you will feel a force pulling you into your seat. The other component, \(a_N\), measures how sharply your direction is changing {\em with respect to time}. So it naturally is related to how sharply the path is curved, measured by \(\kappa\), and also to how fast you are going. Because \(a_N\) includes \(v^2\), note that the effect of speed is magnified; doubling your speed around a curve quadruples the value of \(a_N\). You feel the effect of this as a force pushing you toward the outside of the curve, the "centrifugal force.''

In practice, if want \(a_N\) we would use the formula for \(\kappa\):

$$a_N=\kappa |{\bf v}|^2= {|{\bf r}'\times{\bf r}''|\over |{\bf r}'|^3}|{\bf r}'|^2={|{\bf r}'\times{\bf r}''|\over|{\bf r}'|}.$$

To compute \(a_T\) we can project \({\bf a}\) onto \({\bf v}\):

$$a_T={{\bf v}\cdot{\bf a}\over|{\bf v}|}={{\bf r}'\cdot{\bf r}''\over |{\bf r}'|}.$$

Example 13.4.2

Suppose \({\bf r}=\langle t,t^2,t^3\rangle\). Compute \({\bf v}\), \({\bf a}\), \(a_T\), and \(a_N\).

**Solution**

Taking derivatives we get \({\bf v}=\langle 1,2t,3t^2\rangle\) and \({\bf a}=\langle 0,2,6t\rangle\). Then $$a_T={4t+18t^3\over \sqrt{1+4t^2+9t^4}} \quad\hbox{and}\quad a_N={\sqrt{4+36t^2+36t^4}\over\sqrt{1+4t^2+9t^4}}.$$