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# 13.4: Motion Along a Curve

• • Contributed by David Guichard
• Professor (Mathematics) at Whitman College
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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 $${\bf v}(t)$$. 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

$$a_T={4t+18t^3\over \sqrt{1+4t^2+9t^4}} \quad\hbox{and}\quad a_N={\sqrt{4+36t^2+36t^4}\over\sqrt{1+4t^2+9t^4}}.$$

## Contributors

• Integrated by Justin Marshall.