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 $(0,0,1)$. $\square$

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 \eqalignno{{\bf a}&=|{\bf v}|'{\bf T}+|{\bf v}|{\bf T}'.& (13.4.1)\cr } 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 \eqalignno{{\bf a}&=|{\bf v}|'{\bf T}+\kappa|{\bf v}|^2{\bf N}.& (13.4.2)\cr } 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 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 $a_N$.

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

Exercises 13.4

Here we show how to compute components of acceleration with Sage.

Ex 13.4.1 Let ${\bf r}=\langle \cos t,\sin t,t\rangle$. Compute ${\bf v}$, ${\bf a}$, $a_T$, and $a_N$. (answer)

Ex 13.4.2 Let ${\bf r}=\langle \cos t,\sin t,t^2\rangle$. Compute ${\bf v}$, ${\bf a}$, $a_T$, and $a_N$. (answer)

Ex 13.4.3 Let ${\bf r}=\langle \cos t,\sin t,e^t\rangle$. Compute ${\bf v}$, ${\bf a}$, $a_T$, and $a_N$. (answer)

Ex 13.4.4 Let ${\bf r}=\langle t^2,2t-3,3t^2-3t\rangle$. Compute ${\bf v}$, ${\bf a}$, $a_T$, and $a_N$. (answer)

Ex 13.4.5 Let ${\bf r}=\langle e^t,\sin t,e^t\rangle$. Compute ${\bf v}$, ${\bf a}$, $a_T$, and $a_N$. (answer)

Ex 13.4.6 Suppose an object moves so that its acceleration is given by ${\bf a}=\langle -3\cos t,-2\sin t,0\rangle$. At time $t=0$ the object is at $(3,0,0)$ and its velocity vector is $\langle 0,2,0\rangle$. Find ${\bf v}(t)$ and ${\bf r}(t)$ for the object. (answer)

Ex 13.4.7 Suppose an object moves so that its acceleration is given by ${\bf a}=\langle -3\cos t,-2\sin t,0\rangle$. At time $t=0$ the object is at $(3,0,0)$ and its velocity vector is $\langle 0,2.1,0\rangle$. Find ${\bf v}(t)$ and ${\bf r}(t)$ for the object. (answer)

Ex 13.4.8 Suppose an object moves so that its acceleration is given by ${\bf a}=\langle -3\cos t,-2\sin t,0\rangle$. At time $t=0$ the object is at $(3,0,0)$ and its velocity vector is $\langle 0,2,1\rangle$. Find ${\bf v}(t)$ and ${\bf r}(t)$ for the object. (answer)

Ex 13.4.9 Suppose an object moves so that its acceleration is given by ${\bf a}=\langle -3\cos t,-2\sin t,0\rangle$. At time $t=0$ the object is at $(3,0,0)$ and its velocity vector is $\langle 0,2.1,1\rangle$. Find ${\bf v}(t)$ and ${\bf r}(t)$ for the object. (answer)

Ex 13.4.10 Describe a situation in which the normal component of acceleration is 0 and the tangential component of acceleration is non-zero. Is it possible for the tangential component of acceleration to be 0 while the normal component of acceleration is non-zero? Explain. Finally, is it possible for an object to move (not be stationary) so that both the tangential and normal components of acceleration are 0? Explain.