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

Recall that the unit tangent vector is given by $T(t)= v(t)/|v(t)|$, so $v=|v|T$. If we take the derivative of both sides of this equation we get $$a=|v|'T+|v|T'. \eqrdef{eq:acceleration decomposition initial} \eqno{(\xrefn{eq:acceleration decomposition initial})}$$ Also recall the definition of the curvature, $\kappa=|T'|/|v|$, or $|T'|=\kappa|v|$. Finally, recall that we defined the unit normal vector as $N=T'/|T'|$, so $T'=|T'|N= \kappa|v|N$. Substituting into equation MISSING XREFN(eq:acceleration decomposition initial( we get $$a=|v|'T+\kappa|v|^2N. \eqrdef{eq:acceleration decomposition final} \eqno{(\xrefn{eq:acceleration decomposition final})}$$ The quantity $|v(t)|$ is the speed of the object, often written as $v(t)$; $|v(t)|'$ is the rate at which the speed is changing, or the scalar acceleration of the object, $a(t)$. Rewriting equation MISSING XREFN(eq:acceleration decomposition final( with these gives us $$a=aT+\kappa v^2N= a_{T}T+a_{N}N;$$ $a_T$ 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 $\kappa$: $$a_N=\kappa |v|^2= {|r'\timesr"|\over |r'|^3}|r'|^2={|r'\timesr"|\over|r'|}.$$ To compute $a_T$ we can project $a$ onto $v$: $$a_T={v\cdota\over|v|}={r'\cdotr"\over |r'|}.$$

Exercises 13.4

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

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

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

Ex 13.4.4 Let $r=\langle e^t,\sin t,e^t\rangle$. Compute $v$, $a$, $a_T$, and $a_N$. (answer)

Ex 13.4.5 Suppose an object moves so that its acceleration is given by $a=\langle -3\cos t,-2\sin t,0\rangle$. At time $t=0$ the object is at $(3,0,0)$ and its velocity vector is $\langle 0,2,0\rangle$. Find $v(t)$ and $r(t)$ for the object. (answer)

Ex 13.4.6 Suppose an object moves so that its acceleration is given by $a=\langle -3\cos t,-2\sin t,0\rangle$. At time $t=0$ the object is at $(3,0,0)$ and its velocity vector is $\langle 0,2.1,0\rangle$. Find $v(t)$ and $r(t)$ for the object. (answer)

Ex 13.4.7 Suppose an object moves so that its acceleration is given by $a=\langle -3\cos t,-2\sin t,0\rangle$. At time $t=0$ the object is at $(3,0,0)$ and its velocity vector is $\langle 0,2,1\rangle$. Find $v(t)$ and $r(t)$ for the object. (answer)

Ex 13.4.8 Suppose an object moves so that its acceleration is given by $a=\langle -3\cos t,-2\sin t,0\rangle$. At time $t=0$ the object is at $(3,0,0)$ and its velocity vector is $\langle 0,2.1,1\rangle$. Find $v(t)$ and $r(t)$ for the object. (answer)

Ex 13.4.9 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.