13.4: Motion Along a Curve
( \newcommand{\kernel}{\mathrm{null}\,}\)
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.
Suppose r(t)=⟨cost,sint,1⟩. Then v(t)=⟨−sint,cost,0⟩ and a(t)=⟨−cost,−sint,0⟩. 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 a⋅v=0, so v and a are perpendicular. In fact, it is not hard to see that 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 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′.
Also recall the definition of the curvature, κ=|T′|/|v|, or |T′|=κ|v|. Finally, recall that we defined the unit normal vector as N=T′/|T′|, so T′=|T′|N=κ|v|N. Substituting into equation 13.4.1 we get
\]{\bf a}=|{\bf v}|'{\bf T}+\kappa|{\bf v}|^2{\bf N}.\]
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 13.4.2 with these gives us
a=aT+κv2N=aTT+aNN;
aT is the tangential component of acceleration and aN is the normal component of acceleration.
We have already seen that aT measures how the speed is changing; if you are riding in a vehicle with large aT you will feel a force pulling you into your seat. The other component, aN, 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 κ, and also to how fast you are going. Because aN includes v2, note that the effect of speed is magnified; doubling your speed around a curve quadruples the value of aN. You feel the effect of this as a force pushing you toward the outside of the curve, the "centrifugal force.''
In practice, if want aN we would use the formula for κ:
\]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 aT we can project a onto v:
aT=v⋅a|v|=r′⋅r″|r′|.
Suppose r=⟨t,t2,t3⟩. Compute v, a, aT, and aN.
Solution
Taking derivatives we get v=⟨1,2t,3t2⟩ and a=⟨0,2,6t⟩. Then
aT=4t+18t3√1+4t2+9t4andaN=√4+36t2+36t4√1+4t2+9t4.
Contributors
Integrated by Justin Marshall.