Calculus- Early Transcendentals, 2021a

530 Vector Functions
To remove the dependence on time, we use the arc length parameterization. If a curve is given by r(s),
then the first derivative r ′ (s) is a unit vector, that is, r ′ (s)=T(s). We now compute the second derivative
r ′′ (s)=T ′ (s) and use |T ′ (s)| as the “official” measure of curvature, usually denoted κ.
Example 15.15
We have seen that the arc length parameterization of a particular helix is
r(s)=〈cos(s/ √ 2),sin(s/ √ 2),s/ √ 2〉
Computing the second derivative gives r ′′ (s)=〈−cos(s/ √ 2)/2,−sin(s/ √ 2)/2,0〉 with length 1/2.
What if we are given a curve as a vector function r(t),wheret is not arc length? We have seen that arc
length can be difficult to compute; fortunately, we do not need to convert to the arc length parameterization
to compute curvature. Instead, let us imagine that we have done this, so we have found t = g(s) and then
formed ˆr(s)=r(g(s)). The first derivative ˆr ′ (s) is a unit tangent vector, so it is the same as the unit tangent
vector T(t)=T(g(s)). Taking the derivative of this we get
d
ds T(g(s)) = T′ (g(s))g ′ (s)=T ′ (t) dt
ds .
The curvature is the length of this vector:
κ = |T ′ (t)|| dt
ds | = |T′ (t)|
|ds/dt| = |T′ (t)|
|r ′ (t)| .
(Recall that we have seen that ds/dt = |r ′ (t)|.) Thus we can compute the curvature by computing only
derivatives with respect to t; we do not need to do the conversion to arc length.
Example 15.16
Returning to the helix, suppose we start with the parameterization r(t)=〈cost,sint,t〉. Calculate
the curvature κ.
Solution. First r ′ (t)=〈−sint,cost,1〉, |r ′ (t)| = √ 2, and T(t)=〈−sint,cost,1〉/ √ 2. Then
T ′ (t)=〈−cost,−sint,0〉/ √ 2
Finally, κ = 1/ √ 2/ √ 2 = 1/2, as before.
|T ′ (t)| = 1/ √ 2
♣
Example 15.17
Consider this circle of radius a: r(t)=〈acost,asint,1〉. Find the curvature, κ.
Solution. First r ′ (t)=〈−asint,acost,0〉. Then|r ′ (t)| =
T(t)=〈−asint,acost,0〉/a. Now
T ′ (t)=〈−acost,−asint,0〉/a
√
a 2 sin 2 t + a 2 cos 2 t = a, and