Calculus- Early Transcendentals, 2021a

15.4. Curvature 531
and |T ′ (t)| = 1. Finally,
κ = 1/a
thecurvatureofacircleiseverywheretheinverseofthe radius. It is sometimes useful to think of curvature
as describing what circle a curve most resembles at a point. The curvature of the helix in the previous
example is 1/2; this means that a small piece of the helix looks very much like a circle of radius 2, as
showninFigure15.9.
♣
Figure 15.9: A circle with the same curvature as the helix.
Example 15.18
Consider r(t)=〈cost,sint,cos2t〉, as shown in Figure 15.6. Compute the curvature, κ.
Solution. First r ′ (t)=〈−sint,cost,−2sin(2t)〉 and |r ′ (t)| =
〈
−sint cost
T(t)= √
, √
,
1 + 4sin 2 (2t) 1 + 4sin 2 (2t)
√
1 + 4sin 2 (2t), so
−2sin2t
√
1 + 4sin 2 (2t)
Computing the derivative of this and then the length of the resulting vector is possible but unpleasant.
Fortunately, there is an alternate formula for the curvature that is often simpler than the one we have:
κ = |r′ (t) × r ′′ (t)|
|r ′ (t)| 3 .
We compute the second derivative r ′′ (t)=〈−cost,−sint,−4cos(2t)〉. Then the cross product r ′ (t) ×
r ′′ (t) is
〈−4cost cos2t − 2sint sin2t,2cost sin2t − 4sint cos2t,1〉.
Computing the length of this vector and dividing by |r ′ (t)| 3 is still a bit tedious. With the aid of a computer
we get
κ =
√
48cos 4 t − 48cos 2 t + 17
(−16cos 4 t + 16cos 2 t + 1) 3/2 .
〉
.