This is interpreted as showing the vector ˆet rotates about a line through eˆb with
angular velocity ω. If the curvature κ= 0 , then ω is also zero and so the tangent
vector ˆetis not rotating and consequently the curve is a straight line. Similarly, one
can write
dˆeb
dt=dˆeb
dsds
dt=−τˆends
dt=−τds
dtˆeb׈et=τds
dtˆet׈eb=ω׈eb where ω=τds
dtˆetand this result is interpreted as meaning the vector ˆeb is rotating about the ˆet
direction with angular velocity ω. If the torsion τ = 0 , then there is no rotation
of the binormal vector and so the curve remains a plane curve in the plane of the
normal and tangent vectors.
It is left as an exercise to give a physical interpretation to the derivative d
ˆen
dt.
Example 7-4. Determine how to calculate the curvature κof a space curve.
Solution Use the fact
dˆetds =κeˆn so that
∣∣dˆet
ds∣∣=κ, since ˆen is a unit vector. Let
r =r (t) = x(t)ˆe 1 +y(t)ˆe 2 +z(t)ˆe 3 denote the position vector to a point on the space
curve where trepresents some parameter. If the arc length parameter sis used, then
dr
ds=dr
dtdt
ds=^1
|drdt|dr
dt=ˆetis a unit tangent vector to the curve and the derivative of this vector with respect
to arc length sgives
d^2 r
ds^2=dˆet
ds=κˆenTaking the dot product of this vector with itself gives
d^2 r
ds^2 ·d^2 r
ds^2 = (κˆen)·(κˆen) = κ(^2) = [x′′(s)] (^2) + [y′′(s)] (^2) + [z′′(s)] 2
where
x′(t) =dx
dsds
dt =x′(s)ds
dt =⇒ x′(s) = x′(t)
ds
dt
x′′(t) = x′(s)d^2 s
dt^2 +d
dtx′(s)ds
dtx′′(t) = x′(s)d^2 s
dt^2 +x′′(s)(
ds
dt) 2and solving for x′′(s)one finds x′′(s) =
x′′(t)−x′(s)ddt^2 s 2
(ds
dt