Example 6-18.
Consider the space curve defined by the po-
sition vector
r =r (t) = cos teˆ 1 + sin tˆe 2 +tˆe 3.This curve sweeps out a spiral called a helix^5.
The projection of the position vector r on the
plane z= 0 generates a circle with unit radius
about the origin.
The first and second derivatives of the position vector with respect to the pa-
rameter tare
dr
dt =−sin teˆ^1 + cos tˆe^2 +ˆe^3
d^2 r
dt^2=−cos tˆe 1 −sin tˆe 2.The vector drdt is tangent to the curve at the point (cos t,sin t, t)for any fixed value of
the parameter t.
Tangent Vector to Curve
Let sdenote the distance along a curve mea-
sured from some fixed point on the curve and let
the position vector of a point P on the curve be
represented as a function of this distance. If the
position vector is given by
r =r (s) = x(s)ˆe 1 +y(s)ˆe 2 +z(s)ˆe 3then the derivative with respect to arc length s
is defined
dr
ds = lim∆s→ 0∆r
∆s= lim∆s→ 0r (s+ ∆ s)−r (s)
∆sThis limiting statement can be interpreted by the illustration above with the vector
r (s)pointing to some point P and the vector r (s+ ∆ s)pointing to some near point
(^5) The given equation sweeps out a right-handed helix. Can you determine the equation for a left-handed helix?