Multivariate Calculus - Bruce E. Shapiro

42 LECTURE 5. VELOCITY, ACCELERATION, AND CURVATURE
Example 5.5 Find the curvature and radius of curvature of the curve r = (5 cos t, 5 sin t, 6t).
Solution. The velocity is
hence the speed is
The unit tangent vector is then
The curvature is
κ(t) =
v(t) = (−5 sin t, 5 cos t, 6)
‖v(t)‖ = √ (−5 sin t) 2 + (5 cos t) 2 + 6
√
2
= 25(sin 2 t + cos 2 t) + 36
ˆT(t) =
= √ 25 + 36 = √ 61
v(t)
‖v(t)‖ = √ 1 (−5 sin t, 5 cos t, 6)
61
1
‖v(t)‖
∥
dT
dt
∥
= 1 ‖(−5 cos t, −5 sin t, 0)‖
61
=
61√ 1 (−5 cos t) 2 + (−5 sin t) 2
= 1 √
25(sin 2 t + cos
61
2 t) = 5 61 ≈ 0.08197
The corresponding radius of curvature is R = 1/κ = 61/5.
The Acceleration Vector
Theorem 5.2 The vector dT / ds is normal to the curve.
Proof. By the product rule for derivatives,
( )
d
dT dT
(T · T) = T ·
ds ds + · T = 2T · dT
ds
ds
But since T = v / ‖v‖ (see equation (5.12)),
T · T = v · v
‖v‖ 2 = ‖v‖2
‖v‖ 2 = 1
d
(T · T) = 0
ds
Hence
T · dT
ds = 0
which means that dT / ds is perpendicular to the tangent vector T. Thus dT / ds is
also normal to the curve.
Revised December 6, 2006. Math 250, Fall 2006