Multivariate Calculus - Bruce E. Shapiro

44 LECTURE 5. VELOCITY, ACCELERATION, AND CURVATURE
Figure 5.2: Components of the acceleration.
Example 5.6 Find the normal tangent and perpendicular components of the acceleration
of the curve r(t) = (7 + 21t, 14 − 42t, 28 sin t) at t = π/3.
Solution.Differentiating r gives the velocity
v(t) = r ′ (t) = (21, −42, 28 cos t) = 7(3, −6, 4 cos t)
and the speed as
ds
dt = ‖v‖ = 7√ 45 + 16 cos 2 t
Thus the tangent component of the acceleration is
hence
The acceleration vector is
hence
a ‖ (t) = d2 s
dt = 7 d √
45 + 16 cos
dt
2 112 cos t sin t
t = −√ 45 + 16 cos 2 t
a(t) = dv
dt
a ‖ (π/3) = −4 √ 3
= (0, 0, −28 sin t)
‖a(π/3)‖ = 28 sin(π/3) = 14 √ 3
Since ‖a‖ 2 = a 2 ‖ + a2 ⊥
, the square of the normal component at t = π/3 is
a 2 ⊥
(14 = √ ) 2 (
3 − 4 √ 2
3)
= 540
and consequently
a ⊥ = √ 540 = 6 √ 15.
Revised December 6, 2006. Math 250, Fall 2006