Page 373 - Advanced engineering mathematics

11.2 Velocity and Curvature 353

where
dv
a T = tangential component of the acceleration =
dt
and
2
a N = normal component of the acceleration = v(t) κ(t).
To verify this decomposition of a(t), begin with
1 1

T(t) = F (t) = v(t).

F (t) v(t)
Then
v = vT,

so
d dv
a = v = T + vT
dt dt
dv ds dT
= T + v
dt dt ds
dv 2

= T + v T (s)
dt
dv
2
= T + v κN.
dt
Because T and N are orthogonal, then
2
a = a · a = (a T T + a N N) · (a T T + a N N)
2
2
= a T · T + 2a T a N T · N + a N · N
T
N
2
2
= a + a .
N
T
This means that, whenever two of a , a T , and a N are known, we can compute the third quantity.
If a N is known, it is sometimes convenient to compute the curvature κ(t) as
a N
κ(t) = .
v 2
EXAMPLE 11.4
√
Let F(t) be as in Example 11.3. There we computed v(t) = 5t. Therefore,
dv √
a T = = 5.
dt
The acceleration is
a = v = F (t) =[cos(t) − t sin(t)]i +[sin(t) + t cos(t)]j + 2k.

Then
√
2
a = 5 + t .
so
2
2
2
2
2
a = a −a = 5 + t − 5 = t .
N T
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October 15, 2010 16:11 THM/NEIL Page-353 27410_11_ch11_p343-366