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11.2 Velocity and Curvature 349

1. [F(t) + G(t)] = F (t) + G (t).

2. (αF) (t) = αF (t).

3. [ f (t)F(t)] = f (t)F(t) + f (t)F (t).

4. [F(t) · G(t)] = F (t) · G(t) + F(t) · G (t).

5. [F(t) × G(t)] = F (t) × G(t) + F(t) × G (t).

6. [F( f (t))] = f (t)F ( f (t)).

Rules (3), (4), and (5) are product rules, reminiscent of the rule for differentiating a product
of functions of one variable. In rule (4), the order of the factors is important, since the cross
product is anti-commutative. Rule (6) is a chain rule for vector differentiation.
SECTION 11.1 PROBLEMS
2
In each of Problems 1 through 8, compute the requested 8. F(t)=−4cos(t)k,G(t)=−t i+4sin(t)k;(d/dt)[F(t)·
derivative in two ways, first by using rules (1) through (6) G(t)]
as appropriate, and second by carrying out the vector oper-
ation and then differentiating the resulting vector or scalar In each of Problems 9, 10, and 11, (a) write the position
function. vector and tangent vector for the curve whose parametric
equations are given, (b) find the length function s(t) for
2
1. F(t)=i+3t j+2tk, f (t)=4cos(3t);(d/dt)[ f (t)F(t)] the curve, (c) write the position vector as a function of s,
2 and (d) verify by differentiation that this position vector in
2. F(t)=ti−3t k,G(t)=i+cos(t)k;(d/dt)[F(t)·G(t)]
terms of s is a unit tangent to the curve.
3. F(t) = ti + j + 4k,G(t) = i − cos(t)j + tk;(d/dt)
[F(t) × G(t)]
9. x = sin(t), y = cos(t), z = 45t;0 ≤ t ≤ 2π
2
2
4. F(t) = sinh(t)j − tk,G(t) = ti + t j − t k;(d/dt) 3
10. x = y = z = t ;−1 ≤ t ≤ 1
[F(t) × G(t)]
2
2
2
11. x = 2t , y = 3t , z = 4t ;1 ≤ t ≤ 3
t 3
5. F(t)=ti−cosh(t)j+e k, f (t)=1−2t ;(d/dt)[ f (t)F(t)]
12. Suppose F(t) = x(t)i + y(t)j + z(t)k is the position
2
6. F(t) = ti − tj + t k,G(t) = sin(t)i − 4tj + vector for a particle moving along a curve in 3-space.
3
t k;(d/dt)[F(t) · G(t)]
Suppose that F×F =O. Show that the particle always
2 2 t
7. F(t) =−9i + t j + t k,G(t) = e i;(d/dt)[F(t) × G(t)] moves in the same direction.
11.2 Velocity and Curvature
Imagine a particle or object moving along a path C having the position vector F(t) = x(t)i +
y(t)j+ z(t)k,as t varies from a to b. We want to relate F to the dynamics of the particle. Assume
that the coordinate functions are twice differentiable.
Define the velocity v(t) of the particle at time t to be

v(t) = F (t).
The speed v(t) is the magnitude of the velocity:
v(t) = v(t) .

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October 15, 2010 16:11 THM/NEIL Page-349 27410_11_ch11_p343-366