f( a
y
+ h)
0
x
Figure 1.8.2
Tangent vector f ′( a) and tangent line L = f( a) + sf′( a)
Example 1.36. Let f( t) = (cos t,sin t, t). Then f′( t) = (−sin t,cos t,1) for all t. The tangent line L to the curve at f(2 π) = (1,0,2 π) is L = f(2 π) + s f′(2 π) = (1,0,2 π) + s(0,1,1), or in parametric form: x = 1, y = s, z = 2 π + s for −∞ < s < ∞.
1.8 Vector-Valued Functions
53
A scalar function is a real-valued function. Note that if u( t) is a scalar function and
f( t) is a vector-valued function, then their product, defined by ( u f)( t) = u( t)f( t) for all t, is a vector-valued function (since the product of a scalar with a vector is a vector).
The basic properties of derivatives of vector-valued functions are summarized in the fol-
lowing theorem.
Theorem 1.20. Let f( t) and g( t) be differentiable vector-valued functions, let u( t) be a
differentiable scalar function, let k be a scalar, and let c be a constant vector. Then
d
(a)
(c) = 0
dt
d
df
(b)
( kf) = k
dt
dt
d
df
dg
(c)
(f + g) =
+
dt
dt
dt
d
df
dg
(d)
(f − g) =
−
dt
dt
dt
d
du
df
(e)
( u f) =
f + u
dt
dt
dt
d
df
dg
(f)
(f · g) =
· g + f ·
dt
dt
dt
d
df
dg
(g)
(f × g) =
× g + f ×
dt
dt
dt
Proof: The proofs of parts (a)-(e) follow easily by differentiating the component functions
and using the rules for derivatives from single-variable calculus. We will prove part (f),
and leave the proof of part (g) as an exercise for the reader.
(f) Write f( t) = ( f ( t), f ( t), f ( t)) and g( t)
( t), g ( t), g ( t)), where the component functions
1
2
3
= ( g 1
2
3
f ( t), f ( t), f ( t), g ( t), g ( t), g ( t) are all differentiable real-valued functions. Then 1
2
3
1
2
3
d
d
(f( t) · g( t)) =
( f ( t) g ( t) + f ( t) g ( t) + f ( t) g ( t))
dt
dt
1
1
2
2
3
3
d
d
d
=
( f ( t) g ( t)) +
( f ( t) g ( t)) +
( f ( t) g ( t))
dt
1
1
dt
2
2
dt
3
3
d f
d g
d f
d g
d f
d g
=
1 ( t) g ( t) + f ( t)
1 ( t) +
2 ( t) g ( t) + f ( t)
2 ( t) +
3 ( t) g ( t) + f ( t)
3 ( t)
dt
1
1
dt
dt
2
2
dt
dt
3
3
dt
d f
d f
d f
=
1 ( t),
2 ( t),
3 ( t) · ( g ( t), g ( t), g ( t))
dt
dt
dt
1
2
3
d g
d g
d g
+ ( f ( t), f ( t), f ( t))
1 ( t),
2 ( t),
3 ( t)
1
2
3
· dt
dt
dt
df
dg
=
( t) · g( t) + f( t) ·
( t) for all t.
QED
dt
dt
54
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Example 1.37. Suppose f( t) is differentiable. Find the derivative of f( t) .
Solution: Since f( t) is a real-valued function of t, then by the Chain Rule for real-valued
d
d
functions, we know that
f( t) 2 = 2 f( t)
f( t) .
dt
dt
d
d
But f( t) 2 = f( t) · f( t), so
f( t) 2 =
(f( t) · f( t)). Hence, we have
dt
dt
d
d
2 f( t)
f( t) =
(f( t) · f( t)) = f′( t) · f( t) + f( t) · f′( t) by Theorem 1.20(f), so dt
dt
= 2f′( t) · f( t) , so if f( t) = 0 then
d
f ′( t) · f( t)
f( t) =
.
dt
f( t)
d
We know that f( t) is constant if and only if
f( t) = 0 for all t. Also, f( t) ⊥ f′( t) if and
dt
only if f ′( t) · f( t) = 0. Thus, the above example shows this important fact:
If f( t) = 0, then f( t) is constant if and only if f( t) ⊥ f′( t) for all t.
This means that if a curve lies completely on a sphere (or circle) centered at the origin, then
the tangent vector f ′( t) is always perpendicular to the position vector f( t).
cos t
sin t
− at
Example 1.38. The spherical spiral f( t) =
,
,
, for a = 0.
1 + a 2 t 2
1 + a 2 t 2
1 + a 2 t 2
Figure 1.8.3 shows the graph of the curve when a = 0.2. In the exercises, the reader will be
asked to show that this curve lies on the sphere x 2 + y 2 + z 2 = 1 and to verify directly that
f ′( t) · f( t) = 0 for all t.
1
0.8
0.6
0.4
z
0.2
0
-0.2
-0.4
-0.6
-1
-0.8
-0.8
-0.6
-0.4
-1
-0.2
-1 -0.8
0
-0.6 -0.4
0.2
x
-0.2
0
0.4
0.2
0.6
0.4
y
0.6
0.8
0.8
1
1
Figure 1.8.3
Spherical spiral with a = 0.2
1.8 Vector-Valued Functions
55
Just as in single-variable calculus, higher-order derivatives of vector-valued functions are
obtained by repeatedly differentiating the (first) derivative of the function:
d
d
dnf
d dn−1f
f ′′( t) =
f ′( t) ,
f ′′′( t) =
f ′′( t) ,
. . .
,
=
(for n = 2,3,4,...)
dt
dt
dtn
dt dtn−1
We can use vector-valued functions to represent physical quantities, such as velocity, ac-
celeration, force, momentum, etc. For example, let the real variable t represent time elapsed
from some initial time ( t = 0), and suppose that an object of constant mass m is subjected
to some force so that it moves in space, with its position ( x, y, z) at time t a function of
t. That is, x = x( t), y = y( t), z = z( t) for some real-valued functions x( t), y( t), z( t). Call r( t) = ( x( t), y( t), z( t)) the position vector of the object. We can define various physical quantities associated with the object as follows:14
position: r( t) = ( x( t), y( t), z( t))
dr
velocity: v( t) = ˙r( t) = r′( t) = dt
= ( x′( t), y′( t), z′( t))
dv
acceleration: a( t) = ˙v( t) = v′( t) = dt
d 2r
= ¨r( t) = r′′( t) = dt 2
= ( x′′( t), y′′( t), z′′( t))
momentum: p( t) = mv( t)
dp
force: F( t) = ˙p( t) = p′( t) =
(Newton’s Second Law of Motion)
dt
The magnitude v( t) of the velocity vector is called the speed of the object. Note that since
the mass m is a constant, the force equation becomes the familiar F( t) = ma( t).
Example 1.39. Let r( t) = (5cos t,3sin t,4sin t) be the position vector of an object at time t ≥ 0.
Find its (a) velocity and (b) acceleration vectors.
Solution: (a) v( t) = ˙r( t) = (−5sin t,3cos t,4cos t)
(b) a( t) = ˙v( t) = (−5cos t,−3sin t,−4sin t)
Note that r( t) =
25 cos2 t + 25sin2 t = 5 for all t, so by Example 1.37 we know that r( t) ·
˙r( t) = 0 for all t (which we can verify from part (a)). In fact, v( t) = 5 for all t also. And not
only does r( t) lie on the sphere of radius 5 centered at the origin, but perhaps not so obvious
is that it lies completely within a circle of radius 5 centered at the origin. Also, note that
a( t) = −r( t). It turns out (see Exercise 16) that whenever an object moves in a circle with
constant speed, the acceleration vector will point in the opposite direction of the position
vector (i.e. towards the center of the circle).
14We will often use the older dot notation for derivatives when physics is involved.
56
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Recall from Section 1.5 that if r , r are position vectors to distinct points then r
)
1
2
1 + t(r2 −r1
represents a line through those two points as t varies over all real numbers. That vector
sum can be written as (1 − t)r
. So the function l( t)
is a line through
1 + tr2
= (1 − t)r1 + tr2
the terminal points of r and r , and when t is restricted to the interval
1
2
[0, 1] it is the line
segment between the points, with l(0) = r and l(1)
.
1
= r2
In general, a function of the form f( t) = ( a t
, a t
, a t
) represents a line in
1
+ b 1 2 + b 2 3 + b 3
R3. A
function of the form f( t) = ( a t 2
t
, a t 2
t
, a t 2
t
) represents a (possibly
1
+ b 1 + c 1 2 + b 2 + c 2 3 + b 3 + c 3
degenerate) parabola in R3.
Example 1.40. Bézier curves are used in Computer Aided Design (CAD) to approximate
the shape of a polygonal path in space (called the Bézier polygon or control polygon). For
instance, given three points (or position vectors) b , b , b in
0
1
2
R3, define
b1( t) = (1 − t)b
0
0 + tb1
b1( t) = (1 − t)b
1
1 + tb2
b2( t) = (1 − t)b1( t) + tb1( t)
0
0
1
= (1 − t)2b0 + 2 t(1 − t)b1 + t 2b2
for all real t. For t in the interval [0, 1], we see that