Chapter 3
Vector Functions
3.1
Vector Functions and Space Curves
In general, a function is a rule that assigns to each element in the domain an element in the range. A vector-valued function, or vector function, is simply a function whose domain is a set of real numbers and whose range is a set of vectors.
= , , ℎ( )
= + + ℎ
EXAMPLE 1
If r t =< , ln 3 − , >
Then the component functions are
= , = ln 3 − , ℎ =
By our usual convention, the domain of consists of all values of for which the expression for ( ) is defined. The expressions , ln 3 − , , are all defined when 3 − > 0 and ≥ 0. Therefore the domain of is the interval [0,3).
EXAMPLE 2 Find lim
→ , ℎ = (1 + ) + + ( ) . SOLUTION According to Definition 1, the limit of r is the vector whose components are the limits of the component functions of r:
lim→ = lim
→ (1 + ) + lim
→ + lim
→
sin ( )
lim→ = + .
EXAMPLE 4 Sketch the curve whose vector equation is
= + sin + .
SOLUTION The parametric equations for this curve are
= , = sin , =
3.2 Derivatives and Integrals of Vector Functions
Derivatives
The derivative of a vector function is defined in much the same way as for real valued functions:
Theorem If = , , ℎ( ) = + + ℎ , where , , ℎ are differentiable functions, then
′ = , , ℎ = ′ + ′ + ℎ′
Unit tangent vector ( ) = ′
′
EXAMPLE 1
(a) Find the derivative = (1 + ) + + sin 2 . (b) Find the unit tangent vector at the point where = 0.
SOLUTION
(a) We differentiate each component of r:
′ = 3 + ( − ) + 2 2 .
(b) Since 0 = and ′ 0 = + 2 . The unit tangent vector at the point (1,0,0) is
= ′ 0
′ 0 = + 2
1 + 4 = 1
5 + 2 5
Exercise 18: Find the unit tangent vector ( ) at the point with the given value of the parameter .
=< + 3 , + 1, 3 + 4 > , = 1 . SOLUTION We get ′ as follows
′ =< 3 + 3, 2 , 3 >
′ = 1 =< 6, 2, 3 >
′(1) = 36 + 4 + 9 = 49 = 7
= 1 = 1
1 = 6 + 2 + 3
7 = 6
7 + 2
7 + 3 7 .
Differentiation Rules
Theorem: Suppose and are differentiable vector functions, is a scalar, and is a real-valued function. Then
EXAMPLE 4 Show that if = ( a constant ), then ′( ) is orthogonal to ( ) for all .
SOLUTION Since
Then
Thus ′ . = 0, which says that is ′( ) orthogonal to ( ).
Integrals
The definite integral of a continuous vector function ( ) can be defined in much the same way as for real-valued functions except that the integral is a vector. But then we can express the integral of r in terms of the integrals of its component functions f, g, and h as follows.
EXAMPLE 5 If = 2 + sin + 2 . Then
= 2 + sin + 2
= 2 − + + where C is a vector constant of integration, and
= 2 − + = 2 + +
4
3.3 Arc Length and Curvature
The length of a space curve is defined in exactly the same way (see Figure 1).
Suppose that the curve has the vector equation = , , ℎ( ) , ≤
≤ , or, equivalently, the parametric equations x = , = , = ℎ , where , ℎ′ are continuous. If the curve is traversed exactly once as increases from a to b, then it can be shown that its length is
EXAMPLE 1 Find the length of the arc of the circular helix with vector equation = + sin + . from the point 1,0,0 to the point
1,0,2 .
SOLUTION Since = − + cos + , we have
′ = − + cos + 1 = 2
The arc from 1,0,0 to 1,0,2 is described by the parameter interval 0 ≤
≤ 2 , and so we have
= ′ = 2 = 2 2
Exercise 4: Find the length of the curve
= + sin + ln cos , 0 ≤ ≤ .
SOLUTION
′ = − + cos −
′ = − + cos +
= 1 + = = sec t.
Then
= ′ = sec = ln(sec + tan )
= ln 2 + 1 − ln 1 = ln 2 + 1
Curvature
= ′
′
EXAMPLE 3 Show that the curvature of a circle of radius is .
SOLUTION We can take the circle to have center the origin, and then a parametrization is
= + sin Then
= − + and
′ =
So
= = − +
= − + and
′ = − − This gives
= 1 we have
= ′
′ = 1
Theorem
The curvature of the curve given by the vector function is
=
×EXAMPLE 4 Find the curvature of the twisted cubic =< , , > at a general point and at (0,0,0).
SOLUTION We first compute the required ingredients:
′ =< 1,2 , 3 >
′′ =< 0,2,6 >
′ = 1 + 4 + 9
× = 1 2 3
0 2 6
= 6 − 6 + 2
× = 36 + 36 + 4 = 2 9 + 9 + 1
=
×=
2 9 + 9 + 11 + 4 + 9
At the origin, where = 0, the curvature is = 0 = 2.
=
′′( )1 +
For the special case of a plane curve with equation = ( ).
EXAMPLE 5 Find the curvature of the parabola = at the points 0,0 , 1,1 (2,4).
SOLUTION Since ′ = 2 , and ′′ = 2, then
=
′′1 + ′
= 2
1 + 4 The curvature
0 = 2, 1 =
,2 =
The Normal and Binormal Vectors
At a given point on a smooth space curve ( ), there are many vectors that are orthogonal to the unit tangent vector ( ).
But at any point where we can define the principal unit normal vector ( ) (or simply unit normal) as
=
The vector = × ( ) is called the binormal vector.
It is perpendicular to both and and is also a unit vector.
= , = , = ×
EXAMPLE 6 Find the unit normal and binormal vectors for the circular helix
= cos + sin + .
SOLUTION We first compute the ingredients needed for the unit normal vector:
= − sin + cos +
= sin + cos + 1 = 2
Unit tangent vector
= = 1
2 − sin + cos +
= 1
2 − cos − sin
= 1
2 cos + sin = 1
2
Normal vector
= =
1
2 − cos − sin 1
2
= − cos − sin
Binormal vector
= × = 1
2 − sin cos 1
− cos − sin 0
= 1 2
cos 1
− sin 0 − − sin 1
− cos 0 + − sin cos
− cos − sin
= 1
2 sin − cos + (sin + cos )
= 1
2 sin − cos +
