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. We are most inter- ested in vector functions whose values are three-dimensional vectors. This means that for every number in the domain of there is a unique vector in denoted by . If , , and are the components of the vector , then , , and are real-valued functions called the component functions of and we can write
We use the letter to denote the independent variable because it represents time in most applications of vector functions.
EXAMPLE 1 If
then the component functions are
By our usual convention, the domain of consists of all values of for which the expression for is defined. The expressions , , and are all defined when and . Therefore, the domain of is the interval . ■
The limit of a vector function is defined by taking the limits of its component functions as follows.
If , then
provided the limits of the component functions exist.
Equivalently, we could have used an definition (see Exercise 68). Limits of vec- tor functions obey the same rules as limits of real-valued functions (see Exercise 67).
EXAMPLE 2 Find , where .
SOLUTION According to Definition 1, the limit of r is the vector whose components are the limits of the component functions of r:
(by Equation 1.4.5) ■
i k
limtl 0rt
limtl 01 t3
i
limtl 0tet
jlimtl 0 sin tt k
rt 1 t3 i tetj sin t t k limtl 0rt
-
limtl art
limtl aft, limtl att, limtl ahtrt f t, tt, ht
1
r
0, 3
r t 0
3 t 0rt r t3 ln3 t stt
ht st tt ln3 t
ft t3
rt
t3, ln3 t, stt
rt f t, tt, ht f t i tt j ht k r
t h f rt
ht
tt
ft
rt t r V3
r
10.7
■ If , this definition is
equivalent to saying that the length and direction of the vector approach the length and direction of the vector .L
rt
limtlart L
A vector function is continuous at a if
In view of Definition 1, we see that is continuous at if and only if its component functions , , and are continuous at .
There is a close connection between continuous vector functions and space curves.
Suppose that , , and are continuous real-valued functions on an interval . Then the set of all points in space, where
and varies throughout the interval , is called a space curve. The equations in (2) are called parametric equations of C and is called a parameter. We can think of as being traced out by a moving particle whose position at time is . If we now consider the vector function , then is the position vector of the point on . Thus any continuous vector function defines a space curve that is traced out by the tip of the moving vector , as shown in Figure 1.
EXAMPLE 3 Describe the curve defined by the vector function
SOLUTION The corresponding parametric equations are
which we recognize from Equations 10.5.2 as parametric equations of a line passing through the point and parallel to the vector . Alternatively, we could observe that the function can be written as , where
and , and this is the vector equation of a line as given by Equation 10.5.1.
■ Plane curves can also be represented in vector notation. For instance, the curve given by the parametric equations and (see Example 1 in Sec- tion 9.1) could also be described by the vector equation
where and .
EXAMPLE 4 Sketch the curve whose vector equation is
SOLUTION The parametric equations for this curve are
Since , the curve must lie on the circular cylinder . The point lies directly above the point , which moves counterclockwise around the circle in the xy-plane. (See Example 2 in Section 9.1.) Since , the curve spirals upward around the cylinder as increases. The curve, shown in Figure 2, is called a helix. ■
z tx, y, z x2 y2 1 x, y, 0 t x2 yx22 1 y2 cos2t sin2t 1
z t y sin t
x cos t
rt cos t i sin t j t k
V
j 0, 1
i 1, 0
rt t2 2t, t 1 t2 2t i t 1 j y t 1 x t2 2t
v 1, 5, 61, 2, 1 r r01, 5, 6 tv r0 1, 2, 1
z 1 6t y 2 5t
x 1 t
rt 1 t, 2 5t, 1 6t
V
rt
C
r C
P
(
ft, tt, ht)
rt f t, tt, ht t r(
tft, tt, ht)
C t
I t
z ht
y tt
x f t
2
x, y, z
C
I t h
f
a t h
f
a r
limtl art ra
r
FIGURE 1
C is traced out by the tip of a moving position vector r(t).
C
0 z
x y
P { f(t), g(t), h(t)}
r(t)=kf(t), g(t), h(t)l
Visual 10.7A shows several curves being traced out by position vectors, including those in Figures 1 and 2.
FIGURE 2
”0, 1, ’π2
(1, 0, 0) z
x
y
The corkscrew shape of the helix in Example 4 is familiar from its occurrence in coiled springs. It also occurs in the model of DNA (deoxyribonucleic acid, the genetic material of living cells). In 1953 James Watson and Francis Crick showed that the structure of the DNA molecule is that of two linked, parallel helixes that are inter- twined as in Figure 3.
In Examples 3 and 4 we were given vector equations of curves and asked for a geo- metric description or sketch. In the next two examples we are given a geometric descrip- tion of a curve and are asked to find parametric equations for the curve.
EXAMPLE 5 Find a vector equation and parametric equations for the line segment that joins the point to the point .
SOLUTION In Section 10.5 we found a vector equation for the line segment that joins the tip of the vector to the tip of the vector :
(See Equation 10.5.4.) Here we take and to obtain
a vector equation of the line segment from to :
or
The corresponding parametric equations are
■ EXAMPLE 6 Find a vector function that represents the curve of intersection of
the cylinder and the plane .
SOLUTION Figure 5 shows how the plane and the cylinder intersect, and Figure 6 shows the curve of intersection C, which is an ellipse.
FIGURE 5 FIGURE 6
C (0, _1, 3)
(1, 0, 2)
(_1, 0, 2)
(0, 1, 1) y+z=2
≈+¥=1
z
y 0
x z
x y
y z 2 x2 y2 1
V
0 t 1 z 2 5t
y 3 4t x 1 t
0 t 1 rt 1 t, 3 4t, 2 5t
0 t 1 rt 1 t1, 3, 2 t2, 1, 3
Q P
r1 2, 1, 3
r0 1, 3, 2
0 t 1 rt 1 tr0 tr1
r1
r0
Q2, 1, 3
P1, 3, 2
FIGURE 3
■ Figure 4 shows the line segment in Example 5.
PQ
FIGURE 4 Q(2, _1, 3)
P(1, 3, _2) z
x y
The projection of C onto the xy-plane is the circle . So we know from Example 2 in Section 9.1 that we can write
From the equation of the plane, we have
So we can write parametric equations for C as
The corresponding vector equation is
This equation is called a parametrization of the curve C. The arrows in Figure 6 indicate the direction in which C is traced as the parameter t increases. ■ USING COMPUTERS TO DRAW SPACE CURVES
Space curves are inherently more difficult to draw by hand than plane curves; for an accurate representation we need to use technology. For instance, Figure 7 shows a computer-generated graph of the curve with parametric equations
It’s called a toroidal spiral because it lies on a torus. Another interesting curve, the trefoil knot, with equations
is graphed in Figure 8. It wouldn’t be easy to plot either of these curves by hand.
Even when a computer is used to draw a space curve, optical illusions make it dif- ficult to get a good impression of what the curve really looks like. (This is especially true in Figure 8.) The next example shows how to cope with this problem.
EXAMPLE 7 Use a computer to draw the curve with vector equation This curve is called a twisted cubic.
SOLUTION We start by using the computer to plot the curve with parametric equa- tions , , for . The result is shown in Figure 9(a), but it’s hard to see the true nature of the curve from that graph alone. Most three-
2 t 2 z t3
y t2 x t
rt t, t2, t3.
FIGURE 7 A toroidal spiral FIGURE 8 A trefoil knot z
x
y
z
x y
z sin 1.5t y 2 cos 1.5t sin t
x 2 cos 1.5t cos t
z cos 20t y 4 sin 20t sin t
x 4 sin 20t cos t
0 t 2
rt cos t i sin t j 2 sin t k
0 t 2
z 2 sin t y sin t
x cos t
z 2 y 2 sin t
0 t 2
y sin t x cos t
x2 y2 1, z 0
dimensional computer graphing programs allow the user to enclose a curve or sur- face in a box instead of displaying the coordinate axes. When we look at the same curve in a box in Figure 9(b), we have a much clearer picture of the curve. We can see that it climbs from a lower corner of the box to the upper corner nearest us, and it twists as it climbs. We get an even better idea of the curve when we view it from different vantage points. Part (c) shows the result of rotating the box to give another viewpoint.
■
DERIVATIVES
The derivative of a vector function is defined in much the same way as for real- valued functions:
if this limit exists. The geometric significance of this definition is shown in Figure 10.
If the points and have position vectors and , then PQlrepresents the vector , which can therefore be regarded as a secant vector. If ,
the scalar multiple has the same direction as .
As , it appears that this vector approaches a vector that lies on the tangent line.
For this reason, the vector is called the tangent vector to the curve defined by at the point , provided that exists and . The tangent line to at is defined to be the line through parallel to the tangent vector . We will also have occasion to consider the unit tangent vector, which is
(b) The tangent vector (a) The secant vector
0 P
C
Q r(t+h)-r(t)
r(t) r(t+h)
r(t+h)-r(t) h
0 C
P Q
r(t+h) r(t)
rª(t)
y z
x x
z
y FIGURE 10
Tt rt
rtrt
P
P C rt 0
rt
P
r rt
hl 0rt h rtP Q1hrt h rt rt rt h rt h rth 0 dr
dt rt limhl 0 rt h rt
3 h
r r
FIGURE 9
Views of the twisted cubic x
z
y 2
_2
2 4 6
_6
4 2 0
2 0 _6
_2 6
0
y z
x
_6 6
z 0
2 4
0 2 0
_2
y x
(a) (b) (c)
In Visual 10.7B you can rotate the box in Figure 9 to see the curve from any viewpoint.
Visual 10.7C shows an anima- tion of Figure 10.
The following theorem gives us a convenient method for computing the derivative of a vector function : just differentiate each component of .
THEOREM If , where ,
, and are differentiable functions, then
PROOF
■
EXAMPLE 8
(a) Find the derivative of .
(b) Find the unit tangent vector at the point where . SOLUTION
(a) According to Theorem 4, we differentiate each component of r:
(b) Since and , the unit tangent vector at the point is
■
EXAMPLE 9 For the curve , find and sketch the position vector and the tangent vector .
SOLUTION We have
The curve is a plane curve and elimination of the parameter from the equations , gives , . In Figure 11 we draw the position vec- tor starting at the origin and the tangent vector starting at the corre- sponding pointr1 i jy 2 t1, 1. y 2 x2 x 0 r1 ■ x st
r1 1 2 i j and
rt 1 2st i j
r1
r1 rt st i 2 t j rt
T0 r0
r0 s1 4j 2k s51 j s52 k1, 0, 0
r0 j 2k r0 i
rt 3t2i 1 tetj 2 cos 2t k t 0 rt 1 t3 i tetj sin 2t k
V
f t, tt, ht
t l 0lim ft t f tt , lim
t l 0
tt t tt
t , lim
t l 0
ht t ht
t
lim
t l 0
ft t f tt , tt t tt
t ,ht t ht
t
lim
t l 0
1
t f t t, tt t, ht t f t, tt, ht
rt lim
t l 0
1
t rt t rt
rt f t, tt, ht f t i tt j ht k
t4 h rt f t, tt, ht f t i tt j ht k f r
r
r(1) rª(1) (1, 1)
FIGURE 11 0 y 2
1 x
EXAMPLE 10 Find parametric equations for the tangent line to the helix with parametric equations
at the point .
SOLUTION The vector equation of the helix is , so
The parameter value corresponding to the point is , so the tan- gent vector there is . The tangent line is the line through
parallel to the vector , so by Equations 10.5.2 its parametric equations are
■
Just as for real-valued functions, the second derivative of a vector function r is the derivative of , that is, . For instance, the second derivative of the function in Example 10 is
A curve given by a vector function on an interval is called smooth if is continuous and (except possibly at any endpoints of ). For instance, the helix in Example 10 is smooth because is never 0.
EXAMPLE 11 Determine whether the semicubical parabola is smooth.
SOLUTION Since
we have and, therefore, the curve is not smooth. The point that corresponds to is (1, 0), and we see from the graph in Figure 13 that there is a sharp corner, called a cusp, at (1, 0). Any curve with this type of behavior—an
abrupt change in direction—is not smooth. ■
A curve, such as the semicubical parabola, that is made up of a finite number of smooth pieces is called piecewise smooth.
t 0 r0 0, 0 0
rt 3t2, 2t
rt 1 t3, t2
rt I
rt 0 rt I r
rt 2 cos t, sin t, 0
r r
r
FIGURE 12
z
0 12
0 1
_1 2
0 _2
y x
8 4
_0.5 0.5
z 2 t y 1
x 2t
2, 0, 1
0, 1, 2 r2 2, 0, 1 0, 1, 2 t2 rt 2 sin t, cos t, 1
rt 2 cos t, sin t, t
0, 1, 2
z t y sin t
x 2 cos t
V
■ The helix and the tangent line in Example 10 are shown in Figure 12.
y
1 x 0
cusp
FIGURE 13
The curve r(t)=k1+t # , t@l is not smooth.
■ In Section 10.9 we will see how and can be interpreted as the veloc- ity and acceleration vectors of a particle moving through space with position vector at rt time t.
rt
rt
