Smith R., Minton R. Calculus
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CONFIRMING PAGES
11-57 SECTION 11.6
..
Surfaces in Space 743
Name Generic Equation Graph
Hyperboloid of one sheet ax
2
+ by
2
− cz
2
= 1
(a, b, c > 0)
Hyperboloid of two sheets ax
2
− by
2
− cz
2
= 1
(a, b, c > 0)
EXERCISES 11.6
WRITING EXERCISES
1. In the text, different hints were given for graphing cylinders as
opposedto quadric surfaces.Explain howto tell from the equa-
tion whether you have a cylinder, a quadric surface, a plane or
some other surface.
2. Thefirststepingraphinga quadric surfaceisidentifyingtraces.
Given the traces, explain how to tell whether you have an el-
lipsoid, elliptical cone, paraboloid or hyperboloid. (Hint: For
a paraboloid, how many traces are parabolas?)
3. Suppose you have identified that a given equation represents
a hyperboloid. Explain how to determine whether the hyper-
boloid has one sheet or two sheets.
4. Circular paraboloids have a bowl-like shape. However, the
paraboloids z = x
2
+ y
2
, z = 4 − x
2
− y
2
, y = x
2
+ z
2
and
x = y
2
+ z
2
all open up in different directions. Explain why
these paraboloids are different and how to determine in which
direction a paraboloid opens.
In exercises 1–40, sketch the appropriate traces, and then sketch
and identify the surface.
1. z = x
2
2. z = 4 − y
2
3. x
2
+
y
2
9
+
z
2
4
= 1 4.
x
2
4
+
y
2
4
+
z
2
9
= 1
5. z = 4x
2
+ 4y
2
6. z = x
2
+ 4y
2
7. z
2
= 4x
2
+ y
2
8. z
2
=
x
2
4
+
y
2
9
9. z = x
2
− y
2
10. z = y
2
− x
2
11. x
2
− y
2
+ z
2
= 1 12. x
2
+
y
2
4
− z
2
= 1
13. x
2
−
y
2
9
− z
2
= 1 14. x
2
− y
2
−
z
2
4
= 1
15. z = cos x 16. z =
x
2
+ 4y
2
17. z = 4 − x
2
− y
2
18. x = y
2
+ z
2
19. z = y
3
20. z = 4 − x
2
21. z =
x
2
+ y
2
22. z = sin y
23. y = x
2
24. x = 2 − y
2
25. y = x
2
+ z
2
26. z = 9 − x
2
− y
2
27. x
2
+ 4y
2
+ 16z
2
= 16 28. 2x − z = 4
29. 4x
2
− y − z
2
= 0 30. −x
2
− y
2
+ 9z
2
= 9
31. 4x
2
+ y
2
− z
2
= 4 32. x
2
+ 9y
2
− z
2
= 9
33. −4x
2
+ y
2
− z
2
= 4 34. x
2
− 4y
2
+ z = 0
35. x + y = 1 36. 9x
2
+ y
2
+ 9z
2
= 9
37. x
2
+ y
2
= 4 38. 9x
2
+ z
2
= 9
39. x
2
− y + z
2
= 4 40. x + y
2
+ z
2
= 2
............................................................
In exercises 41–44, sketch graphs for c −1, c 0, c 1 and
other values of c. Describe the effect that the value of c has on
the surface.
41. x
2
+ cy
2
+ z
2
= 1 42. x
2
+ cy
2
− z
2
= 1
43. x
2
+ cy
2
− z = 0 44. x
2
− y
2
+ z
2
= c
............................................................
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CONFIRMING PAGES
744 CHAPTER 11
..
Vectors and the Geometry of Space 11-58
In exercises 45 and 46, match the surfaces with
the equations −x
2
2y
2
z
2
0, −x
2
2y
2
z
2
1,
−x 2y
2
z
2
0, −x
2
2y
2
− z
2
1, −x
2
2y
2
1 and
−x
2
− 2y
2
z
2
1.
45. (a)
y
z
x
22
21
1
1.5
2
1
(b)
y
z
x
2
2
4
2
(c)
y
z
x
24
4
2
22
24
2
4
46. (a)
y
z
x
22
2
2
2
1
(b)
y
25
5
5
10
5
10
z
x
(c)
y
z
x
22
2
2
2
............................................................
Exercises 47–56 involve parametric equations for quadratic
surfaces.
47. If x = a sins cost, y = b sin s sint and z = c coss, show that
(x, y, z) lies on the ellipsoid
x
2
a
2
+
y
2
b
2
+
z
2
c
2
= 1.
48. If x = as cos t, y = bs sin t and z = s
2
, show that (x, y, z) lies
on the paraboloid z =
x
2
a
2
+
y
2
b
2
.
49. If x = as cos t, y = bs sin t and z = s, show that (x, y, z) lies
on the cone z
2
=
x
2
a
2
+
y
2
b
2
.
50. If x = a coss cosht, y = b sin s cosht and z = c sinht, show
that (x, y, z) lies on the hyperboloid of one sheet
x
2
a
2
+
y
2
b
2
−
z
2
c
2
= 1.
51. If x = a coshs, y = b sinhs cost and z = c sinhs sint, show
that (x, y, z) lies on the front half (if a > 0) or back half
(ifa < 0)ofthehyperboloidoftwosheets
x
2
a
2
−
y
2
b
2
−
z
2
c
2
= 1.
52. Show that x = a
t
|t|
coshs, y = b sinhs cost and z =
c sinh s sint, −2π ≤ t ≤ 2π, are parametric equations for
the hyperboloid of two sheets
x
2
a
2
−
y
2
b
2
−
z
2
c
2
= 1.
53. Find parametric equations as in exercises 47–52 for the
surfaces in exercises 3, 5 and 7. Use a CAS to graph the
parametric surfaces.
54. Find parametric equations as in exercises 47–52 for the
surfaces in exercises 11 and 13. Use a CAS to graph the
parametric surfaces.
55. Find parametric equations for the surface in exercise 17.
56. Find parametric equations for the surface in exercise 33.
APPLICATIONS
57. Hyperbolic paraboloids are sometimes called “saddle”graphs.
The architect of the Saddle Dome in the Canadian city of
Calgary used this shape to create an attractive and symboli-
cally meaningful structure.
One issue in using this shape is water drainage from the
roof. If the Saddle Dome roof is described by z = x
2
− y
2
,
−1 ≤ x ≤ 1, −1 ≤ y ≤ 1, in which direction would the water
drain? First, consider traces for which y is constant. Show that
the trace has a minimum at x = 0. Identify the plane x = 0in
the picture. Next, show that the trace at x = 0 has an absolute
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CONFIRMING PAGES
11-59 CHAPTER 11
..
Review Exercises 745
maximum at y = 0. Use this information to identify the two
primary points at which the water would drain.
58. Cooling towers for nuclear reactors are often constructed as
hyperboloids of one sheet because of the structural stability of
that surface. (See the accompanying photo.) Suppose all hor-
izontal cross sections are circular, with a minimum radius of
200 feet occurring at a height of 600 feet. The tower is to be
800feettallwith a maximum cross-sectionalradiusof300feet.
Findanequation forthestructure.(Hint:Theinformationgiven
does not completely determine the shape of the structure, or
the equation.)
59. You can improve the appearance of a wireframe graph by
carefully choosing the viewing window. We commented on
the curved edge in Figure 11.57c. Graph this function with
domain −5 ≤ x ≤ 5 and −5 ≤ y ≤ 5, but limit the z-range to
−1 ≤ z ≤ 20. Does this look more like Figure 11.57b?
EXPLORATORY EXERCISE
1. Golf club manufacturers use ellipsoids (called inertia ellip-
soids) to visualize important characteristics of golf clubs.
A three-dimensional coordinate system is set up as shown
in the figure. The (second) moments of inertia are then
computed for the clubhead about each coordinate axis. The
inertia ellipsoid is defined as I
xx
x
2
+ I
yy
y
2
+ I
zz
z
2
+
2I
xy
xy +2I
yz
yz + 2I
xz
xz = 1. The graph of this ellipsoid
provides important information to the club designer. For com-
parison purposes, a homogeneous spherical shell would have
a perfect sphere as its inertia ellipsoid. In Science and Golf II,
the data given here are provided for a 6-iron and driver, re-
spectively. Graph the ellipsoids and compare the shapes. For
the 6-iron, 89.4x
2
+ 195.8y
2
+ 124.9z
2
− 48.6xy − 111.8xz
+ 0.4yz = 1,000,000 and for the driver, 119.3x
2
+ 243.9y
2
+ 139.4z
2
− 1.2xy − 71.4xz − 25.8yz = 1,000,000.
x
y
z
y
x
z
Review Exercises
WRITING EXERCISES
The following list includes terms that are defined and theorems that
arestatedin this chapter. Foreach term or theorem,(1) givea precise
definition or statement, (2) state in general terms what it means and
(3) describe the types of problems with which it is associated.
Vector Scalar Magnitude
Position vector Unit vector Displacement
First octant Sphere vector
Angle between Triangle Inequality Dot product
vectors Cross product Component
Projection Parametric equations Torque
Magnus force of line Symmetric
Parallel planes Orthogonal planes equations of line
Traces Cylinder Skew lines
Circular paraboloid Hyperbolic paraboloid Ellipsoid
Hyperboloid of Hyperboloid Cone
one sheet of two sheets Saddle
TRUE OR FALSE
State whether each statement is true or false and briefly explain
why. If the statement is false, try to “fix it” by modifying the given
statement to a new statement that is true.
1. Two vectors are parallel if one vector equals the other divided
by a constant.
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CONFIRMING PAGES
746 CHAPTER 11
..
Vectors and the Geometry of Space 11-60
Review Exercises
2. For a given vector, there is one unit vector parallel to it.
3. A sphere is the set of all points at a given distance from a fixed
point.
4. The dot product a ·b = 0 implies that either a = 0 or b = 0.
5. If a ·b > 0, then the angle between a and b is less than
π
2
.
6. (a ·b) ·c = a ·(b ·c) for all vectors a, b and c.
7. (a ×b) ×c = a ×(b ×c) for all vectors a, b and c.
8. a ×b istheuniquevectorperpendiculartotheplanecontaining
a and b.
9. The cross product can be used to determine the angle between
vectors.
10. Two planes are parallel if and only if their normal vectors are
parallel.
11. The distance between parallel planes equals the distance
between any two points in the planes.
12. The equation of a hyperboloid of two sheets has two negative
signs in it.
13. In an equation of a quadric surface, if one variable is linear
and the other two are squared, then the surface is a paraboloid
wrapping around the axis corresponding to the linear variable.
In exercises 1–4, compute a
b, 4b and 2b −
−
− a.
1. a =−2, 3, b =1, 0
2. a =−1, −2, b =2, 3
3. a = 10i +2j −2k, b =−4i +3j +2k
4. a =−i −j +2k, b =−i +j −2k
............................................................
In exercises 5–8, determine whether a and b are parallel,
orthogonal or neither.
5. a =2, 3, b =4, 5
6. a = i −2j, b = 2i − j
7. a =−2, 3, 1, b =4, −6, −2
8. a = 2i −j +2k, b = 4i − 2j + k
............................................................
In exercises 9 and 10, find the displacement vector
→
PQ .
9. P = (3, 1, −2), Q = (2, −1, 1) 10. P = (3, 1), Q = (1, 4)
............................................................
In exercises 11–16, find a unit vector in the same direction as
the given vector.
11. 3, 6 12. −2, 3
13. 10i +2j −2k 14. −i −j +2k
15. from (4, 1, 2) to (1, 1, 6) 16. from (2, −1, 0) to (0, 3, −2)
............................................................
Inexercises17 and18, findthe distancebetween thegivenpoints.
17. (0, −2, 2), (3, 4, 1) 18. (3, 1, 0), (1, 4, 1)
............................................................
In exercises 19 and 20, find a vector with the given magnitude
and in the same direction as the given vector.
19. magnitude 2, v = 2i −2j +2k
20. magnitude
1
2
, v =−i −j +k
............................................................
21. The thrust of an airplane’s engine produces a speed of 500 mph
in still air. The wind velocity is given by 20, −80. In what
direction should the plane head to fly due east?
22. Two ropes are attached to a crate. The ropes exert forces of
−160, 120 and 160, 160, respectively. If the crate weighs
300 pounds, what is the net force on the crate?
In exercises 23 and 24, find an equation of the sphere with radius
r and center (a, b, c).
23. r = 6, (a, b, c) = (0, −2, 0)
24. r =
√
3, (a, b, c) = (−3, 1, 2)
............................................................
In exercises 25–28, compute a ·b.
25. a =2, −1, b =2, 4
26. a = i −2j, b = 4i + 2j
27. a = 3i +j −4k, b =−2i +2j +k
28. a = i +3j −2k, b = 2i − 3k
............................................................
In exercises 29 and 30, find the angle between the vectors.
29. 3, 2, 1 and −1, 1, 2 30. 3, 4 and 2, −1
............................................................
In exercises 31 and 32, find comp
b
a and proj
b
a.
31. a = 3i +j −4k, b = i + 2j + k
32. a = i +3j −2k, b = 2i − 3k
............................................................
In exercises 33–36, compute the cross product a ×
×
× b.
33. a =1, −2, 1, b =2, 0, 1
34. a =1, −2, 0, b =1, 0, −2
35. a = 2j +k, b = 4i + 2j − k
36. a = i −2j −3k, b = 2i − j
............................................................
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CONFIRMING PAGES
11-61 CHAPTER 11
..
Review Exercises 747
Review Exercises
In exercises 37 and 38, find two unit vectors orthogonal to both
given vectors.
37. a = 2i +k, b =−i +2j −k
38. a = 3i +j −2k, b = 2i − j
............................................................
39. A force of 40, −30 pounds moves an object in a straight line
from (1, 0) to (60, 22). Compute the work done.
40. Use vectors to find the angles in the triangle with vertices
(0, 0), (3, 1) and (1, 4).
In exercises 41 and 42, find the distance from the point Q to the
given line.
41. Q = (1, −1, 0), line
⎧
⎨
⎩
x = t + 1
y = 2t − 1
z = 3
42. Q = (0, 1, 0), line
⎧
⎨
⎩
x = 2t − 1
y = 4t
z = 3t + 2
............................................................
In exercises 43 and 44, find the indicated area or volume.
43. Area of the parallelogram with adjacent edges formed by
2, 0, 1 and 0, 1, −3
44. Volume of the parallelepiped with threeadjacent edges formed
by 1, −1, 2, 0, 0, 4 and 3, 0, 1
............................................................
45. A force of magnitude 50 pounds is applied at the end of a
6-inch-long wrench at an angle of
π
6
to the wrench. Find the
magnitude of the torque applied to the bolt.
46. Aballisstruckwithbackspin.Find the direction of the Magnus
force and describe the effect on the ball.
In exercises 47–50, find (a) parametric equations and (b) sym-
metric equations of the line.
47. The line through (2, −1, −3) and (0, 2, −3)
48. The line through (−1, 0, 2) and (−3, 0, −2)
49. The line through (2, −1, 1) and parallel to
x−1
2
= 2y =
z+2
−3
50. The line through (0, 2,1) and normal to the plane
2x − 3y + z = 4
............................................................
In exercises 51 and 52, find the angle between the lines.
51.
⎧
⎨
⎩
x = 4 +t
y = 2
z = 3 +2t
and
⎧
⎨
⎩
x = 4 +2s
y = 2 +2s
z = 3 +4s
52.
⎧
⎨
⎩
x = 3 +t
y = 3 +3t
z = 4 −t
and
⎧
⎨
⎩
x = 3 −s
y = 3 −2s
z = 4 +2s
............................................................
In exercises 53 and 54, determine whether the lines are parallel,
skew or intersect.
53.
⎧
⎨
⎩
x = 2t
y = 3 +t
z =−1 +4t
and
⎧
⎨
⎩
x = 4
y = 4 +s
z = 3 +s
54.
⎧
⎨
⎩
x = 1 −t
y = 2t
z = 5 −t
and
⎧
⎨
⎩
x = 3 +3s
y = 2
z = 1 −3s
............................................................
In exercises 55–58, find an equation of the given plane.
55. The plane containing the point (−5, 0, 1) with normal vector
4, 1, −2
56. The plane containing the point (2, −1, 2) with normal vector
3, −1, 0
57. The plane containing the points (2, 1, 3), (2, −1, 2) and (3, 3, 2)
58. The plane containing the points (2, −1, 2), (1, −1, 4) and
(3, −1, 2)
............................................................
In exercises 59–72, sketch and identify the surface.
59. 9x
2
+ y
2
+ z = 9 60. x
2
+ y + z
2
= 1
61. y
2
+ z
2
= 1 62. x
2
+ 4y
2
= 4
63. x
2
− 2x + y
2
+ z
2
= 3 64. x
2
+ (y + 2)
2
+ z
2
= 6
65. y = 2 66. z = 5
67. 2x − y + z = 4 68. 3x + 2y − z = 6
69. x
2
− y
2
+ 4z
2
= 4 70. x
2
− y
2
− z = 1
71. x
2
− y
2
− 4z
2
= 4 72. x
2
+ y
2
− z = 1
............................................................
73. Use the Cauchy-Schwartz Inequality to show that if a
k
≥ 0,
then
n
k=1
√
a
k
k
p
≤
n
k=1
a
k
n
k=1
1
k
2p
.
74. Show that if a
k
≥ 0, p >
1
2
and
∞
k=1
a
k
converges, then
∞
k=1
√
a
k
k
p
converges.
EXPLORATORY EXERCISES
1. Suppose that a piece of pottery is made in the shape of
z = 4 − x
2
− y
2
for z ≥ 0. A light source is placed at
(2, 2, 100). Draw a sketch showing the pottery and the light
source. Based on this picture, which parts of the pottery would
be brightly lit and which parts would be poorly lit? This can
be quantified, as follows. For several points of your choice
on the pottery, find the vector a that connects the point to the
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CONFIRMING PAGES
748 CHAPTER 11
..
Vectors and the Geometry of Space 11-62
Review Exercises
light source and the normal vector n to the tangent plane at
that point, and then find the angle between the vectors. For
points with relatively large angles, are the points well lit or
poorly lit? Develop a rule for using the angle between vectors
to determine the lighting level. Find the point that is best lit
and the point that is most poorly lit.
2. As we focus on three-dimensional geometry throughout
the balance of the book, some projections will be difficult
but important to visualize. In this exercise, we contrast the
curves C
1
and C
2
defined parametrically by
⎧
⎨
⎩
x = cost
y = cos t
z = sin t
and
⎧
⎨
⎩
x = cost
y = cos t
z =
√
2sint
, respectively. If you have access to three-
dimensional graphics, try sketching each curve from a variety
of perspectives. Our question will be whether either curve is a
circle. For both curves, note that x = y. Describe in words and
sketchagraphoftheplane x = y.Next,notethattheprojection
of C
1
back into the yz-plane is a circle (y = cost, z = sint). If
C
1
is actually a circle in the plane x = y, discuss what its pro-
jection (shadow) in the yz-plane would look like. Given this,
explain whether C
1
is actually a circle or an ellipse. Compare
yourdescriptionoftheprojectionofacircleintothe yz-planeto
theprojection of C
2
intothe yz-plane.Tomakethis more quan-
titative, we can use the general rule that for a two-dimensional
region, the area of its projection onto a plane equals the area
of the region multiplied by cos θ , where θ is the angle between
the plane in which the region lies and the plane into which it is
beingprojected.Giventhis,computetheradiusofthecircleC
2
.
3. In the accompanying figure, the circle x
2
+ y
2
= r
2
is shown.
In this exercise, we will compute the time required for an
u
(x, y)
object to travel the length of a chord from the top of the circle
to another point on the circle at an angle of θ from the vertical,
assuming that gravity (acting downward) is the only force.
From our study of projectile motion in section 5.5, recall
that an object traveling with a constant acceleration a covers
a distance d in time
2d
a
. Show that the component of grav-
ity in the direction of the chord is a = g cosθ. If the chord
ends at the point (x, y), show that the length of the chord is
d =
2r
2
− 2ry. Also, show that cos θ =
r−y
d
. Putting this all
together, compute the time it takes to travel the chord. Explain
why it’s surprising that the answer does not depend on the
value of θ. Note that as θ increases, the distance d decreases
but the effectiveness of gravity decreases. Discuss the balance
between these two factors.
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CONFIRMING PAGES
CHAPTER
12
Vector-Valued Functions
RoboCup is the international championship of robot soccer, where the
robots are engineered and programmed to respond automatically to the
positions of the ball, goal and other players. One of the major challenges
for engineers is that the robots are completely on their own to analyze
the field of play and use teamwork to outmaneuver their opponents and
score goals.
In the small size category, the robots’ “vision” is provided by over-
head cameras, with information relayed wirelessly to the robots. With
the formidable sight problem resolved, the focus is on effective move-
ment of the robots and on providing artificial intelligence for teamwork.
The remarkable abilities of these robots are demonstrated by the sequence of
frames shown below, where a robot on the Cornell Big Red team of 2001 hits a
wide-open teammate with a perfect pass leading to a goal.
There is a considerable amount of mathematics behind this play. To determine
whether or not a teammate is truly open, a robot needs to take into account the
positions and velocities of each robot, since opponents and teammates are all in
motion and could move into the way by the time a pass is executed. Both position
and velocity can be described usingvectors, using adifferent vector for each time.
In this chapter, we introduce vector-valued functions, which assign a vector to
each value of the time variable. The calculus introduced in this chapter is essential
background knowledge for the programmers of RoboCup, as well as for the rest
of this book.
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..
Vector-Valued Functions 12-2
12.1 VECTOR-VALUED FUNCTIONS
Forthe circuitous path of the airplane indicated in Figure 12.1a, it turns out to be convenient
to describe the airplane’s location at any giventime by the endpoint of a vectorwhose initial
point is located at the origin (a position vector). (See Figure 12.1b for vectors indicating
the location of the plane at a number of times.) Notice that a function that gives us a vector
in V
3
for each time t would do the job nicely. This is the concept of a vector-valuedfunction,
which we define more precisely in Definition 1.1.
DEFINITION 1.1
A vector-valued function r(t) is a mapping from its domain D ⊂ R to its range
R ⊂ V
3
, so that for each t in D, r(t) = v for exactly one vector v ∈ R.Wecan
always write a vector-valued function as
r(t) = f (t)i + g(t)j + h(t)k, (1.1)
for some scalar functions f, g and h (called the component functions of r).
For each t, we regard r(t) as a position vector. The endpoint of r(t) then can be viewed
as tracing out a curve, as illustrated in Figure 12.1b. Observe that for r(t) as defined
in (1.1), this curve is the same as that described by the parametric equations x = f (t),
y = g(t) and z = h(t). In three dimensions, such a curve is referred to as a space curve.
y
x
z
t
1
t
2
t
3
t
4
O
FIGURE 12.1a FIGURE 12.1b
Airplane’s flight path Vectors indicating plane’s position
at several times
We can likewise define a vector-valued function r(t)inV
2
by
r(t) = f (t)i + g(t)j,
for some scalar functions f and g.
REMARK 1.1
Although any variable would do, we routinely use the variable t to represent the
independent variable for vector-valued functions, since in many applications t
represents time.
EXAMPLE 1.1 Sketching the Curve Defined
by a Vector-Valued Function
Sketch a graph of the curve traced out by the endpoint of the two-dimensional
vector-valued function
r(t) = (t + 1)i +(t
2
− 2)j.
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Vector-Valued Functions 751
Solution Substituting some values for t,wehaver(0) = i − 2j =1, −2,
r(2) = 3i + 2j =3, 2 and r(−2) =−1, 2. We plot these in Figure 12.2a. The
endpoints of all position vectors r(t) lie on the curve C, described parametrically by
C: x = t +1, y = t
2
− 2, t ∈ R.
We can eliminate the parameter by solving for t in terms of x:
t = x − 1.
The curve is then given by
y = t
2
− 2 = (x −1)
2
− 2.
Notice that the graph of this is a parabola opening up, with vertex at the point (1, −2),
as seen in Figure 12.2b. The small arrows marked on the graph indicate the orientation,
that is, the direction of increasing values of t. If the curve describes the path of an
object, then the orientation indicates the direction in which the object traverses the path.
In this case, we can easily determine the orientation from the parametric representation
of the curve. Since x = t + 1, observe that x increases as t increases.
y
x
1
O
1
1
1
2
1, 2
3, 2
1, 2
2
23
r(2)
r(0)
r(2)
FIGURE 12.2a
Some values of
r(t) = (t + 1)i +(t
2
− 2)j
y
x
1
O
1
1
1
2
1, 2
3, 2
1, 2
2
23
r(2)
r(0)
r(2)
FIGURE 12.2b
Curve defined by
r(t) = (t + 1)i +(t
2
− 2)j
You may recall from your experience with parametric equations in Chapter 10 that
eliminating the parameter from the parametric representation of a curve is not always so
easy as it was in example 1.1. We illustrate this in example 1.2.
EXAMPLE 1.2 A Vector-Valued Function Defining an Ellipse
Sketch a graph of the curve traced out by the endpoint of the vector-valued function
r(t) = 4costi −3sintj, t ∈ R.
Solution In this case, the curve can be written parametrically as
x = 4cost, y =−3sint, t ∈ R.
Instead of solving for the parameter t, it often helps to look for some relationship
between the variables. Here,
x
4
2
+
y
3
2
= cos
2
t +sin
2
t = 1
or
x
4
2
+
y
3
2
= 1,
which is the equation of an ellipse. (See Figure 12.3.) To determine the orientation of
the curve here, you’ll need to look carefully at both parametric equations. First, fix a
starting place on the curve, for convenience, say (4, 0). This corresponds to
t = 0, ±2π, ±4π,....As t increases, notice that cos t (and hence, x) decreases initially,
while sin t increases, so that y =−3sin t decreases (initially). With both x and y
decreasing initially, we get the clockwise orientation indicated in Figure 12.3.
x
y
44
3
3
FIGURE 12.3
Curve defined by
r(t) = 4 cos ti −3 sin tj
Just as the endpoint of a vector-valued function in two dimensions traces out a curve, if
we were to plot the value of r(t) = f (t)i + g(t)j + h(t)k for every value of t, the endpoints
of the vectors would trace out a curve in three dimensions.
EXAMPLE 1.3 A Vector-Valued Function Defining an Elliptical Helix
Plot the curve traced out by the vector-valued function r(t) = sinti − 3costj +2tk,
t ≥ 0.
Solution The curve is given parametrically by
x = sint, y =−3cost, z = 2t, t ≥ 0.
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While most curves in three dimensions are difficult to recognize, you should notice that
there is a relationship between x and y here, namely,
x
2
+
y
3
2
= sin
2
t +cos
2
t = 1. (1.2)
In two dimensions, this is the equation of an ellipse. In three dimensions, since the
equation does not involve z, (1.2) is the equation of an elliptic cylinder whose axis is the
z-axis. This says that every point on the curve defined by r(t) lies on this cylinder. From
the parametric equations for x and y (in two dimensions), the ellipse is traversed in the
counterclockwise direction. This says that the curve will wrap itself around the cylinder
(counterclockwise, as you look down the positive z-axis toward the origin), as t
increases. Finally, since z = 2t, z will increase as t increases and so, the curve will wind
its way up the cylinder, as t increases. We show the curve and the elliptical cylinder in
Figure 12.4a. We call this curve an elliptical helix. In Figure 12.4b, we display a
computer-generated graph of the same helix. There, rather than the usual x-, y- and
z-axes, we show a framed graph, where the values of x, y and z are indicated on three
adjacent edges of a box containing the graph.
x
y
z
FIGURE 12.4a
Elliptical helix:
r(t) = sin ti − 3cos tj + 2tk
⫺2
0
2
0
10
20
30
⫺1
0
1
10
20
30
y
x
z
FIGURE 12.4b
Computer sketch:
r(t) = sin ti − 3cos tj + 2tk
Wecanusevector-valuedfunctionsasaconvenientrepresentationofsomeveryfamiliar
curves, as we see in example 1.4.
EXAMPLE 1.4 A Vector-Valued Function Defining a Line
Plot the curve traced out by the vector-valued function
r(t) =3 +2t, 5 − 3t, 2 −4t, t ∈ R.
Solution Notice that the curve is given parametrically by
x = 3 +2t, y = 5 −3t, z = 2 −4t, t ∈ R.
You should recognize these equations as parametric equations for the straight line
parallel to the vector 2, −3, −4 and passing through the point (3, 5, 2), as seen in
Figure 12.5, where we also note the orientation.
y
z
x
5
5
5
O
(3, 5, 2)
FIGURE 12.5
Straight line:
r(t) =3 +2t, 5 − 3t, 2 −4t
Most three-dimensional graphs are very challenging to sketch by hand. Although you
may want to use computer-generatedgraphics for most sketches, you will need to be knowl-
edgeable enough to know how to adjust such a graph to uncover any hidden features. You
should be able to draw several basic curves by hand, like those in examples 1.3 and 1.4.
More importantly, you should be able to recognize the effects various components have on
the graph of a three-dimensional curve.In example 1.5, we walk you through matching four
vector-valued functions with their computer-generated graphs.
EXAMPLE 1.5 Matching a Vector-Valued Function to Its Graph
Match each of the vector-valued functions f
1
(t) =cos t, lnt, sint,
f
2
(t) =t cost, t sint, t, f
3
(t) =3 sin 2t, t, t and f
4
(t) =5 sin
3
t, 5cos
3
t, t with the
corresponding computer-generated graph (on the following page).
Solution First, realize that there is no single, correct procedure for solving this
problem. Look for familiar functions and match them with familiar graphical properties.
From example 1.3, recall that certain combinations of sines and cosines will produce
curves that lie on a cylinder. Notice that for the function f
1
(t), x = cost and z = sint,
so that
x
2
+ z
2
= cos
2
t +sin
2
t = 1.
This says that every point on the curve lies on the cylinder x
2
+ z
2
= 1 (the right
circular cylinder of radius 1 whose axis is the y-axis). Further, the function y = lnt
tends rapidly to −∞as t → 0
+
and increases slowly as t increases beyond t = 1. Notice
that the curve in Graph B appears to lie on a right circular cylinder and that the spirals