Section 12.4 The Cross Product 819
So the associative law for multiplication does not usually hold; that is, in general,
sa 3 bd 3 c ± a 3 sb 3 cd
However, some of the usual laws of algebra do hold for cross products. The following
theo rem summarizes the properties of vector products.
11 Properties of the Cross Product If a, b, and c are vectors and c is a
scalar, then
1. a 3 b − 2b 3 a
2. scad 3 b − csa 3 bd − a 3 scbd
3. a 3 sb 1 cd − a 3 b 1 a 3 c
4. sa 1 bd 3 c − a 3 c 1 b 3 c
5. a ? sb 3 cd − sa 3 bd ? c
6. a 3 sb 3 cd − sa ? cdb 2 sa ? bdc
These properties can be proved by writing the vectors in terms of their components
and using the definition of a cross product. We give the proof of Property 5 and leave the
remaining proofs as exercises.
Proof of Property 5 If a − k a1, a2, a3l, b − k b1, b2, b3l, and c − k c1, c2, c3l, then
12 a ? sb 3 cd − a1sb2c3 2 b3c2d 1 a2sb3c1 2 b1c3d 1 a3sb1c2 2 b2c1d
− a1b2c3 2 a1b3c2 1 a2b3c1 2 a2b1c3 1 a3b1c2 2 a3b2c1
− sa2b3 2 a3b2 dc1 1 sa3b1 2 a1b3 dc2 1 sa1b2 2 a2b1dc3
− sa 3 bd ? c ■
Triple Products
The product a ? sb 3 cd that occurs in Property 5 is called the scalar triple product of
the vectors a, b, and c. Notice from Equation 12 that we can write the scalar triple prod
uct as a determinant: Z Za1 a2 a3
a ? sb 3 cd − b1 b2 b3
13
c1 c2 c3
bxc b The geometric significance of the scalar triple product can be seen by considering the
h ¨a
c | || | | | | |parallelepiped determined by the vectors a, b, and c. (See Figure 3.) The area of the base
FIGURE 3 parallelogram is A − b 3 c . If is the angle between a and b 3 c, then the height h
of the parallelepiped is h − a cos . (We must use cos instead of cos in case
. y2.) Therefore the volume of the parallelepiped is
V − Ah − | b 3 c | | a | | cos | − | a ? sb 3 cd |
Thus we have proved the following formula.
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820 Chapter 12 Vectors and the Geometry of Space
14 The volume of the parallelepiped determined by the vectors a, b, and c is the
magnitude of their scalar triple product:
V − | a ? sb 3 cd |
If we use the formula in (14) and discover that the volume of the parallelepiped
determined by a, b, and c is 0, then the vectors must lie in the same plane; that is, they
are coplanar.
Example 5 Use the scalar triple product to show that the vectors a − k1, 4, 27 l,
b − k2, 21, 4l, and c − k0, 29, 18l are coplanar.
SOLUTION We use Equation 13 to compute their scalar triple product:
Z Z1 4 27
a ? sb 3 cd − 2 21 4
0 29 18
Z Z Z Z Z Z−
1 21 4 24 2 4 27 2 21
29 18 0 18 0 29
− 1s18d 2 4s36d 2 7s218d − 0
Therefore, by (14), the volume of the parallelepiped determined by a, b, and c is 0.
This means that a, b, and c are coplanar. ■
The product a 3 sb 3 cd that occurs in Property 6 is called the vector triple product
of a, b, and c. Property 6 will be used to derive Kepler’s First Law of planetary motion
in Chapter 13. Its proof is left as Exercise 50.
Torque
r ¨ The idea of a cross product occurs often in physics. In particular, we consider a force F
F acting on a rigid body at a point given by a position vector r. (For instance, if we tighten
FIGURE 4 a bolt by applying a force to a wrench as in Figure 4, we produce a turning effect.) The
torque t (relative to the origin) is defined to be the cross product of the position and
0.25 m 75° force vectors
40 N
t−r3F
and measures the tendency of the body to rotate about the origin. The direction of the
torque vector indicates the axis of rotation. According to Theorem 9, the magnitude of
the torque vector is
| t | − | r 3 F | − | r | | F | sin
| |where is the angle between the position and force vectors. Observe that the only com-
ponent of F that can cause a rotation is the one perpendicular to r, that is, F sin . The
magnitude of the torque is equal to the area of the parallelogram determined by r and F.
FIGURE 5 Example 6 A bolt is tightened by applying a 40-N force to a 0.25-m wrench as
shown in Figure 5. Find the magnitude of the torque about the center of the bolt.
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Section 12.4 The Cross Product 821
SOLUTION The magnitude of the torque vector is ■
| t | − | r 3 F | − | r | | F | sin 75° − s0.25ds40d sin 75°
− 10 sin 75° < 9.66 N ∙m
If the bolt is right-threaded, then the torque vector itself is
t − | t | n < 9.66 n
where n is a unit vector directed down into the page (by the right-hand rule).
1–7 Find the cross product a 3 b and verify that it is orthogonal to (b) Use the right-hand rule to decide whether the components
both a and b. of a 3 b are positive, negative, or 0.
z
1. a − k2, 3, 0l, b − k1, 0, 5l
b
2. a − k4, 3, 22l, b − k2, 21, 1l
x ay
3. a − 2 j 2 4 k, b − 2i 1 3 j 1 k
17. If a − k 2, 21, 3 l and b − k 4, 2, 1l, find a 3 b and b 3 a.
4. a − 3 i 1 3 j 2 3 k, b − 3 i 2 3 j 1 3 k 18. If a − k 1, 0, 1l, b − k 2, 1, 21 l , and c − k 0, 1, 3 l, show that
5. a − 1 i 1 1 j 1 1 k, b − i 1 2j 2 3k a 3 sb 3 cd ± sa 3 bd 3 c.
2 3 4 19. Find two unit vectors orthogonal to both k 3, 2, 1 l and
6. a − t i 1 cos t j 1 sin t k, b − i 2 sin t j 1 cos t k k 21, 1, 0 l.
20. Find two unit vectors orthogonal to both j 2 k and i 1 j.
7. a − k t, 1, 1ytl, b − k t 2, t 2, 1 l 21. Show that 0 3 a − 0 − a 3 0 for any vector a in V3.
22. Show that sa 3 bd ? b − 0 for all vectors a and b in V3.
8. If a − i 2 2 k and b − j 1 k, find a 3 b. Sketch a, b, and 23–26 Prove the property of cross products (Theorem 11).
a 3 b as vectors starting at the origin. 23. Property 1: a 3 b − 2b 3 a
24. Property 2: scad 3 b − csa 3 bd − a 3 scbd
9–12 Find the vector, not with determinants, but by using 25. Property 3: a 3 sb 1 cd − a 3 b 1 a 3 c
properties of cross products. 26. Property 4: sa 1 bd 3 c − a 3 c 1 b 3 c
9. si 3 jd 3 k 10. k 3 si 2 2 jd
11. s j 2 kd 3 sk 2 id 12. si 1 jd 3 si 2 jd 27. Find the area of the parallelogram with vertices As23, 0d,
Bs21, 3d, Cs5, 2d, and Ds3, 21d.
13. S tate whether each expression is meaningful. If not, explain
28. Find the area of the parallelogram with vertices Ps1, 0, 2d,
why. If so, state whether it is a vector or a scalar. Qs3, 3, 3d, Rs7, 5, 8d, and Ss5, 2, 7d.
(a) a ? sb 3 cd (b) a 3 sb ? cd
(c) a 3 sb 3 cd (d) a ? sb ? cd 29–32 (a) Find a nonzero vector orthogonal to the plane through
(e) sa ? bd 3 sc ? dd (f) sa 3 bd ? sc 3 dd the points P, Q, and R, and (b) find the area of triangle PQR.
29. Ps1, 0, 1d, Qs22, 1, 3d, Rs4, 2, 5d
| |14–15 Find u 3 v and determine whether u 3 v is directed into 30. Ps0, 0, 23d, Qs4, 2, 0d, Rs3, 3, 1d
31. Ps0, 22, 0d, Qs4, 1, 22d, Rs5, 3, 1d
the page or out of the page.
14. |v|=5 15. | v |=16
45° | u |=12 120°
| u |=4
16. T he figure shows a vector a in the xy-plane and a vector b in
| | | |the direction of k. Their lengths are a − 3 and b − 2.
| | (a) Find a 3 b .
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822 Chapter 12 Vectors and the Geometry of Space
32. Ps2, 23, 4d, Qs21, 22, 2d, Rs3, 1, 23d 41. A wrench 30 cm long lies along the positive y-axis and grips a
bolt at the origin. A force is applied in the direction k0, 3, 24 l
33–34 Find the volume of the parallelepiped determined by the at the end of the wrench. Find the magnitude of the force
vectors a, b, and c. needed to supply 100 N ∙ m of torque to the bolt.
33. a − k1, 2, 3 l, b − k21, 1, 2 l, c − k 2, 1, 4 l
34. a − i 1 j , b − j 1 k, c − i 1 j 1 k 42. Let v − 5 j and let u be a vector with length 3 that starts at
the origin and rotates in the xy -plane. Find the maximum and
35–36 Find the volume of the parallelepiped with adjacent edges minimum values of the length of the vector u 3 v. In what
PQ, PR, and PS. direction does u 3 v point?
35. Ps22, 1, 0d, Qs2, 3, 2d, Rs1, 4, 21d, Ss3, 6, 1d
36. Ps3, 0, 1d, Qs21, 2, 5d, Rs5, 1, 21d, Ss0, 4, 2d 43. If a ? b − s3 and a 3 b − k1, 2, 2 l, find the angle between a
and b.
37. Use the scalar triple product to verify that the vectors
u − i 1 5 j 2 2 k, v − 3 i 2 j, and w − 5 i 1 9 j 2 4 k 44. (a) Find all vectors v such that
are coplanar.
k1, 2, 1 l 3 v − k 3, 1, 25 l
38. Use the scalar triple product to determine whether the points
As1, 3, 2d, Bs3, 21, 6d, Cs5, 2, 0d, and Ds3, 6, 24d lie in the (b) Explain why there is no vector v such that
same plane.
k1, 2, 1 l 3 v − k3, 1, 5 l
39. A bicycle pedal is pushed by a foot with a 60-N force as
shown. The shaft of the pedal is 18 cm long. Find the magni- 45. (a) Let P be a point not on the line L that passes through the
tude of the torque about P. points Q and R. Show that the distance d from the point P
to the line L is
d− | a3b |
|a|
where a − QlR and b − QlP.
(b) Use the formula in part (a) to find the distance from
the point Ps1, 1, 1d to the line through Qs0, 6, 8d and
Rs21, 4, 7d.
60 N 70° 46. (a) Let P be a point not on the plane that passes through the
10°
points Q, R, and S. Show that the distance d from P to the
plane is | a ? sb 3 cd |
|a 3 b|
P d −
40. (a) A horizontal force of 20 lb is applied to the handle of a where a − QlR, b − QlS, and c − QlP.
gearshift lever as shown. Find the magnitude of the torque (b) Use the formula in part (a) to find the distance from the
about the pivot point P.
point Ps2, 1, 4d to the plane through the points Qs1, 0, 0d,
(b) Find the magnitude of the torque about P if the same force Rs0, 2, 0d, and Ss0, 0, 3d.
is applied at the elbow Q of the lever.
| | | | | | 47. Show that a 3 b 2 − a 2 b 2 2 sa ? bd2.
20 lb
48. If a 1 b 1 c − 0, show that
a3b−b3c−c3a
49. Prove that sa 2 bd 3 sa 1 bd − 2sa 3 bd.
2 ft 50. Prove Property 6 of cross products, that is,
a 3 sb 3 cd − sa ? cdb 2 sa ? bdc
Q 51. Use Exercise 50 to prove that
0.6 ft a 3 sb 3 cd 1 b 3 sc 3 ad 1 c 3 sa 3 bd − 0
P 0.6 ft Z Z 52. Prove that a? c b?c
1 ft sa 3 bd ? sc 3 dd − a?d b?d
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Section 12.5 Equations of Lines and Planes 823
53. Suppose that a ± 0. (These vectors occur in the study of crystallography. Vectors
(a) If a ? b − a ? c, does it follow that b − c?
(b) If a 3 b − a 3 c, does it follow that b − c? of the form n1 v1 1 n2 v2 1 n3 v3 , where each ni is an integer,
(c) If a ? b − a ? c and a 3 b − a 3 c, does it follow form a lattice for a crystal. Vectors written similarly in terms of
that b − c? k1, k2, and k3 form the reciprocal lattice.)
(a) Show that ki is perpendicular to vj if i ± j.
54. If v1, v2, and v3 are noncoplanar vectors, let (b) Show that ki ? vi − 1 for i − 1, 2, 3.
k1 − v2 3 v3 k − v3 3 v1 (c) Show that k1 ? sk2 3 k3d − v1 ? 1 v3 d .
? sv2 3 v3 ? sv2 3 v3 d sv2 3
v1 d 2 v1
k3 − v1 v1 3 v2
? sv2 3 v3 d
discovery Project The geometry of a tetrahedron
P A tetrahedron is a solid with four vertices, P, Q, R, and S, and four triangular faces, as shown in
the figure.
S
QR 1. L et v1, v2, v3, and v4 be vectors with lengths equal to the areas of the faces opposite the
vertices P, Q, R, and S, respectively, and directions perpendicular to the respective faces and
pointing outward. Show that
v1 1 v2 1 v3 1 v4 − 0
2. T he volume V of a tetrahedron is one-third the distance from a vertex to the opposite face,
times the area of that face.
(a) Find a formula for the volume of a tetrahedron in terms of the coordinates of its vertices
P, Q, R, and S.
(b) Find the volume of the tetrahedron whose vertices are Ps1, 1, 1d, Qs1, 2, 3d, Rs1, 1, 2d,
and Ss3, 21, 2d.
3. S uppose the tetrahedron in the figure has a trirectangular vertex S. (This means that the three
angles at S are all right angles.) Let A, B, and C be the areas of the three faces that meet at S,
and let D be the area of the opposite face PQR. Using the result of Problem 1, or otherwise,
show that
D 2 − A2 1 B 2 1 C 2
(This is a three-dimensional version of the Pythagorean Theorem.)
Lines
z A line in the xy-plane is determined when a point on the line and the direction of the line
P¸(x¸, y¸, z¸) (its slope or angle of inclination) are given. The equation of the line can then be written
a using the point-slope form.
L P(x, y, z) Likewise, a line L in three-dimensional space is determined when we know a point
r¸ r P0sx0, y0, z0d on L and the direction of L. In three dimensions the direction of a line is
Ov conv eniently described by a vector, so we let v be a vector parallel to L. Let Psx, y, zd be
OAL Pa¸ndanldetOrl0Pa).nIdf r
x y an arbitrary point on a be the position vectors of P0 and P (that is, they
have representations is the vector with representation PA¸P, as in Fig-
FIGURE 1
ure 1, then the Triangle Law for vector addition gives r − r0 1 a. But, since a and v are
parallel vectors, there is a scalar t such that a − tv. Thus
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824 Chapter 12 Vectors and the Geometry of Space
z 1 r − r0 1 t v
t=0 t>0 which is a vector equation of L. Each value of the parameter t gives the position vector r
t<0 of a point on L. In other words, as t varies, the line is traced out by the tip of the vector r. As
L Figure 2 indicates, positive values of t correspond to points on L that lie on one side
of P0, whereas negative values of t correspond to points that lie on the other side of P0.
r¸
If the vector v that gives the direction of the line L is written in component form as
x y v − ka, b, c l, then we have tv − kta, tb, tcl . We can also write r − k x, y, z l and
r0 − kx0, y0, z0l , so the vector equation (1) becomes
FIGURE 2
k x, y, z l − k x0 1 ta, y0 1 tb, z0 1 tc l
Two vectors are equal if and only if corresponding components are equal. Therefore we
have the three scalar equations:
x − x0 1 at y − y0 1 bt z − z0 1 ct
where t [ R. These equations are called parametric equations of the line L through the
point P0sx0, y0, z0d and parallel to the vector v − ka, b, cl. Each value of the parameter t
gives a point sx, y, zd on L.
2 Parametric equations for a line through the point sx0, y0, z0d and parallel to the
direction vector ka, b, cl are
x − x0 1 at y − y0 1 bt z − z0 1 ct
Figure 3 shows the line L in Examp le 1 Example 1
and its relation to the given point and to
the vector that gives its direction. (a) Find a vector equation and parametric equations for the line that passes through the
point s5, 1, 3d and is parallel to the vector i 1 4 j 2 2k.
z (b) Find two other points on the line.
L r¸ SOLUTION
(5, 1, 3) (a) Here r0 − k 5, 1, 3 l − 5i 1 j 1 3k and v − i 1 4 j 2 2k, so the vector equa-
tion (1) becomes
v=i+4j-2k y
r − s5i 1 j 1 3kd 1 tsi 1 4 j 2 2kd
x
or r − s5 1 td i 1 s1 1 4td j 1 s3 2 2td k
FIGURE 3 Parametric equations are
x − 5 1 t y − 1 1 4t z − 3 2 2t
(b) Choosing the parameter value t − 1 gives x − 6, y − 5, and z − 1, so s6, 5, 1d is
a point on the line. Similarly, t − 21 gives the point s4, 23, 5d.
■
The vector equation and parametric equations of a line are not unique. If we change
the point or the parameter or choose a different parallel vector, then the equations change.
For instance, if, instead of s5, 1, 3d, we choose the point s6, 5, 1d in Example 1, then the
parametric equations of the line become
x − 6 1 t y − 5 1 4t z − 1 2 2t
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Section 12.5 Equations of Lines and Planes 825
Or, if we stay with the point s5, 1, 3d but choose the parallel vector 2i 1 8j 2 4k, we
arrive at the equations
x − 5 1 2t y − 1 1 8t z − 3 2 4t
In general, if a vector v − ka, b, c l is used to describe the direction of a line L, then
the numbers a, b, and c are called direction numbers of L. Since any vector parallel to
v could also be used, we see that any three numbers proportional to a, b, and c could also
be used as a set of direction numbers for L.
Another way of describing a line L is to eliminate the parameter t from Equations 2.
If none of a, b, or c is 0, we can solve each of these equations for t:
t − x 2 x0 t − y 2 y0 t − z 2 z0
a b c
Equating the results, we obtain
3 x 2 x0 − y 2 y0 − z 2 z0
a b c
These equations are called symmetric equations of L. Notice that the numbers a, b, and
c that appear in the denominators of Equations 3 are direction numbers of L, that is,
comp onents of a vector parallel to L. If one of a, b, or c is 0, we can still eliminate t. For
instance, if a − 0, we could write the equations of L as
x − x0 y 2 y0 − z 2 z0
b c
This means that L lies in the vertical plane x − x0.
Figure 4 shows the line L in Example 2 Example 2
and the point P where it intersects the
xy-plane. (a) Find parametric equations and symmetric equations of the line that passes through
the points As2, 4, 23d and Bs3, 21, 1d.
z (b) At what point does this line intersect the xy-plane?
1 SOLUTION
B1 24 (wai)t hWreeparreesennottateixopnliAcliBtlyisgpiavreanllaelvteocttohre parallel to the line, but observe that the vector v
x P _1 y line and
L v − k3 2 2, 21 2 4, 1 2 s23d l − k1, 25, 4 l
A Thus direction numbers are a − 1, b − 25, and c − 4. Taking the point s2, 4, 23d as
P0, we see that parametric equations (2) are
FIGURE 4 x − 2 1 t y − 4 2 5t z − 23 1 4t
and symmetric equations (3) are
x 2 2 − y24 − z 1 3
1 25 4
(b) The line intersects the xy-plane when z − 0, so we put z − 0 in the symmetric
equations and obtain
x 2 2 − y24 − 3
1 25 4
This gives x − 11 and y − 14, so the line intersects the xy-plane at the point (11 , 1 , 0). ■
4 4 4
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826 Chapter 12 Vectors and the Geometry of Space
In general, the procedure of Example 2 shows that direction numbers of the line L
through the points P0sx0, y0, z0 d and P1sx1, y1, z1d are x1 2 x0, y1 2 y0, and z1 2 z0 and so
symmetr ic equations of L are
x 2 x0 − y 2 y0 − z 2 z0
x1 2 x0 y1 2 y0 z1 2 z0
Often, we need a description, not of an entire line, but of just a line segment. How, for
instance, could we describe the line segment AB in Example 2? If we put t − 0 in the
parametric equations in Example 2(a), we get the point s2, 4, 23d and if we put t − 1 we
get s3, 21, 1d. So the line segment AB is described by the parametric equations
x − 2 1 t y − 4 2 5t z − 23 1 4t 0 < t < 1
or by the corresponding vector equation
rstd − k2 1 t, 4 2 5t, 23 1 4tl 0 < t < 1
In general, we know from Equation 1 that the vector equation of a line through the (tip
of the) vector r0 in the direction of a vector v is r − r0 1 tv. If the line also passes
through (the tip of) r1, then we can take v − r1 2 r0 and so its vector equation is
r − r0 1 t sr1 2 r0d − s1 2 tdr0 1 t r1
The line segment from r0 to r1 is given by the parameter interval 0 < t < 1.
4 The line segment from r0 to r1 is given by the vector equation
rstd − s1 2 tdr0 1 t r1 0 < t < 1
The lines L1 and L 2 in Example 3, Example 3 Show that the lines L1 and L 2 with parametric equations
shown in Figure 5, are skew lines.
L1: x − 1 1 t y − 22 1 3t z−42t
z L2: x − 2s y−31s z − 23 1 4s
L¡5 L™ are skew lines; that is, they do not intersect and are not parallel (and therefore do not
lie in the same plane).
5 5 SOLUTION The lines are not parallel because the corresponding direction vectors
x 10 k 1, 3, 21 l and k 2, 1, 4 l are not parallel. (Their components are not proportional.) If L1
y and L 2 had a point of intersection, there would be values of t and s such that
_5
1 1 t − 2s
FIGURE 5 22 1 3t − 3 1 s
4 2 t − 23 1 4s
But if we solve the first two equations, we get t − 11 and s − 8 , and these values don’t
5 5
satisfy the third equation. Therefore there are no values of t and s that satisfy the three
equations, so L1 and L 2 do not intersect. Thus L1 and L 2 are skew lines. ■
Planes
Although a line in space is determined by a point and a direction, a plane in space is
more difficult to describe. A single vector parallel to a plane is not enough to convey the
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Section 12.5 Equations of Lines and Planes 827
z “direction” of the plane, but a vector perpendicular to the plane does completely specify
its direction. Thus a plane in space is determined by a point P0sx0, y0, z0d in the plane and
n a vector n that is orthogonal to the plane. This orthogonal vector n is called a normal
vector. Let Psx, y, zd be an arbitrary point in the plane, and let r0 and r be the position
P( x, y, z) vectors of P0 and P. Then the vector r 2 r0 is represented by PA¸P. (See Figure 6.) The
normal vector n is orthogonal to every vector in the given plane. In particular, n is
r r-r¸ P¸(x¸, y¸, z¸) orthogonal to r 2 r0 and so we have
0 r¸ y
x 5 n ? sr 2 r0 d − 0
FIGURE 6 which can be rewritten as
6 n ? r − n ? r0
Either Equation 5 or Equation 6 is called a vector equation of the plane.
To obtain a scalar equation for the plane, we write n − k a, b, c l, r − k x, y, z l, and
r0 − k x0, y0, z0l . Then the vector equation (5) becomes
k a, b, c l ? k x 2 x0, y 2 y0, z 2 z0l − 0
or
asx 2 x0 d 1 bs y 2 y0 d 1 csz 2 z0 d − 0
7 A scalar equation of the plane through point P0sx0, y0, z0 d with normal
vector n − k a, b, c l is
asx 2 x0 d 1 bs y 2 y0 d 1 csz 2 z0 d − 0
z Example 4 Find an equation of the plane through the point s2, 4, 21d with normal
(0, 0, 3)
vector n − k2, 3, 4 l . Find the intercepts and sketch the plane.
SOLUTION Putting a − 2, b − 3, c − 4, x0 − 2, y0 − 4, and z0 − 21 in Equation 7,
we see that an equation of the plane is
(0, 4, 0) 2sx 2 2d 1 3sy 2 4d 1 4sz 1 1d − 0
y or 2x 1 3y 1 4z − 12
(6, 0, 0) To find the x-intercept we set y − z − 0 in this equation and obtain x − 6. Similarly, the
x
y-intercept is 4 and the z-intercept is 3. This enables us to sketch the portion of the plane
FIGURE 7
that lies in the first octant (see Figure 7). ■
By collecting terms in Equation 7 as we did in Example 4, we can rewrite the equation
of a plane as
8 ax 1 by 1 cz 1 d − 0
where d − 2sax0 1 by0 1 cz0 d. Equation 8 is called a linear equation in x, y, and z.
Conversely, it can be shown that if a, b, and c are not all 0, then the linear equation (8)
represents a plane with normal vector ka, b, c l . (See Exercise 83.)
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828 Chapter 12 Vectors and the Geometry of Space
Figure 8 shows the portion of the Example 5 Find an equation of the plane that passes through the points Ps1, 3, 2d,
plane in Example 5 that is enclosed by
triangle PQR. Qs3, 21, 6d, and Rs5, 2, 0d.
SOLUTION The vectors a and b corresponding to PlQ and PlR are
z
a − k 2, 24, 4 l b − k4, 21, 22 l
Q(3, _1, 6)
Since both a and b lie in the plane, their cross product a 3 b is orthogonal to the plane
P(1, 3, 2) Z Zand can be taken as the normal vector. Thus
y ij k
n − a 3 b − 2 24 4 − 12 i 1 20 j 1 14 k
x 4 21 22
R(5, 2, 0)
With the point Ps1, 3, 2d and the normal vector n, an equation of the plane is
FIGURE 8
12sx 2 1d 1 20sy 2 3d 1 14sz 2 2d − 0
or 6x 1 10y 1 7z − 50 ■
Example 6 Find the point at which the line with parametric equations x − 2 1 3t,
y − 24t, z − 5 1 t intersects the plane 4x 1 5y 2 2z − 18.
SOLUTION We substitute the expressions for x, y, and z from the parametric equations
into the equation of the plane:
4s2 1 3td 1 5s24td 2 2s5 1 td − 18
This simplifies to 210t − 20, so t − 22. Therefore the point of intersection occurs
when the parameter value is t − 22. Then x − 2 1 3s22d − 24, y − 24s22d − 8,
z − 5 2 2 − 3 and so the point of intersection is s24, 8, 3d.
■
n™ ¨ n¡ ¨ Two planes are parallel if their normal vectors are parallel. For instance, the planes
FIGURE 9 x 1 2y 2 3z − 4 and 2x 1 4y 2 6z − 3 are parallel because their normal vectors are
n1 − k1, 2, 23 l and n2 − k 2, 4, 26 l and n2 − 2 n1. If two planes are not parallel, then
they intersect in a straight line and the angle between the two planes is defined as the
acute angle between their normal vectors (see angle in Figure 9).
Figure 10 shows the planes in Example Example 7
7 and their line of intersection L.
(a) Find the angle between the planes x 1 y 1 z − 1 and x 2 2y 1 3z − 1.
(b) Find symmetric equations for the line of intersection L of these two planes.
x+y+z=1 x-2y+3z=1 SOLUTION
(a) The normal vectors of these planes are
6 L n1 − k 1, 1, 1l n2 − k 1, 22, 3l
4 2 0x _2
2 and so, if is the angle between the planes, Corollary 12.3.6 gives
z0
_2 n1 ? n2 1s1d 1 1s22d 1 1s3d 2
_4 n1 n2 s1 1 1 1 1 s1 1 4 1 9 s42
| || |cos − − −
_2 0
y 2 S D − cos212 < 72°
s42
FIGURE 10
(b) We first need to find a point on L. For instance, we can find the point where the line
intersects the xy-plane by setting z − 0 in the equations of both planes. This gives the
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Section 12.5 Equations of Lines and Planes 829
equations x 1 y − 1 and x 2 2y − 1, whose solution is x − 1, y − 0. So the point
s1, 0, 0d lies on L.
Now we observe that, since L lies in both planes, it is perpendicular to both of the
Z Znormal vectors. Thus a vector v parallel to L is given by the cross product
Another way to find the line of inter- i jk
section is to solve the equations of the v − n1 3 n2 − 1 1 1 − 5i 2 2j 2 3k
planes for two of the variables in terms
of the third, which can be taken as the 1 22 3
parameter.
and so the symmetric equations of L can be written as
x 2 1 − y − z ■
5 22 23
2 x-1 = y Note Since a linear equation in x, y, and z represents a plane and two nonparallel
5 _2
planes intersect in a line, it follows that two linear equations can represent a line. The points
1 L sx, y, zd that satisfy both a1 x 1 b1 y 1 c1z 1 d1 − 0 and a2 x 1 b2 y 1 c2 z 1 d2 − 0 lie
on both of these planes, and so the pair of linear equations represents the line of inter-
z0 y = z
_1 2 3 section of the planes (if they are not parallel). For instance, in Example 7 the line L was
given as the line of intersection of the planes x 1 y 1 z − 1 and x 2 2y 1 3z − 1. The
symmetric equations that we found for L could be written as
_2 x 2 1 − y and y − z
5 22 22 23
_1 y 0 0 _1 _2
x
1 21 which is again a pair of linear equations. They exhibit L as the line of intersection of the
planes sx 2 1dy5 − yys22d and yys22d − zys23d. (See Figure 11.)
FIGURE 11
In general, when we write the equations of a line in the symmetric form
Figure 11 shows how the line L in
Example 7 can also be regarded as the x 2 x0 − y 2 y0 − z 2 z0
line of intersection of planes derived a b c
from its symmetric equations.
we can regard the line as the line of intersection of the two planes
x 2 x0 − y 2 y0 and y 2 y0 − z 2 z0
a b b c
Distances
Example 8 Find a formula for the distance D from a point P1sx1, y1, z1d to the
plane ax 1 by 1 cz 1 d − 0.
SOLUTION Let P0sx0, y0, z0 d be any point in the given plane and let b be the vector
corresponding to PA¸P¡. Then
b − k x1 2 x0, y1 2 y0, z1 2 z0l
P¡ From Figure 12 you can see that the distance D from P1 to the plane is equal to the
absolute value of the scalar projection of b onto the normal vector n − ka, b, c l . (See
¨
Section 12.3.) Thus | |
D n b
b D − | compn b | − ?
n | n |
| |−
P¸ asx1 2 x0 d 1 bs y1 2 y0 d 1 csz1 2 z0 d
sa 2 1 b 2 1 c 2
FIGURE 12
| |−
sax1 1 by1 1 cz1d 2 sax0 1 by0 1 cz0 d
sa 2 1 b 2 1 c 2
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830 Chapter 12 Vectors and the Geometry of Space
Since P0 lies in the plane, its coordinates satisfy the equation of the plane and so we
have ax0 1 by0 1 cz0 1 d − 0. Thus the formula for D can be written as
| |9
D− ax1 1 by1 1 cz1 1 d ■
sa 2 1 b 2 1 c 2
Example 9 Find the distance between the parallel planes 10x 1 2y 2 2z − 5 and
5x 1 y 2 z − 1.
SOLUTION First we note that the planes are parallel because their normal vectors
k10, 2, 22 l and k5, 1, 21 l are parallel. To find the distance D between the planes, we
choose any point on one plane and calculate its distance to the other plane. In par-
ticular, if we put y − z − 0 in the equation of the first plane, we get (1120,x0−, 0)5aannddtshoe
( 1 0) in this plane. By Formula 9, the distance between
2 , 0, is a point
plane 5x 1 y 2 z 2 1 − 0 is
| |D −5(1)1 1s0d 2 1s0d 2 1 − 3 − s3
2 2 6
s5 2 1 12 1 s21d2 3s3
So the distance between the planes is s3y6. ■
Example 10 In Example 3 we showed that the lines
L1: x − 1 1 t y − 22 1 3t z − 4 2 t
L2: x − 2s y−31s z − 23 1 4s
are skew. Find the distance between them.
FIGURE 13 SOLUTION Since the two lines L1 and L2 are skew, they can be viewed as lying on two
Skew lines, like those in Example 10, parallel planes P1 and P2. The distance between L1 and L2 is the same as the distance
between P1 and P2, which can be computed as in Example 9. The common normal vec-
always lie on (nonidentical) parallel tor to both planes must be orthogonal to both v1 − k 1, 3, 21l (the direction of L1) and
v2 − k 2, 1, 4l (the direction of L2). So a normal vector is
planes.
Z Zi j k
n − v1 3 v2 − 1 3 21 − 13 i 2 6 j 2 5 k
21 4
If we put s − 0 in the equations of L2, we get the point s0, 3, 23d on L2 and so an
equation for P2 is
13sx 2 0d 2 6sy 2 3d 2 5sz 1 3d − 0 or 13x 2 6y 2 5z 1 3 − 0
If we now set t − 0 in the equations for L1, we get the point s1, 22, 4d on P1. So
the distance between L1 and L2 is the same as the distance from s1, 22, 4d to
13x 2 6y 2 5z 1 3 − 0. By Formula 9, this distance is
| |
D− 13s1d 2 6s22d 2 5s4d 1 3 − 8 < 0.53 ■
s13 2 1 s26d2 1 s25d2 s230
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Section 12.5 Equations of Lines and Planes 831
1. Determine whether each statement is true or false in R3. (b) In what points does this line intersect the coordinate
(a) Two lines parallel to a third line are parallel. planes?
(b) Two lines perpendicular to a third line are parallel.
(c) Two planes parallel to a third plane are parallel. 17. Find a vector equation for the line segment from s6, 21, 9d
(d) Two planes perpendicular to a third plane are parallel. to s7, 6, 0d.
(e) Two lines parallel to a plane are parallel.
(f) Two lines perpendicular to a plane are parallel. 18. Find parametric equations for the line segment from
(g) Two planes parallel to a line are parallel. s22, 18, 31d to s11, 24, 48d.
(h) Two planes perpendicular to a line are parallel.
(i) Two planes either intersect or are parallel. 19–22 Determine whether the lines L1 and L2 are parallel, skew, or
( j) Two lines either intersect or are parallel. intersecting. If they intersect, find the point of intersection.
(k) A plane and a line either intersect or are parallel.
19. L1: x − 3 1 2t, y − 4 2 t, z − 1 1 3t
2–5 Find a vector equation and parametric equations for the line.
L2: x − 1 1 4s, y − 3 2 2s, z − 4 1 5s
2. Tv ehcetolirnke1t,h3ro, u2g23hlthe point s6, 25, 2d and parallel to the
20. L1: x − 5 2 12t, y − 3 1 9t, z − 1 2 3t
3. T he line through the point s2, 2.4, 3.5d and parallel to the
vector 3 i 1 2 j 2 k L2: x − 3 1 8s, y − 26s, z − 7 1 2s
4. T he line through the point s0, 14, 210d and parallel to the line 21. L1: x 2 2 − y23 − z21
x − 21 1 2t, y − 6 2 3t, z − 3 1 9t 1 22 23
5. T he line through the point (1, 0, 6) and perpendicular to the L2: x 2 3 − y 1 4 − z22
plane x 1 3y 1 z − 5 1 3 27
22. L1: x − y21 − z 2 2
1 21 3
L2: x 2 2 − y23 − z
2 22 7
6–12 Find parametric equations and symmetric equations for the
line. 23–40 Find an equation of the plane.
6. The line through the origin and the point s4, 3, 21d 23. The plane through the origin and perpendicular to the
vector k 1, 22, 5 l
7. The line through the points s0, 1 , 1d and s2, 1, 23d
2
8. T he line through the points s1, 2.4, 4.6d and s2.6, 1.2, 0.3d 24. The plane through the point s5, 3, 5d and with normal
9. The line through the points s28, 1, 4d and s3, 22, 4d vector 2 i 1 j 2 k
10. The line through s2, 1, 0d and perpendicular to both i 1 j 25. The plane through the point s21, 1 , 3d and with normal
and j 1 k 2
vector i 1 4 j 1 k
11. The line through s26, 2, 3d and parallel to the line 26. The plane through the point s2, 0, 1d and perpendicular to the
1 1 line x − 3t, y − 2 2 t, z − 3 1 4t
2 x − 3 y − z 1 1
12. The line of intersection of the planes x 1 2y 1 3z − 1 27. The plane through the point s1, 21, 21d and parallel to the
and x 2 y 1 z − 1 plane 5x 2 y 2 z − 6
13. Is the line through s24, 26, 1d and s22, 0, 23d parallel to the 28. The plane through the point s3, 22, 8d and parallel to the
line through s10, 18, 4d and s5, 3, 14d?
plane z − x 1 y
14. I s the line through s22, 4, 0d and s1, 1, 1d perpendicular to the
line through s2, 3, 4d and s3, 21, 28d? 29. The plane through the point s1, 1 , 1 d and parallel to the plane
2 3
15. (a) Find symmetric equations for the line that passes x1y1z−0
through the point s1, 25, 6d and is parallel to the vector
k21, 2, 23 l. 30. The plane that contains the line x − 1 1 t, y − 2 2 t,
z − 4 2 3t and is parallel to the plane 5x 1 2y 1 z − 1
(b) Find the points in which the required line in part (a) inter-
sects the coordinate planes. 31. The plane through the points s0, 1, 1d, s1, 0, 1d, and s1, 1, 0d
16. (a) Find parametric equations for the line through s2, 4, 6d that 32. T he plane through the origin and the points s3, 22, 1d
is perpendicular to the plane x 2 y 1 3z − 7. and s1, 1, 1d
33. The plane through the points s2, 1, 2d, s3, 28, 6d, and
s22, 23, 1d
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832 Chapter 12 Vectors and the Geometry of Space
34. The plane through the points s3, 0, 21d, s22, 22, 3d, and 57–58 (a) Find parametric equations for the line of intersection of
s7, 1, 24d the planes and (b) find the angle between the planes.
35. T he plane that passes through the point s3, 5, 21d and con- 57. x 1 y 1 z − 1, x 1 2y 1 2z − 1
tains the line x − 4 2 t, y − 2t 2 1, z − 23t
58. 3x 2 2y 1 z − 1, 2x 1 y 2 3z − 3
36. The plane that passes through the point s6, 21, 3d and
contains the line with symmetric equations 59–60 Find symmetric equations for the line of intersection of the
xy3 − y 1 4 − zy2 planes.
37. T he plane that passes through the point s3, 1, 4d and contains 59. 5x 2 2y 2 2z − 1, 4x 1 y 1 z − 6
the line of intersection of the planes x 1 2y 1 3z − 1 and
2 x 2 y 1 z − 23 60. z − 2x 2 y 2 5, z − 4x 1 3y 2 5
38. The plane that passes through the points s0, 22, 5d and 61. F ind an equation for the plane consisting of all points that are
s21, 3, 1d and is perpendicular to the plane 2z − 5x 1 4y equidistant from the points s1, 0, 22d and s3, 4, 0d.
39. T he plane that passes through the point s1, 5, 1d and is perpen- 62. Find an equation for the plane consisting of all points that are
dicular to the planes 2x 1 y 2 2z − 2 and x 1 3z − 4 equidistant from the points s2, 5, 5d and s26, 3, 1d.
40. T he plane that passes through the line of intersection of the 63. F ind an equation of the plane with x-intercept a, y-intercept b,
planes x 2 z − 1 and y 1 2z − 3 and is perpendicular to the and z-intercept c.
plane x 1 y 2 2z − 1
64. (a) Find the point at which the given lines intersect:
41–44 Use intercepts to help sketch the plane.
41. 2x 1 5y 1 z − 10 42. 3x 1 y 1 2z − 6 r − k1, 1, 0 l 1 t k1, 21, 2 l
r − k2, 0, 2 l 1 sk21, 1, 0 l
43. 6x 2 3y 1 4z − 6 44. 6x 1 5y 2 3z − 15
(b) Find an equation of the plane that contains these lines.
45–47 Find the point at which the line intersects the given plane.
45. x − 2 2 2t, y − 3t, z − 1 1 t; x 1 2y 2 z − 7 65. Find parametric equations for the line through the point
s0, 1, 2d that is parallel to the plane x 1 y 1 z − 2 and
46. x − t 2 1, y − 1 1 2t, z − 3 2 t; 3x 2 y 1 2z − 5 perpendicular to the line x − 1 1 t, y − 1 2 t, z − 2t.
47. 5x − yy2 − z 1 2; 10x 2 7y 1 3z 1 24 − 0 66. Find parametric equations for the line through the point
s0, 1, 2d that is perpendicular to the line x − 1 1 t,
48. Where does the line through s23, 1, 0d and s21, 5, 6d intersect y − 1 2 t, z − 2t and intersects this line.
the plane 2x 1 y 2 z − 22?
67. Which of the following four planes are parallel? Are any of
49. Find direction numbers for the line of intersection of the planes them identical?
x 1 y 1 z − 1 and x 1 z − 0.
P1: 3x 1 6y 2 3z − 6 P2: 4x 2 12y 1 8z − 5
50. Find the cosine of the angle between the planes x 1 y 1 z − 0
and x 1 2y 1 3z − 1. P3: 9y − 1 1 3x 1 6z P4: z − x 1 2y 2 2
51–56 Determine whether the planes are parallel, perpendicular, 68. Which of the following four lines are parallel? Are any of them
or neither. If neither, find the angle between them. (Round to one identical?
decimal place.)
51. x 1 4y 2 3z − 1, 23x 1 6y 1 7z − 0 L1: x − 1 1 6t, y − 1 2 3t, z − 12t 1 5
52. 9x 2 3y 1 6z − 2, 2y − 6x 1 4z L2: x − 1 1 2t, y − t, z − 1 1 4t
53. x 1 2y 2 z − 2, 2x 2 2y 1 z − 1 L3: 2x 2 2 − 4 2 4y − z 1 1
54. x 2 y 1 3z − 1, 3x 1 y 2 z − 2 L4: r − k 3, 1, 5 l 1 t k 4, 2, 8 l
55. 2x 2 3y − z, 4x − 3 1 6y 1 2z
56. 5x 1 2y 1 3z − 2, y − 4x 2 6z 69–70 Use the formula in Exercise 12.4.45 to find the distance
from the point to the given line.
69. s4, 1, 22d; x − 1 1 t, y − 3 2 2t, z − 4 2 3t
70. s0, 1, 3d; x − 2t, y − 6 2 2t, z − 3 1 t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
laboratory project Putting 3D in Perspective 833
71–72 Find the distance from the point to the given plane. through the points s3, 2, 21d, s0, 0, 1d, and s1, 2, 1d. Calculate
71. s1, 22, 4d, 3x 1 2y 1 6z − 5 the distance between L1 and L2.
72. s26, 3, 5d, x 2 2y 2 4z − 8
81. Two tanks are participating in a battle simulation. Tank A
73–74 Find the distance between the given parallel planes. is at point s325, 810, 561d and tank B is positioned at point
73. 2x 2 3y 1 z − 4, 4x 2 6y 1 2z − 3 s765, 675, 599d.
74. 6z − 4y 2 2x, 9z − 1 2 3x 1 6y
(a) Find parametric equations for the line of sight between the
75. S how that the distance between the parallel planes tanks.
ax 1 by 1 cz 1 d1 − 0 and ax 1 by 1 cz 1 d2 − 0 is
(b) If we divide the line of sight into 5 equal segments, the
elevations of the terrain at the four intermediate points
from tank A to tank B are 549, 566, 586, and 589. Can the
tanks see each other?
| |D
− d1 2 d2 c2
sa 2 1 b 2 1
76. F ind equations of the planes that are parallel to the plane 82. Give a geometric description of each family of planes.
x 1 2y 2 2z − 1 and two units away from it. (a) x 1 y 1 z − c (b) x 1 y 1 cz − 1
77. Show that the lines with symmetric equations x − y − z and (c) y cos 1 z sin − 1
x 1 1 − yy2 − zy3 are skew, and find the distance between
these lines. 83. If a, b, and c are not all 0, show that the equation
ax 1 by 1 cz 1 d − 0 represents a plane and k a, b, c l is
78. Find the distance between the skew lines with parametric
equations x − 1 1 t, y − 1 1 6t, z − 2t, and x − 1 1 2s, a normal vector to the plane.
y − 5 1 15s, z − 22 1 6s.
Hint: Suppose a ± 0 and rewrite the equation in the form
79. Let L1 be the line through the origin and the point s2, 0, 21d. S Da
Let L2 be the line through the points s1, 21, 1d and s4, 1, 3d. x 1 d 1 bs y 2 0d 1 csz 2 0d − 0
Find the distance between L1 and L2. a
80. Let L1 be the line through the points s1, 2, 6d and s2, 4, 8d.
Let L2 be the line of intersection of the planes P1 and P2,
where P1 is the plane x 2 y 1 2z 1 1 − 0 and P2 is the plane
laboratory Project putting 3d in perspective
Computer graphics programmers face the same challenge as the great painters of the past: how
to represent a three-dimensional scene as a flat image on a two-dimensional plane (a screen or a
canvas). To create the illusion of perspective, in which closer objects appear larger than those far-
ther away, three-dimensional objects in the computer’s memory are projected onto a rectangular
screen window from a viewpoint where the eye, or camera, is located. The viewing volume––the
portion of space that will be visible––is the region contained by the four planes that pass through
the viewpoint and an edge of the screen window. If objects in the scene extend beyond these four
planes, they must be truncated before pixel data are sent to the screen. These planes are therefore
called clipping planes.
1. S uppose the screen is represented by a rectangle in the yz-plane with vertices s0, 6400, 0d
and s0, 6400, 600d, and the camera is placed at s1000, 0, 0d. A line L in the scene passes
through the points s230, 2285, 102d and s860, 105, 264d. At what points should L be clipped
by the clipping planes?
2. I f the clipped line segment is projected onto the screen window, identify the resulting line
segment.
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834 Chapter 12 Vectors and the Geometry of Space
3. U se parametric equations to plot the edges of the screen window, the clipped line segment,
and its projection onto the screen window. Then add sight lines connecting the viewpoint to
each end of the clipped segments to verify that the projection is correct.
4. A rectangle with vertices s621, 2147, 206d, s563, 31, 242d, s657, 2111, 86d, and
s599, 67, 122d is added to the scene. The line L intersects this rectangle. To make the rect-
angle appear opaque, a programmer can use hidden line rendering, which removes portions of
objects that are behind other objects. Identify the portion of L that should be removed.
We have already looked at two special types of surfaces: planes (in Section 12.5) and
spheres (in Section 12.1). Here we investigate two other types of surfaces: cylinders and
quadric surfaces.
In order to sketch the graph of a surface, it is useful to determine the curves of inter-
section of the surface with planes parallel to the coordinate planes. These curves are
called traces (or cross-sections) of the surface.
Cylinders
A cylinder is a surface that consists of all lines (called rulings) that are parallel to a
given line and pass through a given plane curve.
Example 1 Sketch the graph of the surface z − x2.
SOLUTION Notice that the equation of the graph, z − x2, doesn’t involve y. This means
that any vertical plane with equation y − k (parallel to the xz-plane) intersects the
graph in a curve with equation z − x2. So these vertical traces are parabolas. Figure 1
shows how the graph is formed by taking the parabola z − x2 in the xz-plane and mov-
ing it in the direction of the y-axis. The graph is a surface, called a parabolic cylinder,
made up of infinitely many shifted copies of the same parabola. Here the rulings of the
cylinder are parallel to the y-axis.
z
FIGURE 1 0 y ■
The surface z − x 2 is a x
parabolic cylinder.
We noticed that the variable y is missing from the equation of the cylinder in Exam
ple 1. This is typical of a surface whose rulings are parallel to one of the coordinate axes.
If one of the variables x, y, or z is missing from the equation of a surface, then the surface
is a cylinder.
Example 2 Identify and sketch the surfaces.
(a) x 2 1 y 2 − 1 (b) y 2 1 z 2 − 1
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Section 12.6 Cylinders and Quadric Surfaces 835
z SOLUTION z
(a) Since z is missing and the equations x2 1 y2 − 1, z − k represent a circle with
radius 1 in the plane z − k, the surface x2 1 y2 − 1 is a circular cylinder whose axis is
the z-axis. (See Figure 2.) Here the rulings are vertical lines.
0 y (b) In this case x is missing and the surface is a circular cylinder whose axis is the
xy-za-pxlias.n(eSaenedFmigouvrein3g.)itIpt aisraolbyletlaitnoetdhebyx-taaxkiisn.g the circle y2 1 z2 − 1, x − 0 in the ■
x
x
FIGURE 2
x2 1 y2 − 1 note When you are dealing with surfaces, it is important to recognize that an equa-
tion like x2 1 y2 − 1 represents a cylinder and not a circle. The trace of the cylinder
z x 2 1 y 2 − 1 in the xy-plane is the circle with equations x 2 1 y 2 − 1, z − 0.
y Quadric Surfaces
A quadric surface is the graph of a second-degree equation in three variables x, y, and
x z. The most general such equation is
FIGURE 3 Ax2 1 By2 1 Cz2 1 Dxy 1 Eyz 1 Fxz 1 Gx 1 Hy 1 Iz 1 J − 0
y2 1 z2 − 1
where A, B, C, . . . , J are constants, but by translation and rotation it can be brought into
one of the two standard forms
Ax 2 1 By 2 1 Cz2 1 J − 0 or Ax 2 1 By 2 1 Iz − 0
Quadric surfaces are the counterparts in three dimensions of the conic sections in the
plane. (See Section 10.5 for a review of conic sections.)
Example 3 Use traces to sketch the quadric surface with equation
x2 1 y2 1 z2 − 1
9 4
SOLUTION By substituting z − 0, we find that the trace in the xy-plane is
x2 1 y2y9 − 1, which we recognize as an equation of an ellipse. In general, the hori-
zontal trace in the plane z − k is
x2 1 y2 − 1 2 k2 z − k
9 4
z which is an ellipse, provided that k2 , 4, that is, 22 , k , 2.
(0, 0, 2) Similarly, vertical traces parallel to the yz- and xz-planes are also ellipses:
y2 1 z2 − 1 2 k2 x−k sif 21 , k , 1d
9 4
0 (0, 3, 0) x2 1 z2 − 1 2 k2 y−k sif 23 , k , 3d
y 4 9
(1, 0, 0)
x Figure 4 shows how drawing some traces indicates the shape of the surface. It’s called an
FIGURE 4 ellipsoid because all of its traces are ellipses. Notice that it is symmetric with respect to
y2 z2 each coordinate plane; this is a reflection of the fact that its equation involves only even
9 4
The ellipsoid x2 1 1 − 1 powers of x, y, and z. ■
Example 4 Use traces to sketch the surface z − 4x2 1 y2.
SOLUTION If we put x − 0, we get z − y 2, so the yz-plane intersects the surface in a
parabola. If we put x − k (a constant), we get z − y2 1 4k2. This means that if we
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836 Chapter 12 Vectors and the Geometry of Space
z slice the graph with any plane parallel to the yz-plane, we obtain a parabola that opens
upward. Similarly, if y − k, the trace is z − 4x2 1 k2, which is again a parabola that
opens upward. If we put z − k, we get the horizontal traces 4x2 1 y2 − k, which we
recognize as a family of ellipses. Knowing the shapes of the traces, we can sketch the
graph in Figure 5. Because of the elliptical and parabolic traces, the quadric surface
z − 4x2 1 y2 is called an elliptic paraboloid. ■
0 Example 5 Sketch the surface z − y2 2 x2.
xy SOLUTION The traces in the vertical planes x − k are the parabolas z − y 2 2 k 2,
which open upward. The traces in y − k are the parabolas z − 2x2 1 k2, which open
FIGURE 5 downward. The horizontal traces are y2 2 x2 − k, a family of hyperbolas. We draw the
The surface z − 4x 2 1 y 2 is an elliptic families of traces in Figure 6, and we show how the traces appear when placed in their
paraboloid. Horizontal traces are correct planes in Figure 7.
ellipses; vertical traces are parabolas.
zz y1
Ϯ2
0 Ϯ1 _1 _1
0
FIGURE 6 y
Vertical traces are parabolas; Ϯ1 xx
horizontal traces are hyperbolas. 0
All traces are labeled with the Ϯ2
value of k. Traces in x=k are z=¥-k@. Traces in y=k are z=_≈+k@. 1
Traces in z=k are ¥-≈=k.
z zz
1
y y x 0
x _1 x Traces in z=k y
FIGURE 7 0 _1 0 _1
Traces moved to their 1
1
correct planes
Traces in x=k Traces in y=k
TEC In Module 12.6A you can inves- In Figure 8 we fit together the traces from Figure 7 to form the surface z − y2 2 x2,
tigate how traces determine the shape a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles
of a surface. that of a saddle. This surface will be investigated further in Section 14.7 when we
discuss saddle points.
zz
FIGURE 8 0 y 0 y
Two views of the surface z − y 2 2 x 2, x x
■
a hyperbolic paraboloid
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Section 12.6 Cylinders and Quadric Surfaces 837
z Example 6 Sketch the surface x2 1 y2 2 z2 − 1.
4 4
SOLUTION The trace in any horizontal plane z − k is the ellipse
(2, 0, 0) (0, 1, 0) x2 1 y2 − 1 1 k2 z−k
x y 4 4
but the traces in the xz- and yz-planes are the hyperbolas
x2 2 z2 − 1 y−0 and y2 2 z2 − 1 x−0
4 4 4
FIGURE 9 This surface is called a hyperboloid of one sheet and is sketched in Figure 9. ■
The idea of using traces to draw a surface is employed in three-dimensional graphing
software. In most such software, traces in the vertical planes x − k and y − k are drawn
for equally spaced values of k, and parts of the graph are eliminated using hidden line
removal. Table 1 shows computer-drawn graphs of the six basic types of quadric surfaces
in standard form. All surfaces are symmetric with respect to the z-axis. If a quadric sur-
face is symmetric about a different axis, its equation changes accordingly.
Table 1 Graphs of Quadric Surfaces
Surface Equation Surface Equation
Ellipsoid x2 1 y2 1 z2 − 1 Cone z2 x2 y2
a2 b2 c2 c2 − a2 1 b2
z z
z
All traces are ellipses. z Horizontal traces are ellipses.
If a − b − c, the ellipsoid is
z y a sphere. z Vertical traces in the planes
x y
x x y x − k and y − k are hyper-
x bolas if k ± 0 but are pairs of
y lines if k − 0.
x yx y
Elliptic Paraboloid z x2 y2 Hyperboloidzof One Sheet x2 1 y2 2 z2 − 1
z c − a2 1 b2 z a2 b2 c2
z
z Horizontal traces are ellipses. z Horizontal traces are ellipses.
x y Vertical traces are parabolas. xy Vertical traces are hyperbolas.
x y xy
The variable raised to the first xy The axis of symmetry corre-
power indicates the axis of the sponds to the variable whose
paraboloid. coefficient is negative.
xz y z
Hyperbolic Pzaraboloid z x2 y2 Hyperboloidzof Two Sheets x2 y2 z2
c − a2 2 b2 2 a2 2 b2 1 c2 − 1
z
z y Horizontal traces are hyper-
y bolas. xy
x xy Horizontal traces in z − k are
x y Vertical traces are parabolas. xy ellipses if k . c or k , 2c.
x The case where c , 0 is
illustrated. Vertical traces are hyperbolas.
The two minus signs indicate
two sheets.
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838 Chapter 12 Vectors and the Geometry of Space
TEC In Module 12.6B you can see Example 7 Identify and sketch the surface 4x2 2 y2 1 2z2 1 4 − 0.
how changing a, b, and c in Table 1
affects the shape of the quadric SOLUTION Dividing by 24, we first put the equation in standard form:
surface.
2x 2 y2 z2 − 1
1 4 2 2
Comparing this equation with Table 1, we see that it represents a hyperboloid of two
sheets, the only difference being that in this case the axis of the hyperboloid is the
y-axis. The traces in the xy- and yz-planes are the hyperbolas
2x 2 1 y2 − 1 z − 0 and y2 2 z2 − 1 x − 0
4 4 2
z | |The surface has no trace in the xz-plane, but traces in the vertical planes y − k for
(0, _2, 0)
k . 2 are the ellipses
0
x2 1 z2 − k2 2 1 y − k
x (0, 2, 0) 2 4
FIGURE 10 which can be written as
4x2 2 y2 1 2z2 1 4 − 0
y S Dx2 z2
k2 1 k2 − 1 y − k
4 4
2 1 2 2 1
These traces are used to make the sketch in Figure 10. ■
Example 8 Classify the quadric surface x2 1 2z2 2 6x 2 y 1 10 − 0.
SOLUTION By completing the square we rewrite the equation as
y 2 1 − sx 2 3d2 1 2z2
Comparing this equation with Table 1, we see that it represents an elliptic paraboloid.
Here, however, the axis of the paraboloid is parallel to the y-axis, and it has been
shifted so that its vertex is the point s3, 1, 0d. The traces in the plane y − k sk . 1d are
the ellipses
sx 2 3d2 1 2z2 − k 2 1 y − k
The trace in the xy-plane is the parabola with equation y − 1 1 sx 2 3d2, z − 0. The
paraboloid is sketched in Figure 11.
z
FIGURE 11 0 y
x 2 1 2z2 2 6x 2 y 1 10 − 0 x (3, 1, 0)
■
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Section 12.6 Cylinders and Quadric Surfaces 839
Applications of Quadric Surfaces
Examples of quadric surfaces can be found in the world around us. In fact, the world
itself is a good example. Although the earth is commonly modeled as a sphere, a more
accurate model is an ellipsoid because the earth’s rotation has caused a flattening at the
poles. (See Exercise 49.)
Circular paraboloids, obtained by rotating a parabola about its axis, are used to collect
and reflect light, sound, and radio and television signals. In a radio telescope, for instance,
signals from distant stars that strike the bowl are all reflected to the receiver at the focus
and are therefore amplified. (The idea is explained in Problem 22 on page 273.) The
same principle applies to microphones and satellite dishes in the shape of paraboloids.
Cooling towers for nuclear reactors are usually designed in the shape of hyperboloids
of one sheet for reasons of structural stability. Pairs of hyperboloids are used to transmit
rotational motion between skew axes. (The cogs of the gears are the generating lines of
the hyperboloids. See Exercise 51.)
David Frazier / Spirit / Corbis
Mark C. Burnett / Science Source
A satellite dish reflects signals to the Nuclear reactors have cooling towers in Hperboloids produce gear transmission.
focus of a paraboloid. the shape of hyperboloids.
1. (a) What does the equation y − x 2 represent as a curve 5. z − 1 2 y 2 6. y − z 2
in R2?
7. xy − 1 8. z − sin y
(b) What does it represent as a surface in R3?
(c) What does the equation z − y 2 represent? 9. (a) Find and identify the traces of the quadric surface
x 2 1 y2 2 z2 − 1 and explain why the graph looks like
2. (a) Sketch the graph of y − e x as a curve in R2. the graph of the hyperboloid of one sheet in Table 1.
(b) Sketch the graph of y − e x as a surface in R3.
(c) Describe and sketch the surface z − e y. (b) If we change the equation in part (a) to x 2 2 y2 1 z2 − 1,
how is the graph affected?
3–8 Describe and sketch the surface.
(c) What if we change the equation in part (a) to
3. x 2 1 z 2 − 1 4. 4x 2 1 y 2 − 4 x 2 1 y2 1 2y 2 z2 − 0?
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840 Chapter 12 Vectors and the Geometry of Space
10. (a) Find and identify the traces of the quadric surface 29–30 Sketch and identify a quadric surface that could have the
2x 2 2 y2 1 z2 − 1 and explain why the graph looks like traces shown.
the graph of the hyperboloid of two sheets in Table 1.
29. Traces in x − k Traces in y − k
(b) If the equation in part (a) is changed to x 2 2 y2 2 z2 − 1,
what happens to the graph? Sketch the new graph. z z
k=Ϯ2 k=_1
11–20 Use traces to sketch and identify the surface. k=Ϯ1
k=0 k=_2
11. x − y 2 1 4z2 12. 4x 2 1 9y 2 1 9z 2 − 36
13. x 2 − 4y 2 1 z 2 14. z 2 2 4x 2 2 y 2 − 4 k=0
k=1
15. 9y 2 1 4z 2 − x 2 1 36 16. 3x 2 1 y 1 3z 2 − 0 y x
17. x2 1 y2 z2 − 1 18. 3x 2 2 y 2 1 3z 2 − 0
9 25 1 4
19. y − z2 2 x 2 20. x − y 2 2 z2
21–28 Match the equation with its graph (labeled I–VIII). Give 30. Traces in x − k Traces in z − k
reasons for your choice. y
z
21. x 2 1 4y 2 1 9z2 − 1 22. 9x 2 1 4y 2 1 z2 − 1 k=Ϯ2 k=0 k=1 k=2
k=Ϯ1 k=2 x
23. x 2 2 y 2 1 z2 − 1 24. 2x 2 1 y 2 2 z2 − 1
k=0
25. y − 2x 2 1 z2 26. y 2 − x 2 1 2z2
y
27. x 2 1 2z2 − 1 28. y − x 2 2 z2
Iz II z k=0
x y y
III z x
31–38 Reduce the equation to one of the standard forms, classify
x IV z the surface, and sketch it.
Vz
y 31. y2 − x2 1 1 z 2 32. 4x 2 2 y 1 2z 2 − 0
x 9
VI z 33. x 2 1 2y 2 2z2 − 0 34. y 2 − x 2 1 4z 2 1 4
35. x 2 1 y 2 2 2x 2 6y 2 z 1 10 − 0
y 36. x 2 2 y 2 2 z 2 2 4x 2 2z 1 3 − 0
37. x 2 2 y 2 1 z 2 2 4x 2 2z − 0
38. 4x 2 1 y 2 1 z 2 2 24x 2 8y 1 4z 1 55 − 0
x yx y ; 39–42 Use a computer with three-dimensional graphing soft-
VII z VIII z ware to graph the surface. Experiment with viewpoints and with
domains for the variables until you get a good view of the surface.
39. 24x 2 2 y 2 1 z2 − 1 40. x 2 2 y 2 2 z − 0
41. 24x 2 2 y 2 1 z2 − 0 42. x 2 2 6x 1 4y 2 2 z − 0
y 43. S ketch the region bounded by the surfaces z − sx 2 1 y 2
x and x 2 1 y 2 − 1 for 1 < z < 2.
44. Sketch the region bounded by the paraboloids z − x 2 1 y 2
and z − 2 2 x 2 2 y 2.
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chapter 12 Review 841
45. Find an equation for the surface obtained by rotating the 50. A cooling tower for a nuclear reactor is to be constructed in
curve y − sx about the x-axis. the shape of a hyperboloid of one sheet (see the photo on
page 839). The diameter at the base is 280 m and the mini-
46. Find an equation for the surface obtained by rotating the line mum diameter, 500 m above the base, is 200 m. Find an equa-
z − 2y about the z-axis. tion for the tower.
47. Find an equation for the surface consisting of all points that 51. S how that if the point sa, b, cd lies on the hyperbolic parabo-
are equidistant from the point s21, 0, 0d and the plane x − 1. loid z − y 2 2 x 2, then the lines with parametric equations
Identify the surface. x − a 1 t, y − b 1 t, z − c 1 2sb 2 adt and x − a 1 t,
y − b 2 t, z − c 2 2sb 1 adt both lie entirely on this
48. F ind an equation for the surface consisting of all points P for paraboloid. (This shows that the hyperbolic paraboloid is
which the distance from P to the x-axis is twice the distance what is called a ruled surface; that is, it can be generated by
from P to the yz-plane. Identify the surface. the motion of a straight line. In fact, this exercise shows that
through each point on the hyperbolic paraboloid there are
49. Traditionally, the earth’s surface has been modeled as a two generating lines. The only other quadric surfaces that are
sphere, but the World Geodetic System of 1984 (WGS-84) ruled surfaces are cylinders, cones, and hyperboloids of one
uses an ellipsoid as a more accurate model. It places the cen- sheet.)
ter of the earth at the origin and the north pole on the positive
z-axis. The distance from the center to the poles is 6356.523 km 52. Show that the curve of intersection of the surfaces
and the distance to a point on the equator is 6378.137 km. x 2 1 2y 2 2 z2 1 3x − 1 and 2x 2 1 4y 2 2 2z2 2 5y − 0
lies in a plane.
(a) Find an equation of the earth’s surface as used by
WGS-84. | | | |; 53. Graph the surfaces z − x 2 1 y 2 and z − 1 2 y 2 on a com-
mon screen using the domain x < 1.2, y < 1.2 and
(b) Curves of equal latitude are traces in the planes z − k. observe the curve of intersection of these surfaces. Show that
What is the shape of these curves? the projection of this curve onto the xy-plane is an ellipse.
(c) Meridians (curves of equal longitude) are traces in
planes of the form y − mx. What is the shape of these
meridians?
12 Review
CONCEPT CHECK Answers to the Concept Check can be found on the back endpapers.
1. What is the difference between a vector and a scalar? 11. How do you find a vector perpendicular to a plane?
2. H ow do you add two vectors geometrically? How do you add 12. How do you find the angle between two intersecting planes?
them algebraically?
13. W rite a vector equation, parametric equations, and symmetric
3. I f a is a vector and c is a scalar, how is ca related to a equations for a line.
geom etrically? How do you find ca algebraically?
14. Write a vector equation and a scalar equation for a plane.
4. How do you find the vector from one point to another?
15. (a) How do you tell if two vectors are parallel?
5. H ow do you find the dot product a ? b of two vectors if you (b) How do you tell if two vectors are perpendicular?
know their lengths and the angle between them? What if you (c) How do you tell if two planes are parallel?
know their components?
16. (a) Describe a method for determining whether three points
6. How are dot products useful? P, Q, and R lie on the same line.
7. W rite expressions for the scalar and vector projections of b (b) Describe a method for determining whether four points
onto a. Illustrate with diagrams. P, Q, R, and S lie in the same plane.
8. H ow do you find the cross product a 3 b of two vectors if you 17. (a) How do you find the distance from a point to a line?
know their lengths and the angle between them? What if you (b) How do you find the distance from a point to a plane?
know their components? (c) How do you find the distance between two lines?
9. How are cross products useful? 18. What are the traces of a surface? How do you find them?
10. (a) How do you find the area of the parallelogram determined 19. Write equations in standard form of the six types of quadric
by a and b? surfaces.
(b) How do you find the volume of the parallelepiped
determined by a, b, and c?
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842 Chapter 12 Vectors and the Geometry of Space
TRUE-FALSE QUIZ
Determine whether the statement is true or false. If it is true, explain 11. For any vectors u, v, and w in V3,
why. If it is false, explain why or give an example that disproves the
statement. u ? sv 3 wd − su 3 vd ? w
12. For any vectors u, v, and w in V3,
1. If u − k u1, u2l and v − kv1, v2l , then u ? v − k u1v1, u2v2l.
u 3 sv 3 wd − su 3 vd 3 w
2. For any vectors u and v in V3, | u 1 v | − | u | 1 | v |. 13. For any vectors u and v in V3, su 3 vd ? u − 0.
3. For any vectors u and v in V3, | u ? v | − | u | | v |. 14. For any vectors u and v in V3, su 1 vd 3 v − u 3 v.
4. For any vectors u and v in V3, | u 3 v | − | u | | v |. 15. T he vector k 3, 21, 2 l is parallel to the plane
5. For any vectors u and v in V3, u ? v − v ? u. 6x 2 2y 1 4z − 1
6. For any vectors u and v in V3, u 3 v − v 3 u. 16. A linear equation Ax 1 By 1 Cz 1 D − 0 represents a line
in space.
| | | | 7. For any vectors u and v in V3, u 3 v − v 3 u .
| 17. The set of points h sx, y, zd x 2 1 y 2 − 1j is a circle.
8. F or any vectors u and v in V3 and any scalar k,
18. In R3 the graph of y − x 2 is a paraboloid.
ksu ? vd − sk ud ? v
19. If u ? v − 0, then u − 0 or v − 0.
9. For any vectors u and v in V3 and any scalar k,
20. If u 3 v − 0, then u − 0 or v − 0.
ksu 3 vd − sk ud 3 v
21. If u ? v − 0 and u 3 v − 0, then u − 0 or v − 0.
10. F or any vectors u, v, and w in V3,
22. If u and v are in V3, then | u ? v | < | u || v |.
su 1 vd 3 w − u 3 w 1 v 3 w
EXERCISES
1. (a) Find an equation of the sphere that passes through the point 4. Calculate the given quantity if
s6, 22, 3d and has center s21, 2, 1d.
a − i 1 j 2 2k
(b) Find the curve in which this sphere intersects the yz-plane.
(c) Find the center and radius of the sphere b − 3i 2 2j 1 k
x 2 1 y2 1 z2 2 8x 1 2y 1 6z 1 1 − 0 c − j 2 5k
2. C opy the vectors in the figure and use them to draw each of the | | (a) 2 a 1 3 b (b) b
following vectors. | | (c) a ? b (d) a 3 b
(e) b 3 c (f ) a ? sb 3 cd
(a)2 a 1 b (b) a 2 b (c) 212 a (d) 2 a 1 b
(g) c 3 c (h) a 3 sb 3 cd
(i) compa b (j) proja b
(k) The angle between a and b (correct to the nearest degree)
a 5. Find the values of x such that the vectors k 3, 2, x l and k 2x, 4, x l
b are orthogonal.
| | 3. I f u and v are the vectors shown in the figure, find u ? v and 6. Find two unit vectors that are orthogonal to both j 1 2 k
u 3 v . Is u 3 v directed into the page or out of it? and i 2 2 j 1 3 k.
| v |=3 7. Suppose that u ? sv 3 wd − 2. Find
45° (a) su 3 vd ? w (b) u ? sw 3 vd
| u |=2 (c) v ? su 3 wd (d) su 3 vd ? v
8. Show that if a, b, and c are in V3, then
sa 3 bd ? fsb 3 cd 3 sc 3 adg − fa ? sb 3 cdg 2
9. Find the acute angle between two diagonals of a cube.
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chapter 12 Review 843
10. Given the points As1, 0, 1d, Bs2, 3, 0d, Cs21, 1, 4d, and 21. F ind the point in which the line with parametric equa-
Ds0, 3, 2d, find the volume of the parallelepiped with adjacent tions x − 2 2 t, y − 1 1 3t, z − 4t intersects the plane
edges AB, AC, and AD. 2 x 2 y 1 z − 2.
11. (a) Find a vector perpendicular to the plane through the points 22. Find the distance from the origin to the line
As1, 0, 0d, Bs2, 0, 21d, and Cs1, 4, 3d. x − 1 1 t, y − 2 2 t, z − 21 1 2t.
(b) Find the area of triangle ABC. 23. D etermine whether the lines given by the symmetric
equations
12. A constant force F − 3 i 1 5 j 1 10 k moves an object along
the line segment from s1, 0, 2d to s5, 3, 8d. Find the work done x 2 1 − y 2 2 − z 2 3
if the distance is measured in meters and the force in newtons. 2 3 4
13. A boat is pulled onto shore using two ropes, as shown in the and x 1 1 − y23 − z 1 5
diagram. If a force of 255 N is needed, find the magnitude of 6 21 2
the force in each rope.
are parallel, skew, or intersecting.
20° 255 N
30° 24. (a) Show that the planes x 1 y 2 z − 1 and
2x 2 3y 1 4z − 5 are neither parallel nor perpendicular.
(b) Find, correct to the nearest degree, the angle between these
planes.
14. Find the magnitude of the torque about P if a 50-N force is 25. F ind an equation of the plane through the line of intersection of
applied as shown. the planes x 2 z − 1 and y 1 2z − 3 and perpendicular to the
plane x 1 y 2 2z − 1.
50 N
30° 26. (a) Find an equation of the plane that passes through the points
As2, 1, 1d, Bs21, 21, 10d, and Cs1, 3, 24d.
40 cm
(b) Find symmetric equations for the line through B that is
perpendicular to the plane in part (a).
(c) A second plane passes through s2, 0, 4d and has normal
vector k2, 24, 23l. Show that the acute angle between the
planes is approximately 438.
(d) Find parametric equations for the line of intersection of the
two planes.
P 27. F ind the distance between the planes 3x 1 y 2 4z − 2
and 3x 1 y 2 4z − 24.
15–17 Find parametric equations for the line. 28–36 Identify and sketch the graph of each surface.
15. The line through s4, 21, 2d and s1, 1, 5d
28. x − 3 29. x − z
16. The line through s1, 0, 21d and parallel to the line 30. y − z2 31. x 2 − y 2 1 4z2
1 s 4d 1 32. 4x 2 y 1 2z − 4 33. 24x 2 1 y 2 2 4z2 − 4
3 x 2 − 2 y − z 1 2 34. y 2 1 z2 − 1 1 x 2
35. 4x 2 1 4y 2 2 8y 1 z2 − 0
17. T he line through s22, 2, 4d and perpendicular to the 36. x − y2 1 z2 2 2y 2 4z 1 5
plane 2x 2 y 1 5z − 12
18–20 Find an equation of the plane. 37. A n ellipsoid is created by rotating the ellipse 4x 2 1 y 2 − 16
18. The plane through s2, 1, 0d and parallel to x 1 4y 2 3z − 1 about the x-axis. Find an equation of the ellipsoid.
19. T he plane through s3, 21, 1d, s4, 0, 2d, and s6, 3, 1d 38. A surface consists of all points P such that the distance from P
to the plane y − 1 is twice the distance from P to the point
20. The plane through s1, 2, 22d that contains the line s0, 21, 0d. Find an equation for this surface and identify it.
x − 2t, y − 3 2 t, z − 1 1 3t
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Problems Plus 1. E ach edge of a cubical box has length 1 m. The box contains nine spherical balls with the
same radius r. The center of one ball is at the center of the cube and it touches the other
1m eight balls. Each of the other eight balls touches three sides of the box. Thus the balls are
1m 1m tightly packed in the box (see the figure). Find r. (If you have trouble with this problem,
read about the problem-solving strategy entitled Use Analogy on page 71.)
FIGURE for problem 1
2. Let B be a solid box with length L , width W, and height H. Let S be the set of all points that
NF are a distance at most 1 from some point of B. Express the volume of S in terms of L, W,
W and H.
¨ 3. Let L be the line of intersection of the planes cx 1 y 1 z − c and x 2 cy 1 cz − 21,
FIGURE for problem 7 where c is a real number.
844 (a) Find symmetric equations for L.
(b) As the number c varies, the line L sweeps out a surface S. Find an equation for the
curve of intersection of S with the horizontal plane z − t (the trace of S in the plane
z − t).
(c) Find the volume of the solid bounded by S and the planes z − 0 and z − 1.
4. A plane is capable of flying at a speed of 180 kmyh in still air. The pilot takes off from an
airfield and heads due north according to the plane’s compass. After 30 minutes of flight
time, the pilot notices that, due to the wind, the plane has actually traveled 80 km at an
angle 5° east of north.
(a) What is the wind velocity?
(b) In what direction should the pilot have headed to reach the intended destination?
| | | | 5. S uppose v1 and v2 are vectors with v1 − 2, v2 − 3, and v1 ? v2 − 5. Let v3 − projv1v2,
| | v 4 − o
projv2v3, v5 − projv3v4, and so on. Compute ` vn .
n−1
6. F ind an equation of the largest sphere that passes through the point s21, 1, 4d and is such
that each of the points sx, y, zd inside the sphere satisfies the condition
x 2 1 y 2 1 z 2 , 136 1 2sx 1 2y 1 3zd
7. S uppose a block of mass m is placed on an inclined plane, as shown in the figure. The
block’s descent down the plane is slowed by friction; if is not too large, friction will pre-
| | | |vent the block from moving at all. The forces acting on the block are the weight W, where
W − mt (t is the acceleration due to gravity); the normal force N (the normal compo-
nent of the reactionary force of the plane on the block), where N − n; and the force F
due to friction, which acts parallel to the inclined plane, opposing the direction of motion.
| | | |If the block is at rest and is increased, F must also increase until ultimately F reaches
| | | | | |its maximum, beyond which the block begins to slide. At this angle s, it has been observed
that F is proportional to n. Thus, when F is maximal, we can say that F − s n,
where s is called the coefficient of static friction and depends on the materials that are in
contact.
(a) Observe that N 1 F 1 W − 0 and deduce that s − tanssd.
| | (b) Suppose that, for . s, an additional outside force H is applied to the block, horizon
tally from the left, and let H − h. If h is small, the block may still slide down the
| |plane; if h is large enough, the block will move up the plane. Let hmin be the smallest
value of h that allows the block to remain motionless (so that F is maximal).
By choosing the coordinate axes so that F lies along the x-axis, resolve each force
into components parallel and perpendicular to the inclined plane and show that
hmin sin 1 m t cos − n and hmin cos 1 s n − mt sin
(c) Show that hmin − m t tans 2 sd
Does this equation seem reasonable? Does it make sense for − s? Does it make
sense as l 908? Explain.
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(d) Let hmax be the largest value of h that allows the block to remain motionless. (In which
direction is F heading?) Show that
hmax − m t tans 1 sd
Does this equation seem reasonable? Explain.
8. A solid has the following properties. When illuminated by rays parallel to the z-axis, its
shadow is a circular disk. If the rays are parallel to the y-axis, its shadow is a square. If
the rays are parallel to the x-axis, its shadow is an isosceles triangle. (In Exercise 12.1.48
you were asked to describe and sketch an example of such a solid, but there are many such
solids.) Assume that the projection onto the xz-plane is a square whose sides have length 1.
(a) What is the volume of the largest such solid?
(b) Is there a smallest volume?
845
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13 Vector Functions
The paths of objects moving © Natalia Davydenko / Shutterstock.com
through space like the jet
planes pictured here can
be described by vector
functions. In Section 13.1
we will see how to use these
vector functions to determine
whether or not two such
objects will collide.
the functions that we have been using so far have been real-valued functions. We now
study functions whose values are vectors because such functions are needed to describe curves
and surfaces in space. We will also use vector-valued functions to describe the motion of objects
through space. In particular, we will use them to derive Kepler’s laws of planetary motion.
847
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848 Chapter 13 Vector Functions
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 r whose values are three-dimensional vectors. This means that
for every number t in the domain of r there is a unique vector in V3 denoted by rstd. If
f std, tstd, and hstd are the components of the vector rstd, then f , t, and h are real-valued
functions called the component functions of r and we can write
rstd − k f std, tstd, hstdl − f std i 1 tstd j 1 hstd k
We use the letter t to denote the independent variable because it represents time in most
applications of vector functions.
Example 1 If
rstd − kt3, lns3 2 td, st l
then the component functions are
f std − t3 tstd − lns3 2 td hstd − st
By our usual convention, the domain of r consists of all values of t for which the expres-
sion for rstd is defined. The expressions t3, lns3 2 td, and st are all defined when
3 2 t . 0 and t > 0. Therefore the domain of r is the interval f0, 3d.
■
Limits and Continuity
The limit of a vector function r is defined by taking the limits of its component functions
as follows.
If lim t la rstd − L, this definition is 1 If rstd − k f std, tstd, hstdl, then
equivalent to saying that the length and
direction of the vector rstd approach the k llim rstd − lim f std, lim tstd, lim hstd
length and direction of the vector L. tla tla tla tla
provided the limits of the component functions exist.
Equivalently, we could have used an «-␦ definition (see Exercise 54). Limits of vector
functions obey the same rules as limits of real-valued functions (see Exercise 53).
Example 2 Find lim rstd, where rstd − s1 1 t3 d i 1 te2t j 1 sin t k.
tl0 t
SOLUTION According to Definition 1, the limit of r is the vector whose components are
the limits of the component functions of r:
f g f g F Glim rstd − lim s1 1 t3d i 1 lim te2t j 1
lim sin t k
t
tl0 tl0 tl0 tl0
− i 1 k (by Equation 3.3.2) ■
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Section 13.1 Vector Functions and Space Curves 849
A vector function r is continuous at a if
lim rstd − rsad
tla
In view of Definition 1, we see that r is continuous at a if and only if its component func-
tions f , t, and h are continuous at a.
z Space Curves
P{f(t), g(t), h(t)} There is a close connection between continuous vector functions and space curves. Sup-
pose that f , t, and h are continuous real-valued functions on an interval I. Then the set
C C of all points sx, y, zd in space, where
0 r(t)=k f(t), g(t), h(t)l 2 x − f std y − tstd z − hstd
xy and t varies throughout the interval I, is called a space curve. The equations in (2) are
called parametric equations of C and t is called a parameter. We can think of C as
FIGURE 1 being traced out by a moving particle whose position at time t is s f std, tstd, hstdd. If we
C is traced out by the tip of a moving now consider the vector function rstd − k f std, tstd, hstdl, then rstd is the position vector
position vector rstd. of the point Ps f std, tstd, hstdd on C. Thus any continuous vector function r defines a space
curve C that is traced out by the tip of the moving vector rstd, as shown in Figure 1.
TEC Visual 13.1A shows several Example 3 Describe the curve defined by the vector function
curves being traced out by position
vectors, including those in Figures 1 rstd − k1 1 t, 2 1 5t, 21 1 6t l
and 2. SOLUTION The corresponding parametric equations are
x − 1 1 t y − 2 1 5t z − 21 1 6t
which we recognize from Equations 12.5.2 as parametric equations of a line passing
through the point s1, 2, 21d and parallel to the vector k1, 5, 6l. Alternatively, we could
observe that the function can be written as r − r0 1 tv, where r0 − k1, 2, 21l and
v − k1, 5, 6l, and this is the vector equation of a line as given by Equation 12.5.1. ■
Plane curves can also be represented in vector notation. For instance, the curve given
by the parametric equations x − t2 2 2t and y − t 1 1 (see Example 10.1.1) could also
be described by the vector equation
rstd − k t2 2 2t, t 1 1l − st2 2 2td i 1 st 1 1d j
where i − k1, 0l and j − k0, 1l.
z Example 4 Sketch the curve whose vector equation is
x (1, 0, 0) rstd − cos t i 1 sin t j 1 t k
SOLUTION The parametric equations for this curve are
FIGURE 2
x − cos t y − sin t z − t
”0, 1, π ’ Since x2 1 y2 − cos2t 1 sin2t − 1 for all values of t, the curve must lie on the circular
2 cylinder x2 1 y2 − 1. The point sx, y, zd lies directly above the point sx, y, 0d, which
y moves counterclockwise around the circle x2 1 y2 − 1 in the xy-plane. (The projection
of the curve onto the xy-plane has vector equation rstd − k cos t, sin t, 0l. See Example
10.1.2.) Since z − t, the curve spirals upward around the cylinder as t increases. The
curve, shown in Figure 2, is called a helix.
■
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850 Chapter 13 Vector Functions
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 struc-
ture of the DNA molecule is that of two linked, parallel helixes that are intertwined 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.
FIGURE 3 Example 5 Find a vector equation and parametric equations for the line segment that
A double helix
joins the point Ps1, 3, 22d to the point Qs2, 21, 3d.
Figure 4 shows the line segment PQ in SOLUTION In Section 12.5 we found a vector equation for the line segment that joins
Example 5. the tip of the vector r0 to the tip of the vector r1:
z rstd − s1 2 td r0 1 tr1 0 < t < 1
Q(2, _1, 3) (See Equation 12.5.4.) Here we take r0 − k1, 3, 22 l and r1 − k2, 21, 3l to obtain a
vector equation of the line segment from P to Q:
rstd − s1 2 td k1, 3, 22l 1 tk2, 21, 3l 0 < t < 1
y or rstd − k1 1 t, 3 2 4t, 22 1 5tl 0<t<1
x The corresponding parametric equations are
P(1, 3, _2) x − 1 1 t y − 3 2 4t z − 22 1 5t 0 < t < 1 ■
FIGURE 4
Example 6 Find a vector function that represents the curve of intersection of the
cylinder x 2 1 y 2 − 1 and the plane y 1 z − 2.
SOLUTION Figure 5 shows how the plane and the cylinder intersect, and Figure 6
shows the curve of intersection C, which is an ellipse.
z z
y+z=2 (0, _1, 3)
(_1, 0, 2)
C
(1, 0, 2)
(0, 1, 1)
≈+¥=1 0
x yx y
FIGURE 5 FIGURE 6
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Section 13.1 Vector Functions and Space Curves 851
The projection of C onto the xy-plane is the circle x2 1 y2 − 1, z − 0. So we know
from Example 10.1.2 that we can write
x − cos t y − sin t 0 < t < 2
From the equation of the plane, we have
z − 2 2 y − 2 2 sin t
So we can write parametric equations for C as
x − cos t y − sin t z − 2 2 sin t 0 < t < 2
The corresponding vector equation is
rstd − cos t i 1 sin t j 1 s2 2 sin td k 0 < t < 2
This equation is called a parametrization of the curve C. The arrows in Figure 6 indi-
cate 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
x − s4 1 sin 20td cos t y − s4 1 sin 20td sin t z − cos 20t
It’s called a toroidal spiral because it lies on a torus. Another interesting curve, the tre-
foil knot, with equations
x − s2 1 cos 1.5td cos t y − s2 1 cos 1.5td sin t z − sin 1.5t
is graphed in Figure 8. It wouldn’t be easy to plot either of these curves by hand.
zz
yx y
x
FIGURE 7 FIGURE 8
A toroidal spiral A trefoil knot
Even when a computer is used to draw a space curve, optical illusions make it difficult
to get a good impression of what the curve really looks like. (This is especially true in
Figure 8. See Exercise 52.) The next example shows how to cope with this problem.
Example 7 Use a computer to draw the curve with vector equation rstd − kt, t2, t3l.
This curve is called a twisted cubic.
SOLUTION We start by using the computer to plot the curve with parametric equations
x − t, y − t2, z − t3 for 22 < t < 2. 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-dimensional
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
852 Chapter 13 Vector Functions
computer graphing programs allow the user to enclose a curve or surface 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.
z _2 6 6
6 2 z0
_6 z0
x 2 _6
0 _2
4y y2 0x _6 _2
(a) 8 42 0 y2 (c) 4 2 0x
_2 (b)
8
_1 4 4
0x z 0 z0
1 _4 _4
2 _8 1 0 _1 _2 _8 123 4
0 1 234 2 x 0 y
(e) (f )
y
(d) We get an even better idea of the curve when we view it from different vantage
FIGURE 9 Views of the twisted cubic points. Part (c) shows the result of rotating the box to give another viewpoint. Parts (d),
TEC In Visual 13.1B you can rotate (e), and (f ) show the views we get when we look directly at a face of the box. In par-
the box in Figure 9 to see the curve
from any viewpoint. ticular, part (d) shows the view from directly above the box. It is the projection of the
curve onto the xy-plane, namely, the parabola y − x2. Part (e) shows the projection
onto the xz-plane, the cubic curve z − x3. It’s now obvious why the given curve is
called a twisted cubic. ■
Another method of visualizing a space curve is to draw it on a surface. For instance,
the twisted cubic in Example 7 lies on the parabolic cylinder y − x2. (Eliminate the
parameter from the first two parametric equations, x − t and y − t2.) Figure 10 shows
both the cylinder and the twisted cubic, and we see that the curve moves upward from the
origin along the surface of the cylinder. We also used this method in Example 4 to visual-
ize the helix lying on the circular cylinder (see Figure 2).
z
x
y
FIGURE 10
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Section 13.1 Vector Functions and Space Curves 853
A third method for visualizing the twisted cubic is to realize that it also lies on the
cylinder z − x3. So it can be viewed as the curve of intersection of the cylinders y − x2
and z − x3. (See Figure 11.)
TEC Visual 13.1C shows how curves 8
arise as intersections of surfaces. 4
z0
_4
FIGURE 11 _8 1 0 2y 4
_1 x 0
Some computer algebra systems pro- We have seen that an interesting space curve, the helix, occurs in the model of DNA.
vide us with a clearer picture of a space Another notable example of a space curve in science is the trajectory of a positively
curve by enclosing it in a tube. Such charged particle in orthogonally oriented electric and magnetic fields E and B. Depend-
a plot enables us to see whether one ing on the initial velocity given the particle at the origin, the path of the particle is either
part of a curve passes in front of or a space curve whose projection onto the horizontal plane is the cycloid we studied in
behind another part of the curve. For Section 10.1 [Figure 12(a)] or a curve whose projection is the trochoid investigated in
example, Figure 13 shows the curve Exercise 10.1.40 [Figure 12(b)].
of Figure 12(b) as rendered by the
tubeplot command in Maple.
BB
EE
t t
(a) r(t) = kt-sin t, 1-cos t, tl (b) r(t) = kt- 23 sin t, 1- 32 cos t, tl
FIGURE 12 FIGURE 13
Motion of a charged particle in
orthogonally oriented electric and For further details concerning the physics involved and animations of the trajectories
magnetic fields of the particles, see the following websites:
■ www.physics.ucla.edu/plasma-exp/Beam/
■ www.phy.ntnu.edu.tw/ntnujava/index.php?topic=36
1–2 Find the domain of the vector function. 3–6 Find the limit.
K L 1. rstd − S D 3. lim
lnst 1 1d, t , 2 t tl0e23t i t2
s9 2 t 2 1 sin2t j 1 cos 2t k
2. rstd − cos t i 1 ln t j 1 t 1 2 k S D 4. limt2 2 ti 1 st 1 8 j 1 sin t k
2 tl1t21 ln t
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
854 chapter 13 Vector Functions
K L 5. lim 1 1 t 2 1 2 e22t 21. x − t cos t, y − t, z − t sin t, t > 0
tl` 1 2 t 2 t 22. x − cos t, y − sin t, z − 1ys1 1 t 2d
, tan21 t, 23. x − t, y − 1ys1 1 t 2 d, z − t 2
24. x − cos t, y − sin t, z − cos 2t
K L 6. lim te2t, t3 1 t , t sin 1 25. x − cos 8t, y − sin 8t, z − e 0.8t, t > 0
tl` 2t3 2 1 t 26. x − cos2 t, y − sin2 t, z − t
7–14 Sketch the curve with the given vector equation. Indicate 27. S how that the curve with parametric equations x − t cos t,
with an arrow the direction in which t increases. y − t sin t, z − t lies on the cone z2 − x 2 1 y 2, and use this
fact to help sketch the curve.
7. rstd − ksin t, t l 8. rstd − kt 2 2 1, tl
9. rstd − kt, 2 2 t, 2tl 10. rstd − ksin t, t, cos tl 28. Show that the curve with parametric equations x − sin t,
11. rstd − k3, t, 2 2 t 2l y − cos t, z − sin2t is the curve of intersection of the surfaces
12. rstd − 2 cos t i 1 2 sin t j 1 k z − x 2 and x 2 1 y 2 − 1. Use this fact to help sketch the curve.
13. rstd − t 2 i 1 t 4 j 1 t 6 k
14. rstd − cos t i 2 cos t j 1 sin t k 29. Find three different surfaces that contain the curve
rstd − 2t i 1 e t j 1 e 2t k.
15–16 Draw the projections of the curve on the three coordinate
planes. Use these projections to help sketch the curve. 30. Find three different surfaces that contain the curve
rstd − t 2 i 1 ln t j 1 s1ytd k.
15. rstd − kt, sin t, 2 cos tl 16. rstd − kt, t, t 2l
31. At what points does the curve rstd − t i 1 s2t 2 t 2d k inter-
17–20 Find a vector equation and parametric equations for the sect the paraboloid z − x 2 1 y 2?
line segment that joins P to Q.
32. At what points does the helix rstd − ksin t, cos t, tl intersect
17. Ps2, 0, 0d, Qs6, 2, 22d 18. Ps21, 2, 22d, Qs23, 5, 1d the sphere x 2 1 y 2 1 z2 − 5?
20. Psa, b, cd, Qsu, v, wd
19. Ps0, 21, 1d, Q ( 12, 31, 1 ) ; 33–37 Use a computer to graph the curve with the given vector
4 equation. Make sure you choose a parameter domain and view-
points that reveal the true nature of the curve.
21–26 Match the parametric equations with the graphs 33. rstd − kcos t sin 2t, sin t sin 2t, cos 2tl
(labeled I–VI). Give reasons for your choices.
34. rstd − k te t, e2t, tl
Iz II z
k l
35. rstd − sin 3t cos t, 1 t, sin 3t sin t
4
36. rstd − kcoss8 cos td sin t, sins8 cos td sin t, cos tl
yx y 37. rstd − kcos 2t, cos 3t, cos 4tl
x
III z IV z ; 38. Graph the curve with parametric equations x − sin t,
y − sin 2t, z − cos 4 t. Explain its shape by graphing its
x y projections onto the three coordinate planes.
Vz
x y ; 39. Graph the curve with parametric equations
VI x − s1 1 cos 16td cos t
z
y − s1 1 cos 16td sin t
z − 1 1 cos 16t
Explain the appearance of the graph by showing that it lies on
a cone.
; 40. Graph the curve with parametric equations
x − s1 2 0.25 cos 2 10t cos t
xy y − s1 2 0.25 cos 2 10t sin t
xy z − 0.5 cos 10t
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Section 13.2 Derivatives and Integrals of Vector Functions 855
Explain the appearance of the graph by showing that it lies ; 51. (a) Graph the curve with parametric equations
on a sphere.
x − 27 sin 8t 2 8 sin 18t
41. Show that the curve with parametric equations x − t 2, 26 39
y − 1 2 3t, z − 1 1 t 3 passes through the points s1, 4, 0d
and s9, 28, 28d but not through the point s4, 7, 26d. y − 2 27 cos 8t 1 8 cos 18t
26 39
42–46 Find a vector function that represents the curve of
intersection of the two surfaces. z − 144 sin 5t
65
42. The cylinder x 2 1 y 2 − 4 and the surface z − xy
(b) Show that the curve lies on the hyperboloid of one sheet
43. The cone z − sx 2 1 y 2 and the plane z − 1 1 y 144x 2 1 144y 2 2 25z 2 − 100.
44. The paraboloid z − 4x 2 1 y 2 and the parabolic
52. The view of the trefoil knot shown in Figure 8 is accurate,
cylinder y − x 2 but it doesn’t reveal the whole story. Use the parametric
45. The hyperboloid z − x 2 2 y 2 and the cylinder x 2 1 y 2 − 1 equations
46. The semiellipsoid x 2 1 y 2 1 4z 2 − 4, y > 0, and the x − s2 1 cos 1.5td cos t
cylinder x 2 1 z 2 − 1 y − s2 1 cos 1.5td sin t
; 47. Try to sketch by hand the curve of intersection of the circu- z − sin 1.5t
lar cylinder x 2 1 y 2 − 4 and the parabolic cylinder z − x 2.
Then find parametric equations for this curve and use these to sketch the curve by hand as viewed from above, with
equations and a computer to graph the curve. gaps indicating where the curve passes over itself. Start by
showing that the projection of the curve onto the xy-plane
; 48. T ry to sketch by hand the curve of intersection of the has polar coordinates r − 2 1 cos 1.5t and − t, so r
parabolic cylinder y − x 2 and the top half of the ellipsoid varies between 1 and 3. Then show that z has maximum and
x 2 1 4y 2 1 4z2 − 16. Then find parametric equations for minimum values when the projection is halfway between
this curve and use these equations and a computer to graph r − 1 and r − 3.
the curve.
; When you have finished your sketch, use a computer to
49. If two objects travel through space along two different draw the curve with viewpoint directly above and compare
curves, it’s often important to know whether they will col- with your sketch. Then use the computer to draw the curve
lide. (Will a missile hit its moving target? Will two aircraft from several other viewpoints. You can get a better impres-
collide?) The curves might intersect, but we need to know sion of the curve if you plot a tube with radius 0.2 around
whether the objects are in the same position at the same the curve. (Use the tubeplot command in Maple or the
time. Suppose the trajectories of two particles are given by tubecurve or Tube command in Mathematica.)
the vector functions
53. Suppose u and v are vector functions that possess limits as
r1 std − kt 2, 7t 2 12, t 2l r2 std − k4t 2 3, t 2, 5t 2 6l
t l a and let c be a constant. Prove the following prope rties
for t > 0. Do the particles collide?
of limits.
50. Two particles travel along the space curves
(a) lim fustd 1 vstdg − lim ustd 1 lim vstd
r1 std − kt, t 2, t 3l r2 std − k1 1 2t, 1 1 6t, 1 1 14tl t la t la t la
Do the particles collide? Do their paths intersect? (b) lim custd − c lim ustd
tla tla
(c) lim fustd ? vstdg − lim ustd ? lim vstd
tla tla tla
(d) lim fustd 3 vstdg − lim ustd 3 lim vstd
tla tla tla
54. Show that lim t l a rstd − b if and only if for every « . 0
there is a number ␦ . 0 such that
if 0 , | t 2 a | , |␦ then rstd 2 b | , «
Later in this chapter we are going to use vector functions to describe the motion of plan-
ets and other objects through space. Here we prepare the way by developing the calculus
of vector functions.
Derivatives
The derivative r9 of a vector function r is defined in much the same way as for real-
valued functions:
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856 Chapter 13 Vector Functions
dr − r9std − lim rst 1 hd 2 rstd
1 dt h
hl0
if this limit exists. The geometric significance of this definition PilsQsrheopwrenseinntsFtihgeurveec1-.
If the points P and Q have position vectors rstd and rst 1 hd, then
tor rst 1 hd 2 rstd, which can therefore be regarded as a secant vector. If h . 0, the
scalar multiple s1yhdsrst 1 hd 2 rstdd has the same direction as rst 1 hd 2 rstd. As
Notice that when 0 , h , 1, h l 0, it appears that this vector approaches a vector that lies on the tangent line. For
multiplying the secant vector by 1yh
stretches the vector, as shown this reason, the vector r9std is called the tangent vector to the curve defined by r at the
in Figure 1(b). point P, provided that r9std exists and r9std ± 0. The tangent line to C at P is defined to
be the line through P parallel to the tangent vector r9std. We will also have occasion to
TEC Visual 13.2 shows an animation consider the unit tangent vector, which is
of Figure 1.
Tstd − r9std
| r9std|
z r(t+h)-r(t) z r ( t+h )-r ( t )
PQ rª(t) h
r(t) PQ
r(t+h)
r(t)
C r(t+h)
0
C
0
FIGURE 1 xy xy
(a) The secant vector PQ (b) The tangent vector rª(t)
The following theorem gives us a convenient method for computing the derivative of
a vector function r: just differentiate each component of r.
2 Theorem If rstd − k f std, tstd, hstdl − f std i 1 tstd j 1 hstd k, where f , t,
and h are differentiable functions, then
r9std − k f 9std, t9std, h9stdl − f 9std i 1 t9std j 1 h9std k
Proof
r9std − lim 1 frst 1 Dtd 2 rstdg
Dt
Dt l 0
− lim 1 fk f st 1 Dtd, tst 1 Dtd, hst 1 Dtdl 2 k f std, tstd, hstdlg
Dt
Dt l 0
K L− lim f st 1 Dtd 2 f std , tst 1 Dtd 2 tstd , hst 1 Dtd 2 hstd
Dt l 0 Dt Dt Dt
K L− lim f st 1 Dtd 2 f std , lim tst 1 Dtd 2 tstd , lim hst 1 Dtd 2 hstd
Dt Dt Dt
Dt l 0 Dt l 0 Dt l 0
− k f 9std, t9std, h9stdl ■
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Section 13.2 Derivatives and Integrals of Vector functions 857
Example 1
(a) Find the derivative of rstd − s1 1 t3 d i 1 te2t j 1 sin 2t k.
(b) Find the unit tangent vector at the point where t − 0.
SOLUTION
(a) According to Theorem 2, we differentiate each component of r:
r9std − 3t 2 i 1 s1 2 tde2t j 1 2 cos 2t k
(b) Since rs0d − i and r9s0d − j 1 2k, the unit tangent vector at the point s1, 0, 0d is
Ts0d − r9s0d j 1 2k 1 j1 2 k
| | r9s0d − s1 1 4 − s5 s5 ■
y Example 2 For the curve rstd − st i 1 s2 2 td j, find r9std and sketch the position
2
vector rs1d and the tangent vector r9s1d.
(1, 1) SOLUTION We have
r(1) rª(1)
r9std − 1 i 2 j and r9s1d − 1 i2j
2 st 2
01 x The curve is a plane curve and elimination of the parameter from the equations
FIGURE 2 x − st , y − 2 2 t gives y − 2 2 x2, x > 0. In Figure 2 we draw the position vector
rs1d − i 1 j starting at the origin and the tangent vector r9s1d starting at the correspond-
Notice from Figure 2 that the tangent ing point s1, 1d.
vector points in the direction of ■
increasing t. (See Exercise 58.)
Example 3 Find parametric equations for the tangent line to the helix with para-
metric equations
x − 2 cos t y − sin t z − t
at the point s0, 1, y2d.
SOLUTION The vector equation of the helix is rstd − k2 cos t, sin t, tl, so
r9std − k22 sin t, cos t, 1l
The parameter value corresponding to the point s0, 1, y2d is t − y2, so the tangent
vector there is r9sy2d − k22, 0, 1l. The tangent line is the line through s0, 1, y2d
parallel to the vector k22, 0, 1l, so by Equations 12.5.2 its parametric equations are
x − 22t y − 1 z − 1 t ■
2
12
The helix and the tangent line in 8
Example 3 are shown in Figure 3. z
FIGURE 3 4
_01 _0.5 0 _2
y 0.5 1 2 0 x
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858 Chapter 13 Vector Functions
In Section 13.4 we will see how r9std Just as for real-valued functions, the second derivative of a vector function r is the
and r0std can be interpreted as the derivative of r9, that is, r0 − sr9d9. For instance, the second derivative of the function in
velocity and acceleration vectors of Example 3 is
a particle moving through space with r0std − k22 cos t, 2sin t, 0l
position vector rstd at time t.
Differentiation Rules
The next theorem shows that the differentiation formulas for real-valued functions have
their counterparts for vector-valued functions.
3 Theorem Suppose u and v are differentiable vector functions, c is a scalar,
and f is a real-valued function. Then
1. d fustd 1 vstdg − u9std 1 v9std
dt
2. d fcustdg − cu9std
dt
3. d f f std ustdg − f 9std ustd 1 f std u9std
dt
4. d fustd ? vstdg − u9std ? vstd 1 ustd ? v9std
dt
5. d fustd 3 vstdg − u9std 3 vstd 1 ustd 3 v9std
dt
6. d fus f stddg − f 9stdu9s f stdd (Chain Rule)
dt
This theorem can be proved either directly from Definition 1 or by using Theorem 2
and the corresponding differentiation formulas for real-valued functions. The proof of
Formula 4 follows; the remaining formulas are left as exercises.
Proof of Formula 4 Let
ustd − k f1std, f2std, f3stdl vstd − k t1std, t2std, t3stdl
Then o3
ustd ? vstd − f1std t1std 1 f2std t2std 1 f3std t3std − fistd tistd
i−1
so the ordinary Product Rule gives
d 3 3d
dt i−1 dt
fistd tistd
i−1
o odfustd ? vstdg − − f fistd tistdg
dt
o3
− f fi9std tistd 1 fistd t9istdg
i−1
o o3 3
− fi9std tistd 1 fistd t9istd
i−1 i−1
− u9std ? vstd 1 ustd ? v9std ■
| |Example 4 Show that if rstd − c (a constant), then r9std is orthogonal to rstd for
all t.
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Section 13.2 Derivatives and Integrals of Vector Functions 859
z SOLUTION Since | |rstd ? rstd − rstd 2 − c2
r(t) rª(t) and c2 is a constant, Formula 4 of Theorem 3 gives
y
0 − d frstd ? rstdg − r9std ? rstd 1 rstd ? r9std − 2r9std ? rstd
x dt
FIGURE 4 Thus r9std ? rstd − 0, which says that r9std is orthogonal to rstd.
Geometrically, this result says that if a curve lies on a sphere with center the origin,
then the tangent vector r9std is always perpendicular to the position vector rstd. (See
Figure 4.) ■
Integrals
The definite integral of a continuous vector function rstd 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 , t, and h
as follows. (We use the notation of Chapter 5.)
y ob rstd dt − lim n rsti*d Dt
a i−1
nl `
FS D S D S D G− lim
nl `
on i1 on j1 on k
f sti*d Dt tsti*d Dt hsti*d Dt
i−1 i−1 i−1
and so
S D S D S Dy y y yb rstd dt − b f std dt i 1 b tstd dt j 1 b hstd dt k
aa a a
This means that we can evaluate an integral of a vector function by integrating each
component function.
We can extend the Fundamental Theorem of Calculus to continuous vector functions
as follows:
y gbrstd dt − Rstd b − Rsbd 2 Rsad
a a
where R is an antiderivative of r, that is, R9std − rstd. We use the notation y rstd dt for
indefinite integrals (antiderivatives).
Example 5 If rstd − 2 cos t i 1 sin t j 1 2t k, then
S D S D S Dy rstd dt − y 2 cos t dt i 1 y sin t dt j 1 y 2t dt k
− 2 sin t i 2 cos t j 1 t2 k 1 C
where C is a vector constant of integration, and
y f g y2 2
y2 rstd dt − 2 sin t i 2 cos t j 1 t 2 k 0 − 2 i 1 j 1 4 k ■
0
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860 Chapter 13 Vector Functions
1. The figure shows a curve C given by a vector function rstd. 13. rstd − t sin t i 1 e t cos t j 1 sin t cos t k
(a) Draw the vectors rs4.5d 2 rs4d and rs4.2d 2 rs4d. 14. rstd − sin2 at i 1 te bt j 1 cos2ct k
(b) Draw the vectors 15. rstd − a 1 t b 1 t 2 c
16. rstd − t a 3 sb 1 t cd
rs4.5d 2 rs4d and rs4.2d 2 rs4d
0.5 0.2
(c) Write expressions for r9s4d and the unit tangent 17–20 Find the unit tangent vector Tstd at the point with the
vector Ts4d. given value of the parameter t.
(d) Draw the vector Ts4d.
y 17. rstd − kt2 2 2t, 1 1 3t, 1 t 3 1 l1t2 , t−2
CR 3
2
r(4.5) 18. rstd − ktan21 t, 2e 2t, 8te t l, t − 0
1 Q 19. rstd − cos t i 1 3t j 1 2 sin 2t k, t − 0
r(4.2) 20. rstd − sin2t i 1 cos2t j 1 tan2 t k, t − y4
P 21. If rstd − kt, t 2, t 3l, find r9std, Ts1d, r0std, and r9std 3 r0std.
r(4) 22. If rstd − ke 2t, e22t, te 2t l, find Ts0d, r0s0d, and r9std ? r0std.
0 1x 23–26 Find parametric equations for the tangent line to the curve
with the given parametric equations at the specified point.
2. (a) Make a large sketch of the curve described by the vector 23. x − t 2 1 1, y − 4st , z − e t 22t; s2, 4, 1d
function rstd − kt 2, t l, 0 < t < 2, and draw the vectors 24. x − lnst 1 1d, y − t cos 2t, z − 2t; s0, 0, 1d
rs1d, rs1.1d, and rs1.1d 2 rs1d. 25. x − e2t cos t, y − e2t sin t, z − e2t; s1, 0, 1d
26. x − st 2 1 3 , y − lnst 2 1 3d, z − t; s2, ln 4, 1d
(b) Draw the vector r9s1d starting at (1, 1), and compare it
with the vector 27. Find a vector equation for the tangent line to the curve of
rs1.1d 2 rs1d intersection of the cylinders x 2 1 y 2 − 25 and y 2 1 z 2 − 20
0.1 at the point s3, 4, 2d.
Explain why these vectors are so close to each other in 28. Find the point on the curve rstd − k2 cos t, 2 sin t, e t l,
length and direction. 0 < t < , where the tangent line is parallel to the plane
s3 x 1 y − 1.
3–8
(a) Sketch the plane curve with the given vector equation. CAS 29–31 Find parametric equations for the tangent line to the curve
(b) Find r9std. with the given parametric equations at the specified point. Illus-
(c) Sketch the position vector rstd and the tangent vector r9std for trate by graphing both the curve and the tangent line on a common
screen.
the given value of t.
29. x − t, y − e2t, z − 2t 2 t 2; s0, 1, 0d
3. rstd − kt 2 2, t 2 1 1l, t − 21
30. x − 2 cos t, y − 2 sin t, z − 4 cos 2t; ss3 , 1, 2d
4. rstd − kt 2, t 3l, t − 1
31. x − t cos t, y − t, z − t sin t; s2, , 0d
5. rstd − e2t i 1 et j, t − 0
6. rstd − e t i 1 2t j, t − 0
7. rstd − 4 sin t i 2 2 cos t j, t − 3y4
8. rstd − scos t 1 1d i 1 ssin t 2 1d j, t − 2y3
9–16 Find the derivative of the vector function. 32. (a) Find the point of intersection of the tangent lines to the
curve rstd − ksin t, 2 sin t, cos tl at the points
9. rstd − kst 2 2 , 3, 1yt 2l where t − 0 and t − 0.5.
10. rstd − ke2t, t 2 t 3, ln tl ; (b) Illustrate by graphing the curve and both tangent lines.
11. rstd − t 2 i 1 cosst 2d j 1 sin2t k 33. The curves r1std − kt, t 2, t 3l and r2std − ksin t, sin 2t, tl
intersect at the origin. Find their angle of intersection correct
12. rstd − 1 1 t i 1 1 t t j 1 1 t2 t k to the nearest degree.
1 1 1
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Section 13.3 Arc Length and Curvature 861
34. At what point do the curves r1std − kt, 1 2 t, 3 1 t 2l and 48. I f u and v are the vector functions in Exercise 47, use For-
r2ssd − k3 2 s, s 2 2, s 2l intersect? Find their angle of
intersection correct to the nearest degree. mula 5 of Theorem 3 to find
d fustd 3 vstdg
dt
35–40 Evaluate the integral.
49. Find f 9s2d, where f std − ustd ? vstd, us2d − k1, 2, 21l,
y 35. 2 st i 2 t 3 j 1 3t 5 kd dt u9s2d − k3, 0, 4l, and vstd − kt, t 2, t 3l.
0
y 36. 4 (2t 3y2 i 1 st 1 1d st k) dt 50. I f rstd − ustd 3 vstd, where u and v are the vector functions
1 in Exercise 49, find r9s2d.
S Dy 37.
1 t 1 1 i 1 t2 1 1 j 1 t2 t 1 k dt 51. If rstd − a cos t 1 b sin t, where a and b are constant
0 1 1 1 vectors, show that rstd 3 r9std − a 3 b.
y 38. y4 ssec t tan t i 1 t cos 2t j 1 sin2 2t cos 2t kd dt 52. If r is the vector function in Exercise 51, show that
0 r 0std 1 2rstd − 0.
y 39. ssec2 t i 1 tst 2 1 1d3 j 1 t 2 ln t kd dt 53. Show that if r is a vector function such that r0 exists, then
S Dy 40. t 1 d frstd 3 r9stdg − rstd 3 r0std
2 s1 2 dt
te 2t i 1 j 1 k dt
1 t t2 d
54. Find an expression for dt fustd ? svstd 3 wstddg.
| | | | 55. d 1
41. Find rstd if r9std − 2t i 1 3t 2 j 1 st k and rs1d − i 1 j. If rstd ± 0, show that dt rstd − rstd rstd ? r9std.
42. Find rstd if r9std − t i 1 e t j 1 te t k and rs0d − i 1 j 1 k. | | [Hint: rstd 2 − rstd ? rstd]
43. Prove Formula 1 of Theorem 3. 56. If a curve has the property that the position vector rstd is
always perpendicular to the tangent vector r9std, show that
44. Prove Formula 3 of Theorem 3. the curve lies on a sphere with center the origin.
45. Prove Formula 5 of Theorem 3. 57. If ustd − rstd ? fr9std 3 r0stdg, show that
46. Prove Formula 6 of Theorem 3. u9std − rstd ? fr9std 3 r-stdg
47. If ustd − ksin t, cos t, tl and vstd − kt, cos t, sin tl, use 58. S how that the tangent vector to a curve defined by a vector
Formula 4 of Theorem 3 to find function rstd points in the direction of increasing t.
[Hint: Refer to Figure 1 and consider the cases h . 0 and
d fustd ? vstdg
dt h , 0 separately.]
z Length of a Curve
0 In Section 10.2 we defined the length of a plane curve with parametric equations x − f std,
y y − tstd, a < t < b, as the limit of lengths of inscribed polygons and, for the case where
f 9 and t9 are continuous, we arrived at the formula
x
y ÎS D S Dy1 dx 2 dy 2
FIGURE 1 L − b sf f 9stdg2 1 ft9stdg2 dt − b dt dt
The length of a space curve is the limit 1 dt
of lengths of inscribed polygons. aa
The length of a space curve is defined in exactly the same way (see Figure 1). Suppose
that the curve has the vector equation rstd − k f std, tstd, hstdl, a < t < b, or, equivalently,
the parametric equations x − f std, y − tstd, z − hstd, where f 9, t9, and h9 are continu-
ous. If the curve is traversed exactly once as t increases from a to b, then it can be shown
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862 Chapter 13 Vector Functions
that its length is
y ÎS D S D S Dy2 L − b sf f 9stdg2 1 ft9stdg2 1 fh9stdg2 dt
a
b dx 2 dy 2 dz 2
dt 1 dt 1 dt
− dt
a
Notice that both of the arc length formulas (1) and (2) can be put into the more com-
pact form
In the next section we will see that if | |y3 L − b r9std dt
rstd is the position vector of a moving a
| |object at time t, then r9std is the velocity because, for plane curves rstd − f std i 1 tstd j,
vector and r9std is the speed. Thus | | | |r9std − f 9std i 1 t9std j − sf f 9stdg2 1 ft9stdg2
Equation 3 says that to compute dis-
tance traveled, we integrate speed.
and for space curves rstd − f std i 1 tstd j 1 hstd k,
| | | |r9std − f 9std i 1 t9std j 1 h9std k − sf f 9stdg2 1 ft9stdg2 1 fh9stdg2
Figure 2 shows the arc of the helix Example 1 Find the length of the arc of the circular helix with vector equation
whose length is computed in
Example 1. rstd − cos t i 1 sin t j 1 t k from the point s1, 0, 0d to the point s1, 0, 2d.
z SOLUTION Since r9std − 2sin t i 1 cos t j 1 k, we have
| |r9std − ss2sin td2 1 cos2t 1 1 − s2
(1, 0, 2π) The arc from s1, 0, 0d to s1, 0, 2d is described by the parameter interval 0 < t < 2
and so, from Formula 3, we have
(1, 0, 0)
x | |y y
L − 2 r9std dt − 2 s2 dt − 2 s2 ■
FIGURE 2 00
y A single curve C can be represented by more than one vector function. For instance,
the twisted cubic
4 r1std − kt, t 2, t 3l 1 < t < 2
could also be represented by the function
5 r2sud − ke u, e 2u, e 3ul 0 < u < ln 2
where the connection between the parameters t and u is given by t − eu. We say that
Equations 4 and 5 are parametrizations of the curve C. If we were to use Equation 3 to
comp ute the length of C using Equations 4 and 5, we would get the same answer. In
general, it can be shown that when Equation 3 is used to compute arc length, the answer
is independent of the parametrization that is used.
The Arc Length Function
Now we suppose that C is a curve given by a vector function
rstd − f stdi 1 tstd j 1 hstdk a < t < b
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Section 13.3 Arc Length and Curvature 863
z where r9 is continuous and C is traversed exactly once as t increases from a to b. We
s(t)
define its arc length function s by
r(t) ÎS D S D S Dy | | yC
r(a) 6 sstd − t r9sud du − t dx 2 dy 2 dz 2
du 1 du 1 du
0 aa du
x
Thus sstd is the length of the part of C between rsad and rstd. (See Figure 3.) If we dif-
FIGURE 3
ferentiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus,
y we obtain
7 ds − | r9std |
dt
It is often useful to parametrize a curve with respect to arc length because arc
length arises naturally from the shape of the curve and does not depend on a particular
coordinate system. If a curve rstd is already given in terms of a parameter t and sstd is the
arc length function given by Equation 6, then we may be able to solve for t as a function
of s: t − tssd. Then the curve can be reparametrized in terms of s by substituting for t:
r − rstssdd. Thus, if s − 3 for instance, rsts3dd is the position vector of the point 3 units
of length along the curve from its starting point.
Example 2 Reparametrize the helix rstd − cos t i 1 sin t j 1 t k with respect to arc
length measured from s1, 0, 0d in the direction of increasing t.
SOLUTION The initial point s1, 0, 0d corresponds to the parameter value t − 0. From
Example 1 we have
ds − | r9std | − s2
dt
and so | |y ys − sstd − t r9sud du − t s2 du − s2 t
00
Therefore t − sys2 and the required reparametrization is obtained by substituting for t:
rstssdd − cosssys2 d i 1 sinssys2 d j 1 ssys2 d k ■
Curvature
TEC Visual 13.3A shows animated A parametrization rstd is called smooth on an interval I if r9 is continuous and r9std ± 0
unit tangent vectors, like those in on I. A curve is called smooth if it has a smooth parametrization. A smooth curve has no
Figure 4, for a variety of plane curves
and space curves. sharp corners or cusps; when the tangent vector turns, it does so continuously.
If C is a smooth curve defined by the vector function r, recall that the unit tangent
z
vector Tstd is given by
Tstd − r9std
| r9std |
0C y and indicates the direction of the curve. From Figure 4 you can see that Tstd changes
x direction very slowly when C is fairly straight, but it changes direction more quickly
FIGURE 4 when C bends or twists more sharply.
Unit tangent vectors at equally spaced
The curvature of C at a given point is a measure of how quickly the curve changes
points on C
direction at that point. Specifically, we define it to be the magnitude of the rate of change
of the unit tangent vector with respect to arc length. (We use arc length so that the curva-
ture will be independent of the parametrization.) Because the unit tangent vector has
constant length, only changes in direction contribute to the rate of change of T.
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864 Chapter 13 Vector Functions
Z Z8 Definition The curvature of a curve is−dT
ds
where T is the unit tangent vector.
The curvature is easier to compute if it is expressed in terms of the parameter t instead
of s, so we use the Chain Rule (Theorem 13.2.3, Formula 6) to write
Z Z Z ZdTdTds dT d Tydt
ds dt ds dsydt
dt
− and − −
| |But dsydt − r9std from Equation 7, so
9 std − ||Tr99ssttdd||
Example 3 Show that the curvature of a circle of radius a is 1ya.
SOLUTION We can take the circle to have center the origin, and then a parametrization
is
rstd − a cos t i 1 a sin t j
Therefore | |r9std − 2a sin t i 1 a cos t j and r9std − a
Tstd − r9std − 2sin t i 1 cos t j
| |so r9std
and T9std − 2cos t i 2 sin t j
| |This gives T9std − 1, so using Formula 9, we have
std − ||Tr99ssttdd|| − 1 ■
a
The result of Example 3 shows that small circles have large curvature and large circles
have small curvature, in accordance with our intuition. We can see directly from the defi
nition of curvature that the curvature of a straight line is always 0 because the tangent
vect or is constant.
Although Formula 9 can be used in all cases to compute the curvature, the formula
given by the following theorem is often more convenient to apply.
10 Theorem The curvature of the curve given by the vector function r is
std − | r9std 3 r0std |
| r9std |3
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Section 13.3 Arc Length and Curvature 865
| | | |Proof Since T − r9y r9 and r9 − dsydt, we have
r9 − | r9|T − ds T
dt
so the Product Rule (Theorem 13.2.3, Formula 3) gives
r0 − d 2s T 1 ds T9
dt 2 dt
S DUsing the fact that T 3 T − 0 (see Example 12.4.2), we have
r9 3 r0 − ds 2
dt
sT 3 T9d
| |Now Tstd − 1 for all t, so T and T9 are orthogonal by Example 13.2.4. Therefore, by
S D S D S DTheorem 12.4.9,
| r9 3 r0 | − ds 2 | T 3 T9 | − ds 2 | T | | T9 | − ds 2 | T9 |
dt dt dt
Thus | T9 | − | r9 3 r0 | − | r9 3 r0 |
| r9|2
sdsydtd2
and − | T9 | − | r9 3r 0 | ■
| r9|
| |r9 3
Example 4 Find the curvature of the twisted cubic rstd − kt, t2, t3l at a general point
and at s0, 0, 0d.
SOLUTION We first compute the required ingredients:
r9std − k1, 2t, 3t2l r0std − k0, 2, 6tl
Z Z| |r9std − s1 1 4t2 1 9t4
ij k
r9std 3 r 0std − 1 2t 3t 2 − 6t 2 i 2 6t j 1 2 k
0 2 6t
| |r9std 3 r0std − s36t 4 1 36t2 1 4 − 2 s9t 4 1 9t2 1 1
Theorem 10 then gives
| | | |std −
r9std 3 r0std − 2 s1 1 9t 2 1 9t4
r9std 3 s1 1 4t 2 1 9t 4 d3y2
At the origin, where t − 0, the curvature is s0d − 2. ■
For the special case of a plane curve with equation y − f sxd, we choose x as the
parameter and write rsxd − x i 1 f sxd j. Then r9sxd − i 1 f 9sxd j and r0sxd − f 0sxd j.
| |Since i 3 j − k and j 3 j − 0, it follows that r9sxd 3 r0sxd − f 0sxd k. We also have
r9sxd − s1 1 f f 9sxdg2 and so, by Theorem 10,
| |11
sxd − f1 1 f 0sxd
s f 9sxdd2 g3y2
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866 Chapter 13 Vector Functions
y Example 5 Find the curvature of the parabola y − x2 at the points s0, 0d, s1, 1d,
2 y=≈
and s2, 4d.
y=k(x)
0 1x SOLUTION Since y9 − 2 x and y0 − 2, Formula 11 gives
FIGURE 5 | |sxd − 2
The parabola y − x2 and its curvature y0 − s1 1 4x 2 d3y2
function f1 1 s y9d2 g3y2
T( t ) The curvature at s0, 0d is s0d − 2. At s1, 1d it is s1d − 2y53y2 < 0.18. At s2, 4d it is
B( t ) s2d − 2y173y2 < 0.03. Observe from the expression for sxd or the graph of in
Figure 5 that sxd l 0 as x l 6`. This corresponds to the fact that the parabola
N( t ) appears to become flatter as x l 6`.
■
FIGURE 6
The Normal and Binormal Vectors
| |At a given point on a smooth space curve rstd, there are many vectors that are orthogonal
to the unit tangent vector Tstd. We single out one by observing that, because Tstd − 1
for all t, we have Tstd ? T9std − 0 by Example 13.2.4, so T9std is orthogonal to Tstd. Note
that, typically, T9std is itself not a unit vector. But at any point where ± 0 we can define
the principal unit normal vector Nstd (or simply unit normal) as
Nstd − T9std
| T9std |
We can think of the unit normal vector as indicating the direction in which the curve is
turning at each point. The vector Bstd − Tstd 3 Nstd is called the binormal vector. It is
perpendicular to both T and N and is also a unit vector. (See Figure 6.)
Figure 7 illustrates Example 6 by Example 6 Find the unit normal and binormal vectors for the circular helix
showing the vectors T, N, and B at
two locations on the helix. In general, rstd − cos t i 1 sin t j 1 t k
the vectors T, N, and B, starti ng at
the various points on a curve, form a SOLUTION We first compute the ingredients needed for the unit normal vector:
set of orthogonal vectors, called the
TNB frame, that moves along the | | r9std − 2sin t i 1 cos t j 1 k r9std − s2
curve as t varies. This TNB frame
plays an important role in the branch Tstd − r9std 1 s2sin t i 1 cos t j 1 kd
of mathematics known as differential | | r9std − s2
geometry and in its applications to the
motion of spacecraft. | | 1 1
T9std − s2 s2cos t i 2 sin t jd T9std − s2
z
T | |Nstd −T9std − 2cos t i 2 sin t j − k2cos t, 2sin t, 0l
T9std
B
This shows that the normal vector at any point on the helix is horizontal and points
N
Ftoward the z-axis. The binormal vector isBstd−Tstd3Nstd−1i j Gk
BT s2 2sin t cos t
2cos t 2sin t 1
N 0
y
x − 1 ksin t, 2cos t, 1l ■
s2
FIGURE 7
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Section 13.3 Arc Length and Curvature 867
TEC Visual 13.3B shows how the The plane determined by the normal and binormal vectors N and B at a point P on a
TNB frame moves along several curve C is called the normal plane of C at P. It consists of all lines that are orthogonal
to the tangent vector T. The plane determined by the vectors T and N is called the oscu-
curves. lating plane of C at P. The name comes from the Latin osculum, meaning “kiss.” It is the
plane that comes closest to containing the part of the curve near P. (For a plane curve, the
osculating plane is simply the plane that contains the curve.)
The circle that lies in the osculating plane of C at P, has the same tangent as C at P,
lies on the concave side of C (toward which N points), and has radius − 1y (the recip-
rocal of the curvature) is called the osculating circle (or the circle of curvature) of C at
P. It is the circle that best describes how C behaves near P; it shares the same tangent,
normal, and curvature at P.
Example 7 Find equations of the normal plane and osculating plane of the helix in
Example 6 at the point Ps0, 1, y2d.
Figure 8 shows the helix and the SOLUTION The point P corresponds to t − y2 and the normal plane there has normal
osculating plane in Example 7. vector r9sy2d − k 21, 0, 1l, so an equation is
S D21sx 2 0d 1 0sy 2 1d 1 1
z z 2 − 0 or z − x 1
2 2
z=_x+π2
The osculating plane at P contains the vectors T and N, so its normal vector is
P S D K LT 3 N − B. From Example 6 we have
x y Bstd − 1 ksin t, 2cos t, 1 l B − 1 , 0, 1
s2 2 s2 s2
FIGURE 8
A simpler normal vector is k1, 0, 1 l, so an equation of the osculating plane is
S D
1sx 2 0d 1 0sy 2 1d 1 1 z 2 − 0 or z − 2x 1 ■
2 2
y y=≈ Example 8 Find and graph the osculating circle of the parabola y − x2 at the origin.
osculating
SOLUTION From Example 5, the curvature of the parabola at the origin is s0d − 2.
circle s0, d.
S o the radius of the osculating circle at the origin is 1y − 1 and its center is 1 Its
2 2
equation is therefore
s dx2 1 y 2 1 2 − 1
2 4
1
2 For the graph in Figure 9 we use parametric equations of this circle:
0 1x x − 1 cos t y − 1 1 1 sin t ■
2 2 2
FIGURE 9
Notice that the circle and the parabola We summarize here the formulas for unit tangent, unit normal and binormal vectors,
appear to bend similarly at the origin. and curvature.
r9std Nstd − T9std Bstd − Tstd 3 Nstd
TEC Visual 13.3C shows how the | | | |Tstd − r9std T9std
osculating circle changes as a point
moves along a curve. Z Z −dT − | T9std | − | r9std 3 r0std |
ds | r9std | | r9std |3
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
868 Chapter 13 Vector Functions
1–6 Find the length of the curve. 17–20
(a) Find the unit tangent and unit normal vectors Tstd and Nstd.
1. rstd − k t, 3 cos t, 3 sin t l, 25 < t < 5 (b) Use Formula 9 to find the curvature.
2. rstd − k 2 t, t 2, tl13 , 0 < t < 1 17. rstd − k t, 3 cos t, 3 sin t l
3
3. rstd − s2 t i 1 e t j 1 e2t k, 0 < t < 1 18. rstd − kt 2, sin t 2 t cos t, cos t 1 t sin t l, t . 0
4. rstd − cos t i 1 sin t j 1 ln cos t k, 0 < t < y4 19. rstd − k s2 t, e t, le2t
5. rstd − i 1 t 2 j 1 t 3 k, 0 < t < 1 20. rstd k t, tl12, t 2
6. rstd − t 2 i 1 9t j 1 4t 3y2 k, 1 < t < 4 − 2
7–9 Find the length of the curve correct to four decimal places. 21–23 Use Theorem 10 to find the curvature.
(Use a calculator to approximate the integral.) 21. rstd − t 3 j 1 t 2 k
22. rstd − t i 1 t 2 j 1 e t k
7. rstd − k t 2, t 3, t 4 l, 0 < t < 2 23. rstd − s6 t 2 i 1 2t j 1 2t 3 k
8. rstd − k t, e2t, te2t l, 1 < t < 3 24. Find the curvature of rstd − k t 2, ln t, t ln t l at the
point s1, 0, 0d.
9. rstd − kcos t, 2t, sin 2tl, from s1, 0, 0d to s1, 4, 0d
25. Find the curvature of rstd − k t, t 2, t 3 l at the point (1, 1, 1).
; 10. Graph the curve with parametric equations x − sin t, ; 26. G raph the curve with parametric equations x − cos t,
y − sin 2t, z − sin 3t. Find the total length of this curve
correct to four decimal places. y − sin t, z − sin 5t and find the curvature at the
point s1, 0, 0d.
11. Let C be the curve of intersection of the parabolic cylinder 27–29 Use Formula 11 to find the curvature.
x 2 − 2y and the surface 3z − xy. Find the exact length of C 27. y − x 4 28. y − tan x 29. y − xe x
from the origin to the point s6, 18, 36d.
30–31 At what point does the curve have maximum curvature?
12. F ind, correct to four decimal places, the length of the curve What happens to the curvature as x l `?
of intersection of the cylinder 4x 2 1 y 2 − 4 and the plane
x 1 y 1 z − 2. 30. y − ln x 31. y − e x
13–14 (a) Find the arc length function for the curve measured 32. Find an equation of a parabola that has curvature 4 at the
from the point P in the direction of increasing t and then origin.
reparametrize the curve with respect to arc length starting from
P, and (b) find the point 4 units along the curve (in the direction 33. (a) Is the curvature of the curve C shown in the figure
of increasing t) from P. greater at P or at Q? Explain.
13. rstd − s5 2 td i 1 s4t 2 3d j 1 3t k, Ps4, 1, 3d (b) Estimate the curvature at P and at Q by sketching the
osculating circles at those points.
14. rstd − e t sin t i 1 e t cos t j 1 s2 e t k, P(0, 1, s2 )
15. Suppose you start at the point s0, 0, 3d and move 5 units
along the curve x − 3 sin t, y − 4t, z − 3 cos t in the posi-
tive direction. Where are you now?
16. Reparametrize the curve yP
S Drstd −
2 2 1 i 1 2t j C
1 t2 1
t2 1 1
w ith respect to arc length measured from the point (1, 0) 1 Q
in the direction of increasing t. Express the reparametriza- 01 x
tion in its simplest form. What can you conclude about the
curve?
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