132
Vci.,iurcc,-su
i,,IIAVICISS
nJlr\T
Equations of Lines and Planes
9.5
1. (a) True; each of the first two lines has a direction ‘vector parallel to the direction xector of the third line, so these vectors are
each scalar multiples of the third direction vector. Then the first two direction ectors are also scalar multiples of each
other, so these vectors, and hence the
: for example. the
Fai
e
(h> 5
i-
two
lines, are parallel.
and p-axes are both perpendicular to the c-axis, yet the a’- and p-axes are not parallel.
(c) True: each of the first txso planes has a normal Vector parallel to the normal vector of the third plane, so these tsso normal
vectors are parallel to each other and tile planes are parallel.
(d) I alse: (hr example. the a y- and pc-planes are not parallel, yet they are both perpendicular to the cc-plane.
(e) False: tile
i-
1.
and p-axes are not parallel, yet they are both parallel to the plane c
(f) True: if each line is perpendicular to a plane. then the lines’ direction ectors are both parallel to a nornlal xeetorfortlle
plane. Tllus, the direction vectors are parallel to each other and the lines are parallel.
(g) False;
tile
planes
1 and a
ii
=
1 are not parallel, yet they are both parallel to tile .r-axis.
(h) True: if eacil plane is perpendicular to a line, then any nomlal vector for each plane is parallel to a direction vector for
line. Thus,
tile
nomlal xectors are parallel to each other and the planes are parallel.
tile
(i) True: see Figure 9 and the accompanying discussion.
(j)
False; tiley can be skew, as in Example 3
(k) irue. Consider any normal vector for tile plane and any direction vector for the line. If tile normal vector is perpendicular
to tile
direction ector. the line and plane are parallel. Other ise. the vectors meet at
Irne v. ill intersect the plane at an angle 90
6i
2. For this line, we have ro
0 +tv
r— r
(61
—
—
—
angle 0. 0 < 0 < 90-. and the
0.
1
5] + 2k and v
5j-t 2k)— t(i)-3j
an
k, so a vector equation is
3]
±
(6-f I)i+(—5m3t)j+ (2—1)kandparametricequationsare
k)
.=6.y=—.5-t-3t.crr2-/.
3. Forthisiine.wehaxe r
0
r
=rO
=
21
-,-
2.4j
-‘-
3..5kandv
x=2—(-3t,y—2.1+2t,a
3.5
I.
21
4. This line has the same direction as the given line. v
r
=
(14]
p
=
1-1— 3/, z= 10a 9/,
—
10k)
-
(2-i-3t) i+(2.4±2f)j + (3.5— f)kandparametricequationsare
k)
(2i— 2.Jj -3,5k) -r ((31 ±2]
-1-/v
3i --2j —k,soavectorequation is
_—
(i —3]-- 9k
=
2/i
+
—
0
3] a- 9k. Here r
=
14]
—
10k. so a vector equation is
(14— 3i’)j —(—10 ± 9t)k and parametric equations are.i’
—
2/,
5. A line perpendicular to the given plane has the same direction as a normal vector to the plane, such as
i’s
0
(1.3,1). So r
=
j -F- 6k. and we can take v
=
i
—
3] 1-k. Then a vector equation is
r ii-- 6kH-t(i±3j÷k)(1--t)i±3tj-r-(6+L)k,andparametricequationsare.r= 1-t-t.y=3t.c-=6-i-t.
6.
V
Z
(2
--
6.1
—
1.5
—
(—3j’
=
0
—4. 3. S. and letting P
3 + St. while symmetric equations are
=
=
--
(6. 1. —3). parametric equations are a’
—
6
—
-it. p
=
1
—
3t.
7. v
(2
0. 1
a.
—3
-
i)
3
,
(2.
0
1). and letting P
.
4E, while symmetric equations are
r
9
(2.1.
3), parametric equations are a
z—3
y.-l
—j-—
—i—
or
r-2
2y
—--—
2
2t, y
-
1 t
2
ijk
8. v
(i
--j)
x
(j ±
k)
1
—
1
0
0 1
1
I
—
—
j
k is the direction of the line perpendicular to both I -+
2
With P
0 — (2, 1. 0). parametric equations are a.
on’
2.— 1
1
t. a.
I and symmetric equations are a.
2
p=
9. The line has direction v.- (1. 2. 1
and smmetnic equations are a.
10. Setting
t, p
andj + k
j
(1.
Letting P
0
1
1. 1). parametnc equations arex
1 + t, p
—
—1
-
21, z
1 -1-
1.
a.
= 0 we see that (1.0.0) satisfies the equations of both planes, so they do in fact have a line of intersection.
The line is perpendicular to the normal sectors of both planes. so a direction vector for the line is
V
Th ><
p
21, z
n
(1.2,3)
<
(1.
(5.2
1, 1?
(5
. Taking the point (1.0,0) as P
, parametric equations area.
0
1 -r 51,
—31. and symmetric equations are
11. Direction vectors of the lines are v
1 — (—2
V2
3
10,3
18. 14
(
4)
15, 10
5,
12. Direction vectors ofthe lines are v
4),0
(
,
(—6),
and since
2.4.4) and v
(
3
1)
(2,6.
4) and
the direction sectors and thus the lines are parallel.
V2
1 v
(8, —1, 1, Since v
16
4 4 16
0. the vectors and
thus the lines are not perpendicular.
5.6) and a direction sector for the line is
13. (a) The line passes through the point (1.
the line are
a.
1
3
(b) The line intersects the a’y-plane when a. = 0. so we need
- 2
we need .r
.r—1
—“i—-
p-f-S
——
—
0 6
a—i
—— or
= 2
—..——
1. Thus the point of intersection with the .ey-plane is
p
-
Ii
1
the az-plane, we needy
0
a.-1
—i—
-
=
5
—1,
1.—i. 0). Similarly for the pa-plane.
3. Thus the line intersects the pa-plane at (0. —3,3). For
3, z
p
2
(
c
z-6
——
a. .
—. So the line intersects the .rz-plane
.3
—,
a.
).
at(—.0,
14. (a) A vector normal to the plane a
p
-
Ta. = 7 is n
(1,
1.3). and since the line is to be perpendicular to the plane, n is
2 4 t, y = 4 — t, a
also a direction sector for the line. Thus parametric equations of the line are a.
(b) On the ap-plane, a
cc-
31 = 0
0. So a = 6
-
01
a.
cl2
.1
plane. p
U
\\
niLls
lilipilLS
1
6 ± 31.
I = —2 in the parametric equations of the line, and therefore a. = 0
and p = 6, giving the point of intersection (0. 6, 0). For the pa-plane, a.
(0. U, U). 1
3), so symmetric equations for
6
z
II-f
2
1
1, 2,
(
i,
s
U
L
OCc,cL..m, IlRc,h’ Rccrccd \i, rcta .a.cncd cc ccd cr5 pcc acccj
c
auU
0 so we get the same point of interesection;
Ic
_cjccaphlccc cc ±
c
1
cd
ccc
L1c
dr
I
ccc
Olih üi ifliLl coLa
cnn un
c i
(
,
.
15. From Equation 4 the line segment from r
21
=
—
j
-
1
4k to r
=
4i
—
(ij
—
k is
3k).01<1.
r(t)=(1-t)ro—trj=(—1)i—j1k)—/(1i—6j -k)---(2i--j-1k)--t(2i—Tj
.
r(t)
-
1) r ÷ I ri
(1
(1
=
=(10i+3j-)-k)±t(-5i-3j- 1k).
The corresponding parametric equations are a:
17. Since the direction vectors are
Vj =
-
0
1<1
10
51. p
2
—6. 9. —3 and v
—
—
6j
—
3k is
3k)
1)(10 i I 3j -i k) + 1(51 -r 6j
-
Si
1
k to r
101 -1- 3j
16, From Equation 4 the line segment from ro
31.
3
-=
1
—
11. 0 < I
1
2. —3. 1,. we have v
=
1
<—
2 so the lines are parallel
—:3v
--
1 and 1 1 :3 aren’t parallel. For the lines to Intersect we must
18. The lines aren’t parallel since the direction vectors .2.3,
able to find one value of I and one value of a that produce the same point from the respectie parametric equations. Thus ssc
need to satisfy the following three equations: 1
we get I
=
--
a, 31
1
21
4 + a. 2
1
1
Solving the first two equation
3s.
1 and L
2 aren’t parallel
11 and checking, we see that these alues don’t satisfy the thIrd equation. Thus L
6, a
and don’t intersect, so they must be skew lines.
1.3. 2, are not scalar multtples of each other, the lines are not parallel, so we
19. Since the direction vectors (1.2.3, and
1 : a:
check to see if the lines intersect. The parametric equations of the lines are L
a:
=
2
4s, p
3
1
3s, z
—
±
1, p
1
—
—
2/,
—
2
±
:
2
3/ and L
2s. For the lines to intersect, we must he able to find one value of I and one value of s th
produce the same point from the respective parametric equations. Thus we need to satisfy the following three equations:
t
=
3
—
21
4s, 1
=
3a, 2 +- 31
2
—
1 a 2+. Solving the first two equations we get I
=
I and checking. we sec
--1, a
fv the third equation. Thus the lines aren’t parallel and don’t intersect, so they must be skew line
5
that these values don’t sati
1) and ‘1. —1. 3 aren’t parallel, the lines aren’t parallel. Here the parametric equations a
20, Since the direction vectors (2. 2.
=
1 + 21, p
:3 -+ 21,
=
-=
equations 1 + 21
=
2 —)- a, 3 + 21
equations gives I
=
1, a
t
1 and a
=
2
2 — I and La: a’
a, and 2
6
a,
=
(5
2 —j--3s. Thus, for the lines to intersect, the th
a.
2 -+•3s must be satisfied simultaneously. Solving the first two
/
1 and, checking, we see that these values do satisfy the third equation, so the lines intersect whet
1, that is, at the point (3.5,1).
21, Since the plane is perpendicular to the vector /—2. 1.5). we can take (—2.1. 5 as a normal vector to the plane.
(6,3,2) is a point on the plane. so setting
- 6)
22,
+ 2k
Xp—
l(y
—
3)
5(z — 2)
=
u —
0 or
-2,5
2+
p
=
+
5z
=
(0,1,2) iS a normal sector to the plane and (4.0.
zg
,0
--
3inEquation7gi9es0(a
.5 and
1. c
.I
=
6.
q
=
3,
=
2 in Equation 7 gives
ito be an equation of the plane
3) is a point on the plane, so setting o
4) - 1(p—0)-f 2[z—( 3)’
0, b
Oorp-t 2zr--
1,
c
— 2.
-6tobeanequation
the plane.
.
Smee the two planes are parallel, they will have the same normal vectors. So we can take
=
(3. 0. — 7). and an equation
C
0_( 7(z —3j =Oor3.r—7z=—9.
2
Planeis3(4)-r
_
1
)
vector for theplane 5+ + 2
1 is n
(1 1, 3), ar
(5.2. 1. A direction vector for the line isv
‘e $t6w the line is perpendicular to n and hence parallel to the plane. Thus, there is a parallel plane which
-
-
a1Ofltt,a.t,,p
-.
‘.‘‘
1
0, we ow that the point 11.2.1) is on the line and hence the new plane. We can use the
.
.-‘.‘d.Lqua1iOlOi
I
dilvIsO.L_1)±2t]2)_j)4)UUt3t_2y
a
a,,
a, a pub’ j,a,,, Ha
,fr p,
,-
sH-a or
a,
pro1
:=1
FUN
25. Heretheectorsa
a
‘1
0.0
1.1
-
1’
/1
1.0. andb
.1.
. b is a normal xector to the plane. Thus. se can Lake n
=
a x b
(0.1.1).anequationoftheplaneisli.r—0)-—1(y-—1H-1(:
26. Here the sectors a
n
a x b
18(.r
‘2. —-1.6/ and b
—
12
=-
0) -21(y-- 0) + 22(c
=
UUAF UNS U- LiNSS AND (‘LANES
1
1.0
0.0
:1
1.0. -i\ lieintheplane.so
1.0
—t-
135
1.1. li. lff
3 is the point
1
1)=0or.r—--z=2.
(5.1. 3 lie in the plane, so
6,2 -1- 20
6.30
0.1
a
18.21.22’ is a normal sector to the plane and an equation of the plane is
0)
Oor
18r
24y±22z=0.
-
27. If \S e first find tss o nonparallel vectors in the plane. their cross product ss ill be a normal sector to the plane Since the given
—
line lies in the plane. its direction vector a
/ —2, 5.4 is one vector in the plane. We can verify that the given point (6.0. —2.
does not lie on this line, so to find another nonparallel sector b sshich lies in the plane. we can pick any point on the line and
find a vector connecting the points. lfwe put /
b
(6- 4.0
2 —7)
3,
equation of’ the plane H.
26. Since the liner
2y
(2
10(9
jfr
3z, or .r
—
9 andn
3.
6
71r
The point (0.0.0) is on the line (put /
=
b
--
=
4.
4(.r—0)—4(
—
1.—i
)
H. on the line, so
axb= —151 12,8
0)
( 2)
1
0 or
a
1N.6
J-
10
1 0i
-
/
1.
100
4-
lies in the plane, its direction vector a
0>. and n e can serify that the gisen point (1.
— 1. 1
(--33,---10, —4). Thus, an
-
1.
. ) is parallel to the plane.
1. 1) in the plane is not on the line.
is therefore parallel to the plane. but not parallel to 1. 2, 3). Then
,
4. 4. —)() is a normal vector to the plane. and an equation of the plane is
=
(z- 0)
0)
—
13’
The sector connecting these two p0mb. b
a
0. we see that (4.3.
Oor5r-4g
29. A direction vector for the line of intersection isa
9z0.
n
‘1, 1.
>< fl2
1)
(2.
1.3)
—
(2.— 5—3), and a is parallel to the
desired plane. Another vector parallel to the plane is the sector connecting any point on the line of intersection to the gisen
point (—-1.2,1) in the plane. Setting a
simultaneous solution p
—1.—
.
=
4
and a
=
0. the equations of the planes reduce toy
—
So a point on the line is (0.
.
—k). Then a normal veetorto the plane is n
theplanejs_2(rt 1)
2)- sc
L4(/J
2.
=
1) —Oorr
5.
4)
=
—
—
Anormal vector to the desired plane is n
16.
Ting (r
. yn. a)
0
=
—10
= V X V2 =
-r-
(—.2.1. —8) and an equation of
1— 1.5, —4) connecting them is parallel to
—
0 and for perpendicular planes,
2z
2) is also parallel to the desired plane.
(5. 1.
20— 2.—I
(0. —2. .5). the equation se are looking for is 6(.r
1 with
1.
5.r + ly or 5i .4— 4y
Ia normal vector for one plane is parallel to the other plane, so v
2
2 and —y -I- 3a
—4. —
iL The points (0. —2, 5) and (—-1, 3. 1) lie in the desired plane, so the sector v
1
the plane. The desired plane is perpendicular to the plane 2 a
—
and another vector parallel to the plane is
/_1.
3
- -Ia
—
.
a
-
0)
—
—
6. —22, —29.
25)
(y ÷ 2)
22
—
29(a
— 5)
0 or
2
9
z- —101.
j1hme is perpendicular to tss o other planes. its normal vector is perpendicular to the normal vectors of the other two
planes.
S1. —2
A
1,0.3
lies on the plane
=
(3
—
0. —2
eqi
—
i
6.0
3(1
—
1
1)
=
(3. -8. —1) is a nonnal vector to the desired plane. The point
8(//
5
(
1
0
ni
3a
8ij
-
38
SECTION 9.5
38. A direction vector fIr the line through (1.0. 1) and (4. —2. 2) is v
equations for the line are .r
1 F It. p
r-
planegivesi f3t—2t—tl+1---6
of intersection is (7.
3
fl r
21
-
21
—
40. The normal ‘,ectors are n
1
Furthermore. n
1 ro
cosU
rit’fl2
-
2 —r- 1
‘
112
1
.
1121
--
fl
-=
1.andz 1—F-2— 3sothepoint
-
3. 6. 7, so the normals (and thus the planes) aren’t parallel.
*
1
1
2. —1.2). The normals are not parallel, so neither are the planes.
14
cos
63.6
..
(r)
a
1. 1. 1) and na
—
=
-
1. —1. 1. The normals are not parallel, so neither are the planes.
0. so the planes aren’t perpendicular. The angle between them is given by
1
3
v3
2
U
9
0—cor
42. Normal vectors for the planes are n
1
planes) are
2(2)—
0. so the planes aren’t perpendicular. The angle between them is given b
=—
..
fl11
4
—
1
2
11
fli’
cosU=,
(1.4. —3, and
4
-
1
1
—
—
=-
41. Normal vectors for the planes are n
1
ni
1. 0. 1). parametric
t. Substitution of the parametric equations into the equation of the
t=2.Then.t —1÷3(2) 7,y=
—
(1.2.2. and
//
Ii
Furthermore,
—
-=
0, so the normals (and thus the planes) are perpendicular.
=
2
—
=
—
1
3. —2. 1 and. tallng P
0
=
137
4.3).
39. Normal vectors for the planes are n
1
But n
—2t, z
r-
EQUATIONS OF LINES AND PLANES
( —1,4.
—1
1\
—
(-)11l
.
0
.D.
—
2 and n2
12.6). Since n
2
(3,
=
=
3nj, the normals (and thus the
parallel.
43. (a) To find a point on the line of intersection. set one of the variables equal to a constant, sai a
intersection does not cross the .ey-plane in that case, try setting
toe ± p
—
1 andi
4
0. (This will fail if the line of
p equal to 0.) The equations of the two planes reduce
t or
1. Solving these two equations gies a’
2y
—
1, p
—
-
0. Thus a point on the line is (1,0.0).
A vector v in the direction of this intersecting line is perpendicular to the normal \ectors of both planes. so we can take
V = fli < fla =
linearex=1,y
(1.1. 1) x i. 2. 2
-
=
(2 —2.1
—
2.2
it
—
—
t.
t,z
(bi The angle between the planes satisfies cos U
1±2±2
fll’fl2
=
=
,
_
gives .r
—
0 then the equations of the planes reduce to 3.r
1, p
Setting z
=
=
.5
Therefore 6?
3/)
—.
1 and 2
+
=
= cos
-1/
I
5
—
NI
inS
3 and solving these two equations
1. Thus a point on the line of intcrscction is (1.1.0). A vector v in the direction of this intersecting line
Equations 2, parametric equations for the line are a’
i’fl
2p
-
is perpendicular to the normal vectors of both planes, so let v
b cos e
=
—
9
7
3y
111)112)
44. (a) lfwe set z
‘0. —1. 14. By Equations 2 .parametric equations forthe
623
1
=
1
U=cos
0, the equations of the two planes become 5
-‘-
=
St. p
n
=
X 2 =
3. —2. 1) y2. 1. —-3
1 + 111,
=
(5. 11.7). By
7L
()8u.9.
—
2p
=
1 and Ir ±
=
6. Solving these two equations gives
l,y
2so a point on the line of intersection is (1,2.0). A vector v in the direction of this intersecting line is
endieular to the normal vectors of both planes.
So we can use V
flj X 112 = (5, —2. —2, >< /4, 1.1) = (0, —13. 13) or
—
‘Cl
1
1
-
9
f
1
l’9’.
1j..’:i
-
RIr’ ir.’rC,,ra M
C.”
hr
ord er drpl..r’.rd
‘C
po rdr,
5
’,
11
hr’
l
—1
,.rSNC r’r’,r”r ,r, s’.’rohr
C,,, rzr
1
I IUI
COUMI IUN Oh LN
ANU HLANh
119
parallel. ote that L
2 and L
2 are not parallel.) L
1 contains the point (1. 1.5). hut this point does not lie on L
. so they’re not
3
identical. (3. 1.5) lies on L
4 and also on L
2 (fort
53. Let
Q
P
(1.3,4) and R
=
(2, 1. 1). points on the line corresponding to t
(4.1. —2). Then a
=
OR
a
d
Q
54. Let
ci
bI
.
a
—
2
—
(0, 6. 3) and R
=
—
d—
ax b
a
(2.—2.lt x (0.—SO)
(2. 2,1,.
.
56. By Equation 9, the distance is D
=
OP
=
—
1 -+ by
Io
1
--
(—3) —1>
(_3)2
ç_2)2
—
—
\12
-—
52
—
6)
,/12
—
2(3)
—
-
\
6(1)
—
—-10
-
\/ii
(4)2
5\/
x’125
—
3
—
i
5
/2+62
1(5) — 8
(2)2
—
61
14’
—
14
-—
1. Let
=-
(_10)2
2)2m12
21 2)
3(1)
-
1(
=
—
—
0 and I
02_
‘22*(
—
4
1
(Z
1. Let
=
(0 —5.0 The distance is
—-
(5.0.—iD)
2,- 2.1
—
0 and t
63. —2. -—6. The distance is
=
=
-
z
OP
\6 —3 4
ti. —2.—3
—
2. —2.1 and b
=
55. B Equation 9, the distance is D
57, Put y
=
=
(2. 4, 1), points on the line corresponding to 1
(0.1.3). Then a-— QR
—
(1. —2. —3), b
=
3 x (3 —2 —6
(1—2. —3
P
—
1), so L
2 and L
1 are the same line.
=
i
,,,/
—
—
7
40
=
v’21
0 in the equation of the first plane to get the point (2.0,0) on the plane Because the planes are parallel, the
distanceD betsseen them is the distance from (2.0,0)10 the second plane. By Equation 9,
D
4(2)—6(0)-t-2(0)—3
5
5
\/4
(_6)2(2)a
-- /ii2\/i
—
--
5L Put.r
p
=
—
2s
Din the equation of the first plane to get the point (0.0,0) on the plane Because the planes are parallel the
distance D between them is the distance from (0. 0, 0)10 the second plane 3r
D=
3
1
3
6(D)-r9(D)—i:
32
± (_6)2
fig + 9z
1
—
0. By Equation 9,
11
92
—
ii26
—
3
The distance between two parallel planes is the same as the distance between a point on one of the planes and the other
plane.
) be a point on the plane given by
0
(o.yn, z
by
ni
dtance between P
0 and the plane given by
54
D
0 + ago + czo ± d-,
ax
b2+c2
=
rz + 4,
+d
1 ±2
c1
d
.
x/o2 2
tb ±c
2
=
0 is.
-
2)
=
from
OTr,
hyo
UI Rr R,d Ua
0 is such a plane, then for some I
1 —
=
7 and .r
b
+ d,
0 and the
— d’
\/12
1anes have equations x
-- 2i, — 2z
czo
Equation 9.
(1,21. —21). So this plane is given by the equation x + 2y — 2z + h’
iSe59,thedistance betweentheplanes is 2
+
\52 (.2
—
iplaxies must have parallel non-nal
vectors, so if ax + by + (‘z -- 4
1<, 2,
0 Then
+ 22 — (_2)2
2,
2z
rId
r
-
V•VJ
6
=
1— k
==
0, where 5’
k
5
r’i:-
=
,
=
0,
d/t. By
Tor —5. So the
SECTiON 9.6
11. z
=
6
-
3r
—
2yor3rt2g
z
12.:
6. aplane with
=
145
FUNCTIONS AND SURFACES
coJ,a’\2\e.’
intercepts 2, 3. and 6.
13. z
=
er
A + 1. a parabolic cylind
1
14. (at The traces in .zthe traces in z
-=
r—
A are parabolas of the form :
2 ±
/- are circles r
2 =
2+
A
y. the traces in y
=
A are parabolas of the form :
A A > 0.
Combining these traces e form the graph.
All Rt&oRes,ã Ma0 oat ho saoaood. tape-i or doroarod or pt-oLd to a
puNch acearbia toSs to
it
oh a or to pot-
:2
2 and
b-
CHAPTERS vC utn RINU
i
146
-iC rCUJ!C Y ui ai-sw,i
(c)
b)
hC.
g(s’ y) is the graph of f(x. y) reflected in tile
eg-plane. (Note that g .r. y)
y;.]
f(s.
15. All si graphs have different traces in the planes a
(a)
-
(b) f(i’.
.ij)
p1.
+
f(s.y)
The trace ins
The trace in
—
1
(c) f(J.u
1 ±i’
i
y) is the graph of g(s. y) shifted upward 3 units
0 is:
0 is :
=
=
ft and p
0 is:
=
0 is:
,
0. so it
—
so it must be graph VI.
must
1
,,.andin;=Ots:
1 + jj
be graph V.
1
.
.
Thetraceina
0. so we investigate these fur each function.
y, and in p
0. and in p
-
=
Ois:=
+
=
1
..
.
Inaddition.weeanseetharJ is
-.
—
close to 0 for large values of a and p. so this is graph I.
f(r )
(d)
(2
—
j2)2
notice the trace in :
(e)
fis.y)
(a
—
IC. p)
sin
0 is 0
=
= (.z2
(i’
0 is 0
—
+
p1).
—
(a
0 is
=
—
. The trace ins
2
p)
-
notice the trace in :
(f)
The trace in .r
=
. and in p
1
p
2)
p
p
0 is:
2
p)
—
z =
=
.
The trace in a
0 is
=
=
=
0 is
=
s’. Both graph IT and
graph IV seem plausible;
±s. so it must be graph IV
. and in y
2
p
0 is :
=
=
.r. Both graph II and graph IV seem plausible:
p
a, so it must be graph 11.
z
sin y, and in p
=
0 is:
sin
‘l.
In addition, notice that the
oscillating nature of the graph is characteristic of trigonometric functions. So this is graph 111.
16. The equation of tile graph is :
.2
2 +
+16p
—
‘
16
—
2
a
16,: >0. Traces ins
z2
—
=
2 or equivalently
16y
kare 16p
2 +z
2
16
z > 0, a family of ellipses where here we hae only the upper halves. Traces in
p
—
in :
P are
=
.12
:2 =
—
P. 1.’
16
—
, z > 0. again a family of half-ellipses. Traces
2
16k
0. are another family of ellipses.
Note that the equation can be written as
2
2
16y
1,
±p
2+
16
=
.
2
P
—
1, ti
0, which we
z
recognize as the top half of an ellipsoid with intercepts +4. ±1, and -1.
17. The equation of tile graph is:
0. Traces in
S-=
P are
:2
\/452
2
p
=
2 or equivalently
±p
,a
2
4k
we have only the upper branch. Traces in p
=
again a famila of half-hyperbolas. Traces in:
or .r
2
+
written as
=
2
.
±
_
2 = :2
422
0, a family of hyperbolas where
k are
=
z2
k, P
—
42
0. are
=
2
p
42
0,
+2
2
p
a family of ellipses. Note that the original equation can be
-,
2
> 0, which we recognize as the upper half of an
ciliptical Cone.
Ol{rrr,’-rr Lrrrr mr ‘1] P’ht. P., ,rrd kIr m’tS,
2
rr-,d, r pd
rd
rprrd ror prrb c’
rrI—it, rr
‘.
,.irrrrrn
SECTION 9.6
18. The equation of the graph is a
—
22
2
a
—
02.
02.
zj2
2
The traces in a
FUNCTIONS AND SURFACES
147
A are
a family of paraholas opening downward. In p
A, sse hase
a family of parabolas opening upward. The traces in a
=
A are
A, a family of hyperbolas. The surface is a hyperbolic paraboloid
with saddle point (0. 0. 0).
j.
19. p
=
A are A
The traces in a
= 2
i.2
A are the parabolas
/
a
-
A-
—
.r., Thus.
a
p
.-
A are the parabolas
is a hyperbolic paraboloid.
—
2 i—.tbetraccini
p
20, For.r
the traces in
which are hyperbolas (note the hyperbolas are oriented
differently for A > 0 than for A < O; and the traces in a
.
2.
—
hase a fttmily of ellipses. When A
2
arey
4,2
I,WhenA
:5
Osse
Owe has ejust a point at the origin, and
the trace is empty for 0 <0. The traces in p
A are .r
2 f k, a
4a
family of parabolas opening in the positise a-direction. Similarly, the traccs
in a
A are .r
2 4i
IA-. a family of paraholas opening in the positive
.i’-direction. We recognize the graph as an elliptic paraboloid ss ith avis the
i-axis and sertcx the origin.
21. Completing squares in p and a gises
.12.2+(y_2)2T
1(a
3)2
22. Completing squares in p and a gives
2)2÷12 2)2 .r -Oor
ci
br
2
(y_9)
(a
.
3)2
=
I,
1. an ellipsoid with
—
center (0.2,3).
(p
—
2)
(.
9)2
.an elliptic paraboloid with
sertex (0.2.2) and axis the horizontal line p
-
2, a
0.4.3
0,2,2,
a) in R
2
2
(b) In
the equation doesn’t ins olve a, which means
that any horizontal plane a
A intersects the surface in a circle
1, z
A. Thus the surthce is a circular cylinder, made up of infinitely many shifted copies the
of
circle
1, with axis the a-axis.
,
c) In a,
=
1 represents a circle of radius 1 centered at the origin.
2
1 t
a
s
n r
-a
‘.115 l, R
d M r
‘
I
..
drdrI’..t .1 op )
1
-
r.b’.,R
‘:
,bI,
.r.
—
2.
148
L7
CTORS AND
CHAPTER 9 VE
= .2
2
ces of z
24. (a) The tra
are
aces in z = P
Tr
2
,t’
OF
THE GEOMETRY
P are
2 in ,r
±p
SPACE
e traces in p
hyperbolas, as ar
A’ a family of
2
2
=
2
k, z
72
—
per half of
+y2 is the up
p) = /T
.r,
f(
of
h
ap
gr
e
Th
(c)
h of
(h), and the grap
the cone in part
half
2 is the lower
p
s.
2 a family of circle
p
is.
h axis the c-ax
neit
is a circular co
(h) The surface
g(i y) =
2
- \/12
hyperbolas.
A’ a family of
2
perbolas are
(Note that the hy
, a similar
2
1 — k
2
z
= 1
2
P are p
e £‘ —
c r— 1 in ,r —
— 2
2
p
t,r
ces in p = P ar
2
of
ces
k > 1,) The tra
or
25. (a) The tra
—1
<
rk
trace in the
= 0, the
P < 1 than fo
ntly for 1 <
of circles For A’
re
ily
ffe
m
di
fa
a
ed
2
nt
,
2
k
ie
or
2
+
with the
+
r- f are
vior, combined
rcle, This beha
The traces in z
.
ci
e
as
th
ol
rb
of
pe
us
hy
di
ra
es the
family of
increases, so do
2.
radius 1. As P
of
is
le
rc
ci
e
e sheet in Table
th
i’p-plane,
perholoid of on
hy
e
th
of
h
ap
s the gr
cal traces, give
hyperbolic verti
d is
t the hyperboloi
is unchanged, bu
ce
rfa
su
e
th
of
les.
(b) The shape
p A’ are circ
p-axis. Traces in
e
th
is
is
ax
its
rotated so that
.
A’ are hyperbolas
dz
2 = P an
in
s
ce
tra
le
whi
:2
1. The
1)2
,,2
(p ±
gives
p
in
re
ua
sq
e
th
g
(c) Completin
rt (a) but shifted
to the one in pa
al
tic
en
id
d
oi
ol
rb
surface is a hype
gative p-direction.
one unit in the ne
p
are the traces in
hyperbolas, as
of
ily
m
fa
a
P
.
e radii
= 1 -‘- 2
—p -increases, th
1 in i = A’ are 2
2
r P > 1. As JP
2 + z
p
fo
s
le
rc
ci
of
s
of
ce
aph of
1, a family
26. (a) The tra
2
P
2
gives the gr
2 +p
k are .r
a
vertical traces,
e
th
ith
. The traces in
2
w
d
1 +P
ne
. combi
1, This beha ior
s are empty for fa <
ce
tra
e
th
;
se
ea
cr
of the circles in
Table 2.
d of two sheets in
the hyperboloi
d so
rt (a) but is rotate
the hyperbolnid in pa
as
e
ap
sh
e
m
sa
e
th
has
(b) The graph
s, shile traces
k. kI > 1, are circle
.r
in
es
ac
Tr
lie -e
- I
that it
e hyperbolas
— P ar
in p = P and a
—
C
All Rh-
\4,nt
-had,,
‘
4h2’,
,
mh’k
SECTION 9.6
FUNCTiONS AND SURFACES
149
27. Graph Ill has these traces. One indication is found by noting that the higher c-values occur for negative values of p in the
traces in .r
—
1 and
.1
2, and for positie alues oft in the traces in p
—
-
1 andy
=
—2. Thus the graph should have a
“hill” over the fourth quadrant of the ag-plane. Suilarly. ve should expect a “valley” corresponding to the second quadrant of
the ag-plane.
28.
f(r.
y)
-,
=
9
.rge
r
1
j
1
-
0
Three-dimensional x iex
Side view
The function does have a maximum xalue. which it appears to achie e at two different points (the to highest “hilltops”).
From the side xie graph, xe can estimate the maximum alue to be approximately 0.011. These same two points can also he
considered local maximum points along Gth the two loer hilltops The graph touches the .r- and y-axes. along the “alles”
between the hills, and rises as we move away from the axes, so we might regard points on the
.t’-
and p-axes as local minimum
points.
29. f(i-. p1
=
a
2
—
V
Three-dimensional xiex
Front view
It doe, appear that the function has a maximum value, at the
higher of the two “hilltops.” From the front view aph, s.x e can
$tmate the maximum value to be approximately 0044. Both hilltops
could be considered local maximum points, as the
ies of f there are larger than
at the neighboring points. Similarly, the vo “valley bottoms” visible in the graph can be
dered local minimum points, as all the neighboring
points give greater values of f. (And
lower valley dip. I
‘.11 R,,±9 Rr
‘.i.,
-rd
,.
a’.cL,
f achieves a minimum value
Li
178
VECTOR FUNCTIONS
CHAPTER 10
From the projection onto the iy-plane
plane p
=
ve see that the cune lies on the \ertical
The other two projections show that the curve is a parabola contained
i.
in this plane.
0
15. Taking r
11,2.3
1
O. 0,01 and r
=
r(t)=(1—t)ru In =(1
i’
rQ)
(1
=
I) ro + I ri
—
Parametric equations are
17. Takingro
—
(I.
=
—
1.2.3,0<t<1 or r(t)(t.2/ .3/,, 0<1<1,
3/. 0
<.
/ < 1.
/) (1.0.1) ft (2.3, 1), 0 <
(1
1 f t, p
1)1.
1
=1
1.2
31. p
2.1.0 andr
1
21.
1
-
2
-
5/. 0 < / < 1.
1.2 . v.e haie
:6.
19.
=
—2
=
-i-
20. x
=
8/, p
tcost. p =1, a —/sint. 1>0.
cost, p
=
sin/, a
=
on a circular cylinder a
2 -i p
2
—
,2
=
1/(1 +
1
=
4
—-
orn(/)/2aS1.1_5/.21,.0</<1.
51.
—
2/. 0 < / < 1.
2
Atanvpoint(.r.g.a)onthecurve..r
2
curve lies on the circular cone a
2
l2t245t),0.a’I1.
<lorr(t)—Li.—3t.
-tiI.1.7’.0
r(t)(1—1)rotrl=(1—t)/-2.1.0,-.1/6.
Parametric equations are .o
/. 3t. 1). 0 < I < 1.
(1
1. 0 < I < 1
31, a
—
<1 or r 1)
1,1 7 ,vehave
1,2 andri
r(/)=(1—t)ru*iri—(
18. Takingro
9.5.4
(2.3. 1), we have from Fquation 9 5.4
,i
Parametric equations arei
21. a
/, p
1
(1,0. 1) and r
=-
ve hiue from Equation
t)0.O.O +
Parametric equations are r
16, Taking
.
±:2
=/2
2 so the
p
with axis the p-axis. Also notice that p > 0: the graph is II.
2
At any point on the curve ve have a
2
12),
l+t
=t°cos
s
2
/
in
ith axis the a-axis. Notice that 0 < a < 1 and a
1, so the curve lies
2 1
+ sin
=
1 only for 1
0. A point (a. p.
:)
on
the cune lies directly abo\e the point (a. gO), which moves counterclockwise around the unit circle in the ag-plane as I
increases, and a
2t. .r
—
1, p - 17(1 +
0 as I
p2),
±. The graph must be VI.
a
—
12.
At any point on the curve we have a
, so the curve lies on a parabolic cylinder parallel
2
a
to the p-axis. Notice that 0 < p < 1 and a > 0. Also the curxe passes through (0,1,0) when 1
22.
I
F, so the graph must be V.
i’
cost. p
sin t, a
=
co 21.
.2
—
=
2
cos
—i- sin
21
=
0 and p
0. a
as
1. so the curve lies on a circular cylinder with axis the
a-axis. A point (a, y. a) on the curve lies directly above or belo (.r. p.0). which moves around the unit circle in the
.ry plane
with penod 2ir. At the same time, the a-value of the point (a.
p. a) oscillates with a period of a’. So the curve repeats itself and
me graph is 1.
2)15
A)) R,5, Rrd M.’.
“A
h ‘rd
pd ord.phA.d
A’
r’,h:),
)‘
A’’
,
.rrnpA
SECTION 10.1
23..r
=
cos Nt, y
=
sin bt. z
axis the c-axis. A point
t > O.a’
2 -i-
,
cos t -t sin
2 t
VECTOR FUNCTIONS AND SPACE CURVES
179
1. so the curse lies on a circular cylinder with
y. z) on the cure lies directly above the point (a. gO). which moves counterclockw
ise around the
unit circle in the .rg-plane as increases. The cur’e starts at (1, 0. 1), when
t
0. and z
(at an increasing rate) as
c, so the graph is IV
t
(i.
—
24. a
=
2 t, y
cos
= sin
2
1, z
=
1.
a
2I
cos
g
--
2
sin
I
1.so the curve lies in the ertical plane
—
.r and g are periodic, both with period r. and z increases as
-
25,1I=tcos1, y
.t.then.r 2
.
Ismt,
so the curve lies on the cone z
2
2
9
)
9.)
Since
1. the
6
9
y—tcost -tsint
* g
-
increases, so the graph is Ill.
t—z,
cuie is a spiral on
this cone.
26. Here
2,2 =
and
=
2
s’
-
2
sin
2
cos
1
1. so the
curve is contained in the intersection of the parabolic cylinder
,r2
2
with the circular cylinder
2
y
—
1. We get the complete
intersection for 0 < I < 2ir
27. Parametric
gives 2t
equations for
=-
the
2
t
curve are .r
212
21
1, y
-
1
=
=
0,
—
. Substituting into the equation of the paraboloid
2
t
21
0, 1. Since r(0)
=
0 and r(1)
i
-
k. the points of intersection
are (0, 0, 0) and (1,0.1).
28, Parametric equations for the helix are x
2 cos
siti
t+1
2
r( 2)
-
5
(sin(2).cos(—2).
(sin(—2,), cos(—2). —2)
29, r(t)
1
(cost sin 2t. sin I
t, y
cost, z
1
=
t. Substituting into the equation of the sphere gives
±2. Sincer(2)
=
s1
iii2,cos2,2 and
2. the points of intersection are (sin2.cos2.2)
(—0.909,
sin
sin
± 2 =
0.416, —2).
21. cos 2t.
Wê’inc1ude both a regular plot and a plot
howing a tube of radius 0.08 around
the
iR.h,Rr-.,d 5I,b
,r,d
(0.909. —0.416.2 and
C))
U)
0
F
C-)
U)
CD
U)
U
0
IC)
LU
>
U-
Cl)
o
-J
U)
CD
LU
CD
U)
LU
-
U
o
U
U
CD
U
U
U
C)
;;
*
—
U
CD
CU)
U)
CD
U)
(0
U
CD
U
o
CD
-
=
C_f
U)
—
U
—
U
C_f
—
U
—
U)
U
C-)
U)
—
C-f)
C-C
U
U
_U)
-.
4
U
0
0
U
U)
—
Cl
-
C)
-
U
-U)
-4
CD
U)
CD
U
C
U
>
C)
4
CD
0
II
-
*
C/)
U
CD
CD
U)
t
U)
C)
4
U
-
0CC
-
—U
U)
—
0
C)
U
-U)
C)
U)
CD
N
*t•
N
lCD
U)
U
—
NS
VECTOR FUNCTIO
CHAPTER 10
188
6. Since y
curve
e
1•
(a), (c)
the
=
t.
i—c
t.
j.
-.-j
r’(O)=i
perbola
is part of the hy
1
=
=
(b)r(t)=c
—.
1)
> ft
Note that.r > 0. y
(b) r’(t)
7. Since g
=
(3f
=
()3
=
(a). (c)
, the
3
.r
3 i ± 3c j.
c
=
r’(O)
=
i — 3j
ote
a cubic cove. N
cure is part of
that here ..r > 0.
0
(b) r’(/)
s/,
8.i=I- co
—
1)2 f
curse is a
(a). (c)
’--sintso
/1= 2—
costj,
-
_.J
1
r
6
2
2’
.3,
circle.
2t
F
i sini.
9. r’(t)
tcos2t)
,
=
(tees /
10. r(t)
-
land the
2)
(q
sin j
=
sin 1.21. (‘OS
r’(tj
‘sec /.
=
secttant,
2
i
r’(t)=2tf
3/)k
(1
n=
)
jI
(t
—
r
i
11, 2
t) 2 ± cos2t)
sint, 2t, t(— sin2
‘teost
2/sin 2/,
2/
l/t2
(tan/sect.
=
)k
3cos /( sint
tj ±c 2
sn
oi
cs
2
3
/
b
1+
t1
3sin3/) -aeos3
12. r’(t) = [at(—
3rcos I sintk
stj 2
3hsin / co
i 2
l)
n3
si
3d
(a cos 3/
rem 3.
1 and 3 of Theo
2/c by formulas
b
+
c
21
13. r’(t) = 0 + b
(a x c), so
2
t(a>. b) -I- t
+ / c)
lb
x
/a
=
t)
r(
st expand
14. To find r’(t), we fir
. So r’(O)
r’(0) = /l.2.2
r)
2c
,
2
t
1
0
_
/(
(
1
t
± e.2
15. r’(t
.
—
-
—
T(0)—
=
99
=
x c).
a x b ± 2t(a
+2
+22
=
=
3 and
=
—
‘
Ir’(O)I
j1.k,Thus
r’(l)=2i±2
+k
2tj
16.r’(t)j+
T(1)
r’(/)
r’(l)
ir’(l)j
1
25iO,
‘-Lr
cpL
to
te oh,to
uH,J o,tb
-‘
tt5...o °r°
P°”
SECTION 10.2
30, (a) The tangent line at
0 is the line through the point v ith position sector r(0)
the direction of the tangent sector. r’
p. z
(i’.
—
r()
r’()
r(O)
=
DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
Ltr’(O)
=
(0— mj.0
-
sin.2sin.eos)
rrcos .2cos.
‘rr ens 0. 2rr ens 0.
=
—
(sisi 0.2 sinO. cos Oi
r. 2ir. 0
(0. 0. 1), and in
So an equation of the line is
çu.2u. 1).
1.2,0).
rrsin
-
2u, 1,
rr sin 0,
—
191
(b)
=
So the equation of the second line is
=1.2.0 =u(0.0.-
‘.nij.
‘1.2.t.
The lines intersect sshere —
1
u, 2ru I
r-.
(1. 2.
‘‘
so the point of intersection is (1 2. 1).
31. The angle of intersection of the tsso curves is the angle hetncen the tao tangent sectors to the curses at the point of
intersection, Since r (t)
r(!)
=
1, 2t.
3[2
and t
cost. 2eos2t. 1 and sincerl(OI
Oat (0.0.0), ri(0)
0. 0.0
=
hetaeen these two tangent sectors, then cos 0
,
ri (0)
- a,
I
I
Thus the point of
2,3 1- t
2
a
2
(1.0.0 is a tangent sector to r at (0.0.0). Similarly.
(1.2.1 is a tangent vectorto r
2 at (0.0.0). IfO is the anole
(1,0,0)
(1. 2. 1
.
32. To find the point of intersection, we must find the values oft and
—3
o-
.
and 0
1
ens
()
66
a hich satisfy the follos ing three eqLlations simultaneousl\
Solsing the last two equations gives t
1, a
2 (check these in the first equation)
intersection is Il. 0.-il. To find the angle 0 of intersection, we proceed as in Exercise 31. The tangent
vectors to the respective curses at (1.0.4) are r(1)
tl,
1.2 and r(2)
-
(—1.1, 1,. So
1—S)=—=—andO=eos i(+).5.5
cDs0=_Tl._(_-l
Note’ In Exercise 31, the curves intersect when the value of both parameters is zero. however, as
seen in this exercise, it is not
necessary for the parameters to be of equal value at the point of intersection.
3. j (16I’ I
— 9t
j
2
÷
25t k) dt
=
—
f
i±
k) dt-
(i
3 dL) i
lOt
4
E1t
[4tan
(J
j + (j 2511 di) k
912 di)
[3tj-i- 15[0]k=1i_3ji.5k
i
tij
-4 1n(1 +t2)k]
=
[4tan’ lj -41112k]
—
E4tan’ Oj +1111k]
=1(-)j—1n2k—0j—0k=rrj-r-1u2k
fq
2
(
3sin t C05t1
3sin I cos
2 Ii
(f/23.2
jsin
t
3
]
2
t eostdi) i
E—cos
I
3
’
2sint costk) dt
(J
2
2
3sint cos
i dt)j
2
[sint] k
i2
=
O tososo L,om,,, sil 5,,’’, R,or,d 1.1.,.
rot b,. ,oor.od optd. cr5 pocotod
(f22sint
costdt) k
(1—0)1 + (0+ 1)j + (1— 0)k
sd too pohLiy
thir ,,obo”rrnhoi
r-r’
=
1±j +k
Cl
.
i
ci
—
ci
V
ci
Cl
>
-
hl
-cci
c>
I
c
H
Li-
>
cc
+
2—i
H
-
1—
c
--
t
,
L>
II
-c
ccc
P
ci
-
—
cc)
C)
-c
—
-
C__
ci
ci
-:
-
-
1
-,
IcN
H
ci
±
Cl
h
—
-
H
ft
1—
—
—
*
cii
sI-ft
ci
Cl
1.
I -I
-
I
cci
±
*
II
L-’
cc
-z
-
±
—
i—
-
-r
—
c-c
-1--
LU
DC
CD
C)
-‘C
CD
I—
C)
Cl
CD
Cl
-‘
“>
-
-J
DC
LU
C’]
C)
C:
.
8
t
if]
Cl
8
+
‘
I-’
C
Cl
C
C::i
o
DC
C’)
DC
0
C
it
C’I
C’]
C)
if:
C)
C)
I]
C
+
C:
“IC’
‘C)
C
±
Cl
Cl
Ci
i
>
C’]
ii
‘;:‘
ifE’
if]
1,1,
.2
-2
C’]
.
+
-
C’1
C:—’
—
C’]
—
Cl
—‘
c;Cl
‘
,-‘
C:
CI
—_- t
‘>
—
‘C:
8
—
CD
—
C)
C:,
C)
-
2
I:
C
if
C)
:
CD
Cl
C
C)
C)
—
2
8
-
ff
C
H
1-2
CC
:
‘:;::-
-
-t
C
CD
2
c
8
U
‘C:)
-
-I-)t
-
2
-
C:
It)
—
-
-
C)
-
-
Cl
Cl
iH
-i--
-H
2
C)]
C)
C)
—
C))
-
C)]
2
!
‘ft
.!
:
CCl
-
.
ii]
±
Cl
C:
-I’)
Cl
Cl
C
-
C)
-‘
‘, -j
-8
—
Cl
C
]D
C-
C)
C)
II
H
-
-2
2
-2
‘— ‘
C)
-
C
t I
IC]
CI
CI
ii
C)
C:
Ci
if>
Cli’)
CC
8
Z
H1 !f!;
C’)
o
>
w
a—
a-
0
a
C)
—
A
C)
,
aD
3
C)
—
—
ft
—
1
>
-
.
F
—
3
!H
C)
aD
.
a
—
H
—-
F
I
>lft)
-
.—
C)
d
H
£4
0
CC
-
a-’
-a-
a
-—a-
—
C)
-
__c_
H
H
-
a
H
F
ft
-H
H
C)
+
1
-H
C)
4
Cal
F
—
±
C)
p
H
C)
-3
i
>
ft
—
-
E-4
-
-
—
d
c’1
--
—
I.
-IC’]
¶A
rH
C”
•
:
-
-
‘—,
-H
--
-H
-4-
—
H
(‘I
B
i—
1+
—
-ft
-H
+
ftO
—
SECTION 10.3
T’(t)
rr’
\/(1
— 3
t -+ t
j
k
2
r’(/)
(b) rh)
21.
r(t)
r’i)
(1 —
512
j
2
3t
21k,
r”(t)!
r’(ti
r”(t)
612.
6/j
+
2k,
r’(l)
t)2÷(22
V
-
ti 12j+((k
2
/‘
2
T”j’2
‘
r’(t)
r’(tJxr”//) =
i—21j—’k.
r’(t)
/“2t
(l)2
‘21—2)’
/1/1
Then
h
r (/)
t)
><
r (1)
(4/-
6/2
4j2
4 in tj
3/i
23, r(t)
-.
r’(t)
256 r lilcos
/
2
3i
4 os
j
x r°(/)
—
2k,
1
°
-
—lsin tj
1)
16i 12costj
1
t k,
l2sintk.
r’(L) xr”(t)
20. Then t(t)
o
20
d
/j2,
24. r(t)
tot
—
mt. Lint)
1, and r’(l) = 2. 1. 1),
Ir’(l) x r”1)
25. r(t)
(1,
r’11)
(r’(l)
r’(t)
.
2, ti)
O
(
r’(t)
4)2
-
(1,21,
in /,
vü
3/2
,
2
.
-.
1i/2,1 1;.
(2,
r°(1) = 2. —1. 1..
or
(_)3
r”(1) = (0.2.6.
1, and r’(l)
r’(lj> r”(l(
-
r’(l)xr°(1)
Then
r°(1) a-s 2.0.
r’(l)
jr°(1)
Then n(1)
1
_.i
The point (lU. 0) corresponds
The point (1.1,1) correspondsto /
r”(t) s-s 0. 2
.ö/i
r°(1)I = 36+4
r°(t)
\_12=12 = \/6.
r’(l)
4 ±9
—
1
v’
X
(2/. 1//.1
4/2)32
(91
\(t15)t1.
2
2
L0 1k,
i”(/)
j
1
r
1i
2)r
8t42
(1
t
2
llIsin
t
(21
t
v(’
[21)
2
\‘9+16
jr’(t) = t
co +16in
/ = 9— 165.
2
Ir’(t) x r°()
r”(t)
>‘
1
5)c-
4/2
I cos 1k
r’(/)
)3
2j—
k
t
.
22\ 2
(2/—2
2
c
t
( c2’t_
81
1
r”()
\/TTi2,
6/2
= jr’(/) x r”t)
Thenrr(/)
(V9t1
22. r(1)
/6.
(1,2.3)
6.2, so
1
Note that e get the complete curse for 0 < / < 2ir.
rh)
((‘Os
=-
r°(t)
—
t.sint.sin .5t
(-- cost
corresponds to 1
r’(O)
-i
0. and r’(O)
52
=
UI (lagS, Rats-, ad
(— sint. cost. 5cos5t).
r’(t) =
The point (1,0.0)
(0. 1.5)
r°(0) = (—1.0.0),
,
r’(O) x r°(0
1 = 0. —5.1)
/biT)E5)2E’i2
the point (1.0.0) is h’(D) =
carmaig
=
sin /. -25 sin 5/.
12
—
[r’(O) x r”(O)
35! (cnaac
199
5L)
1
r”(/) = 2
—6t
i
.
ARC LENGTH AND CURVATURE
is S s-ar-red arrpted ord 1
-‘
,.satod
it,
po aid
i,
The
=
r’(O) x r”(0’
3
i(fl3
a pabiIt :,ccaraihS arar
tm
total
‘
curvature at
1
=
is ta parr
-
1,.
VECTOR FUNCTIONS
CHAPTER 10
200
27.
f(r)
28.
f()
f’(c)
,r.
=
f”(a’)l
=
[1 +
(f(a5))213
xe’.
f’(a)
-
29. [(a’)
=
)
(1
—
2 ,r tan.c,
2 sec
2e’,
-
+ 2e
±e)
’e
t1(a 2
2
I())23 2
f1+
[1
tanr
r) 2
5
sec 2
4
,c ‘ 21
(1crc)a]
3
1+
ac’
h\.L)—
=
2
c
2
2sc
.ccr
f”(a)
c,
a
(ff())2]3 2
2 sec x sec tan x
ec a’ tanil
2
I2s
l3/2
(sec j)2
2
2
=
12752,
=
f”(x)
2 a’,
sec
f(.r)
tan a,
=
f”(r)
4a.
=
(
33,
11 16°)
(43)2’3 2
=
2
34,
1
1
30.
17(1)
11
2
(r))2
3
(y1
(1 +
(72
1)3/2
(a’s
2
1/12)32
2
+ 1)3
±
r
5
(
i)
2
—
i()(a
2
f(
0
and e’(a’) < 0 for ,z’ >
Since lim
L
,
‘2
1)
2
(a
—
attains its maximum at a
(r’)
0.
=
q
-—
,
Since
0 forD <
e’(i’)
,1
,ln
c.
36
C
(y (a’)
S
31, Stnce q
=
re occurs at
Thus. the maximum curvatu
.
approaches 0 as .r
(i’)
thecurvaturels17(a)
S
c’
(2
a’
number in the domain is
0, so the onl critical
212
1
=4-
t
32
35.
212
1
]
2
3r
r ± 1)
(.’
i)
[
2
(2 + 1)3
1)3,2
(12
ical numbers of tt(a’):
vature. ne first find the crit
To find the n3aximum cur
1)2(2I)
[slncea’ >01.
(2 )3 2
1
1
1
V” (c)
=
S
(1 +
(gI(I))2]
2
’
3
[1
e (1
‘
rn
3
)
1
3
c2
2
/
critical numbers of tt(x):
vature, we first find the
To find the maximum cur
=
=
for a’ >
t72)
e2(
0
when
—
I
-
22
(—)(i ±
0.so
212w
2r
=
3/22E2.)
32
7
and 1
’ >0 forx < —ln2
2
2c
—1n2. And since 1—
or.r
re is aained at the point
in 2, the rnaxin3urn curvatu
+
)
7
(1
2
eI
1
Since urn r
+2
(—
in 2,
1n2;, 2)
=
(—
In 2,
a’’—’.
=0.n(,r)approaches0as
equation is
opening upward. so the
‘ertex at the origin and
32. We can take the parabola as having its
“(1’)
Fquation 11,
it
J$(T(
2
2 and the equation is y
’ <0
2
2c
=
7 23101
‘1
-
)
2
,’
1
2
(9
3
’
=
a’2)3’
2
4a2
’
(1
thus ta(O)
=
f(.r)
=
nY
ox. a> 0. The
2u We wanta(O)
212,
g0
L’o,’,o 0,j3 R,oh 117
r17
Olo’
,,ol
7, ,,,o’,c’J
72
c,
. ,r,”csl,d
o17l
5
dp
,
0,
,,b.,_l
eb
.
5
o
io
,ho7 or
4.so
__
__
__
__
__
_
__
__
__
ature to be greater at P.
would expect the curs
se
so
n
tha
P
Q.
at
y
more quickl
he changing direction
33. (a) C appeais to
the
cles at P and Q. Using
roximate osculating cir
(b) First we sketch app
of the osculating circle
sse measure the radius
axes scale as a guide,
ly 0.8 units, thus p
at P to be approximate
=
ate the radius of the
1.3. Similarly, we estim
=
osculating circle at
Q to be
1.4 units, son
0.7.
1.4
p
to
34,p=.r2a
—
2 and
.
”=1
2
y
1
.
x
.r
4
y’4
—
3
1
.
n (I) =
)
2
4e
- ]
(it
3
,
[i
()2]
The graph of the
1
e graph of p
at we ssould expect. Th
cursalure here is wh
gin and near.,
g most sharply at the ori
appears to be bendin
35
2
=
P
-
2a
‘
,
p’
—
6a
o.
-
(
2i
‘,
—
2,2
1.
=
1
and
ti
‘
—
1)2132
r)
3
a’(1 L
3
4
g. hut it is
haps a little surprisin
ps in this graph is per
hum
two
the
of
e
anc
The appear
both
lly at the origin from
2
increases asymptotica
t
p
tha
t
fac
the
by
explained
) is undefined.]
d there. [Note that n(U
ben
le
litt
y
ser
has
ph
directions, and so its gra
a isO or near 0 where 6 is
tuming more sharply. and
is
6
se
cur
ich
wh
at
e i-values
o is highest for the sam
36. Notice that the curve
•fLr).
the graph of p
of p --- a(i). and 6 is
ph
gra
the
be
st
mu
,
o
nearly straight. So
ect the curvature to
straight. We ould exp
ost
alm
s
ear
app
ph
which the gra
two inflection points at
n the graph
37. Notice that the curve 6 has
of q = f(.r) rather tha
us. o must he the graph
Th
re.
the
0
r
nea
’t
isn
, but the curve o
be 0 or nearly 0 at these values
of p
of curvature, and 6 is the graph
e
0 < t < 2r. Curvatur
(a) The complete curse is gisen by
te) maximum at 6
appears to have a local (or absolu
curse appears to turn
points. (Look at points where the
more sharply.)
hi.bt
hc:ç
SECTION 10.3 ARC LENGTH AND CURVATURE
44. f(x)
f’(.r)
—
=
(0)
[I
(1
J”(.a)
.
2
r
2
(f’(.r))
U
e
2
c
2
2
(.2)3
3
c
numberwhen2c
(1
L
(.2)3
—n
=0
and negatE e elsess here, so
-.
2]2
1
c’)
2(9.)
1) corresponds tot
(2)
c(2
=
(212 + 1)
=
1.
2
1)
0orc.
f(.r)
(2(2
(2t. 2t
. 1)
2
2g2
2
2.
2(2/2 ± 1)
2
c
2(2)2
+ It. —4t
(2t) +
and jr’(t)
1
r(!)
T(/)
=
(— sin!, cost
(—[(sin!)(— sint)
Finally, B(0)
=
T(0) . N
47. (0, r, -2) corresponds to t
/
=
T(7rj —
planeis
r’(t)
r’çt)j
=
< / g
,
71.
r(t)
2(2, 21, 2!)
(1
+it48(2
and then sect > 0
-(-1,0,1)=(
‘0.1.0 x
=
.0 < en
so the members
=
.
.
;.
1’
2!)
(1
.(
=‘
sint)
cost)
=
and
, 2!, _2t)
2
2t
1 4 2(2
,
and in Exercise 4 we found that r’(t)
2 I.
(— sint cost, cos
r’/
sect.
T(0)
(0.1.0).
(
sin 1, coSt. — tan t
! — cos
2
(sin
2 /. —2sintcost. — cost). so
,0.
‘—-4 0
/—--.o.
2/’
(2 sin 3!, (.2 (‘05 3!)
6cos3!,1,—6sin3/
1
—
=
2 31 + 1 + 36 sin
ens
2 3!
.
bcos3t. 1. —6s]n3t...
‘-4 (—6, 1, 0) is a nonnal vector for the normal plane, and so
6, 1, 0) is also normal. Thus an equation for the
6(a’—O)-e 1(y—a)--i-0(z+2)=Oorg—6r=ir.
(—l8sin3t,0,
NU)
(cost. sin 1. in cost
tan!)
-- 2(2.2!.
.—(—-—).i--
(cos/)(cos/). 2(cost)(— sint).
1.0,-1)
N(0)=
Tt()
r(t)
0.
sect . Here we can assume -
=
. \Ve ha\e a critical
2
. In either case. U 0)
y2
2/2/2
2!)
N)=-..-—4)andB(1)=TU)’n N(1)=
46. (1.0,0) corresponds to t
=
—
(
.2)
[by I ormula 3 of Theorem 10.23]
2(2, 2!. -21)
(1
2\/(
)
2
(2_r
(1
:t. 2t
. 1
2
.soT(1)
212
1
2 + W -r 1
•e1t
3(3
j3
Then
C
2(. 212. 1
1) ‘(2.11.0)
0 is
=
f’(c)ispositiveforc<
‘2 or — s
and f(.r)
2)3 2’
— 3c -
3
)
2
c
(1
=
r’(t)
‘(1)
T(t)
2(2(2 H- 1)
T’(t)
T’,
N(t)
(1 +
)
2
2c(1—c
J’ achieves its maximum alue when c —
=
2
4t(21
2
(1—c
=
of the family ssith the largest value of n(0) are
T’(t) —
1(c)
=
(1
.
so the curvature at z’
2
To determine the maximum salue for K(O). let
)3 2
2(1
45. (1.
Using I om2ula II e hae
‘.
3
’
2
(cc’)
=
l9cos3t)
=n
T’(!)
(—sisi3t.0.—cos3t). So N(71)
/n23t+182cos23t
0.0. 1 and B(71)
=
) is a normal to the
71
ceB(
osculating plane. so is ‘1.6,0
equation for the plane is 1Cr
—0) 6(y — 71) -- 0(z -+- 2)
0 or
-n-
C ‘Ptfl Ltr CII R Sn Pn-nd Can
nt
203
a ncnntJ
p
1
dnpl
n.t
a
71 -i--
(6.1.0) x (0.0.1)
=
6y
=
=
671.
,pn an, patch c,.cttib’n tn-kIln in nh tintt r
(1.6.0.