132 i,,IIAVICISS Vci.,iurcc,-su nJlr\T

9.5 Equations of Lines and Planes

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 two lines, are parallel.

(h> Fai5e: for example. the i- 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- and p-axes are not parallel, yet they are both parallel to the plane c 1.

(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 ii 1 and a = 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 tile

line. Thus, tile nomlal xectors are parallel to each other and the planes are parallel.

(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 an angle 0. 0 < 0 < 90-. and the

Irne v. ill intersect the plane at an angle 90 — 0.

2. For this line, we have ro 6 i — 5] + 2k and v 1 ± 3] k, so a vector equation is

r— r0 +tv (61 — 5j-t 2k)— t(i)-3j k) (6-f I)i+(—5m3t)j+ (2—1)kandparametricequationsare

.=6.y=—.5-t-3t.crr2-/.

3. Forthisiine.wehaxe r0 = 21 -,- 2.4j -‘- 3..5kandv _— 3i --2j —k,soavectorequation is

r =rO -1-/v (2i— 2.Jj -3,5k) -r ((31 ±2] k) (2-i-3t) i+(2.4±2f)j + (3.5— f)kandparametricequationsare

x=2—(-3t,y—2.1+2t,a 3.5 I.

4. This line has the same direction as the given line. v 2 1 — 3] a- 9k. Here r0 = 14] — 10k. so a vector equation is

r = (14] — 10k) - (i —3]-- 9k = 2/i + (14— 3i’)j —(—10 ± 9t)k and parametric equations are.i’ — 2/,

p = 1-1— 3/, z= 10a 9/,

5. A line perpendicular to the given plane has the same direction as a normal vector to the plane, such as

i’s (1.3,1). So r0 = 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 (2-- 6.1 — 1.5 — (—3j’ = —4. 3. S. and letting P0 = (6. 1. —3). parametric equations are a’ — 6 — -it. p = 1 — 3t.

Z 3 + St. while symmetric equations are= --

7. v (2 0. 1 , 3 i) (2. . 1). and letting P0 (2.1. 3), parametric equations are a 2 - 2t, y 1 t

r 9 y.-l z—3 r-2a. —3 - 4E, while symmetric equations are

—j-——i— or

—--—

2y 2

ijk

8. v (i --j) x (j ± k) — 1 1 0 — I— j k is the direction of the line perpendicular to both I -+ j andj + k

0 1 1

With P0 — (2, 1. 0). parametric equations are a. 2 t, p 1 t. a. I and symmetric equations are a. 2

on’ 2.— 1 p =

9. The line has direction v.- (1. 2. 1 Letting P0 (1. 1. 1). parametnc equations arex 1 + t, p — —1 - 21, z 1 -1-

and smmetnic equations are a. 1 a. 1.

10. Setting = 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 >< n (1.2,3) < (1. 1, 1? (5.2 3 . Taking the point (1.0,0) as P0, parametric equations area. 1 -r 51,

p 21, z —31. and symmetric equations are

11. Direction vectors of the lines are v1 — (—2 ( 4),0 (—6), 3 1) (2,6. 4) and

V2 (5 10,3 18. 14 4) ( 5, 15, 10 , and since V2 the direction sectors and thus the lines are parallel.

12. Direction vectors ofthe lines are v ( 2.4.4) and v (8, —1, 1, Since v1 v 16 4 4 16 0. the vectors and

thus the lines are not perpendicular.

13. (a) The line passes through the point (1. 5.6) and a direction sector for the line is ( 1, 2, 3), so symmetric equations for

a. 1 II-f z 6the line are

1 2 3

.r—1 p-f-S 0 6 a—i(b) The line intersects the a’y-plane when a. = 0. so we need —“i—- —— — —— or

—..——= 2 c —1,

- 2-

p 1. Thus the point of intersection with the .ey-plane is ( 1.—i. 0). Similarly for the pa-plane.

we need .r Ii 12

p 3, z 3. Thus the line intersects the pa-plane at (0. —3,3). For

a.-1 5 z-6 .3the az-plane, we needy 0 - —i— = —— a. . —, a. —. So the line intersects the .rz-plane

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

also a direction sector for the line. Thus parametric equations of the line are a. 2 4 t, y = 4 — t, a 6 ± 31.

(b) On the ap-plane, a cc- 0. So a = 6 31 = 0 - 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 so we get the same point of interesection;

(0. U, U). 1 01 cl2 .1 plane. p U \\ niLls lilipilLS 1 i, s L U auU Ic cd L1c I Olih üi ifliLl coLa c i ( , .

a. OCc,cL..m, IlRc,h’ Rccrccd \i, rcta .a.cncd cc ccd cr5 pcc acccj c _cjccaphlccc cc ±1c ccc dr ccc cnn un

15. From Equation 4 the line segment from r = 21—

j - 4k to r1 = 4 i — (ij — k is

r(t)=(1-t)ro—trj=(—1)i—j1k)—/(1i—6j -k)---(2i--j-1k)--t(2i—Tj 3k).01<1.

16, From Equation 4 . the line segment from ro 101 -1- 3j k to r1 Si — 6j — 3k is

r(t) - (1 1) r ÷ I ri = (1 - 1)(10 i I 3j -i k) + 1(51 -r 6j 3k)

=(10i+3j-)-k)±t(-5i-3j- 1k). 0 1<1

The corresponding parametric equations are a: - 10 51. p — 3 31. -= 1 — 11. 0 < I <— 1

17. Since the direction vectors are Vj = —6. 9. —3 and v2 = 2. —3. 1,. we have v1 -- —:3v2 so the lines are parallel

18. The lines aren’t parallel since the direction vectors .2.3, 1 and 1 1 :3 aren’t parallel. For the lines to Intersect we must

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 -- 21 1 a, 31 4 + a. 2 1 1 3s. Solving the first two equation

we get I = 6, a 11 and checking, we see that these alues don’t satisfy the thIrd equation. Thus L1 and L2 aren’t parallel

and don’t intersect, so they must be skew lines.

19. Since the direction vectors (1.2.3, and 1.3. 2, are not scalar multtples of each other, the lines are not parallel, so we

check to see if the lines intersect. The parametric equations of the lines are L1 : a: 1, p — 1 — 2/, — 2 ± 3/ and L2:

a: = 3 4s, p 2 — 3s, z 1 ± 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 — 4s, 1 21 = 2 3a, 2 +- 31 — 1 a 2+. Solving the first two equations we get I = --1, a I and checking. we sec

that these values don’t sati5fv the third equation. Thus the lines aren’t parallel and don’t intersect, so they must be skew line

20, Since the direction vectors (2. 2. 1) and ‘1. —1. 3 aren’t parallel, the lines aren’t parallel. Here the parametric equations a

= 1 + 21, p = :3 -+ 21, -= 2 — I and La: a’ 2 a, = (5 a. 2 —j--3s. Thus, for the lines to intersect, the th

equations 1 + 21 = 2 —)- a, 3 + 21 6 a, and 2 / 2 -+•3s must be satisfied simultaneously. Solving the first two

equations gives I = 1, a = 1 and, checking, we see that these values do satisfy the third equation, so the lines intersect whet

t 1 and a 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 u — -2,5 = 1. c .5 and .I = 6. q = 3, = 2 in Equation 7 gives

- 6) l(y— 3) 5(z — 2) = 0 or 2+ p + 5z = ito be an equation of the plane

22, + 2k (0,1,2) iS a normal sector to the plane and (4.0. 3) is a point on the plane, so setting o 0, b 1, c — 2.

Xp— 0,zg -- 3inEquation7gi9es0(a 4) - 1(p—0)-f 2[z—( 3)’ Oorp-t 2zr-- -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

Planeis3(4)-r0_(2)1_7(z —3j =Oor3.r—7z=—9.

vector for theplane 5+ + 2 1 is n (5.2. 1. A direction vector for the line isv (1 1, 3), ar

‘e $t6w the line is perpendicular to n and hence parallel to the plane. Thus, there is a parallel plane which

- 0, we ow that the point 11.2.1) is on the line and hence the new plane. We can use thea1Ofltt,a.t,,p

‘.‘‘ 1 - . - -- -. .-‘.‘d.Lqua1iOlOi I dilvIsO.L_1)±2t]2)_j)4)UUt3t_2y :=1

a a,, a, a pub’ j,a,,, Ha ,fr p, ,- sH-a or a, pro1

FUN a UUAF UNS U- LiNSS AND (‘LANES 135

25. Heretheectorsa ‘1 0.0 1.1 1’ - .1. 1.0. andb /1 0.1 1.0 1 1.0. -i\ lieintheplane.so

a . b is a normal xector to the plane. Thus. se can Lake n = a x b = :1 0.0 —t- 1.0 1 1.1. li. lff3 is the point

(0.1.1).anequationoftheplaneisli.r—0)-—1(y-—1H-1(: 1)=0or.r—--z=2.

26. Here the sectors a — ‘2. —-1.6/ and b (5.1. 3 lie in the plane, so

n a x b =- 12 6.30 6,2 -1- 20 18.21.22’ is a normal sector to the plane and an equation of the plane is

18(.r 0) -21(y-- 0) + 22(c 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 / 0. we see that (4.3. H. on the line, so

b (6- 4.0 3, 2 —7) (2 3. 9 andn axb= —151 12,8 1N.6 10 - (--33,---10, —4). Thus, an

equation of’ the plane H. 71r 6 10(9 — 0) 1 ( 2) 0 or a J- 1 0i 4- 100

26. Since the liner 2y — 3z, or .r jfr 13’lies in the plane, its direction vector a 1. . ) is parallel to the plane.

The point (0.0.0) is on the line (put / = 0>. and n e can serify that the gisen point (1. 1. 1) in the plane is not on the line.

The sector connecting these two p0mb. b - / 1. — 1. 1 , is therefore parallel to the plane. but not parallel to 1. 2, 3). Then

a b = -- 4. — 1.—i ) = 4. 4. —)() is a normal vector to the plane. and an equation of the plane is

4(.r—0)—4( 0) (z- 0) Oor5r-4g 9z0.

29. A direction vector for the line of intersection isa n >< fl2 ‘1, 1. 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 — 0. the equations of the planes reduce toy a — 2 and —y -I- 3a 1 with

simultaneous solution p = 4 and a = . So a point on the line is (0.. 4) and another vector parallel to the plane is

—1.— . —k). Then a normal veetorto the plane is n = 2. 5. 3 /_1. —4. — (—.2.1. —8) and an equation of

theplanejs_2(rt 1) L4(/J 2)- sc 1) —Oorr — - -Ia = 1.

iL The points (0. —2, 5) and (—-1, 3. 1) lie in the desired plane, so the sector v1 1— 1.5, —4) connecting them is parallel to

the plane. The desired plane is perpendicular to the plane 2 a — 5.r + ly or 5i .4— 4y — 2z 0 and for perpendicular planes,

Ia normal vector for one plane is parallel to the other plane, so v2 — (5. 1. 2) is also parallel to the desired plane.

Anormal vector to the desired plane is n = V X V2 = —10 -r- 16. 20— 2.—I — 25) 6. —22, —29.

Ting (r0. yn. a) = (0. —2. .5). the equation se are looking for is 6(.r - 0) — 22(y ÷ 2) — 29(a — 5) 0 or

29z- —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 = (3 — 0. —2 — 6.0 — 1) = (3. -8. —1) is a nonnal vector to the desired plane. The point

lies on the plane eqi i 3(1 1 8(// 5 ( 1 0 ni 3 a 8ij - 38

SECTION 9.5 EQUATIONS OF LINES AND PLANES 137

38. A direction vector fIr the line through (1.0. 1) and (4. —2. 2) is v = 3. —2. 1 and. tallng P0 -= 1. 0. 1). parametric

equations for the line are .r r- 1 F It. p r- —2t, z 1 — t. Substitution of the parametric equations into the equation of the

planegivesi f3t—2t—tl+1---6 — t=2.Then.t —1÷3(2) 7,y= 2(2)— - 1.andz 1—F-2— 3sothepoint

of intersection is (7. 4.3).

39. Normal vectors for the planes are n1 — (1.4. —3, and fl*

3. 6. 7, so the normals (and thus the planes) aren’t parallel.

But n fl r 3 - 21 — 21 0, so the normals (and thus the planes) are perpendicular.

40. The normal ‘,ectors are n1 = (1.2.2. and 2 -= 2. —1.2). The normals are not parallel, so neither are the planes.

Furthermore. n1 ro — 2 - 2 —r- 1 =- 4 0. so the planes aren’t perpendicular. The angle between them is given b

rit’fl2 1 4 14 .. -cosU — = -— U cos (r) 63.6

Ii // 9 a

41. Normal vectors for the planes are n1 — 1. 1. 1) and na = 1. —1. 1. The normals are not parallel, so neither are the planes.

Furthermore, ni ‘ 112 — 1 1--

1 1 0. so the planes aren’t perpendicular. The angle between them is given by

fli’112 1 1 —1 1\ — —cosU=, . .. =— 0—cor (-)11l0..D.fl11 1121 v3 3

42. Normal vectors for the planes are n1 ( —1,4. 2 and n2 = (3, 12.6). Since n2 = 3nj, the normals (and thus the

planes) are parallel.

43. (a) To find a point on the line of intersection. set one of the variables equal to a constant, sai a — 0. (This will fail if the line of

intersection does not cross the .ey-plane in that case, try setting t or p equal to 0.) The equations of the two planes reduce

toe ± p — 1 andi 4 2y 1. Solving these two equations gies a’ — 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 = (1.1. 1) x i. 2. 2 = (2 —2.1 — 2.2 — it — ‘0. —1. 14. By Equations 2 .parametric equations forthe

linearex=1,y - t,z t.

fll’fl2 1±2±2 .5 -1/ 5 N(bi The angle between the planes satisfies cos U = , = _ — = —. Therefore 6? = cos I — I inS111)112) 3y79 3/)

44. (a) lfwe set z 0 then the equations of the planes reduce to 3.r-

2p = 1 and 2 + = 3 and solving these two equations

gives .r — 1, p 1. Thus a point on the line of intcrscction is (1.1.0). A vector v in the direction of this intersecting line

is perpendicular to the normal vectors of both planes, so let v = n X 2 = 3. —2. 1) y2. 1. —-3 = (5. 11.7). By

Equations 2, parametric equations for the line are a’ = 1 -‘- St. p = 1 + 111, 7L

b e i’fl 623 1

_

cos U=cos ()8u.9.

Setting z = 0, the equations of the two planes become 5—

2p = 1 and Ir ± = 6. Solving these two equations givesl,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 91fl’9’. 1j..’:i l-

- —1 1

RIr’ ir.’rC,,ra M C.” hr ord er drpl..r’.rd ‘C po rdr, hr’11’,5,.rSNC r’r’,r”r ,r, s’.’rohr C,,, rzr

I IUI COUMI IUN Oh LN ANU HLANh 119

parallel. ote that L2 and L2 are not parallel.) L1 contains the point (1. 1.5). hut this point does not lie on L3. so they’re not

identical. (3. 1.5) lies on L4 and also on L2 (fort = 1), so L2 and L1 are the same line.

53. Let Q (1.3,4) and R = (2, 1. 1). points on the line corresponding to t = 0 and t = 1. Let

P (4.1. —2). Then a = OR = (1. —2. —3), b = OP = 63. —2. -—6. The distance is

da . bI ci 2 3 x (3 —2 —6 \6 —3 4 — — (—3) —1>

— — 61— a — (1—2. —3 — ti. —2.—3 -— \12 — ç_2)2 — (_3)2

-— 14 - \ 14’

54. Let Q = (0, 6. 3) and R — (2. 4, 1), points on the line corresponding to 1 0 and I =- 1. Let

P (0.1.3). Then a-— QR = 2. —2.1 and b = OP—- (0 —5.0 The distance is

d—ax b (2.—2.lt x (0.—SO) — (5.0.—iD) 52 02_ (_10)2 x’125 5\/

— a — (2. 2,1,. — 2,- 2.1 — ‘22*( 2)2m12 — — 3

Io1 -+ by1 -- (Z1 4 3(1) 21 2) — 6(1) 5 i i55. B Equation 9, the distance is D= - /2+62

— ,,,/ — 7

- . 1( 6) — 2(3) — 1(5) — 8 —-10 4056. By Equation 9, the distance is D = - =- ,/12 — (2)2 (4)2 \/ii v’21

57, Put y z = 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,

D4(2)—6(0)-t-2(0)—3 — 5 — 5

-- \/4 (_6)2(2)a -- /ii2\/i 2s

5L Put.r p = 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 fig + 9z 1 — 0. By Equation 9,

D=1-33 6(D)-r9(D)—i: 1132 ± (_6)2 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.

(o.yn, z0) be a point on the plane given by ni by rz + 4, 0 Then OTr, hyo + czo + d, 0 and the

dtance between P0 and the plane given by + d = 0 is. from Equation 9.54

D ax0 + ago + czo ± d-, c11 ± d2. — d’b2+c2 = x/o2 tb2 ±c2 —

\52 (.2

iplaxies must have parallel non-nal vectors, so if ax + by + (‘z -- 4 0 is such a plane, then for some I 0,

-1<, 2, 2)

= (1,21. —21). So this plane is given by the equation x + 2y — 2z + h’ = 0, where 5’ d/t. By

iSe59,thedistance betweentheplanes is 2= 1 —

6 = 1— k k = Tor —5. So the\/12 + 22 — (_2)2

1anes have equations x -- 2i, — 2z 7 and .r 2, 2z - 5

UI Rr R,d Ua b rId r V•VJ r’i:- == ,

SECTiON 9.6 FUNCTIONS AND SURFACES 145

11. z = 6 - 3r — 2yor3rt2g z 6. aplane with

intercepts 2, 3. and 6.

13. z = A + 1. a parabolic cylinder

12.: = coJ,a’\2\e.’

14. (at The traces in .z- -= A are parabolas of the form : A2 +y. the traces in y = A are parabolas of the form : :2 b-2 and

the traces in z r— /- are circles r2 ± 2 = A A > 0.

Combining these traces e form the graph.

1

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-

146 i CHAPTERS vC utn RINU -iC rCUJ!C Y ui ai-sw,i

b) (c)

g(s’ y) is the graph of f(x. y) reflected in tile hC. y) is the graph of g(s. y) shifted upward 3 units

eg-plane. (Note that g .r. y) f(s. y;.]

15. All si graphs have different traces in the planes a = ft and p = 0. so we investigate these fur each function.

(a) f(s.y)-

+ p1. The trace ins 0 is: = y, and in p 0 is: , so it must be graph VI.

(b) f(i’. .ij) — The trace in i 0 is : - 0. and in p = 0 is: — 0. so it must be graph V.

1. 1 . 1 .. -.(c) f(J.u Thetraceina Ois:= ,,.andin;=Ots: =

. Inaddition.weeanseetharJ is1 ±i’ + 1 + jj 1 —

close to 0 for large values of a and p. so this is graph I.

(d) f(r ) — (2 j2)2 The trace in .r = 0 is z = p1. and in p = 0 is = s’. Both graph IT and graph IV seem plausible;

notice the trace in : = 0 is 0 = (.z2—

p2 ) p = ±s. so it must be graph IV

(e) fis.y) — (a-

p)2. The trace ins = 0 is: = p2. and in y = 0 is : = .r. Both graph II and graph IV seem plausible:

notice the trace in : — 0 is 0 — (a—

p)2. p a, so it must be graph 11.

(f) IC. p) sin (i’ + p1). The trace in a = 0 is 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 : — ‘ 16 — a2—

16y2 or equivalently

.2 +16p2 + z2 16,: >0. Traces ins = kare 16p2 +z2 16

z > 0, a family of ellipses where here we hae only the upper halves. Traces in

p — P are .12 — :2 = 16 — 16k2, z > 0. again a family of half-ellipses. Traces

in : = P. 1.’ 0. are another family of ellipses. 2 16y2 = 16 — P2.

Note that the equation can be written as ± p2 + 1, z 0, which we

recognize as the top half of an ellipsoid with intercepts +4. ±1, and -1.

17. The equation of tile graph is: \/452 ± p2 or equivalently 422 2 = :2

0. Traces in S-= P are :2 p2 = 4k2, a 0, a family of hyperbolas where

we have only the upper branch. Traces in p = k are z2 — 42= p2 0,

again a famila of half-hyperbolas. Traces in: = k, P 0. are 42 + 2 p2

or .r2 + = . a family of ellipses. Note that the original equation can be

written as 2± _

-, 2 > 0, which we recognize as the upper half of an

ciliptical Cone.

1, ti

2Ol{rrr,’-rr Lrrrr mr ‘1] P’ht. P., ,rrd kIr m’tS, rr-,d, r pd rd rprrd ror prrb c’ rrI—it, rr ‘. ,.irrrrrn

SECTION 9.6 FUNCTIONS AND SURFACES 147

18. The equation of the graph is a 2 The traces in a A are

— 02. a family of paraholas opening downward. In p A, sse hase22 — 02. a family of parabolas opening upward. The traces in a = A are

a2 zj2 A, a family of hyperbolas. The surface is a hyperbolic paraboloid

with saddle point (0. 0. 0).

19. p j. The traces in a A are the parabolas / 2. the traces in

= A are A = 2 i.2 which are hyperbolas (note the hyperbolas are oriented

differently for A > 0 than for A < O; and the traces in a A are the parabolas

. , p aa - A- — .r. Thus.

.- —

is a hyperbolic paraboloid.

20, For.r p2 i—.tbetraccini — arey2 4,2 I,WhenA Osse

hase a fttmily of ellipses. When A Owe has ejust a point at the origin, and

the trace is empty for 0 <0. The traces in p A are .r 4a2 f k, a

family of parabolas opening in the positise a-direction. Similarly, the traccs

in a A are .r i2 4- 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 22. Completing squares in p and a gives.12.2+(y_2)2T 1(a 3)2 br

ci 2)2÷12 2)2 .r -Oor(y_9)2

I, (. 9)2(a . 3)2

= 1. an ellipsoid with— (p — 2) .an elliptic paraboloid with

center (0.2,3). sertex (0.2.2) and axis the horizontal line p - 2, a — 2.

a) in R2 2= 1 represents a circle of radius 1 centered at the origin.

(b) In , the equation doesn’t ins olve a, which means that any horizontal plane a A intersects the surface in a circle1, z A. Thus the surthce is a circular cylinder, made up of infinitely many shifted copies of the circle1, with axis the a-axis.

c) In a, 21 atsn r -a

‘ .. 1 - ‘:

:5

0.4.30,2,2,

‘.115 l, R d M r I drdrI’..t .1 op ) r.b’.,R ,bI, .r.

148 L7 CHAPTER 9 VECTORS AND THE GEOMETRY OF SPACE

24. (a) The traces of z2 = .2 ± p2 in ,r P are =2 A’2 a family of hyperbolas, as are traces in p k, z2

72—

Traces in z = P are ,t’2p2 a family of circles.

(h) The surface is a circular coneith axis the c-axis. (c) The graph of f(.r, p) = /T+y2 is the upper half of

the cone in part (h), and the graph of

g(i y) =- \/12 p2 is the lower half

25. (a) The traces of,r2 t- p2 — c2 r— 1 in ,r — P are p22

= 1 A’2 a family of hyperbolas. (Note that the hyperbolas are

oriented differently for 1 < P < 1 than fork < —1 ork > 1,) The traces in p = P are £‘ — z2 1 — k2, a similar

family of hyperbolas. The traces in z r- f are 2 +2 + k2, a family of circles For A’ = 0, the trace in the

i’p-plane, the circle is of radius 1. As P increases, so does the radius of the circle, This behavior, combined with the

hyperbolic vertical traces, gives the graph of the hyperholoid of one sheet in Table 2.

(b) The shape of the surface is unchanged, but the hyperboloid is

rotated so that its axis is the p-axis. Traces in p A’ are circles.

while traces in 2 = P and z A’ are hyperbolas.

(c) Completing the square in p gives ,,2 (p ± 1)2 :2 1. The

surface is a hyperboloid identical to the one in part (a) but shifted

one unit in the negative p-direction.

26. (a) The traces of p2 + z2 1 in i = A’ are —p2 -- = 1 -‘- P2. a family of hyperbolas, as are the traces in p

— 1 + P2. The traces in a k are .r2 + p2 P2 1, a family of circles for P > 1. As JP increases, the radii

of the circles increase; the traces are empty for fa < 1, This beha ior. combined with the vertical traces, gives the graph of

the hyperboloid of two sheets in Table 2.

(b) The graph has the same shape as the hyperbolnid in part (a) but is rotated so

that it - I lie -e Traces in .r k. kI > 1, are circles, shile traces

in p = P and a — P are hyperbolas

CAll 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.

-, 9 1r 1j28. f(r. y) = .rge -

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 = —2a

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 f achieves a minimum value

lower valley dip. I

Three-dimensional x iex

0

Side view

V

‘.11 R,,±9 Rr ‘.i., -rd,.

a’.cL,

178 Li CHAPTER 10 VECTOR FUNCTIONS

From the projection onto the iy-plane ve see that the cune lies on the \ertical

plane p = i. The other two projections show that the curve is a parabola contained

in this plane.

15. Taking r0 = O. 0,01 and r1 11,2.3. ve hiue from Equation 9.5.4

r(t)=(1—t)ru In =(1 t)0.O.O + 1.2.3,0<t<1 or r(t)(t.2/ .3/,, 0<1<1,

Parametric equations are r /, p 21. a 3/. 0 <. / < 1.

16, Taking i’ =- (1,0. 1) and r1 (2.3. 1), we have from Fquation 9 5.4

rQ) = (1 I) ro + I ri — (1 /) (1.0.1) ft (2.3, 1), 0 < <1 or r 1) (1 /. 3t. 1). 0 < I < 1.

Parametric equations are ,i 1 f t, p — 31, a 1. 0 < I < 1

17. Takingro—

(I. 1,2 andri 1,1 7 ,vehave

r(/)=(1—t)ru*iri—( 1)1. 1.2 -tiI.1.7’.0 <lorr(t)—Li.—3t. l2t245t),0.a’I1.

Parametric equations arei =1 1 31. p 1 21. - 2 - 5/. 0 < / < 1.

18. Takingro = — 2.1.0 andr1 :6. 1.2 . v.e haie

r(t)(1—1)rotrl=(1—t)/-2.1.0,-.1/6. orn(/)/2aS1.1_5/.21,.0</<1.

Parametric equations are .o = —2 -i- 8/, p = 4 —- 51. — 2/. 0 < / < 1.

19. = tcost. p =1, a —/sint. 1>0. Atanvpoint(.r.g.a)onthecurve..r2±:2 =t°cos2l+t2sin2/=/2 p2 so the

curve lies on the circular cone a2 2=

,2 with axis the p-axis. Also notice that p > 0: the graph is II.

20. x = cost, p = sin/, a = 1/(1 + 12), At any point on the curve ve have a2 2 + sin2 1 1, so the curve lies

on a circular cylinder a2 -i p2 — 1 ith axis the a-axis. Notice that 0 < a < 1 and a = 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 — 0 as I ±. The graph must be VI.

2t. .r 1, p - 17(1 + p2), a — 12. At any point on the curve we have a a2, so the curve lies on a parabolic cylinder parallel

to the p-axis. Notice that 0 < p < 1 and a > 0. Also the curxe passes through (0,1,0) when 1 0 and p 0. a as

I F, so the graph must be V.

22. i’ cost. p sin t, a = co 21. .2— = cos2 —i- sin2 1 = 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 VECTOR FUNCTIONS AND SPACE CURVES 179

23..r = cos Nt, y = sin bt. z , t > O.a’2 -i- cos t -t sin2 t 1. so the curse lies on a circular cylinder withaxis the c-axis. A point (i. y. z) on the cure lies directly above the point (a. gO). which moves counterclockwise around theunit circle in the .rg-plane as increases. The cur’e starts at (1, 0. 1), when t 0. and z (at an increasing rate) ast — c, so the graph is IV

24. a = cos2 t, y = sin2 1, z = 1. a--

g cos2 I sin2 I — 1.so the curve lies in the ertical plane- 6

.r and g are periodic, both with period r. and z increases as increases, so the graph is Ill.

-. 2 9 ) 9.) 925,1I=tcos1, y Ismt, .t.then.r y—tcost -tsint t—z,

so the curve lies on the cone z2 2 * g Since 1. the cuie is a spiral on

this cone.

26. Here 2,2 = = and s’2 - sin2 cos2 1 1. so the

curve is contained in the intersection of the parabolic cylinder

,r2 with the circular cylinder 2y2 — 1. We get the complete

intersection for 0 < I < 2ir

27. Parametric equations for the curve are .r - 1, y = 0, — 21 t2. Substituting into the equation of the paraboloidgives 2t =- t2 21 212 1 = 0, 1. Since r(0) = 0 and r(1) i - k. the points of intersectionare (0, 0, 0) and (1,0.1).

28, Parametric equations for the helix are x sin t, y cost, z = t. Substituting into the equation of the sphere givessiti2cos2t+12- 5 1 ± 2

= 1 ±2. Sincer(2) =1siii2,cos2,2 andr( 2) (sin(2).cos(—2). 2. the points of intersection are (sin2.cos2.2) (0.909. —0.416.2 and(sin(—2,), cos(—2). —2) (—0.909, 0.416, —2).

29, r(t) (cost sin 2t. sin I sin 21. cos 2t.

Wê’inc1ude both a regular plot and a plot

howing a tube of radius 0.08 around the

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7. Since g= (3f = ()3 = .r3, the

cure is part of a cubic cove. Note

that here ..r > 0.

9. r’(t)=

i sini. Ft2 , tcos2t) = ‘teost sint, 2t, t(— sin2t) 2 ± cos2t)

(tees / sin 1.21. (‘OS 2/ 2/sin 2/,

10. r(t) (tan/sect. l/t2 r’(tj = ‘sec /. secttant, 2

11, r(t)=2i—jIn(1 3/)k r’(t)=2tfi

12. r’(t) = [at(—3sin3/) -aeos3t1 1+ b . 3sin2/costj ±c 3cos2 /( sint)k

(a cos 3/ — 3d sin3l) i -3hsin2 / costj —

3rcos2 I sintk

13. r’(t) = 0 + b 21 c b + 2/c by formulas 1 and 3 of Theorem 3.

14. To find r’(t), we first expand r(t) = /a x lb + / c) t(a>. b) -I- t2(a x c), so r’(/) = a x b ± 2t(a x c).

15. r’(t(_10t ± e.2/(1 t2, 2cr) r’(0) = /l.2.2. So r’(O) + 2 +22 = = 3 and

T(0)—99 —

= =

Ir’(O)I‘

16.r’(t)j+2tj+k r’(l)=2i±2j1.k,Thus

T(1)r’(l) 1

ir’(l)j

188 CHAPTER 10 VECTOR FUNCTIONS

6. Since y = e =the

curve is part of the hyperbola

1= —. Note that.r > 0. y > ft

(a), (c)

1• t. t.

(b)r(t)=c i—c j.

r’(O)=i -.-j

1)

(a). (c)

(b) r’(t) = c3 i ± 3c j.

r’(O) = i — 3j

0

8.i=I- cos/, /1= 2—’--sintso

—1)2 f (q 2) land the

curse is a circle.

(a). (c)

.3,

(b) r’(/) = sin j - costj,

_.J-

r61 2’ 2

25iO,‘-Lr cpL

to uH,J o,tbte oh,to -‘ tt5...o °r° P°”

SECTION 10.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS 191

30, (a) The tangent line at 0 is the line through the point v ith position sector r(0) — (sisi 0.2 sinO. cos Oi (0. 0. 1), and in

the direction of the tangent sector. r’ ‘rr ens 0. 2rr ens 0. rr sin 0, — r. 2ir. 0 So an equation of the line is

(i’. p. z = r(O) Ltr’(O) = (0— mj.0 2u, 1, = çu.2u. 1).

r() — sin.2sin.eos) - 1.2,0).(b)

r’() rrcos .2cos. - rrsin =

So the equation of the second line is

‘.nij. =1.2.0 =u(0.0.- ‘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) 1, 2t. 3[2 and t Oat (0.0.0), ri(0) o- (1.0.0 is a tangent sector to r at (0.0.0). Similarly.

r(!) = cost. 2eos2t. 1 and sincerl(OI = 0. 0.0 , ri (0) (1.2.1 is a tangent vectorto r2 at (0.0.0). IfO is the anole

hetaeen these two tangent sectors, then cos 0 (1,0,0) . (1. 2. 1 and 0 ens 1 () 66

32. To find the point of intersection, we must find the values oft and . a hich satisfy the follos ing three eqLlations simultaneousl\

—3 - a, I I a 2,3 1- t2 2 Solsing the last two equations gives t 1, a 2 (check these in the first equation)

Thus the point of 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

cDs0=_Tl._(_-l 1—S)=—=—andO=eos i(+).5.5

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 — 9t2j ÷ 25t k) dt= (i lOt3 dL) i (J 912 di) j + (j 2511 di) k

— E1t4 i [3tj-i- 15[0]k=1i_3ji.5k

f i ± k) dt- [4tan 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 cos2 Ii 2sint costk) dt

(f/23.2t eostdi) i (J 2

3sint cos2idt)j (f22sint costdt) k

jsin3t] 2 E—cos3I’ 2i [sint]2 k = (1—0)1 + (0+ 1)j + (1— 0)k = 1±j +k

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SECTION 10.3 ARC LENGTH AND CURVATURE 199

(b) rh)T’(t) \/(1 5L)

_________

rr’ 1 512 (1 —

21. r(t) — t3j -+ t2k r’(/) 3t2j 21k, r”(t) 6/j + 2k, r’(l) V t)2÷(22 \/TTi2,

r’i) r”(/) = —6t2i. r’(ti r”(t)! 612. Thenrr(/)= jr’(/) x r”t) 6/2 6/2

(V9t1 4j2 )3 (91 4/2)32

22. r(1) - ti 12j+((k r’(t) i—21j—’k. r”() 2j—tk.

r’(t) /‘‘22T”j’2 /“2t r’(/) >‘ r”(t) (21 2)r1 i r1j — 2k,

r’(tJxr”//) = ‘21—2)’ (l)2 22\ (2/—2(2c2tc2’t_ \(t15)t1.

/1/1r (/) >< r (1) (4/- 81 5)c- 1 v(’t- 8t- 1Then h t)1 4/2 [21) (1 42 2 2 -

23, r(t) 3/i 4 in tj I cos 1k r’(t) 3 i 4 os j L0 1k, ° 1) —lsin tj 1 o t k,

jr’(t) = \‘9+16co2t+16in2/= 9— 165. i”(/) x r°(/) 16i 12costj l2sintk.

Ir’(t) x r°() -. 256 r lilcos2/t llIsin2t 20. Then t(t)r’(L) xr”(t) 20 1

d _.i

24. r(t) /j2, mt. Lint) . r’(t) (2/. 1//.1 in /, r°(t) (2, 1i/2,1 1;. The point (lU. 0) corresponds

tot — 1, and r’(l) = 2. 1. 1), r’(l) \_12=12 = \/6. r°(1) = 2. —1. 1.. r’(l) r°(1) a-s 2.0. - 1,.

Ir’(l) x r”1) O ( 4)2- vü 2 . Then n(1)

jr°(1)(_)3 or

25. r(t) (1, 2, ti) r’(t) (1,21, 3/2, The point (1.1,1) correspondsto / 1, and r’(l) (1,2.3)

r’11) v’1—4±9 r”(t) s-s 0. 2.ö/i r”(1) = (0.2.6. r’(lj> r”(l( /6. 6.2, so

(r’(l) X r°(1)I = 36+4 -. Then r’(l)xr°(1) - 1

Note that e get the complete curse for 0 < / < 2ir.

rh) =- ((‘Os t.sint.sin .5t = r’(t) = (— sint. cost. 5cos5t).

r°(t)— (-- cost sin /. -25 sin 5/. The point (1,0.0)

corresponds to 1 0. and r’(O) (0. 1.5)

r’(O) — 12 52 =, r°(0) = (—1.0.0),

-i r’(O) x r°(01 = 0. —5.1)

[r’(O) x r”(O) /biT)E5)2E’i2 = The curvature at

r’(O) x r”(0’ 1the point (1.0.0) is h’(D) = i(fl33 ‘ =

35! (cnaac carmaig UI (lagS, Rats-, ad is S s-ar-red arrpted ord‘1-,.satod it, po aid i, a pabiIt :,ccaraihS arar tm total is ta parr

200 CHAPTER 10 VECTOR FUNCTIONS

27. f(r) = ,r. f’(c) = 4a. f”(r) = 12752, ) (ff())2]3 2= [1 (43)2’3 2 11 16°) 2

28. f() = tan a, f(.r) sec2 a’, f”(x) = 2 sec x sec tan x 2 sec2 ,r tan.c,

f”(a’)l I2sec2 a’ tanil 2scc2 tanr

= [1 + (f(a5))213 2 (sec2 j)2l3/2— (1 sec4 5r)2 2

29. [(a’) - xe’. f’(a) = a c, f”(a) .ccr- 2e’,

ac’ + 2e ,c ‘ 21

h\.L)— f1+ I())23 2 t1(a’e±e)22 = 1+(1crc)a]32

1 130.

V” (c) 1 1 1(2 )3 2

17(1)(y1(r))23

2 11 (1 + 1/12)32 2 (a’s 1)3/2 (72 + 1)3 2 (12 1)3,2[slncea’ >01.

To find the n3aximum curvature. ne first find the critical numbers of tt(a’):

± i)2

— i()(a t 1)2(2I) (a2 i)’2[(.r2± 1) 3r2] 1 212

(5r f(2 32(2 + 1)3

(2

0 =4- 1 212 0, so the onl critical number in the domain is a’ = , Since e’(i’) 0 forD < ,1

and e’(a’) < 0 for ,z’ > , (r’) attains its maximum at a . Thus. the maximum curvature occurs at ,ln

Since limL 1) ‘2

— 0. (i’) approaches 0 as .r c.

SS

(y (a’) C ‘ 3

31, Stnce q = q -— c’ thecurvaturels17(a) S =e (1 rn )

[1 (gI(I))2]3’2(1 + c213/2

To find the maximum curvature, we first find the critical numbers of tt(x):

= e2( t72) 22 (—)(i ±32 3/22E2.) 7

= 0 when I - 212w 0.so 2r = or.r —1n2. And since 1— 2c2’ >0 forx < —ln2 and 1 2c2’ <0

for a’ > —in 2, the rnaxin3urn curvature is aained at the point (— in 2, 1n2;, 2)

= (— In 2,

Since urn eI(1+172r) +2 =0.n(,r)approaches0asa’’—’.

32. We can take the parabola as having its ‘ertex at the origin and opening upward. so the equation is f(.r) = ox. a> 0. Then Y

Fquation 11, J$(T(“(1’)

2 ‘1 - (9,’)2’312 = (1 4a2a’2)3’2’thus ta(O) = 2u We wanta(O) 4.so

it 2 and the equation is y = 212,

7 23101 g0 L’o,’,o 0,j3 R,oh 117 r17 Olo’ ,,ol 7, ,,,o’,c’J 72 c, dpl5,o17. ,r,”csl,d 0, ,,b.,_l eb 5o. io ,ho7 or

33, (

34,

35.

36

33. (a) C appeais to he changing direction more quickly at P than Q. so se would expect the curs ature to be greater at P.

(b) First we sketch approximate osculating circles at P and Q. Using the

axes scale as a guide, sse measure the radius of the osculating circle

at P to be approximately 0.8 units, thus p =

= 1.3. Similarly, we estimate the radius of the

osculating circle at Q to be 1.4 units, son 0.7.p 1.4

34,p=.r2a —y’4x3—4.r. y”=1221. and

1

n (I) =,

. The graph of the

[i()2] (it3-- 4e)2] -

cursalure here is what we ssould expect. The graph of p — 2,2

appears to be bending most sharply at the origin and near., = 1.

35 =2

‘2a

, p’ — 6a ‘, and

o. ‘ ti

_______

__________

—

_________

-( 2i

1)2132 a’(1 Lr)3

The appearance of the two humps in this graph is perhaps a little surprising. hut it is

explained by the fact that p2 increases asymptotically at the origin from both

directions, and so its graph has sery little bend there. [Note that n(U) is undefined.]

to

P

1

1

43

36. Notice that the curve o is highest for the same i-values at which curse 6 is tuming more sharply. and a isO or near 0 where 6 is

nearly straight. So, o must be the graph of p --- a(i). and 6 is the graph of p •fLr).

37. Notice that the curve 6 has two inflection points at which the graph appears almost straight. We ould expect the curvature to

be 0 or nearly 0 at these values, but the curve o isn’t near 0 there. Thus. o must he the graph of q = f(.r) rather than the graph

of curvature, and 6 is the graph of p

(a) The complete curse is gisen by 0 < t < 2r. Curvature

appears to have a local (or absolute) maximum at 6

points. (Look at points where the curse appears to turn

more sharply.)

hi.bt hc:ç

SECTION 10.3 ARC LENGTH AND CURVATURE 203

44. f(x) — f’(.r) . J”(.a) c2e ‘. Using I om2ula II e hae

= [I (f’(.r))2r 2 U (cc’)2’3 2 2so the curvature at z’ = 0 is

(0)(1 )3 2

To determine the maximum salue for K(O). let 1(c)= (1 + 2)3 2’

Then

2(1 (.2)3 2 (1 c’)1 2(9.) (1—c 2 2c(1—c2)— 3c (2_r2)-. = -

___________

. \Ve ha\e a critical(1 L (.2)3 2]2 (1 c2)3 (1 .2) 2

numberwhen2c c3 =0 —n c(2 (2)= 0orc. f’(c)ispositiveforc< .0 < en

and negatE e elsess here, so J’ achieves its maximum alue when c — ‘2 or — s . In either case. U 0) = so the members

of the family ssith the largest value of n(0) are f(.r) c and f(.r) C

r’(t) 2(. 212. 1 :t. 2t2. 145. (1. . 1) corresponds tot = 1. T(t)= ‘(1) •e1t2 + W -r 1 212 1

.soT(1) = .. ;.

T’(t) — 4t(212 1) 2 (2t. 2t2. 1) (2(2 1) ‘(2.11.0) [by I ormula 3 of Theorem 10.23]

= (212 + 1)2g2 2 2. j3 3(3

+ It. —4t = 2/2/2 y 2 -- 2(2.2!. 2!)

N(t)T’(t) 2(2(2 H- 1) 2 (1 2(2, 2!. -21) — (1 2(2, 21, 2!) (1 2t2, 2!, _2t)T’, 2(2/2 ± 1) 2\/( 2(2)2 (2t) + ( 2!) +it48(2 1 4 2(2

N)=-..-—4)andB(1)=TU)’n N(1)= .—(—-—).i-- 1’ .(,

46. (1.0,0) corresponds to t 0. r(t) (cost. sin 1. in cost , and in Exercise 4 we found that r’(t) ( sin 1, coSt. — tan t

and jr’(t)1 = sect . Here we can assume - < / g and then sect > 0 =‘ r’/ sect.

T(/)r(!) (— sin!, cost tan!)

(— sint cost, cos2 I. sint) and T(0) (0.1.0).

= (—[(sin!)(— sint) (cos/)(cos/). 2(cost)(— sint). cost) = (sin2! — cos2 /. —2sintcost. — cost). so

N(0)= 1.0,-1) -(-1,0,1)=( ,0.

Finally, B(0) = T(0) . N ‘0.1.0 x /—--.o. ‘—-4 02/’

47. (0, r, -2) corresponds to t 71. r(t) (2 sin 3!, (.2 (‘05 3!)

/ r’(t) 6cos3!,1,—6sin3/ 1 .Tt() = =

____________________

— bcos3t. 1. —6s]n3t...r’çt)j ens2 31 + 1 + 36 sin2 3!

T(7rj — ‘-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

planeis 6(a’—O)-e 1(y—a)--i-0(z+2)=Oorg—6r=ir.

(—l8sin3t,0, l9cos3t) =n T’(!)/n23t+182cos23t =

NU)= = (—sisi3t.0.—cos3t). So N(71) = 0.0. 1 and B(71)

=(6.1.0) x (0.0.1) (1.6.0.

ceB(71)is a normal to the osculating plane. so is ‘1.6,0equation for the plane is 1Cr —0) -n- 6(y — 71) -- 0(z -+- 2) 0 or 71 -i-- 6y = 671.

C ‘Ptfl Ltr CII R Sn Pn-nd Can nt a ncnntJ p 1 dnpl n.t a ,pn an, patch c,.cttib’n tn-kIln in nh tintt r