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Unformatted text preview: 4 D VECTOR FUNCTIONS Vector Functions and Space Curves ctions V4 ~ t2, e’“, and ln(t — , the domain ofris (~1,2]. 4.1
20 => + 1) are all deﬁned when 4 — t2 t+1>0 => t> t ~ 2 , sint, and ln(9 ~ it?) are all deﬁned when t gé ~2and9~t2>0 :> ~23t_<_2and lim —t = 0 so the domain 0
i . lnt 1/t
z ‘ t.— 0~1l' 't: 0:0,l‘ tltzl' ———:l' ————=
3. $131+ cos cos , $151+ sm sm t:er n 110+ 1/1: L35: _1/t2 HM
[by l’lIospital’s Rule]. Thus lim (cost,sint,tln t) = < lim cost, lim sint, lim tlnt>
t0+ t—vO‘l' t—+O+ t—+0+
/ilim 9::  11111 at — 1 [usin l’Hos ital’s Rule]
t—~ t T t—vo 1 ' g p ’
/ ._ 1/ — \/1 t
hle :1im,lit._—l._/ii—1— :1gm/l’ :— in _,3._ :3_
mo t we t t/1+t+1 t~>0\/1+t+1 2t~01+t
Thus the given limit equals (1, %,3>.
. _ t2 1 1 1 1
511m63t=60:1,11m :hm’f'i—St—‘ﬁ’. ﬂi’EZEZ1
H0 t—>0 sm t t—*0 srn t hm sm t 1. sint 1
t2 Ho 162 113% t
and lira?) cos 2t : cos 0 = 1 Thus the given limit equals 1 + _] + k
. l t . 1 t
6. 11m arctant : %, lim 8‘” = 0, lim 3— : 11m —— = 0 [by l’Hospital sRule]
t—voo t—aoo twoo t t—voo 1
Thus lim arctant 6'” 13: =<1 0 0)
taco , 7 t 2) 7
7. The corresponding parametric equations for this curve are m ~ sm t, y — t
We can make a table of values, or we can eliminate the parameter: t = y :>
ection in a: : sin y, with y E R. By comparing different values of t, we ﬁnd the dir which t increases as indicated in the graph. ~3<t<3, 285 286 :1 CHAPTER 14 thlUK ruwununc u. mm. ._.. . ~
, CL [4—3 ( 8. The corresponding parametric equations for this curve are m : t3, y = t2. y
We can make a table of values, or we can eliminate the parameter:
aszt3 :> t: {3/5 => y=t2=(€/§)2:m2/3,
0 x with t E R => an e R. By comparing different values oft, we ﬁnd the direction in which if increases as indicated in the graph. ,xi“
MT he corresponding parametric equations are a: = t, y : cos 2t, z = sin 2t. Note that y2 + z2 = cos2 2t + sin2 2t 2 1, so the curve lies on the circular cylinder y2 + 22 : 1. Since :r = t, the curve is a helix. f” t #10.) The corresponding parametric equations are cc : 1 + t, y : 3t, z : k, which are parametric equations of a line through the point (1, 07 0) and with direction vector (1, 3,  1). 11. The corresponding parametric equations are m = 1, y 2 cost, 2 = 2 sin t. Eliminating the parameter in y and z gives 342 + (2/2)2 : cos2 t + sinzt = 1 or y2 + z2/4 7— 1. Since so = 1, the curve is an ellipse centered at (1, O7 0) in the plane as = 1. / . . .
1127 The parametric equations are an : t2, y = t, z = 2, so we have m z y2 With z = 2. \J
(o, 0, 2) Thus the curve is a parabola in the plane 2 = 2 with vertex (0, 07 2). g;
S:
g.
E
.3. SECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES ET SECTION 13.1 [parametric equations are a: = 12, y = if“, z = t6. These are positive f 0 and 0 when t = 0. So the curve lies entirely in the ﬁrst quadrant. projection of the graph onto the my—plane is y = :52, y > 0, a half parabola. :jourthe mzplane z = 303, z > 0, a half cubic, and the yzplane, y3 : 22. “I Ifg; = cost, y 2 —cost, 2 = sint, then $2 + Z2 = 1 and y2 + 22 = 1,
so the curve is contained in the intersection of circular cylinders along the
z and yaxes. Furthermore, y = —m, so the curve is an ellipse in the plane y = —a:, centered at the origin. 15. Taking r0 = (0,0,0) and r1 = (1, 2, 3), we have from Equation 13.5.4 [ET 12.5.4]
r(t) = (1—t)ro+tr1 = (1—t) (0,0,0) +t(1,2,3),0§ t S 1 or r(t) : <t,2t,3t),0§ t _<_ 1.
Parametric equations are a: = t, y 2 2t, 2 2 3t, 0 g t g 1.
16. Taking r0 = (1,0,1) and r1 = (2,3,1), we have from Equation 13.5.4 [ET 12.5.4]
r(t) : (1 —t)ro+tr1 = (1—t) (1,0,1) +t(2,3,1),0 gtg 1 or r(t) : (1+t,3t,1),0_<_t g 1.
Parametric equations are m = 1 + t, y 2 3t, z = 1, 0 S t S 1.
17. Taking r0 = (1, —1,2) and r1 = (4, 1,7),we have
r(t) = (1 —t)r0+tr1 = (1—t)(1,—1,2)+t<4,1,7),0gtg 1 or r(t) = (1+3t,—1+2t,2+5t),0 gt 3 1.
Parametric equations are at = 1 + 3t, y = —1 + 2t, 2 = 2 + 5t, 0 g t S 1. 18. Taking r0 = (—2,4,0) and r1 = (6, —1, 2), we have r(t) = (1—t)rg+tr1 (1—t)(—2,4,0)+t<6,—1,2),0§tg 1 or r(t) = (—2+8t,4—5t,2t),ogtg 1. ll Parametric equations are a: = —2 + 8t, y = 4 — 5t, 2 2 2t, 0 S t g 1.
.75) [ 191x 2 cos 4t, y = t, z : sin 4t. At any point (9:,y, z) on the curve, 952 + 22 = cos2 4t + sin2 4t : 1. So the curve lies on a x/ circular cylinder with axis the y—axis. Since y = 75, this is a helix. So the graph is Vl. z t, y 2 t2, z = 6“. At any point on the curve, y = $2. So the curve lies on the parabolic cylinder y = 932. Note that y
and z are positive for all t, and the point (0,0, 1) is on the curve (when t : 0). As t —> 00, (3:, y, z) —> (oo, oo, 0), while as t —> —00, (117,31, 2) —> (—oo, oo, 00), so the graph must be 11. 21. a: : t, y = 1/(1 + t2), 2 = t2. Note that y and z are positive for all t. The curve passes through (0, 1,0) when t = 0. Ast —> 00, (ac, y, z) ——> (00,0,00), and as t —> —oo, (x,y,z) —+ (—oo,0,oo). So the graph is IV. mm”, a” ,, ﬂ ‘W'Mém*mﬂw~vwmemwwwe , W. 288 CHAPTER14 VECTOR FUNCTIONS ETCHAPTER 13 < 'L . T‘ . /(;22/ a: : e_t cos 10t, y z 6" Sin 1015, 2 = e". " p
/”“ ‘.‘ :02 + y2 = 6;” cos2 10t + 6'” sin2 10t : e'2t(cos2 1015 + sin2 1016) = e‘2t : 22, so the curve lies on theﬂcpgp" x2 + y2 = Z2. Also, 2 is always positive; the graph must be I. x2 + y2 = cos2 t + sin2 t = 1, so the curve lies on a circular cylinder with axis the 23. ac : cos t, y = sint, z = sin 5t.
t = 27r the curve passes through the same point, so the curve repeats z—axis. Each of a), y and z is periodic, and at t = 0 and itself and the graph is V. 24. m 2 cost, y : sin t, z = In t. :52 + y2 = cos2 t + sin2 t = 1, so the curve lies on a circular cylinder with axis the zaxis. Ast ——> 0, z —> —00, so the graph is III. 25.1fmztcost, =tsint, z=t,thena:2+ 2:t2coszt+tzsin2t2152:22,
y 3/ so the curve lies on the cone 22 = $2 + 3/2. Since 2 : t, the curve is a spiral on this cone. 26. Here $2 = sin2 15 = z and x2 + y2 = sin2 t + cos2 t = 1, so the
curve is contained in the intersection of the parabolic cylinder z = 9:2 with the circular cylinder 302 + y2 = 1. We get the complete intersection for 0 g t g 271'. 27. Parametric equations for the curve are x = t, y : 0, z 2 2t — t2. Substituting into the equation of the paraboloid gives 2t — t2 = t2 :> 2t 2 2t2 => 15 = 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 as = sin t, y : cost, z = 15. Substituting into the equation of the sphere gives it = i2. Since r(2) = (sin 2,cos 2,2) and sin2t+cos2t+t2 = 5 : 1+1:2 : 5 2
) as (0.909, —0.416,2) and r(—2) = (sin(—2), cos(—2), —2), the points of intersection are (sin 2, cos 2, 2 (sin(—2),cos(—2), —2) m (—0.909, —0.416, —2). 29. r(t) = {cos t sin 2t, sin t sin 2t, cos 2t).
We include both a regular plot and a plot
showing a tube of radius 0.08 around the CUI'VC. SECTION 14.1 VECTOR FUNCTIONS AND SPACE CURVES ET SECTION 13.1 E] 291 'cles to collide, we require r1 t) = r2(t <:> t2, 7t ~ 12, t2 2 4t — 3, t2, 5t — 6 . Equating components
,partl 2513—6.Fromtheﬁrstequation,t2 —4t+3:0 <:> (t—3)(t—1)=Osot: 1 5 415—3, 7t— 12 = t2, andt2
s collide when t = 3, at the ingeSﬁLtz
‘ point (9,9, 9). _ t = 1 does not satisfy the other two equations, but t : 3 does. The particle 9 particles collide provided r1(t) = r2 (t) <=> ' (t, t2, t3) = (1 + 2t,1 + (it, 1 + 1415). Equating components gives = 1 + (it, and t3 : 1 + 14t. The ﬁrst equation gives it : —1, but this does not satisfy the other e uations, so
‘1 t = 1 + 2t, t2
ect, we need to ﬁnd a value for t and a value for s where r; (t) the particles do not collide. For the paths to inters : 13(8) <:> (t,t2,t3) : (1 + 25,1 + 63,1 + 143). Equating components, t : 1 + 23, t2 = 1 + 63, and t3 = 1 + 145. Substituting the ﬁrst equation into the second gives (1 + 23)2 = 1 + 63 => 432 — 2s = 0 :> 25(25 — 1) = 0 => 3 = 0 or s 2 we see that both pairs of values satisfy the From the ﬁrst equation, 3 = 0 => t = 1 and s = % :> t = 2. Checking, NIH third equation. Thus the paths intersect twice, at the point (1, 1, 1) when 5 = 0 and t = 1, and at (2, 4, 8) when 5 = and t = 2.
43. (a) lim u(t) + lim v(t) = lim ul (t), tlim U2(t), lim U3 (t) + <lim m (t), lim 122 (t), tlim v3 and the limits of these component functions must each exist since the vector functions both possess limits as t ——> a. Then adding the two vectors and using the addition property of limits for realvalued functions, we have that lim u(t) + tlirn v(t) : lim m (t) + tlim ’U1(t), tlim U205) + tlim 02(t), tlim u3(t) + tlim 113(t) t—va = lim [mm + mm] , gig; [mm + v2 (01 gig; lua(t) + mun) t—aa. : lim (u1(t) + U1 (t), ug (t) + ’02 (t), u3(t) + v3(t)) [using (1) backward] t—>a = lim [u(t) + v(t)] t—>a (b) i133 cu(t) : lim (cu1(t), ma), also» = <31; cu1(t), in}; cuz (t), lip; mm) t—vu = <clim u1(t),cmu2(t),cmu3(t)> = c<li‘n(11u1(t),lﬂu2(t),lhﬂu3(t)> tau. : 0 lim (m (t), U2(t), U3 = c lim u(t) t—wa t——>a <°> 11‘2"“)  122% = <mu1<t>mu2<t>tmust»  two 122w» 122w) : [lim ul vl + [lirn 11,2 v2 + U3 v3 t——>a. t—uz : lim u1(t)v1 (t) + uz(t)v2 (t) + us(t)v3 (t) t—aa. = lim [m (U010) + U2 (07120?) + us(t)v3(t)l = gig}; [u(t) v(t)l t——va ...
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This note was uploaded on 05/02/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas.
 Spring '09
 Chu

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