hw-10-sec-14.1-solns

# hw-10-sec-14.1-solns - 4 D VECTOR FUNCTIONS Vector...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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’l-Iospital’s Rule]. Thus lim (cost,sint,tln t) = < lim cost, lim sint, lim tlnt> t-0+ 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’EZ-EZ1 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 t-woo 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 mz-plane z = 303, z > 0, a half cubic, and the yz-plane, 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 y-axes. 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~vwmemwww-e , 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 z-axis. 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 ingeSﬁLt-z ‘ 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 real-valued 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 ...
View Full Document

## 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.

### Page1 / 5

hw-10-sec-14.1-solns - 4 D VECTOR FUNCTIONS Vector...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online