Homework 2 - Vincent(jmv777 — HW02 — Gilbert w(57495 1...

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Unformatted text preview: Vincent (jmv777) — HW02 — Gilbert w (57495) 1 This print—out should have 19‘ questions. Multiple—choice questions may continue on the next column or page — find all choices before answering. 001 10. 0 points If the constant C is chosen so that the curve given parametrieally by ogt<3, ($132,025), is the arc of the parabola y2 m 8:0 from (0, 0) t0 (2, 4), find theeoordinates of the point P on this are corresponding to t m 2. 1. so 2. on 3- P m o Si 4 (29%) a. so 5.}: m (3, gym: The point P has coordinates 2 2 (gt2lt22’ Ctltmz) 2 (1’20) 8 2 so we need to find 0. But we are told that the graph of 02 2 ( “ti = Ct) passes through (2, 4) when t 2 3. Thus 30 m 4, m, o = Consequently, keywords: parametric curve, parabola 002 10.0 points Determine a Cartesian equation for the curve given in parametric form by fit) m Zet, y(t) = Ste—2*. 2 1. 59— m :2 y 2 2. 35m = 18 y 3. $23; 3 12 correct 4 egg a: 18 5 my? 2 6 cc 6. "if? _" 6 Expianation: We have to eliminate the parameter t from the equations for :c and y. Now from the equation for so it follows that Consequently, . 003 10.0 points Describe the motion of a particle with posi— tion P(33, y) when a; :22 2sint, y = 300325 Vincent (jrnv???) — HWOZ -— Gilbert ~ (57495) 2 as t varies in the interval 0 g t 3 2w. 1. Moves once clockwise along the ellipse 2 2 mas—y— : 1, 4 9 starting and ending at (O, 3). correct 2. Moves once clockwise along the ellipse (237? + (w m 1. starting and ending at (0, 3). 3. Moves once counterclockwise along the ellipse (M2 + (331)2 e 1. starting and ending at (O, 3). 4. Moves along the line a: y W “i=1, 2+3 starting at (2, 0) and ending at (U, 3). 5. Nieves along the line '3 y w hm} 2+3 ’ ‘ starting at (0, 3) and ending at (2, 0). 6. Moves once counterclockwise along the ellipse 2 2 a i- a z 1, 4 9 starting and ending at (0, 3). Explanation: Since cos2 t + sing it = 1 for all t, the particle travels along the curve given in Cartesian form by 2 2 33 y _ __ 3 1- 4 + 9 ’ this is an ellipse centered at the origin, At t m 0, the particle is at (231110, 3cos 0), ale, at the point (0, 3) on the ellipse. Now as t increases from t m 0 to t = n/Z, 3:05) increases from a: m 0 to :3 m 2, While Mt) decreases from y = 3 to y = 0; in particular, the particle moves from a point on the positive y—axis to a point on the positive mwaxis, so it is moving clockwise. In the same way, we see that as it increases from 7r/2 to 72*, the particle rnoves to a point on the negative y—axis, then to a point on the negative m—axis as it increases from 7r to Zia/2, until finally it returns to its starting point on the positive y—axis as t increases from 37r/2 to 271'. Consequently, the particle moves clockwise once around the ellipse starting and ending at (O, 3). keywords: motion on curve, ellipse 604 10.0 points Which one of the following could be the graph of the curve given parametrically by 33(t) 2 722—43, y(t) x 19—213, Where the'arrows indicate the direction of increasing t? C01“ rect Vincent (jmv'???) — HWOZ — Gilbert — {57495) 3 Explanation: All the graphs are symmetric either about the ywaxis or the m—axis. Let’s check which it is for the graph of ($05), y(t)) = (t2 93, t3 .— 2:). and 9(4) a (-03 — 2(4) = “(t3 4 2t) = -y(t), SO (xt-t), 34(4)) = (3005'), WW»- Thus the graph is symmetric about the 3;— axis, eliminating three choices. To decide which one of the remaining three it is, we can check the path traced out as t ranges from moo to +00 by looking at sign charts for fit) and. y(t) because this will tell us in which quadrant the graph lies: 1.x(t)=t2w3: + e w e + t—————+——-~w--+mm~a. «00 _.\/§ x/fi +00 2. y(t) = 25052 m 2): ~0+0~0+ WWW-l woo “Jig \/§ +00 Thus the graph starts in quadrant I, crosses the y—axis into quadrant II, then crosses the mmaxis into quadrant IV, crosses back into quadrant II and so on. Consequently, the graph is Vincent (jaw???) m HWUZ w Gilbert M (57495) " 4 correct keywords: parametric curve, graph, direction, 005 10.0 points Which one of the foliowing could be the graph of the curve givenlearametrically by (36(25): will when the graphs of 33(t) and Mt) are shown in Explanation: Since 33(0) x 10(0) = 0, 33(1) m 0=y(1) : l: the graph passes through the origin and the point (0, 1). This already eliminates two of the graphs. There are two ways of deciding which one of the remaining two it is: (1) since me) =11 yo= :1: the graph must pass also through (1, 1/ Vincent (jmv’???) — Hw02 ~ Gilbert — (57495). 5 (ii) the graph ofwfit) is increasing neart = 0 Find y(t) . at the same rate as it is decreasing near t = i, While the graph of y(t) is increasing faster 1. ya) 2 1 3 near t z: 1 than it is neart = 0. So the graph 5 “3" 3t of (m(t), y(t)) is increasing more slowly near 1 x I a c m 2‘ t 2 t 0 than it IS decreasmg neart 1. y( ) 3 + 5t3 Consequently, the graph of (suit), y(t)) is t 3 t m m ) 5 + 37:3 t 4 t = M l 3 + 57:3 t2 5 y(t) = 3 + 5t?) correct t2 6. t m M ) 5 4» 3t3 Explanation: The line 3; : tic intersects W 3 r 3 W keywords: parametric curve, graph, 3m +99 W my 006 10.0 points When As the graph 33:3 +5t3m3 = tsc2_ After canceling the common factor 9:2 this becomes m(3+5t3) = 1;. Consequently, solving for 33 gives t m m 3 + 5153’ which means that t2 t = M l 3 + 5t3 of the e uation q keywords: 33:3 + 53/3 2 my shows, each line “i m ta: intersects the graph 007 10‘0 pomts at the origin and one other point. So every point P 79 (0, 0) on the graph can be written parametricaily as P m (we), gm). 27(t) 2 t3 — 2t, y(t) m t2—t? Which one of the following could be the graph of the curve given parametricelly by Vincent SUN???) — HW'OQ ~ Gilbert ~ (57495) 6 Explanation: These examples iilustrate the diversity of curves in 2—space. But simple properties such as (i) behaviour as t ——> 00, (ii) an and y—intercepts, (iii) passing through the origin or not, (iv) symmetry (even or oddness), can often be used to determine which graph goes with which function. For instance, three of the graphs above iie within a square cen- tered at the origin, suggesting that the other three are embounded; on the other hand, only three pass through the origin; and some, hut not ali, have various symmetries such as even— ness, oddness, or symmetry about a iine. In the case of the curve given parametrically by m) z: t3~2t, W) m {2—25, we see that he me = the m = 00> while 33(0) 2 y(0) = 0. But oniy two of the graphs have these properties. On the other hand, teem 93“) r ‘00 tiara W) = 00: leaving Vincent (jmv’???) ~ HWOZ — Gilbert — (57495) 7 as the only possible graph. keywords: ZD-graph, parametric function, limit at infinity, symmetry, periodicity 008 10.0 points Find the path (33(t), y(t)) of a particie that moves once counterwclockwise around the curve 232+(ym3)2 = 25, starting at (5, 3). 1. (— 5cost, 3 —— 55int), 0 _<_ t 3 7r 2. (m 500st, 3 -~ 5sint), 0 g t < 27r 3. (Scost, 3 — 5sint), 4. (5 cost, 3 —- Ssint), 5. (5cost, 3+SSint), 6. (5 cost, 3+551nt), 0 g t 3 21’1" correct Explanation: The graph of 332-1-(y—3)2 : 25 is the circie (not drawn to scaie) centered at (0,3) hav— ing radius 5. As the particle moves counter— clockwise around this circie starting at (5, 3), it starts at A, then moves to D, then to C, and finaliy to B. The Pythagorean identity singtil +cos269 =1, suggests setting 1:05): 5cost, y(t) = 3i5sint. For then stew—(me? = 25 for either choice of sign :h. The choice of sign - determines whether the path moves clockwise or counterciockwise. Indeed, suppose we take 32(t) = 500st, y(t) = 3+5sint. Then (33(0), y(0)) 2—" (5, 3), so the path starts at the right point. But as t increases from 0 to ir/Q, e is decreasing while 9 is increasing; in fact, the point (33(t), y(t)) moves from (5, 3) to (0, 8), and it does so counter-ciockwise. On the other hand, if we take 93(t) = 500st, y(t) = 3—5sint. Then (33(0), 14(0)) m (5, 3), so the path starts at the right point. But as it increases from 0 to 7r/ 2, both :6 and y are decreasing; in _ fact, the point (x(t), y(t)) moves from (5, 3) to (0, m2), and it does so ciockwise. Consequently, the path of the particle is described parametricaily by (500st, 3+5sint), 0 S t < 27? 009 10.0 points Let Q, R be the points where the ray of angle (9 intersects circies centered at the origin as Shown in Vincent (jmv’???) ~ HWOZ ~— Gilbert — (57495) 8 . , i q . . u i . . , ,, . . 3 t . . . i and let P be the point of intersection of the vertical line through Q and the horizontal line through R. As 9 varies, P traces out a curve. Write this curve in parametric form ._ (meme), mega. (Hint: find the coordinates of Q and R.) 1. (SCOSQ, 4sin6) - 2. (Attanéi, Seed?) 3. (Ssecfl, 413mm?) 4. (4 cos 9, 8sin8) correct 5. (Ssi‘nfl, 4cost?) 6. (4 secé‘, 8 tan 6) Explanation: From the graph it follows that the inner circle has radius 4, while the outer circle has radius 8. On the other hand, any circle of radius r centered at the origin can be written in parametric form as (rcosG, rsinti), 0 S 6 < 271', Thus Q and R are the points (22(4 (3086, 4sin9), R(88089,881D6). But P,- Q have the same m—coordinate, and P, R have the same y—coordinate. Consequentiy, the curve traced out by P has the parametric form (4 cos 6, 8 $1119) for O S 6 3 2w. Eliminating 9, we see that P traces out the ellipse 010 10,0 points When a mortar sheli is fired with an initial velocity of 00 ft/sec at an angle a above the horizontal, then its position after 15 seconds is given by the parametric equations cc 2 ('00 cos 001:, y m (129 sin 0e)t — 16752. If the mortar shell hits the ground 900 feet from the mortar when or 3 75°, determine '00. 1. 129 m 240 ft/sec correct 2. 129 = 260 ft/sec 3. a) m 270 fig/sec 4. w} m 230 ft/sec Vincent (jinx/7777) ‘ HVJOE — Gilbert — (57495) 9 5. no = 250 ft/sec EXpIanation: The mortar shell hits the ground when ya) m 90(sin75°)t—— 16152 m 0, 216., after t1 = ('09 sin 75") / 16 seconds, and it will then be a distance of 900 : 55(251) __ v3 sin 75" cos 75° _ v8 sin 150° _ 16 “ 32 feet from the mortar since sinflcosfl = $311129. But sin 1500 x 1/2. Consequently, no =— V64>< 900 = 240 ft/sec keywords: trajectory, parabola, distance, range, initial velocity 011 10.0 points 1' BE 2 4—61 2 fig _ 3et(1+t) ' dzt _ 4+et 3 fig __ Bet(1+t) ' dm w 4wet dy Met 4. W = t d2: 3&(1 + t) come 5 dy _ filmet ’ d3: m 3et(l—-t) 6. 2‘52 m 4+6: dm 3€i(1~t) Explanation: Differentiating with respect to t we see that m’Ufi) = 3(el 4» test), y'(t) m 4 + at. Consequently, 933 W 9’05) m “met de “” :3’(t) m" 3e’9(1+t) ' 012 10.0 points Find an equation £01" the tangent line to the curve given parametriealiy by wt mi 3305) m 2tsin(—§—), fit) 2 3teos(w-2~») at the point P(m(2), 1:03)). 1. y §(—l—x‘i)7r El 3 l 2. y W ;(2n~§m) 3 i 3. y s §(1—§m)w 1 4. y = ~3u(——33+21r)correct 7r 2 3 I. 3 I s. y = ;(§mw27r) Explanation: The point slope formula with t m 2 can be used to find an equation for the tangent line at P. ' We first need to find the slope at P. New by the Chain Rule and Product Rule, 93’(t) m 2(sin(Z;—t) + “EECOS(%E)), m m were Vincent (jmv???) w HWOZ - Gilbert w (57495) 10 Thus _d_y m 9’0?) dm 23’ (t) m 3 2cos(7rt/2) — 7rtsin(7rt/2) m g (25in(7rt/2) + 7ft cos(7rt/2)) ’ so the tangent line at P(:z:(2), y(2)) hes slope 2 On'the other hand, PWZL 31(2)) = (0, 6). By the point slope formula, therefore, an equation for the tangent line is m hilt which after simplification becomes 11 2 §(%az+27r) . keywords: parametric curve, tangent line, slope, trig function, 013 10.0 points Find the y-intercept of the tangent line at P(l, 2) to the graph of the curve defined parar metrically by $(t) 2 20052235 y(t) 2 4sint. 1. y—interoept 2 3 correct 2. y—intercept 2 m3 3. ywintercept 2 ~3— 4. y-intercept 2 m6 5. y—intercept 2 :2: 6. ywintercept 2 6 Explanation: The point PU, 2) on the graph of a: 2 2008215, 3; w 4sint is the point at which t 2 7r / 6. On the other hand, :c’(t) m —4sin2t, y’(t) 2 ecost, and so the siope of the tangent line at P is given by yin/wt) = = But then by the pointwslope formula, 21 — 2 e -(~’6 ~ 1} is an equation for the tangent line at P, which after simplification becomes y+m 2 3. Consequently, this tangent line has y—intercept m . 014 100 points Find an equation for the tangent iine at P(2:(t), y(t)) to the cycloid defined parametw rically by ‘ 22(t) 2 5(t—sint), y(t) 2 5(1'w cost). Hint: remember the trig identities 811175 2 ZSln(%)COS(":—) , l w cost 2 28in2(g). 1. y—l—(rz: +5t)ten(—t—) = 10 2 2. y m (st—5t)tan(%)+§.0 t 3. y 2 (mn5t) eot(§)+10 correct 4. y 2 (23+5t)cot(~g)e10 Vincent (jmv777) m HVVO2 w Gilbert m (57495) 11 5. y+(w+5t)cot(%) m 10 6. y ~1~ (a: - 5t) tan(—;—) 3 10 Explanation: Since 33%) m 5(1— cost) m iOsiHZCE) 2 and y’(t) : 53int m iOsin(%)cos(—g), the slope of the tangent line at P is given by _ dy W y’(t) W sin(t/2)coe(t/2) 3 (2012(3) 83?, “"‘ w’(t) “"““ sin2(t/2) 2 By the point~slope formula, therefore, the tan— gent line at P has equation 1; y ~—— 5(1 e cost) = mtg) (:3 —— 503 ~— sin t)) which after simplification becomes t y = (m w 575) c0t(§) + 5(Sint00t(“:-)+{l —— coet)). But by the given trig identities, . _ I t _ 28in(t/2)cos(t/2)cos(t/2) SiniCOL(§) —. Sin(t/2) t 2o, 008 2 While 1. 1w cost = 231n2(—2—) . Consequently, an equation for the tangent line to the cycloid at P is y : (a:-~5t)cot(~;1)+10 . keywords: parametric form, trig function, tangent line, point—slope formula, 015 10.0 points d _ Find mg for the curve given parametically d3 by m) : 2+12, y(t) m t2+2t3, I dy _ 1 d3: ‘- 1+3t dy' 3 2. m = 2 mt dm +2 3. fig 2 2+3t d3: dy 1 4" E}; W 2+6t 5 fig m *3“ da: 2+3t dy 6. W = 1 t dm +6 7. gig = 1+3t correct d3: 8_ Fig m. ,2... dx 2+3t Explanation: Differentiating with respect to t we see that fit) m 2t, y'{t) m 2t + 69. Consequently, 55g W y’(t) W 2t+6t2 012: W $’(t) W 2t 2: 1+3t . 016 10.0 points d2 Find E5 when 23(t) = 3+t2, y(t) = than-2). 1fl_i££§ 'dw2_2t3 Vincent (jmv???) —- HW’OE w Gilbert w (57495) 12 3 1 t 2 g = MZE§ correc 3 fife 2 Jet do:2 t3 4. fl = dag 2t“2 5 die m 1.1}: deg 132 6 fl 2 3:13 ' dis? :53 Explanation: First notice that y(t) w Satin t. After diflerentietion with respect to t, therefore, we see that fit) m 2t, y’(t) m 2(l+int). Th as Fig W l—I—lnt dm W t ' On the other hand, by the Chain Rule, 2 re) R see» m iii-which case day m d(dy) is: it}? — tit da: 0123' But £(dy) W t(1/t)— (l—Hnt) __ _l_n_t dt d3: m t2 m 252' Consequently, e m. 01332 m 2t3 ’ 617 10.0 points Determine all points P at which the tangent line to the curve given parametricaliy by 33(3) m t3- 6t, y m 31:2 is pereliei to the iine (mt, 2t). 1. P = (5, 3), (m4,12) 2. P = (“5, —1), (4,1) 3. P m (5, —1), (m4, 1) 4. P = (5,1), (m4, m1) 5. P m (~5,12), (4, 3) 6. P m (—5, 3), (4, 12) correct Explanation: Since 33’(t) = 3:22” 6, W) = fit, the slope of the tangent line to the curve is given by dy _ 3W) 5‘5 dz: ‘ 57(5 3 site But if this tangent tine is to be parallel to the line (iefined parametriceliy by (wt, 2t), then 61; = “2. sews After cross—muitipiicetion and simplification, this becomes the equation - 6t2+6tm12 x 6(t~1)(t+2) e 0. Consequently, P is the point corresponding to the values if =2 1 and t m —2, in which case P = (“5,3), (4,12) _ 018 10.0 points The region A enclosed by the line y = 9 and the graph of the curve given parametricelly by .7:(t)=t—E, 2 aim/it - t M) +75 is similar to the shaded region in Vincent (jmv'f'f'?) — HW02 ~ Giibert ~ (57495) 13 Where 1 1 m~m(t)—t—E, dxm(1+E§)dt. Thus after a change of variable from 1: to t we reduce the integral describing area(.A)- to Determine the area. of .4. 1‘1 ‘ o 189 6 9 2 1.area(A)=§+61n8 m/to (9*4t“;+E§—Eg)dt 197 ' 9 l ’51 2- fl "_"“ 1 :5: -~ 2 — m» w -— . eree(A) 8 +6 :18 [9t 2t filnt t + t2 L0 4 197 3. area(A) = mg— “ 61n8 But the graph of 4. areai(A) = 3:3 w 61118 correct mg) m t _. Eli, ya) 3 43 in? 181 . 5. area(A) = —g- + 61n8 intersects y w 9 when l l 6. area,(A) m ——:———61n8 Kit—i“? = 9, I t' : Exp mm 1011 . 316" when If the graph of 2 #0:”? y(t)m4t+—t— 4t2~9t+2:(4tw1)(t~—2)m0‘ intersects y = 9 when t m 159. and 1’: 2 t1, then A is similar $20 the shaded region in Thug to m 1/4 Whfle ti 2 2’ SO area(A) m (gfm61n2)w(m£§—3—~6In(Z—:D. Consequently, aree(A) m lgg—(ilnéi . I l I I 1 ! l l I l with a :2 mag) and b = $(t1). In this case 'area(A) m /b (9 w y) d3: keywords: parametric curve, area, 019 10.0 points fwd (9 _ 4t _ dm Which integral represents the arc length of m0) ’ the astroid shown in Vincent (jaw???) — HVVO2 — Gilbert - (57495) 14 and given parametriceliy by $05) m 50053t, y(t) = 58in3t. 271' 1. I = 15/ [costsintldt correct 0 7T 2. I m 15/ |costsint|dt 0 . 21f 3. I m 5/ costsintdt {J 271' 4. I z 15/ costsintdt 0 2?!“ 5. I = 5/ Icostslntldt 0 6.1" W 5/ icostsintldt 0 Explanation: The arc length of the parametric curve (560»): W», is given by the integral 5 I m f x/(w’(t))2+(y’(t))2dt- But when agtgb 220') == 5 008315, y(t) m 5 3111315 we see that 16%) 2 "~15 8111150032 t, y’(t) = 15 008155111213, in which case (68125))2 4“ (9%))2 = (l5costsin t)2(0032 t + sin2 t) = ilScostsintl. Consequently, 211' I m 15/ leostsintldt . 0 keywords: arc length, parametric curve, as— troid trig functions, definite integral ...
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This note was uploaded on 03/24/2010 for the course M 57500 taught by Professor Gilbert during the Fall '09 term at University of Texas.

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Homework 2 - Vincent(jmv777 — HW02 — Gilbert w(57495 1...

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