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hw-6-sec-11.1-solns - 11 Cl PARAMETRIC EQUATIONS AND POLAR...

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Unformatted text preview: 11 Cl PARAMETRIC EQUATIONS AND POLAR COORDINATES 11.1 Curves Defined by Parametric Equations 1.:t:=1+\/1_5, y=t2—4t, 03:35 2.x:2cost, y=t—cost, OStSQW ,rg;:éz:;.‘z‘.-u;;v' a»; .‘1-4 2.47 9.87 ”:35: ‘ tagfflmifh w ‘ ’ 2 yr.) . m.“ ‘ ‘1‘“ A"; it";. L ra‘ g. 844 D CHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES (b)z=3t—5 => 3t=z+5 => t=§(z+5) : y=2.%(m+5)+l,soy=§z+l§. 6.:L‘=1+t, y=5—2t, —2St£3 (a) (b)z=1+t => t=z—1 2 y=5—2($—1), soy=—2a;+7, —IS$S4. (b):c=1+3t : t=§(x—1) 2 y=2—[§(x—1)]2, In 0 1 1.414 1.732 2 1 0 —1 —2 -3 (b)x=¢£ => t=z2 => y=1—t=1—$2.Sincet20,m20. So the curve is the right half of the parabola y = 1 — $2. ® (a) :c:sin0,y=cos0,0§ 0 S711 ar:2+y2 =sin29+cos20= 1. SinceO 30 SW, we have sin0 2 0, so :1: 2 0. Thus, the curve is the right half of the circle 2:2 + y2 = 1. @(a)z=sint,y=csct,0<t<12'—. 1 y=csct= — = LForO <t< §,Wehave sin t a: 0 < z < 1 and y > 1. Thus, the curve is the portion ofthe hyperbola y = l/z with y > 1. (b) y (1,1) SECTION 11.1 tER,yER,$20. 12. (a) m = 4cos€,y = 55in0, —1r/2 g 0 S 7r/2. ($2 +01%)2 = cos2 0 + sin2 0 = 1, which is an ellipse with x-intercepts (21:4, 0) and y-intercepts (0, :l:5). We obtain the portion of the ellipse with '2: Z Osince4cos€ Z 0 for —7r/2 S 6 g 7r/2. (b) 14. (a) m = et —- 1, y = e”. y 2 (at)2 = ('1: + 1)2 and since 1 > —1, we have the right side of the parabola y = (a: + U2. 0)) 16.(a)m=lnt,y=\/Z,t21. m=lnt => t=ez => y=\/E=ex/2,x20. (b) y (0, 1) CURVES DEFINED BY PARAMETRIC EQUATIONS U 845 846 C] CHAPTER 11 PARAMETRlC EQUATIONS AND POLAR COORDINATES 17. (a) :1: = sinh t, y = cosht => 1,:2 — $2 = cosh2 t - sinhzt = 1. Since (b) y = cosht Z 1, we have the upper branch of the hyperbola y2 — x2 = 1. 18. (a)$=2cosht,y=55inht => g=cosht,%=sinht => (b) 53 2 _ 2 fl 2 _ . 2 - 2 - 2 _ (2) —— cosh t, (5) — smh t. Since cosh t — smh t _ 1, we have $2 yz I — 55 = 1, a hyperbola. Because a: 2 2, we have the right branch of the .\ hyperbola. ' 19. a: = 3 + 2cos t, y = 1 + 2sint, 1r/2 g t S 37r/2. By Example 4 with r = 2, h = 3, and k = 1, the motion ofthe particle takes place on a circle centered at (3, 1) with a radius of 2. As t goes from g to 37", the particle starts at the point (3, 3) and moves counterclockwise to (3, —1) [one-half of a circle]. 2 20.x:2sint,y=4+cost : sint=Ec2-,cost=y—4. sin2t+cos2t=1 => (g) +(y—4)2=1.Themotion of the particle takes place on an ellipse centered at (0,4). As t goes from 0 to 37", the particle starts at the point (0, 5) and moves clockwise to (—2, 4) [three-quarters of an ellipse]. 2 2 ®z=5sint,y=2cost 9 sint=§,cost=‘%. sin2t+cos2t=1 => (g) +(%) =1.Themotionofthe particle takes place on an ellipse centered at (0,0). As t goes from —7r to 57r, the particle starts at the point (0, —2) and moves clockwise around the ellipse 3 times. 22. y = cos2 t = 1 — sin2 t = 1 — :52. The motion of the particle takes place on the parabola y = 1 — 9732. As t goes from —27r to —7r, the particle starts at the point (0, 1), moves to (1, 0), and goes back to (O, 1). As t goes from —7r to 0, the particle moves to (—1,0) and goes back to (0, 1). The particle repeats this motion as t goes from 0 to 27r. 23. We must have 1 g a: S 4 and 2 S y g 3. So the graph of the curve must be contained in the rectangle [1, 4] by [2, 3]. 24. (a) From the first graph, we have 1 S :5 S 2. From the second graph, we have —1 g y g 1. The only choice that satisfies either of those conditions is 111. (b) From the first graph, the values of :3 cycle through the values from —2 to 2 four times. From the second graph, the values of y cycle through the values from —2 to 2 six times. Choice 1 satisfies these conditions. (c) From the first graph, the values of x cycle through the values from —2 to 2 three times. From the second graph, we have 0 S y S 2. Choice IV satisfies these conditions. (d) From the rst graph, the values of :3 cycle through the values from -2 to 2 two times. From the second graph, the values of y do the same thing. Choice 11 satisfies these conditions. .....m..ze«...k.rs«:m .n'w.5;.;..m<.,« ‘ ‘ ‘ ; 1+: . . )f SEC110N 11.1 CURVES DEFINED BY PARAMETRIC EQUATIONS El 847 25. Whent = —1, (:12, y) = (0, —1). As t increases to 0, 3: decreases to ——1 and y increases to 0. As t increases from 0 to 1, 2 increases to 0 and 3/ increases to 1. As it increases beyond 1, both a: and y increase. Fort < —1, a: is positive and decreasing and y is negative and increasing. We could achieve greater accuracy by estimating x- and y-values for selected values of t from the given graphs and plotting the corresponding points. 26. For t < —1, z is positive and decreasing, while y is negative and increasing (these points are in Quadrant IV). When t = —1, (cc, y) = (0, 0) and, as t increases from —1 to 0, :5 becomes negative and 3; increases from 0 to 1. At t = O, (z, y) = (0, 1) and, as t increases from O to 1, y decreases from 1 to 0 and a: is positive. At t = 1, (x, y) = (0, 0) again, so the loop is completed. For t > 1, x and y both become large negative. This enables us to draw a rough sketch. We could achieve greater accuracy by estimating w- and y-values for selected values of t from the given graphs and plotting the corresponding points. 27. When t = 0 we see that m = 0 and y 2 0, so the curve starts at the origin. As t increases from O to %, the graphs show that y increases from 0 to 1 while an increases from 0 to 1, decreases to 0 and to —1, then increases back to 0, so we I arrive at the point (0, 1). Similarly, as t increases from é to 1, y decreases from 1 to 0 while 27 repeats its pattern, and we arrive back at the origin. We could achieve greater accuracy by estimating m- and y-values for selected values of t from the given graphs and plotting the corresponding points. .(a)as=t4—t+1 = (t4+1)—t>0 [thinkofthegraphsofy=t4+1andy=t] andyzt2 20,sotheseequations are matched with graph V. (b) y = x/f Z 0. as = t2 — 2t 2 t(t — 2) is negative for 0 < t < 2, so these equations are matched with graph I. (c) m = sin 2t has period 27r/ 2 = 71'. Note that y(t + 271') = sin[t + 271' + sin 2(t + 2%)] = sin(t + 2n + sin 2t) = sin(t + sin 2t) 2 y(t), so g has period 27r. These equations match graph 11 since 2: cycles through the values —1 to 1 twice as y cycles through those values once. (d) a: = cos 5t has period 27r / 5 and y = sin 2t has period 7r, so .2 will take on the values ——1 to 1, and then 1 to —1, before y takes on the values -—1 to 1. Note that when t = 0, (22,31) 2 (1,0). These equations are matched with graph VI. (e) x = t + sin 4t, y = t2 + cos 3t. As t becomes large, t and t2 become the dominant terms in the expressions for z and y, so the graph will look like the graph of y = 232, but with oscillations. These equations are matched with graph IV. sin2t _cos2t 4+t2’ y“ 4+t2' (f) a: 2 As it —+ 00, x and y both approach 0. These equations are matched with graph III. ...
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