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# Review Sheet for MA 154 at IPFW

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COURSE
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TYPE
Class Notes
PAGES
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KARMA
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This 6 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Indiana University taught by a professor in Fall. Since its upload, it has received 27 views.

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Date Created: 02/06/15
1 Review for the MA 154 Final Fall 2008 Date Time Room 639 Write each ofthe complex numbers 2122 and 23 In the 5 form zre rcos6isin6 1 Write 24 and z in rectangular form 2 x yi Exact please 3 Plot and label each number on the complex plane 2 a 214 b 225 75i c 233i d 24 32quotquot 3cos71r i3sin71t 39 e 25 NEZ Z MEcos37 i 2sin37 45 5 21 3 2 14 1 2 3 4 5 5 72 Perform the complex number arithmetic and writeyour 3 answer in rectangular formx yi m 39 1 V a2lt42z7lt371gt b 2067 c i 75 1 43 d Zeasmyn e1i14 Eliminate the parameter and write a formula which only involves x and y Give the domain and range a x 2cost y 73cost3 b x 4cost y 4cos2t c x cos2t y 2cost Hint consider a double angle idmtity Eliminate the parameter and write a formula which only involves x and y axt71ye bxe 5 ye Write implicit formulas for each ofthe conics graphed below Give the center focal points exact and approximate to two decimal places and vertices if any For parts a through c give the parametric formulas as well For part d report the equations of the asymptotes a b E 2 2 4 a 10 gt6 6 d 6 Sketch a graph of each ofthe conics Report the vertices or vertex and focal points and plot them For parts a d give the center For parts e and I give the equation ofthe directrix 2 2 2 2 ax y 1 b amp1 4 16 25 9 2 2 2 2 cyx1 dampamp1 25 100 1 4 e yz8x f x2riy1 7 A parabola has its vertex at 0 0 and its focus at 0 3 Give its equation 8 The ellipse shown has focal points at 1 0 and 9 0 Give its equation 9 For Question 9 a day is a 24 hour period beginning at midnight Two species of plant A and B propagate by dispersing their seeds in the wind during the height of summer On a typical day in the height of summer the wind speed wt measured in miles per hour thours a er midnight is given by the formulawt 76 cos 15 a Species A favors propagation in high winds andwill only release its seeds if the wind speed is no less than twenty miles per hour In the figure the graph ofwt is given and the time interval over which species Awill release its seeds is also indicated Complete the figure by calculating the values of the two endpoints of the interval b The seeds of species B are destroyed in high winds and hence species B has a seed release mechanism which only releases seeds ifthe wind speed is no greater than ten miles per hour For how many hours on a typical day in high summer will species B be releasing seeds 10 Find a possible formula for each a b Additional Problems from Text Section 66 17 19 23 Section 67 37 43 51 Ch 6 Tools 1 3 23 Chapter 6 Review 3 15 4347 odd 61 Section 71 5 11 17 39 Section 72 3 5 1741 45 Section 76 114 1725 Chapter 7 Review 3 5 7 9 Section 101 5 12 Section 102 3 13 14 22 Section 103 12 15 Section 104 1 25 Chapter 10 Review 5 1129 35 Section 111 725 odd 3033 Section 112 3 13 19 23 35 41 Section 113 115 odd Section 114 19 odd 19 Chapter 11 Review 1 3 5 7 9 10 Section 121 111 odd 28 Section 122 113a15 17 Section 123 1 3 17a Section 124 1 3 11a Section 125 123 odd 25 26 27 30 31 Chapter 12 Review 5 9 1719 28 m CAquot go In cr m Answers to Review 5f 21 44cos7ri4sin7r4e r 5f 22 5 7515450 cos7T i 50sin7T 0817m4 3 23 3i 3cos i3sin 38W 239 24 3e 3cos7ir i3sin77r 371 30 73 3 is Mew ea i2 Bi 242i737i84i73i55i 2i37i6i7212 6i 72716i 226i iii iiiil7 i2i771i i i i i2 71 71 gem2 210equotlt5 2gt10 210equotlt25 gt 210cos7257r isin7257r 21071 71024 To find 1 i14 write 1 iin the form re g The angle 0 714 and distance r 12 so 1 1 Egan4 Then ex7r4l4 JE14 ex7r414 212l4ex77f2 27ex77r27 274 7128 y 7x 3 lii Since x 2cos I y 73 cosI 3 solve for cos I Therefore cosIi soy 73cosI 3 73 3 39 3 2 Using a grapher we can see that when I 0 the curve starts at the endpoint 2 0 travels on a line to the endpoint 72 6 and repeats back Note that the domain is restricted to 72 S x S 2 since 72 S 2cos I S 2 and the range is restricted to 0 S y S 6 since 0 S 73cosI 3 S 6 E 5 4 3 2 1 y 4x 7 42 Solve for cos I to get cos I x7 4 and substitute therefore y 4cos I2 4x7 42 which is a parabola Using a grapher we can see that when I 0 the curve starts at the endpoint 5 4 travels on the parabola to the endpoint 3 4 and repeats back The domain is 73 S x S 5 and the range is 0 S y S 4 x yZ 71 We must write cos 2I in terms of cos I which can be done if we use the double angle identity cos 2I 2cos2 I71 Solving y 2cosI for cos I we have cosI and substituting gives x 2Z fl Simplifying gives x 2y4 2 fl 471 The domainis fl SxSl andtherange is 72SyS2 4 Eliminate the parameter and Write a formula Which only involves x and y a y e x1 1 since tx1andye e b xJy or yx2xgt0 since xe 5 yer and xe 5y 5 y 5 a 6 2 1 center 0 0 Vertices r 8 0 The focal points are c unis from the center Where C2 64 716 48 so 6 m z 693 So the focal poinw are ixE 0 or r 693 0 Parametric x 8cos ty 4sin t b x7 32 y 42 9 center 3 74 There are no Vertices or focal points this is a circle Parametric x 3cos t 3y 3sin 7 4 c is 1 center 73 6 Vertices 73 12 and 73 0 We have C2 36 7 9 27 so 6 m z 520 and the focal poinw are on the major axis at 73 6ix or approximately at 73 080 and 73 1120 Parametric x 3cos 7 3y 6sin 6 d Lg 7 1 center 74 4 Vertices 74 1 and 47 The focal poinw are 6 units from the center Where C2 9 4 13 so 6 J z 361 So the focal poinw are 74 445 or 74 039 or 74 761 Asymptotes y74 ix4 ory15x10y715x7 2 6 a center 0 0 b center 1 4 Vertices 0 4 0 74 Vertices 6 74 74 74 focal points 0rxE focal points 73 74 and 5 74 Whichare 0 4346 WehaVe 6425794 so at 1i474 4 3 2 1 7473724 1234557 78 c center 0 0 d center 72 5 asymptotes y ilinxor y ix asymptotes y i2 1D Vertices 0 5 0 75 Vertices 71 5 74 5 focal points mfgE or 0 1118 focal points 7 2i 5 5 or 424 5 024 5 o e The vertex rs atthe ohgh Smceyxs ofsecond degree we expeet the parabolato have a hohzohtal axrs of symmetry The equatroh y 4pxhas afocusp uhrts from the vertex The graph of y xx hasp 2 so the focus rs at 2 0 The dArecmx rs the vertical hhe x 23455 f The vertex rs at 0 71 Srheexrs ofsecond degree we expeet the parabolato have avemcal axrs of symmetry srrmlartoy x2 Here 4p 7 so p 46 The foeus is p e umts from the vertex oh the 4 en 5 n 5 1 vemcal axrs or at 0 7146 or 0 77 The duecmx 15 p 46 uhrts from the vertex so 39D 5 rt has the equanony 771516 7 The focus 0 3 rs oh the axrs of symmetry vertrea1 so the parabola has the equatroh x2 4py wrthp 3 so 12y 4 5 8 eehter 5 0 RUN5 s4 Smce 2 RUN2 7021515 5 7RISE2 2 we can solve for the RISE 4 52 7021515 RISE2 52 74 25716 9 Therefore RISE 3 2 7 a a x 5 y 1 25 9 9 a t 9 76311 14 237 Solve graphmally b 19 526hours Fmdwhere wt10 Thrsrs 52237ahdz2217o3sozrzo19 526 10 a y4srhTx b y225m27x

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