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2 AP#Calculus#BC# CHAPTER(11(WORKSHEET( PARAMETRIC)EQUATIONS)AND)POLAR)COORDINATES) Name Sea# Dae Moion(Problems(in(Parameric(Equaions( 1. on he coordinae plane xy is given by he vecor sec, an. a) Find he velociy vecor a any ime. Use your expression o find he velociy vecor for he paricle a. 6 b) Find he acceleraion vecor a any ime. Use your expression o find he acceleraion vecor for he paricle a. 6 c). 6 d) Find he equaion of he line angen o he moion of he paricle a. 6 e) Se up an inegral expression o find he oal disance raveled by he paricle in he ime inerval,. Use your calculaor o evaluae your expression.. on he coordinae plane xy is given by he vecor, 1. a) Find he velociy vecor of he paricle a ime =. b) Find he acceleraion vecor of he paricle a ime =. c) =. d) Find he equaion of he line angen o he moion of he paricle a. e) Se up an inegral expression o find he oal disance raveled by he paricle in he ime inerval,. Use your calculaor o evaluae your expression. (You)may)use)your)calculaor)o)answer)his)quesion.)). The posiion of a paricle moving on a plane is given by he parameric equaions x sin y cos, for. (. The acceleraion of an objec is he derivaive of is speed. 5. The velociy vecor poins in he direcion of moion. and a) Find an equaion for he normal line o he pah described by he paricle a =. The normal line is perpendicular o he angen line o he curve. b) Find he iniial and final posiion of he paricle for. Find he disance beween hese wo poins. c) Find he oal disance raveled by he paricle for. d) Are your answers o pars (b) and (c) equal? Explain. SEE(OTHER(SIDE(

3 6. The velociy of a moving paricle a ime is given by he vecor,. A ime 1, he 1 posiion vecor of he paricle was,. a) Find he posiion vecor of he paricle a any ime ime. To do so, solve he iniial value dr problem v. d b) Find he velociy vecor a =. Inerpre he values found for boh he horizonal and verical velociy of he paricle a ime =. c) Find he paric 1. d) Find he acceleraion vecor of he paricle a any ime ime. 7. The posiion of a paricle is given by he equaions x sin and y cos, for. a) Find he velociy vecor for he paricle a any ime ime. dr b) For wha value(s) of is d? c) While describing he moion of he paricle, wha is he significance of he value(s) you found in par (b)? Explain. #CALCULUS#BC#(no\$calculaor\$allowed)# 8. A moving paricle has posiion y x, a ime. The posiion of he paricle a ime = 1 is 1 1 (, 6), and he velociy vecor a any ime > is given by 1,. a) Find he acceleraion vecor a ime =. b) Find he posiion of he paricle a any ime. Use your formula o find he posiion of he paricle a ime =. c) For wha ime > does he line angen o he pah of he paricle a x, y have a slope of 8? d) The paricle approaches a line as. Find he slope of his line. Show he work ha leads o your conclusion.

4 AP Calculus BC Review Chaper 11 (Parameric Equaions and Polar Coordinaes) Things o Know and Be Able o Do Undersand he meaning of equaions given in parameric and polar forms, and develop a skech of he appropriae graph Find he slope of a angen line o a paramerically-given curve a a paricular poin using he equaion d =, and undersand how his is derived using he Chain Rule d Find he second derivaive of a paramerically-given curve a a paricular poin using he equaion d d ( ) =, and undersand how his is derived using he Chain Rule d Find he area enclosed by a paramerically-given curve Find he arc lengh of a paramerically-given curve and he area of he surface generaed by revolving a paramerically-given curve Find he area enclosed by a polar curve using he equaion A 1 = r dθ (undersanding ha r f ( θ )) undersand how his is derived. Be very careful wih finding he endpoins for evaluaing he inegral Find he arc lengh of a polar curve = and Pracice Problems These problems should be done wihou a calculaor, wih he excepion of 1, as explained below. The original es, of course, required ha you show relevan work for free-response problems. x = cos 1 For 1, an objec ravels along an ellipical pah given paramerically by. A he poin a which y = sin = 1, he objec leaves he pah and ravels along he line angen o he pah a ha poin. Wha is he slope of he line on which he objec ravels? an1 a b c d e an1 an1 x= The posiion of a paricle moving in he xy-plane is given by he parameric equaions. y = 1 values of is he paricle a res? a 1 only b only c only d 1 and only e 1,, and For wha A curve C is defined by he parameric equaions angen o he graph of C a he poin (,8 )? x= + y = a x = b x = c 8 1. Which of he following is an equaion of he line y = 7 d y= ( x+ ) + e y ( x ) 1 8 =

5 x= A paricle moves so ha is posiion a ime is given by. y = sin cos1 a 8sin1 b 6 5 Which of he following inegrals represens he area shaded in he graph shown a righ? The curve is given by r = sin θ. a sin ( θ ) dθ b 8sin ( θ ) dθ c sin θ dθ d sin ( θ ) dθ e sin θ dθ 6 Which of he following inegrals represens he arc lengh of he polar funcion r = 1+ cosθ from θ? a b 1+ cosθ + sinθ dθ 1+ sin θ dθ c ( + θ ) 1 cos dθ d Wha is he speed of he paricle when =? c ( cos1) + 6 d ( sin1) + 81 e ( cos1) θ 1 cos dθ e ( 1 + cos θ ) sin θ ( 1 + cos θ ) + ( sin θ ) d θ 7 Consider he graph of he vecor funcion r () = 1 +,+. Wha is he value of where x =? a b 8 A paricle moves so ha a ime c 8 d ln,. > is posiion vecor is 8 e 9 a he poin on he graph A ime =, is velociy vecor is a,8 b, c 1,8 d 1, e Consider he curves r 1 = cosθ and r =. a Skech he curves on he axes provided a righ. b Show use of calculus o find he area of he region common o boh graphs. x = 1 Consider he curve given paramerically by. y = 1 a In erms of, find. b Wrie an equaion for he line angen o he curve a he poin a which = 1. c Find he x- and y-coordinaes for each criical poin on he curve and idenify each poin as having a verical or horizonal angen. 16,

6 11 A projecile is launched from he edge of a cliff 19 fee above he ground below. The projecile has an iniial velociy of 18 fee per second and is launched a an angle of o he horizonal. A horizonal wind is also blowing agains he projecile a a velociy of 18 fee per second. The acceleraion of graviy is fee per second per second. a Skech he siuaion and he pah of he projecile. b Wrie a vecor equaion ha represens he pah of he projec ile, using in seconds as he parameer. c When does he projecile reach is maximum heigh, and wha is his maximum heigh? Jusify your answer. d How far, horizonally, does he projecile ravel before landing on he ground? When his es was originally adminisered, he following quesion was o be aken home and compleed wih he use of a calculaor bu wih no oher resources permied o be used, such as oher people, books, compuers, or anyhing else. Sudens had one weekend o complee his. 1 An objec moving along a curve in he xy-plane has posiion ( x, y( )) a ime wih 1 d = and = ln( 1 + ( ) ). A ime =, he objec is a posiion ( ) d 1,5. A ime =, he objec is a poin P wih x- coordinae. a Find he acceleraion vecor and he speed a ime =. b Find he y-coordinae of poin P. c Wrie an equaion for he line angen o he curve a poin P. d For wha value(s) of, if any, is he objec a res? Jusify your answer.

7 AP#Calculus#BC# CHAPTER(11(WORKSHEET( PARAMETRIC)EQUATIONS)AND)POLAR)COORDINATES) ANSWER(KEY( 1. a) v sec an, sec. A v b) a sec an sec, sec an. A a, 1.95, ,.. 5 c) Speed: ; a d d 6 d) A d. Poin: 6, y d e) sec an sec d.8. x. a) v 9, 8 b) a, c) Speed: d) 1 d 1. Poin: 6, 7 y 7 x d e) 1 d a) d sin Slope normal line:. 57. cos 1.97 d.599,.597 y x.599 Poin: b) Iniial:, 1, 1. Final: c) cos sin d Disance: d) No. One is he arc lengh he oher one is a chord (line segmen).

8 . False. The acceleraion is he derivaive of he velociy. 5. True. 6. a) Doing aniderivaives: r C1, C. Using he iniial condiion: 7 r, b) v,. The paricle horizonal velociy is zero: i is no moving horizonally. The paricle verical velociy is : i is moving upwards. c) Speed: 1 1 d) a, 1 7. a) v cos, sin b) d r cos d and sin. Tha happens when and. c) Those are he imes when he #CALCULUS#BC#(no\$calculaor\$allowed)# 8. a) a,. A a, r C1, C 5 1 r,, d d b) Doing aniderivaives: 1 1 r,. A = : 1 1 c) 1, 1 Since 8 d 1 1 d 1 d) lim lim. The slope of he line is Using he iniial condiion:

9 Answers 1 d c a c 5 b 6 a 7 d 8 a 9a see soluions 9b a = y 11 = x ,16, 16 ; 1b 1c horizonal a ( ) and verical a ( 1, 11) and (, ) 11a see soluions 11b r () = 6 18, c 56 f a = s 11d 8 18 f 1a a =, ; v = 1+ ( ln17) ln17 1b c y = ( x ) 1 1d = only Soluions 1 We find = sin and cos. d d = Then (by he Chain Rule) d cos = = =, d sin an. = an1 This is choice d. so a = 1, We wan a ime a which boh = and ; d d = ha is, boh 6= and 6 6 1=. The firs equaion is = 6 or =, and he second equaion is he same as =, which facors o ( )( + 1) =. Therefore, 1 ; he inersecion of he wo soluion ses is = only, choice c. d = for { } Firsly, find ha ( xy, ) = (,8) occurs a. = Now find d = and, d = so d = =. Then d 1 = =. This is undefined, so he angen line mus be verical. Since he poin has an x-value of =, he relevan angen line is x =, choice a. d d The paricle s velociy is given by v () =, =,cos. We are ineresed in v = 6 + cos1 = cos1 + 6, choice c. which is 6,cos 1, 1 5 Recall ha he area enclosed by a polar graph is given by r dθ. Therefore he inegrand mus be 1 1 ( sinθ ) = ( 16) sin θ = 8sin θ. Wihou even having o consider he limis, we alrea know ha he correc choice mus be b. dr r + dθ. Since 6 Recall ha he arc lengh of a curve represened in polar coordinaes is given by dθ dr = sin θ, he inegral mus ake he form ( 1 cos ) ( sin ) d. dθ + θ + θ θ We are given he limis, and he correc answer is a.

10 d d d 7 Recall ha when a curve is given paramerically, =. Therefore we mus find. = d This is. Therefore choice d. d 8 d 8 = = = 5. The poin a which x = is he poin a which = 1, so The velociy vecor is v () =, =,. A =, we have v ( ) =,8, choice a. d d + 9a The firs equaion is a circle of diameer angen o he y-axis and shifed along he posiive x-axis; he second is a circle of diameer cenered a he origin. The circles along wih he region referred o in par b are shown a righ. 9b As shown in he diagram, he circles inersec a he poins ( r θ ) d = = 1, =, 6 and,. Noe ha he region is symmeric abou he line θ =, so we 6 θ = 6 have he luxury of considering only he porion of i wih θ, and hen doubling is area. For θ,, he ouer limi of he region is r =, so ha par s area is dθ =. For θ,, he ouer limi of he region is r1 = cos θ, so ha par s area is 6 1 ( cos θ ) d θ = cos θ d θ. To evaluae his, recall he ideniy 6 6 cos θ = + cos θ. Therefore he ine- 1 gral may also be wrien as ( 1+ cosθ ) dθ θ ] θ , 9 = + sin = + =. The union of hese 6 wo regions is hen + = 7, which is half of he area of he enire region. Therefore he area of 1 he whole region is 7. 6 d 1a The Chain Rule gives =. This is d 6 6 If you are so inclined, you may facor ou a from he numeraor and a 6 from he denominaor, yielding 1., bu such simplificaion is enirely opional. θ = 6 ( 1) 1 9 1b We need = = =. We also mus find x and y values: x( 1) = ( 1) ( 1) = 1 6( 1) 6( 1) 1 = 5 and y( 1) = ( 1) 1( 1) = 11. Therefore an equaion for he angen line is given by, in poin-slope form, y 11 = ( x+ 5 ). 1c Criical poins are hose a which he curve s derivaive is eiher undefined (verical angen line) or zero (horizonal angen line). The derivaive will be undefined iff (if and only if) is denominaor is zero; ha is, =, so we d

11 solve ; = his gives { },1. A, has verical angen lines a ( 1, 11) and so we solve 1= o ge =±. A, he curve has horizonal angen lines a ( 8,16) = xy, =, ; a 1, = xy, = 1, 11. Therefore he curve,. The derivaive is zero iff is numeraor is zero; ha is,, d = xy, = 8,16, and a, xy, =, 16. Therefore = and (, 16 ). = 11a An annoaed hand-skech is presened below. I is very much no o scale, and he pah should be symmeric and parabolic. θ = wind = 18 f s a g = f s 11b The componen of he iniial velociy in he horizonal direcion is 18cos, and he wind slows he projecile x = 18cos 18= In he verical direcion, he down by 18 fee every second. Therefore () iniial velociy s componen is 18sin, he paricle has an iniial heigh of 19 f, and graviy s conribuion is 1 given by a = 1 = 16. The oal is y= 18sin = Therefore he g vecor equaion is r () = 6 18, c We wan o maximize y(), so we find = 6 and se i equal o zero, finding. d gives = 56 f. 11d To begin, find he ime a which he paricle lands by seing y= ; ha is, {,6 }, from which we can immediaely discard = Evaluaing y = This gives = because i is nonsensical o have negaive ime given he siuaion. Therefore he paricle his he ground a = 6 s; we find is horizonal displacemen by finding x ( 6) = ( 6 18)( 6) = 8 18 f. 1a The acceleraion vecor () h = 19 f a is given by a () =, = 1 6,. Evaluaing his a = gives d d 1+ a =, 17. The speed a = is given by he magniude of he velociy vecor a = ; since he velociy vecor is v () =, = 1,ln( 1+ ( ) ), we have v = 1,ln17, so v = 1 + ( ln17) d d 1.. 1b The change in he objec s y-coordinae is given by inegraing wih respec o ime over he ime inerval in d quesion; his is ln ( 1 + ( ) ) d. The inegrand canno be anidiffereniaed wihou he echniques of com-

12 plex analysis, bu he calculaor will (afer raher a while) give he approximaion However, noe ha his is he change in he objec s y-coordinae; o ge he acual y-coordinae, we mus add is iniial y-coordinae of 5, giving ( + ( ) ) ( ) d ln 1 ln 1 + ln17 1c We find he slope a P by = = ; a =, = =. Since we know d he x-coordinae of P (i is given) and he y-coordinae (from par b), we can wrie he equaion for he line in ln17 poin-slope form as y = ( x ). 1 1d The objec is saionary in he x-direcion when ; d = ha is, 1 =, or {, }. The objec is saionary in he y-direcion when = ; ha is, ln( 1 + ( ) ) =, or =. The objec is wholly saionary a he in- d ersecion of he soluion ses, which is =.

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