AP Calculus BC - Parametric equations and vectors Chapter 9- AP Exam Problems solutions
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1 AP Calculus BC - Parameric equaions and vecors Chaper 9- AP Exam Problems soluions. A 5 and 5. B A, C A, ; he poin a is (,). y + ( x ) x e + e D A, slope.5 6 e e e 5. A d hus d d y, ; 4 6. B If x hen y 5. 6 x + y ; () E x e, y sin( ); cos() c os() e e 8. C x, y ; Verical angens a, ( )
2 9...
3 . 97 BC4 A kie flies according o he parameric equaions x, y ( 8) 8 64 where is measured in seconds and < 9. (a) (c) (d) How high is he kie above he ground a ime seconds? A wha rae is he kie rising a seconds? A wha rae is he sring being reeled ou a seconds? A wha ime does he kie sar o lose aliude? (a) y ( 8) 64 A, y ()( 8) ( 96) ( 64 8) 64 A, 9 + ( 96) + 64 (c) Le s be he lengh of he sring. s x + y ds x y ( + ) x + y A, x 4 and y 44. A, and 8. Therefore a, ds (4) + (44)() (6 9) (d) The kie will sar o lose aliude a he insan when This occurs a 64 seconds.
4 . 974 BC5 Given he parameric equaions x (θ sin θ) and y ( cosθ) : (a) Find in erms of θ. Find an equaion of he line angen o he graph a θ π. (c) Find an equaion of he line angen o he graph a θ π. (d) Se up bu do no evaluae an inegral represening he lengh of he curve over he inerval θ π. Express he inegrand as a funcion of θ. 974 BC5 Soluion (a) cosθ, sinθ dθ dθ d sinθ sinθ θ cosθ cosθ dθ A θπ, sin π. This means he angen line is horizonal. cosπ A θπ, y 4, so he equaion of he angen line is y 4. (c) A θπ, sin π cosπ which is no defined. lim sin θ lim cos θ θ θ π cosθ θ π sin This means ha here is a verical angen. A θ π, x 4π, so he equaion of he angen line is x 4π (d) π π Lengh ( cos θ ) + (sin θ) dθ cosθdθ
5 4. 98 BC6 Poin Px, ( y) moves in he xy-plane in such a way ha + and for. (a) Find he coordinaes of P in erms of if, when, x ln and y. Wrie an equaion expressing y in erms of x. (c) Find he average rae of change of y wih respec o x as varies from o 4. (d) Find he insananeous rae of change of y wih respec o x when. 98 BC6 Soluion (a) x ln( + ) + + C ln x() ln + C, so C x ln( + ) y +C y() + C, s o C y x e x +, so y (e ) e x e x (c) y(4) y() 5 ( ) 6 x(4) x () ln 5 ln ln 5 (d) A, 4 + or From, using x ln when e ln ln e 8 4 4
6 BC The pah of a paricle is given for ime > by he parameric equaions x y. + and (a) Find he coordinaes of each poin on he pah where he velociy of he paricle in he x direcion is zero. Find when. (c) Find d y when y. 984 BC Soluion (a) when. The corresponding poin on he pah is (,6). (c) 6 A, 6 d 4 6 d y 6 When y, and so d y
7 BC4 Consider he curve given by he parameric equaions x and y (a) In erms of, find. Wrie an equaion for he line angen o he curve a he poin where. (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. 989 BC4 Soluion (a) 6 6 (+ )( ) ( ) x 5, y 4 y ( x + 5) 4 or 9 y x y+ x 9 (c) ( x,y) ype ( 8,6) horizonal ( ) (, ) ( ), verical verical 4, 6 horizonal
8 7. 99 BC Soluion x y (a) () () x ( ) y ( ) v ( ) ( ) 5 / 6 / ( ) () () (c) y x x x+ x+ ; x + or ; + x x y ; y (x+ ) x + /
9 8. D L D cos, y sin π x for. L π + π π 4 4 L ( cos sin ) + (sin cos ) 9cos sin + 9sin cos C The lengh of his parameric curve is given by ( ) + +.
10 . 97 BC7 Soluion, 6 y 4( x 5 x+ 4) for x 4, so 4(x 5) 5 x 9 y for y 9 (lef half of curve), so 4 9 y Each of he inegrals can be wrien in erms of,, or. (a) b Lengh + a (6 ) 4 + 6(x 5) ( y) Volume π π b y a 6 ((6 )) 4 π 6( x 5x+ 4) 9 π y 9 y b (c) Surface area π yds a 6 π (6 ) +(6 ) 4 4 π 4(x 5 x+4) +6(x 5) 9 4π y + 6(9 y)
11 . 974 BC5 Given he parameric equaions x (θ sin θ) and y ( cosθ) : (a) Find in erms of θ. Find an equaion of he line angen o he graph a θ π. (c) Find an equaion of he line angen o he graph a θ π. (d) Se up bu do no evaluae an inegral represening he lengh of he curve over he inerval θ π. Express he inegrand as a funcion of θ. 974 BC5 Soluion (a) cosθ, sinθ dθ dθ d sinθ sinθ θ cosθ cosθ dθ A θπ, sin π. This means he angen line is horizonal. cosπ A θπ, y 4, so he equaion of he angen line is y 4. (c) A θπ, sin π cosπ which is no defined. lim sin θ lim cos θ θ θ π cosθ θ π sin This means ha here is a verical angen. A θ π, x 4π, so he equaion of he angen line is x 4π (d) π π Lengh ( cos θ ) + (sin θ) dθ cosθdθ
12 . E f () ( e, sin); f ( ) ( e, cos ). 4. D x( ) a() 4 y ) 4 and d x ; ( 6 and d y d x d y, (, ) a() (,) 5. E d x d y 4 x +,, ; y ln(+), ; + ( + ) 6. A + v,4, v ( ) 6,8, E v) ( ( ) and a() 6, 4( ),6( ) ( ) a() ( 6, 4) 8.
13 BC A paricle moves on he circle x + y so ha a ime he posiion is given by he vecor, +. + (a) (c) Find he velociy vecor. Is he paricle ever a res? Jusify your answer. Give he coordinaes of he poin ha he paricle approaches as increases wihou bound. 975 BC Soluion (a) (+ )( ) ( )() 4 (+ ) ( + ) ( + )() () (+ ) ( + ) The velociy vecor is 4, ( + ) (+ ) a and a. Therefore here is no such ha a he same ime. Hence he paricle is never a res. (c) lim x) ( lim lim + + lim y) ( lim + lim + Hence he paricle approaches he poin (,) as increases wihou bound.
14 . 987 BC5 Soluion (a) x sin( ), y cos( ) cos(), sin( ) v (cos( ), sin( )) or v cos( i ) +( sin()) j v cos( ) and sin( ) Therefore cos() and 4sin( )cos( ). The only choice is cos(). π π Therefore he paricle is a res when,. (c) x sin (), y x y cos ( ) sin ( ) (d)
15 . 99 BC A ime, π, he posiion of a paricle moving along a pah in he xy-plane is given by he parameric equaions x e sin and y e cos. (a) Find he slope of he pah of he paricle a ime π. Find he speed of he paricle when. (c) Find he disance raveled by he paricle along he pah from o. 99 BC Soluion (a) e sin + e cos e cos e sin / e / e s + ( cos sin ) ( in cos) π /( ) π / ( + ) π e a, e e sin + speed ( e cos) +( e cos e sin when speed is ( esin+ ecos) +( ecos esin) e ) (c) disance is ( e sin + e cos) +( e cos e sin e sin + cos ( ) e ( e ) ) e
16 . 99 BC Soluion x y (a) () () x ( ) y ( ) v ( ) ( ) 5 / 6 / ( ) () () (c) y x x x+ x+ ; x + or x x ; + y ; y (x+ ) x + /
17 . 994 BC A paricle moves along he graph of y cos x so ha he x-coordinae of acceleraion is always. A ime, he paricle is a he poin ( π, ) and he velociy vecor of he paricle is (,). (a) Find he x- and y-coordinaes of he posiion of he paricle in erms of. Find he speed of he paricle when is posiion is ( 4,cos4 ). 994 BC (a) x () () ( ) x ( ) () +, x( ) π x() + π y() cos + π x +C x C ; x k k ( ) sin ( + π ) s() + ( ( )) ( + π ) ( ) sin π sin when x 4, 4; 4 + π π ( ) ( ) s 4 4 π +4 4 si.4 π n 4
18 BC Soluion, (a) VA, 4 ; VA VB, ; VB, disance 4 (c) Se 4; 4 When 4, he y-coordinaes for A and B are also equal. Paricles collide a (,4) when 4. (d) Viewing Window [ 7,7] [ 5,5]
19 BC Soluion (a) x y.5 cos.5.5 5sin.5 5 y x (c) (d) x( ) sin y() 5 cos v sin,5 cos.75 (e) disance 9 sin 5 cos.5 5.9
20 6. BC{ 999. A paricle moves in he xy{plane so ha is posiion a any ime,, isgiven by x() ; ln( + ) an() sin. (a) Skech he pah of he paricle in he xy{plane below. Indicae he direcion of moion along he pah. A wha ime,, doesx() aain is minimum value? Wha is he posiion (x()y()) of he paricle a his ime? (c) A wha ime, <<, is he paricle on he y{axis? Find he speed and he acceleraion vecor of he paricle a his ime. (a) y ( : graph : direcion O x x () ; + + ; ;+p 5 or :68 in [] x(:68) ;:9 y(:68) :78 (c) x() ; ln( + ) :85 or :86 x () ; y () cos speed p + (x (:86)) +(y (:86)) :96 x () + ( + ) y () ; sin 4 8 >< >: 8 > >< > >: : x () : soluion for : posiion : x() : soluion for : speed : acceleraion vecor acceleraion vecor <x (:86)y (:86) > <:9;:879 >
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22 8. AP CALCULUS BC SCORING GUIDELINES Quesion An objec moving along a curve in he xy-plane has posiion x (),y() a ime wih cos and sin for > >. A ime, he objec is a posiion (4,5). (a) Wrie an equaion for he line angen o he curve a (4,5). Find he speed of he objec a ime. (c) Find he oal disance raveled by he objec over he ime inerval (d) Find he posiion of he objec a ime. > >. (a) sin cos sin cos 5.64 y Г ( x Г4) : angen line Speed cos (8) 9 sin (4).75 : answer (c) Disance cos 9 sin.458 : : disance inegral < Г each inegrand error < Г error in limis : answer (d) cos x () or.954 y () 5 sin : : definie inegral for x : answer for x() : definie inegral for y : answer for y()
23 9. AP CALCULUS BC SCORING GUIDELINES Quesion > J > &IA?@I6DAFIEJE BJDA?HJJEA J IA?@I?>A@AA@FHAJHE?O >O N J J " IEJ O J ГJ Г? IJ MDAHA O HAAIKHA@EAJAHI6DA@AHELJELAIBJDAIABK?JEIHACELA>O N J "?IJ O J ГJIEJ?IJ Г.E@JDAIFABJDAFJDJJEAJ 5DMJDA?FKJJEIJDJA@JOKHIMAH >.E@JDA??AAHJELA?JHBJDA?HJJDAJEAMDAJDA?H\IDHEJFIEJEEIN"?.E@JDAJEA J JMDE?DJDA?HEIJEJINEKDAECDJ@BE@JDAIFAA@EIA?BJDA?HJ J &JDAHAHAJMJEAIJMDE?DJDA?HEIJCHK@ALA O & IE?I Г J N "?I %'!H%'" > NJ J " IEJ " J!$"%&! N J Г!# ' O J #H $ )??AAHJELA?JHEIГ!# ' # HГ!# ' $? OJ ГJIEJ?IJ Г J!!H! "JNEKDAECDJ O $ %H$ & 5FAA@ N OJ N J J " " )LAHCAIFAA@ N J O " "?I J E@AJEBEAI??AAHJELA?JH I@AHELJELABLA?EJOLA?JH?FKJAI??AAHJELA?JH MDAN " IAJIO J! IAA?JIBEHIJJ IFAA@ J J "! EEJI@?IJJ EJACH@
24 4. AP CALCULUS BC SCORING GUIDELINES (Form B) Quesion )FHJE?ALAIEJDANOFAIJDJEJIFIEJEJOJEAJBHГ > J > EICELA>O NJ IE! J 5AJ?DJDAFJDBJDAFHJE?AEJDANOFAFHLE@A@@E?JAJDA@EHA?JEBJE CJDAFJD >.E@JDAHCABNJ@JDAHCAB OJ?.E@JDAIAIJFIEJELALKABJBHMDE?DJDAN?H@EJABJDAFHJE?AEI? IJDA@EIJ?AJHLAA@>OJDAFHJE?ABHJ Г JJ CHAJAHJD # KIJEBO OKHIMAH CHFD JDHAA?O?AIBIEA N >AJMAA > Г > NJ > Г > OJ >? NJ!?I! J!J J $ 5FAA@ '?I! J " )JJ $ 5FAA@ '?I '?I! J "@J Г %'%! #?IA@EJAHLBHNJ?IA@EJAHLBHOJ N J!?I! J! ILAIBHJ IFAA@JIJK@AJ\IJEA EJACHBH@EIJ?A??KIEMEJDKIJEBE?JE
25 4. AP CALCULUS BC SCORING GUIDELINES Quesion A paricle sars a poin A on he posiive x-axis a ime and ravels along he curve from A o B o C o D, as shown above. The coordinaes of he paricle s posiion x( ), y x ( 9cos sin + ) 6 ( ( )) are differeniable funcions of, where and y ( ) is no explicily given. A ime 9, he paricle reaches is final posiion a poin D on he posiive x-axis. (a) A poin C, is posiive? A poin C, is posiive? Give a reason for each answer. The slope of he curve is undefined a poin B. A wha ime is he paricle a poin B? 5 (c) The line angen o he curve a he poin ( x( 8 ), y( 8) ) has equaion y x. Find he 9 velociy vecor and he speed of he paricle a his poin. (d) How far apar are poins A and D, he iniial and final posiions, respecively, of he paricle? (a) A poin C, is no posiive because y() is decreasing along he arc BD as increases. A poin C, is no posiive because x() is decreasing along he arc BD as increases. : : no posiive wih reason : no posiive wih reason ; cos or 6 ( ) sin + + or ; for boh. 6 Paricle is a poin B a. : : ses : (c) ( ) sin (8) 9 cos ( ) 4 9 x y(8) 5 x (8) y(8) x(8) 9 The velociy vecor is < 4.5,.5 >. : : x (8) : y (8) : speed Speed or 5.48 (d) x(9) x() x ( ) The iniial and final posiions are 9.55 apar. : inegral : : answer
26 4. AP CALCULUS BC SCORING GUIDELINES (Form B) Quesion 4 A paricle moves in he xy-plane so ha he posiion of he paricle a any ime is given by x () e 7 + e and ( y ) e e. (a) Find he velociy vecor for he paricle in erms of, and find he speed of he paricle a ime. Find in erms of, and find lim. (c) Find each value a which he line angen o he pah of he paricle is horizonal, or explain why none exiss. (d) Find each value a which he line angen o he pah of he paricle is verical, or explain why none exiss. (a) x() 6e 7e 7 y () 9e + e Velociy vecor is < 6e 7 7e, 9 e + e > : : x( ) : y( ) : speed Speed x() + y() () + 9e + e 7 6e 7e : : in erms of : limi 9e + e 9 lim lim 7 6e 7e 6 (c) Need y(), bu 9e e + > for all, so none exiss. : : considers y( ) : explains why none exiss (d) Need x() and y(). 6e 7e 7 7 e 6 7 ln 6 ( ) : considers x( ) : : soluion
27 4. AP CALCULUS BC 4 SCORING GUIDELINES Quesion An objec moving along a curve in he xy-plane has posiion ( x( ), y() ) a ime wih + cos ( ). The derivaive is no explicily given. A ime, he objec is a posiion (, 8). (a) Find he x-coordinae of he posiion of he objec a ime 4. A ime, he value of is 7. Wrie an equaion for he line angen o he curve a he poin (x(, ) y( ) ). (c) Find he speed of he objec a ime. (d) For, he line angen o he curve a ( x( ), y() ) has a slope of +. Find he acceleraion vecor of he objec a ime 4. x x( ) 4 ( ( )) 4 ( ( )) (a) ( 4) + + cos + + cos 7. or 7. : : 4 ( + cos( )) : handles iniial condiion : answer cos4 y 8.98( x ) : finds : : equaion (c) The speed of he objec a ime is ( x ( ) ) + ( y ( ) ) 7.8 or 7.8. : answer (d) x ( 4). y () ( + y ( 4) 4.8 or 4.84 The acceleraion vecor a 4 is., 4.8 or., ) ( + cos( )) : : x ( 4) : : answer
28 44. AP CALCULUS BC 4 SCORING GUIDELINES (Form B) Quesion A paricle moving along a curve in he plane has posiion ( x( ), y() ) a ime, where and e + 5e for all real values of. A ime, he paricle is a he poin (4, ). (a) Find he speed of he paricle and is acceleraion vecor a ime. Find an equaion of he line angen o he pah of he paricle a ime. (c) Find he oal disance raveled by he paricle over he ime inerval. (d) Find he x-coordinae of he posiion of he paricle a ime. (a) A ime : Speed x () + y () : speed : : acceleraion vecor Acceleraion vecor x (, ) y ( ), y ( ) 7 x ( ) Tangen line is y 7 ( x 4) + : slope : : angen line 4 (c) Disance ( + ( ) 9 + e + 5e ) 45.6 or 45.7 : disance inegral each inegrand error : error in limis : answer 4 (d) x ( ) : inegral : : answer 7.9 or 7.9
29 45. AP CALCULUS BC 5 SCORING GUIDELINES (Form B) Quesion An objec moving along a curve in he xy-plane has posiion ( x( ), y() ) a ime wih and ln + ( 4 ( 4) ). A ime, he objec is a posiion (, 5). A ime, he objec is a poin P wih x-coordinae. (a) Find he acceleraion vecor a ime and he speed a ime. Find he y-coordinae of P. (c) Wrie an equaion for he line angen o he curve a P. (d) For wha value of, if any, is he objec a res? Explain your reasoning. (a) x ( ), y ( ).88 7 a( ),.88 Speed + ( ln ( 7) ).9 or. : acceleraion vecor : : speed ( ) y () y() + ln + ( u 4)4 y ( ) 5 + ln + ( u 4 ) 4 du ( ). du 67 : ( : ln ( u ) ) du : handles iniial condiion : answer ln ( 7) (c) A, slope.6 y.67.6( x ) : slope : : equaion (d) x () if, 4 y () if 4 4 : reason : : answer
30 46. AP CALCULUS BC 6 SCORING GUIDELINES Quesion An objec moving along a curve in he xy-plane is a posiion ( x( ), y() ) a ime, where ( ) sin e and 4 + for. A ime, he objec is a he poin ( 6, ). (Noe: sin x arcsin x ) (a) Find he acceleraion vecor and he speed of he objec a ime. The curve has a verical angen line a one poin. A wha ime is he objec a his poin? (c) Le m() denoe he slope of he line angen o he curve a he poin ( x( ), y() ). Wrie an expression for m () in erms of and use i o evaluae lim m (). (d) The graph of he curve has a horizonal asympoe y c. Wrie, bu do no evaluae, an expression involving an improper inegral ha represens his value c. (a) a( ).95 or.96,.74 or.74 Speed x ( ) + y (). 7 or.8 : { : acceleraion : speed ( ) sin e e ln.69 and when ln : x () : : answer (c) m () lim 4 + sin ( e ) m () 4 lim sin ( e + ) sin () : m() : : limi value (d) Since lim x (), c lim y() : : inegrand : limis : iniial value consisen wih lower limi
31 47. AP CALCULUS BC 6 SCORING GUIDELINES (Form B) Quesion An objec moving along a curve in he xy-plane is a posiion ( x( ), y() ) a ime, where an( e ) for. A ime, he objec is a posiion (, ). ( ) and sec e (a) Wrie an equaion for he line angen o he curve a posiion (, ). Find he acceleraion vecor and he speed of he objec a ime. (c) Find he oal disance raveled by he objec over he ime inerval. (d) Is here a ime a which he objec is on he y-axis? Explain why or why no. (a) sec( e ) an ( e ) sin( e ) sin e.78 or.78 (, ) ( ) y + ( x ) sin ( e ) : : (, ) : equaion of angen line x ().45, y ().596 a( ).4,.5 or.4,.5. : acceleraion vecor : { : speed ( ) ( ( )) speed sec( e ) + an e.8 or.9 (c) ( x () ) + ( y () ).59 : { : inegral : answer (d) x() x() x ( ) > The paricle sars o he righ of he y-axis. Since x ()> for all, he objec is always moving o he righ and hus is never on he y-axis. : : x ( ) expression : x () > : conclusion and reason
32 48. AP CALCULUS BC 7 SCORING GUIDELINES (Form B) Quesion An objec moving along a curve in he xy-plane is a posiion ( x( ), y( )) a ime wih ( ) arcan + and ln ( + ) for. A ime, he objec is a posiion (, 4). (Noe: an x arcan x ) (a) Find he speed of he objec a ime 4. Find he oal disance raveled by he objec over he ime inerval (c) Find x( 4. ) 4. (d) For >, here is a poin on he curve where he line angen o he curve has slope. A wha ime is he objec a his poin? Find he acceleraion vecor a his poin. (a) Speed x ( 4) + y ( 4).9 : speed a 4 Disance : { : inegral : answer (c) x( 4) x( ) 4 + x ( ) : inegrand : : : uses x( ) : answer (d) The slope is, so, or ln ( ) + arcan ( ). Since >, A his ime, he acceleraion is x ( ), y () , : : : -value : values for x and y
33 49. AP CALCULUS BC 8 SCORING GUIDELINES (Form B) Quesion A paricle moving along a curve in he xy-plane has posiion ( x( ), y() ) a ime wih and cos. The paricle is a posiion (, 5) a ime 4. (a) Find he acceleraion vecor a ime 4. Find he y-coordinae of he posiion of he paricle a ime. (c) On he inerval 4, a wha ime does he speed of he paricle firs reach.5? (d) Find he oal disance raveled by he paricle over he ime inerval 4. (a) a( 4) x ( 4), y ( 4).4,.87 : answer y( ) cos.6 or.6 : : inegrand : uses y( 4) 5 : answer (c) Speed ( x () ) + ( y () ) + 9cos.5 : : expression for speed : equaion : answer The paricle firs reaches his speed when.5 or.6. (d) 4 + 9cos.8 : { : inegral : answer
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