3, so θ = arccos

Transcription

1 Mahemaics 210 Professor Alan H Sein Monday, Ocober 1, 2007 SOLUTIONS This problem se is worh 50 poins 1 Find he angle beween he vecors (2, 7, 3) and (5, 2, 4) Soluion: Le θ be he angle (2, 7, 3) (5, 2, 4) (2, 7, 3) (5, 2, 4) cos θ, so 8 62 ( ) cos θ, cos θ 3, so θ arccos Find vecor parameric, scalar parameric and scalar symmeric equaions for he line hrough he poin (5, 3, 1) and orhogonal o he plane 3x 7y + 2z 8 Soluion: The normal o he plane is in he direcion of he line, so he vecor (3, 7, 2) is in he direcion of he line Vecor Equaion: x (5, 3, 1) + (3, 7, 2) Scalar Parameric Equaions: x 5 + 3, y 3 7, z Scalar Symmeric Equaions: x 5 y z Find an equaion of he plane hrough he poins (1, 5, 4), (4, 9, 1) and (5, 17, 5) Exra Credi: Obain he same equaion using an enirely differen mehod Soluion: Le v (4, 9, 1) (1, 5, 4) (3, 4, 3), w (5, 17, 5) (1, 5, 4) (4, 12, 1), i j k n v w (3, 4, 3) (4, 12, 1) (40, 15, 20) 5(8, 3, 4) Since we can use any vecor in he same direcion as a normal o he plane, we redefine n (8, 3, 4) We ge an equaion for he plane n x n (1, 5, 4) Wriing x (x, y, z), we ge (8, 3, 4) (x, y, z) (8, 3, 4) (1, 5, 4) or 8x 3y + 4z 9 Alernaively, we know he hree poins saisfy an equaion of he form ax + by + cz d We may hus ry o solve he following sysem for a, b, c, d a(1) + b(5) + c(4) d a(4) + b(9) + c(1) d a(5) + b(17) + c(5) d Since we have an exra variable, if we so desire we can arbiarily choose a value for one of hem If we le d 9, we will obain a 8, b 3, c 4 If we chose any oher value for d, we would have obained an equivalen equaion Page 1 of 5

2 Page 2 of 5 4 Find he disance beween he poin (1, 2, 3) and he plane 4x + 5y + 6z 7 Exra Credi: Find he disance using an enirely differen mehod Soluion: If we le (x, y, z) represen a vecor whose ip is in he plane, he disance (4, 5, 6) (4, 5, 6) The is he projecion of (x, y, z) (1, 2, 3) ono he uni normal lengh will be (4, 5, 6) ((x, y, z) (1, 2, 3)) x + 5y + 6z ( ) Alernaively, one could find where he line x (1, 2, 3) + (4, 5, 6) inersecs he plane and find he disance beween ha poin and (1, 2, 3) 5 An an walks on he paraboloid z x 2 +y 2 Using he usual orienaion for he coordinae axis, he an sars a he origin and is rising a a rae of 4 unis per second Looking from above, i looks as if he an is spiralling around he z-axis, saring by he x-axis and making a complee revoluion every 5 seconds (a) Find a vecor parameric equaion for he pah of he an Soluion: Clearly, z 4, where represens ime Since x 2 + y 2 z, we have x 2 + y 2 4 If θ is he angle we visualize looking sraigh down beween he x-axis and he line from he origin o he an, x 4 cos θ and y 4 sin θ Since θ sars a 0 and increases by 2π every 5 seconds, θ 2π We hus obain 5 r (2 cos( 2π), 2 ), 4) 5 5 (b) Find he velociy of he an Soluion: v r ( 1 cos( 2π 5 ) 4π 5 5 ), 1 5 ) + 4π (c) Find he acceleraion of he an Soluion: a v ( cos( 2π 5 5 ), 4) 2π cos( ) 4π 5 5 2π sin( ) 8π2 cos( 2π), π sin( 4π 5 2π cos( ) 8π2 ), 0) 5 Represen ime by Each of your conclusions should be in erms of 6 Consider he vecor funcion x (, 3 sin, 4 cos ), 0 (a) Skech he graph 5 ) + (b) Find he lengh of he porion of he curve for which π 2π Give your answer in erms of a definie inegral You do no need o evaluae he inegral bu may do so for exra credi Soluion: Since dx d (1, 3 cos, 4 sin ), he lengh is 2π π cos sin 2 d

3 Page 3 of 5 7 Consider he vecor funcion x (, 2, 3 ), 1 (a) Skech he graph (b) Find he uni angen vecor T a 2 Soluion: v (1, 2, 3 2 ), so T v v (1, 2, 32 ) When 2, T (1, 4, 12) (c) Find he normal vecor N a 2 Soluion: N T T (0, 2, 6) (1, 2, ) T (0, 2, 12) (1, 4, 12) When 2, we ge T 152 2( 76, 143, 54) ( 76, 143, 54) ( 76, 143, 54) Thus N ( 76, 143, 54) (d) Find he binormal vecor B a 2 (1, 4, 12) ( 76, 143, 54) Soluion: B T N i j k (1, 4, 12) ( 76, 143, 54) (1932, 966, ) (1932, 966, ) We hus have B We can simplify his by observing B is a uni vecor in he direcion of (1932, 966, ) (12, 6, 1) (12, 6, 1) (12, 6, 1), so B (12, 6, 1) 181 (e) Verify T, N and B are muually orhogonal Soluion: We calculae each of he do producs Each comes ou o be 0 (f) Find he curvaure κ a 2 Soluion: κ T ds d When 2, T 2( 76, 143, 54), so T Also, when 2, v (1, 4, 12), so ds d v ( 2 ) 181 We hus have κ 2 181

4 Page 4 of 5 8 A jogger runs leisurely a a consan pace of a six minue mile around a quarer mile rack The rack consiss of wo parallel sraighaways conneced by wo semicircles, each 110 yards long (a) Se up an appropriae coordinae sysem Soluion: We se he origin a he cener of he rack The x-axis is parallel o he sraigh porion, oriened in he direcion he jogger sars ou We orien he y-axis so he y-coordinae is negaive when he jogger begins Since each semicircle is of lengh 110 (measured in yards), we effecively have par of a circle wih circumference 220 Leing r be he radius, we have 2πr 220, so r 110 π We will consider he horizonal porion o be going eas-wes, wih he jogger saring ou going o he eas, hen counerclockwise around a semicircle, hen going

5 Page 5 of 5 (c) Find he acceleraion of he runner when he or she is exacly halfway hrough one of he semicircles Exra Credi: Obain he acceleraion again using an enirely differen mehod Soluion: Around he firs curve, r (55, 0) + 110(cos( π + 8π( 3)), sin( π + π π ( ))), so v ( sin( π + 8π( 3)), cos( π + 8π( 3))) 8π and a 3 8 π ( cos( π + 8π( 3)), sin( π + 8π( 3))) ( 8π π )2