Chapter 6 Applications of Trigonometry

Transcription

1 Chter 6 Alictions of Trigonometry Chter 6 Alictions of Trigonometry Section 6. Vectors in the Plne Exlortion. Use the HMT rle, which sttes tht if n rrow hs initil oint x, y nd terminl oint x, y, it reresents the ector x -x, y -y. If the initil oint is (, ) nd the terminl oint is (7, ), the ector is 7-, -,.. Use the HMT rle, which sttes tht if n rrow hs initil oint x, y nd terminl oint x, y, it reresents the ector x -x, y -y. If the initil oint is (, ) nd the terminl oint is x, y, the ector is x -, y -. Using the gien ector, 6, we he x - nd y -6. x - x 0; y -6 y. The terminl oint is (0, ).. Use the HMT rle, which sttes tht if n rrow hs initil oint x, y nd terminl oint x, y, it reresents the ector x -x, y -y. If the initil oint P is (, ) nd the terminl oint Q is x, y, the ector PQ is x -, y -( ). Using the gien ector PQ,, we he x - nd y +. x - x 6; y + y 7. The oint Q is (6, 7).. Use the HMT rle, which sttes tht if n rrow hs initil oint x, y nd terminl oint x, y, it reresents the ector x -x, y -y. If the initil oint P is x, y nd the terminl oint Q is (, ), the ector PQ is -x, -y. Using the gien ector PQ,, we he -x nd -y. -x x ; -y y. The oint P is (, ). Qick Reiew x9 cos 0, y9 sin 0.. x cos 0 7., y sin 0. x7 cos 0.6, y7 sin 0.. x6 cos( 0 ).86, y6 sin( 0 ).60 For 6, se clcltor For 7 0, the ngle determined y P(x, y) inoles tn y>x. Since this will lwys e etween 80 nd+80, yo my need to dd 80 or 60 to t the ngle in the correct qdrnt. 7. tn 9 L tn - 7 L tn L After hors, the shi hs treled ()( sin 0 ) nt mi est nd ()( cos 0 ) nt mi north. Fie hors lter, it is ()( sin 0 )+()( sin ).0 nt mi est nd ()( cos 0 )+()( cos ).99 nt mi north (ot.9 nt mi soth) of Port Norfolk. Bering: 80 +tn L 9.0 Distnce: L. nt mi. Section 6. Exercises For, recll tht two ectors re eqilent if they he the sme mgnitde nd direction. If R hs coordintes (, ) nd S hs coordintes (c, d), then the mgnitde of RS c - + d - RS, the distnce from R to S. The direction of RS is determined y the coordintes (c-, d-).. If R(, 7) nd S(, ), then, sing the HMT rle, RS -( ), -7,. If P(0, 0) nd Q(, ), then, sing the HMT rle, PQ -0, 0,. Both ectors reresent, y the HMT rle.. If R(7, ) nd S(, ), then, sing the HMT rle, RS -7, -( ),. If P(0, 0) nd Q(, ), then, sing the HMT rle, PQ 0, 0,. Both ectors reresent, y the HMT rle.. If R(, ) nd S(0, ), then, sing the HMT rle, RS 0-,,. If P(, ) nd Q(, ), then, sing the HMT rle, PQ -, -,. Both ectors reresent, y the HMT rle.. If R(, ) nd S(, ), then, sing the HMT rle, RS -( ), -( ),. If P(, ) nd Q(, ), then, sing the HMT rle, PQ -( ), -( ),. Both ectors reresent, y the HMT rle.. PQ -( ), + 9

2 Section 6. Vectors in the Plne 6. RS -( ), QR -, PS -( ), 8-, QS -, PR -( ), QR + PS, +, 0, QR PS - PQ, 0 -,, 6, PS - PQ , +,, 7., -,,., -,, 8 6., 6, 7., +,, 9 8., -, 0, 0 9., -,, 8 0., -,, , i+0.89j. + -, i-0.7j w , - 0 w i-0.89j w 0 w i+0.7j For 8, the nit ector in the direction of, is 0 0 +, + + i +. () () 6. () - () - +, +, i + j, i + j + j. 7. () - () - 8. () (), - i + - j i - j cos, 8 sin 6., cos, sin 8.0,.7. 7 cos 08, 7 sin 08.,.70. cos 6, sin 6.7,.9., Åcos +.., Åcos , Å60 -cos , - i + - j - i - j , Å60 -cos Since (7 cos )i+(7 sin )j( cos Å)i +( sin Å)j, 7 nd Å. 8. Since ( cos 60 )i+( sin 60 )j( cos Å)i +( sin Å)j, nd Å60. For 9 0, first find the nit ector in the direction of. Then mltily y the mgnitde of, , , -, , , 0.8.9,.07. A ering of corresonds to direction ngle of. 0 cos, sin.99, 80.. A ering of 70 corresonds to direction ngle of cos ( 80 ), sin ( 80 ) 79.88,.0. () A ering of 0 corresonds to direction ngle of 0. cos 0, sin 0.6, 0.0 () The wind ering of 0 corresonds to direction ngle of 0. The wind ector is w0 cos 0, sin 0.7, 0.6. Actl elocity ector: +w 6.87, 6.0. Actl seed: +w mh. Actl direction: 80 +tn , so the ering is ot 7.8.

3 Chter 6 Alictions of Trigonometry. () A ering of 70 corresonds to direction ngle of 80 : 60 cos ( 80 ), sin ( 80 ) 79.88,.0. () The wind ering of 00 corresonds to direction ngle of 0. The wind ector is w80 cos ( 0 ), sin ( 0 ) 7.6, 7.8. Actl elocity ector: +w., 8.9. Actl seed: +w mh. Actl direction: 80 +tn , so the ering is ot 7... () 0 cos 70, sin 70., 9.0 () The horizontl comonent is the (constnt) horizontl seed of the sketll s it trels towrd the sket. The erticl comonent is the erticl elocity of the sketll, ffected y oth the initil seed nd the downwrd ll of grity. 6. (). cos, sin., 0.6 () The horizontl comonent is the force moing the ox forwrd. The erticl comonent is the force moing the ox wrd ginst the ll of grity. 7. We need to choose w, k cos, sin,so tht k cos( - )k cos 8.. (Redefine horizontl to men the rllel to the inclined lne; then the towing ector mkes n ngle of 8 with the. horizontl. ) Then k.6 l, so tht cos 8 w.0,.. 8. Jn s force cn e reresented y cos 8, sin 8.87, 7., while Diego s force is 7 cos( ), sin ( ) 6.08, Their totl force is therefore 7.9, 0., so Cororl mst e lling with n eql force in the oosite direction: 7.9, 0.. The mgnitde of Cororl s force is ot 7.9 l. 9. F 0 cos, 0 sin + 7 cos ( 0 ), 7 sin ( 0 ) 00.,., so F 00. l nd.. 0. F00 cos 0, sin 0 +0 cos 60, sin cos( 0), sin( 0 ) 9.7, 66., so F.8 l nd.66.. Shi heding: cos 90, sin 90 0, Crrent heding: cos, sin.8,.8 The shi s ctl elocity ector is.8, 9.7, so its seed is mh nd the direction ngle is cos , so the 9.6 ering is ot.86.. Let 0, 0 e the elocity of the ot nd w 8, 0 e the elocity ector of the crrent. If the ot trels t mintes to rech the oosite shore, then its osition, in ector form, mst e 0, 0t + 8t, 0 8t, 0t 8t,. So 0t t 8t0. mi. The ot meets 0 the shore 0. mi downstrem.. Let w e the seed of the shi. The shi s elocity (in still wter) is w cos 70, w sin 70 0, w. Let z e the seed of the crrent. Then, the crrent elocity is z cos, z sin 0.7z, 0.7z. The osition of the shi fter two hors is 0 cos 0, 0 sin 0 0, 7.. Ptting ll this together we he: 0, w + 0.7z, 0.7z 0, 7.,.z, w+.z 0, 7., so z 7.07 nd w.66. The seed of the shi is ot.66 mh, nd the seed of the crrent is ot 7.07 mh.. Let,,,, nd w w, w. () +, +, +, + +, +, +, + () (+)+w +, + + w, w + +w, + +w, + +w, +w +(+w) (c) +0, + 0, 0 +0, +0, (d) +( ), +, +( ), +( ) 0, 0 0 (e) (+) +, + ( + ), ( + ) +, +, +,, +, + (f) (+) (+), (+) +, +, +,, +, + (g) () (), () ( ), ( ), () (h) 0 0, 0 0, 0 0, , 0,0 0, 0 0 (i) () (), (), ( ) ( ), ( ), (j), Tre. Vectors nd he the sme length t oosite directions. Ths, the length of is lso. 6. Flse. / is not ector t ll , The nswer is D , 9-8, , , 9 The nswer is E. 9. The x-comonent is cos 0, nd the y-comonent is sin 0. The nswer is A , so the nit ector is 8 -, 9>0. The nswer is C.

4 Section 6. Dot Prodct of Vectors 6. () Let A e the oint (, ) nd B e the oint (, ). Then OA is the ector, nd OB is the ector,. So, BA -, -, -, OA -OB () x OA +y OB x( + CA )+y( + CB) (from rt ()) x +x CA +y +y CB (x+y) +x CA +y CB +x +y CB CA (since x+y) 0 BC BC@ CA CB 0 CA CA@ +y 0BC 0 CA + CB 0 CA 0 +y +y since xy CA is nit ector, sme direction. 0 BC +y 0BC BC oints in the + CB 0 BC 0 +y ( BC + CB ) 6. () By Exercise 6, OM x OA +y OB, where x+y. Since M is the midoint, ƒ BM ƒ 0 M A 0. We know from Exercise 6, howeer So xy As reslt,. M y. The roof for OM OA + OB. OM nd OM re similr. () OM + OA + OB + OA + OB +. Use the sme method for the other roofs. (c) Prt () imlies tht OM + OM OM + OB + OA. Ech of the three ectors lies long different medin (tht is, if nonzero, the three ectors he three different directions). Hence they cn only e eql if ll re eql to 0. Ths OM OM -, -OA, nd 0 OM 0 OM 0 OM OM so -OB, OA 0 0 OB 0 6. Use the reslt of Exercise 6. First we show tht if C is on the line segment, then there is rel nmer t so 0 BC 0 t tht (Conince yorself tht 0 CA 0 - t. BC t works.) Then t OA + - t OB. BC + CA A similr rgment cn e sed in the cses where B is on the line segment AC or A is on the line segment BC. Sose there is rel nmer t so tht t OA + - t OB. We lso know OB + BC nd OA + AC. So we he t OA + - tob OB + BC nd t OA + - tob OA + AC. Therefore, t OA - OB BC nd t - OA - OB AC. Hence BC nd AC he the sme or oosite directions, so C mst lie on the line L throgh the two oints A nd B. 6. y 0 0 A P Q M R The line segment OM is medin of ^O since M is midoint of. The line segment AQ is medin of ^O since digonls of rllelogrm isect ech other. By the reslt of Exercise 6, since P is the intersection CR oint of medins, Similrly, This RQ PQ.. imlies tht PRRC, so the digonl hs een trisected. Section 6. Dot Prodct of Vectors Exlortion. -x, 0-y -x, y -x, 0-y -x, y. ( -x)(-x)+( y)( y) +x +y +0 Therefore, 90.. Answers will ry. C Qick Reiew cos + sin. cos 7 + sin 7. -( ), -0, 6. -, -0, 7. -, - - 0, ( ), - -0, -,, +, 6 -, -, - + N 0 -, 9 B x

5 6 Chter 6 Alictions of Trigonometry Section 6. Exercises , 6 + 9, + 6, cos L , , + 9, 9 +, cos - 9 L , + 9, , cos L , , + 6, cos , hs direction ngle nd hs direction ngle (which is eqilent to - ), so the ngle etween the ectors is or hs direction ngle nd hs direction ngle, 6 so the ngle etween the ectors is or , , , cos L , , , -69 cos L.0., h + -, - i - 0 Since 0, nd re orthogonl..,, Since 0, nd re orthogonl. For 8, first find roj. Then se the fct tht 0 when nd re orthogonl.. roj -8, -6, - -6, , - -6, , 0-7, + -, roj, -7 -, -6 -, , -6-9, - 9, + 8, - 7. roj 8, -9, - -9, , - 9, , + -, roj -, 8 9, - 9, , - 7 -, 7 -, +, 9. y B(, ) 0 A(, 0) C(, ) x CA CB -, 0- -, -, 9 7, +60, 0 CA , 0 CB , 0 Ccos 8 6 L 7.7 BC BA 7, -( ), 0-7,, -0, 0 BC 0 6, 0 BA 0 0, Bcos A80 -B-C

6 Section 6. Dot Prodct of Vectors 7 0. CA CB -, -( 6) -, -( 6), 7, 0+, Ccos 7 L 7.0 BA BC -, -( ) -, 6-( ) 9,, 6-06, Bcos A80 -B-C For, se the reltionshi cos... A(, ) y C(, 6) x B(, ) 0 CA , 0 CB 0 6 +, 0 BA , 0 BC 0, 8 cos 0 L cos 0 For 8, ectors re orthogonl if 0 nd re rllel if k for some constnt k.. Prllel: - 0 0,, -,. Neither: 0 nd Z 0 0, Z,. Neither: -0 Z 0 nd - -,, -7 Z 6. Orthogonl: Orthogonl: Prllel: - - -,, -7 For 9 rt (), first find the direction(s) of nd then find the nit ectors. Then find P y dding the coordintes of A to the comonents of nit ector. 9. () A is (, 0) nd B is (0, ). () The line is rllel to 0-, -0,, so the direction of is, or,., - or ƒ ƒ, , -,. ƒ ƒ So, P is (.6, 0.8) or (., 0.8). 0. () A is (, 0) nd B is (0, ). () The line is rllel to 0-( ), -0,, so the direction of is, or,. -, , So P is - - (.7, 0.9) or 9, (.6, 0.9). 9, - 9. () A is (7, 0) nd B is (0, ). () The line is rllel to 0-7, -0 7,, so the direction of is, 7 or, 7., -7 or 8, , So P is 7 + ( 7.9, 0.9) or 8, (6.6, 0.9). 8, 7 8. () A is (6, 0) nd B is (0, ). () The line is rllel to 0-6, -0 6,, so the direction of is, 6 or, 6., 6, , So P is 6 +,, - (6., 0.89) or 6 - (., 0.89)., , + 7. Since -, + 9-7, , - +80, , 8 ( -)( -8)0, so or. Therefore,, or.07, 0.6., 8

7 8 Chter 6 Alictions of Trigonometry. +, + 0. Since +, + 0, , 9, , ( +)(9 +8)0, so or. 9 Therefore,, or - 9, ,.79.. (cos 60 )i+(sin 60 )j i+ j F roj F(F ) -60 j i + j -80. i + j -0 i - 0 j The mgnitde of the force is ƒ F ƒ , onds. 6. In this cse, F j nd remins the sme s in Exmle 6. F roj F(F ) 6. (i+j). The mgnitde of the force is F onds. 7. () (cos )i+(sin )j F 000j F roj F(F ) ( 0, 000 cos, sin ) cos, sin ( 000 sin ) cos, sin. Since cos, sin is nit ector,the mgnitde of the force eing extended is F 000 sin.8 onds. () We re looking for the grittionl force exerted erendiclr to the street. A nit ector erendiclr to the street is w cos ( 78 ), sin ( 78 ), so F roj w F(F w)w ( 000 sin ( 78 )) cos ( 78 ), sin ( 78 ) Since cos ( 78 ), sin ( 78 ) is nit ector, the mgnitde of the force erendiclr to the street is 000 sin ( 78 ) 96.0 onds. 8. We wnt to determine how mch of the 60 ond force is rojected long the inclined lne. F60 cos, sin.88, 0.9 nd cos 8, sin 8 0.9, 0..88, , , 0. roj F.8 0.9, 0..7, The mgnitde of this force is 0 F onds. Of note, it is lso ossile to elte this rolem considering the x-xis rllel to the inclined lne nd the y-xis erendiclr to the lne. In this cse F60 cos, sin.8,.6. Since we only wnt the force in the x-direction, we immeditely find or nswer of ot.8 onds. 9. Since the cr weighs 600 onds, the force needed to lift the cr is 0, 600. W F 0, 600 0,.,00 foot-onds 0. Since the ottoes weigh 00 onds, the force needed to lift the ottoes is 0, 00. W F 0, 00 0, 00 foot-onds,. F, 0, 0 8 W F,, 0.7 foot-onds,. F,, 0, W F, 0 80 W F,, foot-onds,. F 0 0,, 0, 0 + Since we wnt to moe feet long the line y x,we sole for x nd y y sing the Pythgoren theorem: x +y, x + 9, x 9, x 6 x, y 6, 6 WF,, foot-onds B, 0 0. F 0,, 0, 0 + Since we wnt to moe the oject feet long the line yx, we sole for x nd y y sing the Pythgoren theorem: x +y, x +x, x, x., y...,.. WF 0,.,. 6 B. foot-onds. WF 0 F cos 00 cos L 6. foot-onds 6., -,, WF 0 F cos 79 cos foot-onds 79 L 0.9

8 Section 6. Dot Prodct of Vectors 9 7. () Let,,,, nd w w, w. 0 0, 0, () (+w), (, + w, w ), +w, +w ( +w )+ ( +w ) + + w + w,, +, w, w + w (c) (+) w(, +, ) w, w +, + w, w ( + ) w +( + )w w + w + w + w, w, w +, w, w w+ w (d) (c) (c, ), c, c, c +c, c, c (c) c( + )c(,, )c( ) 8. () When we elte the rojection of onto we re ctlly trying to determine how mch of is going in the direction of. Using Figre 6.9, imgine tht rns long or x-xis, with the y-xis erendiclr to it. Written in comonent form, ( cos, sin ) nd we see tht the rojection of onto is exctly cos times s nit ector. Ths, roj cos () Recll Figre 6.9 nd let w e PRroj nd w RQ -roj. Then, (-roj ) (roj )w w. Since w nd w re erendiclr, w w As the digrm indictes, the long digonl of the rllelogrm cn e exressed s the ector + while the short digonl cn e exressed s the ector -.The sm of the sqres of the digonls is (+) (+) +(-) (-) , which is the sm of the sqres of the sides. 60. Let,. ( i)i+( j)j (,, 0 )i+(, 0, )j ( )i+( )j i+ j, 6. Flse. If either or is the zero ector, then 0 nd so nd re orthogonl, t they do not cont s erendiclr. 6. Tre , so the ectors re erendiclr. The nswer is D. 6.,, ( )+( )( ) 8+ 7 The nswer is C. 6. roj , 0, 0 k h, 0 i The nswer is A. 66. The nit ector in the direction of is,. The force is reresented y times the nit ector. The nswer is B. 67. () 0+ 0 nd + 00 () -0, 7-, -0, 0-, w roj,, -7, ,, 9 9 w -roj, -, 9 0 -, + 6, (c) w is ector from oint on to oint P. Since w is erendiclr to, w is the shortest distnce from to P. w B ,869 9 B 9 9 (d) Consider Figre 6.9. To find the distnce from oint P to line L, we mst first find roj.in this cse, roj x 0, y 0 -, -, + - x 0 - y 0 +, 9 x 0-0y 0 + 0, -0x 0 + y roj x 0, y 0 - x - 0-0y 0 + 0, -0x 0 + y ( 9x 0, 9 (y 0 -) 9 - x 0-0y 0 +0, 0x 0 +y 0-8 ) x 0 +0y 0-0, 0x 0 +y nd

9 @ 0 Chter 6 Alictions of Trigonometry So, the distnce is the mgnitde of this ector. d 9 x 0 + 0y x 0 + y 0-0 x 0 + y x 0 + y x 0 + y x 0 + y c, -c c x 0, y 0 -, so roj (e) In the generl cse, nd x 0c - cy 0 - c h c, - c i c + - c -x x 0 - y 0 + c 0 + y 0 - c, + + -roj x c 0, y 0 - -x x 0 - y 0 + c 0 + y 0 - c, + + x 0 + y 0 - c, x 0 + y 0 - c + + The mgnitde of this ector, x 0 + y 0 - c the distnce from oint P to L: () Yes, if 0, 0 or tn, nny integer n () Yes, if 0, 0 or t, nodd integer (c) Generlly, no, ecse sin t Z cos t for most t. Excetions, howeer, wold occr when t, + n nny integer, or if 0, 0 nd/or 0, One ossile nswer: If +c+d -c+-d0 (-c)+(-d)0 Since nd re not rllel, the only wy for this eqlity to hold tre for ll ectors nd is if (-c)0 nd (-d)0, which indictes tht c nd d. - Section 6. Prmetric Eqtions nd Motion Exlortion.. 0.(7)+.0, so the oint (7, 0) is on the grh, t 8. 0.( )+. 0, so the oint (, 0) is on the grh, t.. x-t, t-, t -. Alterntiely, -t, so t-.. Choose Tmin nd Tmx so tht Tmin - nd Tmx.. Exlortion. It looks like the line in Figre 6... The grh is erticl line segment tht extends from (00, 0) to (00, 0).. For 9 nd 0, the ll does not cler the fence, s shown elow. 9 : 0 : [ 0, ] y [, ] [0, 0] y [0, 80] [0, 0] y [0, 80] For nd, the ll clers the fence, s shown elow. : [0, 0] y [0, 80] [0, 0] y [0, 80]