Chapter 6 Applications of Trigonometry

Transcription

1 Chapter 6 Applications of Trigonometr Chapter 6 Applications of Trigonometr Section 6. Vectors in the Plane Eploration. Use the HMT rule, which states that if an arrow has initial point (, ) and terminal point (, ), it represents the vector -, -. If the initial point is (, ) and the terminal point is (7, ), the vector is 7-, - =,.. Use the HMT rule, which states that if an arrow has initial point (, ) and terminal point (, ), it represents the vector -, -. If the initial point is (, ) and the terminal point is (, ), the vector is -, -. Using the given vector, 6, we have - = and -=6. -= Q =0; -=6 Q =. The terminal point is (0, ).. Use the HMT rule, which states that if an arrow has initial point (, ) and terminal point (, ), it represents the vector -, -. If the initial point P is (, ) and the terminal point Q is (, ), the vector PQ is -, -( ). Using the given vector PQ,, we have -= and +=. -= Q =6; += Q = 7. The point Q is (6, 7).. Use the HMT rule, which states that if an arrow has initial point (, ) and terminal point (, ), it represents the vector -, -. If the initial point P is (, ) and the terminal point Q is (, ), the vector PQ is -, -. Using the given vector PQ,, we have - = and - =. - = Q =; - = Q =. The point P is (, ). Quick Review =9 cos 0 =, =9 sin 0 =.. = cos 0 = 7., = sin 0 =. =7 cos 0.6, =7 sin 0.. =6 cos ( 0 ).86, =6 sin ( 0 ).60 For # and 6, use a calculator For #7 9, the angle determined b P(, ) involves tan (/). Since this will alwas be between 80 and+80, ou ma need to add 80 or 60 to put the angle in the correct quadrant. 7. =tan a 9 b L =60 +tan a - 7 b L =80 +tan a b L After hours, the ship has traveled ()( sin 0 ) naut mi east and ()( cos 0 ) naut mi north. Five hours later, it is ()( sin 0 )+()( sin ).0 naut mi east and ()( cos 0 )+()( cos ).99 naut mi north (about.9 naut mi south) of Port Norfolk. Bearing: 80 +tan a b L 9.0 Distance: (.0) + (-.99) L. naut mi. Section 6. Eercises For #, recall that two vectors are equivalent if the have the same magnitude and direction. If R has coordinates (a, b) and S has coordinates (c, d), then the magnitude of RS is ƒ RS ƒ = (c - a) + (d - b) = RS, the distance from R to S. The direction of RS is determined b the coordinates (c-a, d-b).. If R=(, 7) and S=(, ), then, using the HMT rule, RS = -( ), -7 =,. If P=(0, 0) and Q=(, ), then, using the HMT rule, PQ = -0, 0 =,. Both vectors represent, b the HMT rule.. If R=(7, ) and S=(, ), then, using the HMT rule, RS = -7, -( ) =,. If P=(0, 0) and Q=(, ), then, using the HMT rule, PQ = 0, 0 =,. Both vectors represent, b the HMT rule.. If R=(, ) and S=(0, ), then, using the HMT rule, RS = 0-, =,. If P=(, ) and Q=(, ), then, using the HMT rule, PQ = -, - =,. Both vectors represent, b the HMT rule.. If R=(, ) and S=(, ), then, using the HMT rule, RS = -( ), -( ) =,. If P=(, ) and Q=(, ), then, using the HMT rule, PQ = -( ), -( ) =,. Both vectors represent, b the HMT rule.. PQ = -( ), - =,, ƒ PQ ƒ = + = 9 6. RS = -( ), 8- =,, ƒ RS ƒ = + (-) = 8 Copright 0 Pearson Education, Inc.

2 Section 6. Vectors in the Plane 7. QR = -, - =,, ƒ QR 6. (a) h - ƒ = (-) + = 6, i 8. PS = -( ), 8- =, 0, (b) - = + (-0) i + j ƒ PS ƒ = 6= 9 9. QS= -, 8- =,, 7. (a) h -, - i ƒ QS ƒ = (-) + (-) = 80 = 0. () PR (b) - = -( ), - = 0,, i + a - bj = - i - j ƒ QR = 0 + () =. QR + PS =, +, 0 =, 7, ƒqr + PS ƒ = (-) + (-7) = 70. PS - PQ =, 0 -, =, 6, ƒ PS - PQ ƒ = (-) + (-6) = 77., +, =, 7., -, =,., -, =, 8 6., = 6, 7., +, =, 9 8., -, = 0, 0 9., -, =, 8 0., -, =, 7.. u ƒ u ƒ = - i + j v ƒ v ƒ ƒ = h = h - (-) +, (-) + i + (-), - + (-) i = i - j 8. (a) (b) h, - i i + a - bj = i - j 9. v= 8 cos, 8 sin 6., v= cos, sin 8.0,.7. v= 7 cos 08, 7 sin 08.,.70. v= cos 6, sin 6.7,.9. u =, Å=cos a + =. b. u =, Å=cos a - (-) + = 6.7 b. u =, Å=60 -cos a + (-) = b u = (-) + (-) =, Å=60 -cos a b 7. Since (7 cos )i+(7 sin )j=( u cos Å)i +( u sin Å)j, u =7 and Å=. 8. Since ( cos 60 )i+( sin 60 )j=( u cos Å)i +( u sin Å)j, u = and Å=60... w - = h ƒ w ƒ (-) + (-), - (-) + (-) i = - i - j w = h ƒ w ƒ +, + i = i + j For #9 and 0, first find the unit vector in the direction of u. Then multipl b the magnitude of v, v. 9. v = ƒ v ƒ # u = h ƒ u ƒ + (-), - + (-) i = h, - i = 8, v = ƒ v ƒ # u ƒ u ƒ = h - (-) + 7, 7 (-) + 7 i For # 8, the unit vector in the direction of v= a, b is ƒ v ƒ a = h a + b, a = a + b i +. (a) h, i (b) i + j b a + b i b a + b j. 0.8, 0.8 =.9,.07. A bearing of corresponds to a direction angle of. v=0 cos, sin.99, A bearing of 70 corresponds to a direction angle of 80. v=60 cos ( 80 ), sin ( 80 ) 79.88,.0.. (a) A bearing of 0 corresponds to a direction angle of 0. v= cos 0, sin 0.6, 0.0. (b) The wind bearing of 0 corresponds to a direction angle of 0. The wind vector is w=0 cos 0, sin 0.7, 0.6. Actual velocit vector: v+w 6.87, 6.0. Actual speed: v+w L 6.8 mph. Copright 0 Pearson Education, Inc.

3 Chapter 6 Applications of Trigonometr Actual direction: =80 +tan a b L So 0t= Q t= 8t=0. mi. The boat meets 0 Q.6, so the bearing is about 7.8. the shore 0. mi downstream.. Let w be the speed of the ship. The ship s velocit (in still water) is w cos 70, w sin 70 = 0, w. Let z be the speed of the current. Then, the current velocit is z cos, z sin 0.7z, 0.7z. The position of the ship after two hours is 0 cos 0, 0 sin 0 0, 7.. Putting all this together we have: 0, w + 0.7z, 0.7z = 0, 7.,.z, w+.z = 0, 7., so z 7.07 and w.66. The speed of the ship is about.66 mph, and the speed of the current is about 7.07 mph.. (a) A bearing of 70 corresponds to a direction angle of 80 : v=60 cos ( 80 ), sin ( 80 ) 79.88,.0. (b) The wind bearing of 00 corresponds to a direction angle of 0. The wind vector is w=80 cos ( 0 ), sin ( 0 ) 7.6, 7.8. Actual velocit vector: v+w., 8.9. Actual speed: v+w L 0.79 mph. Actual direction: =80 +tan a b L 8., so the bearing is about 7... (a) v=0 cos 70, sin 70., 9.0 (b) The horizontal component is the (constant) horizontal speed of the basketball as it travels toward the basket. The vertical component is the vertical velocit of the basketball, affected b both the initial speed and the downward pull of gravit. 6. (a) v=. cos, sin., 0.6 (b) The horizontal component is the force moving the bo forward. The vertical component is the force moving the bo upward against the pull of gravit. 7. We need to choose w= a, b =k cos, sin,so that k cos ( - )=k cos 8 =.. (Redefine horizontal to mean the parallel to the inclined plane; then the towing vector makes an angle of 8 with the. horizontal. ) Then k=.6 lb, so that cos 8 w.0,.. 8. Juana s force can be represented b cos 8, sin 8.87, 7., while Diego s force is 7 cos ( ), sin ( ) 6.08, Their total force is therefore 7.9, 0., so Corporal must be pulling with an equal force in the opposite direction: 7.9, 0.. The magnitude of Corporal s force is about 7.9 lb. 9. F= 0 cos, 0 sin + 7 cos ( 0 ), 7 sin ( 0 ) 00.,., so F 00. lb and.. 0. F=00 cos 0, sin 0 +0 cos 60, sin cos ( 0), sin ( 0 ) 9.7, 66., so F.8 lb and.66.. Ship heading: cos 90, sin 90 = 0, Current heading: cos, sin.8,.8 The ship s actual velocit vector is.8, 9.7, so its speed is (-.8) mph and the direction angle is cos a , so the 9.6 b bearing is about.86.. Let v= 0, 0 be the velocit of the boat and w= 8, 0 be the velocit vector of the current. If the boat travels t minutes to reach the opposite shore, then its position, in vector form, must be 0, 0t + 8t, 0 = 8t, 0t = 8t,.. Let u= u, u, v= v, v, and w= w, w. (a) u+v= u, u + v, v = u +v, u +v = v +u, v +u = v, v + u, u =v+u (b) (u+v)+w= u +v, u +v + w, w = u +v +w, u +v +w = u, u + v +w, v +w =u+(v+w) (c) u+0= u, u + 0, 0 = u +0, u +0 = u, u =u (d) u+( u)= u, u + u, u = u +( u ), u +( u ) = 0, 0 =0 (e) a(u+v)=a u +v, u +v = a(u +v ), a(u +v ) = au +av, au +av = au, au + av, av =a u, u +a v, v =au+av (f) (a+b)u= (a+b)u, (a+b)u = au +bu, au +bu = au, au + bu, bu =a u, u +b u, u =au+bu (g) (ab)u= (ab)u, (ab)u = a(bu ), a(bu ) =a bu, bu =a(bu) (h) a0=a 0, 0 = a0, a0 = 0, 0 =0 0u=0 u, u = 0u,0u = 0, 0 =0 (i) ()u= ()u, ()u = u, u =u ( )u= ( )u, ( )u = u, u = u (j) au = au, au = (au ) + (au ) = a u + a u = a (u + u ) = ƒ a ƒ u + u = a u. True. Vectors u and u have the same length but opposite directions. Thus, the length of u is also. 6. False. /u is not a vector at all. 7. ƒ8,-9ƒ = + (-) = The answer is D. 8. u - v = 8-, 9-8,-9 = 8- -, - (-)9 = 8-6, 9 The answer is E. 9. The -component is cos 0, and the -component is sin 0. The answer is A. 60. ƒ v ƒ = (-) + = 0, so the unit vector is 8-, 9/0. The answer is C. Copright 0 Pearson Education, Inc.

4 Section 6. Dot Product of Vectors 6. (a) Let A be the point (a, a ) and B be the point (b, b ). Then OA is the vector a, a and OB is the vector b, b. So, BA = a -b, a -b = a, a - b, b = OA -OB (b) OA + OB =( OC+ CA )+( OC+ CB ) (from (a)) = OC+ CA + OC+ CB =(+) OC+ CA + CB = OC+ CA + CB (since +=) ƒbcƒ = + ƒ BC ƒ OC # + CB since a = b ƒ CA CA ƒ ƒcaƒ = OC+ a ƒ BC ƒ # CA ƒ CA ƒ + CB b CA is a unit vector, and BC points in the ƒ CA ƒ same direction. = + a ƒ BC ƒ # BC OC ƒ BC + CB b ƒ = OC+ ( BC + CB ) = OC 6. (a) B Eercise 6, OM = OA + OB, where +=. Since M is the midpoint, ƒbm ƒ= ƒ M Aƒ. We know from Eercise 6, however, that ƒbm ƒ So == As a result,. ƒ = = M. Aƒ = The proof for OM OA + OB. OM and OM are similar. (b) OM + OC = a OC OA + OB b + = OA + OB + OC. Use the same method for the other proofs. (c) (b) implies that OM = OM + OB + OC = OM + OA. Each of the three vectors lies along a different median (that is, if nonzero, the three vectors have three different directions). Hence the can onl be equal if all are equal to 0. Thus OM OM = -OA = -OC,, and ƒom ƒ ƒom ƒ ƒom ƒ OM so =. ƒ OB = = -OB, ƒoa = ƒ OC ƒ ƒ ƒ 6. Use the result of Eercise 6. First we show that if C is on the line segment AB, then there is a real number t so ƒ BC ƒ t that (Convince ourself that ƒ CA = ƒ - t. BC t= works.) Then OC= t OA + ( - t) OB. BC + CA A similar argument can be used in the cases where B is on the line segment AC or A is on the line segment BC. Suppose there is a real number t so that OC = t OA + ( - t) OB. We also know OC = OB + BC and OC = OA + AC. So we have t OA + ( - t)ob = OB + BC and t OA + ( - t)ob = OA + AC. Therefore, and (t - )(OA - OB t (OA - OB ) = BC ) = AC. Hence BC and AC have the same or opposite directions, so C must lie on the line L through the two points A and B A P Q M R The line segment OM is a median of ^ABO since M is a midpoint of AB. The line segment AQ is a median of ^ABO since diagonals of a parallelogram bisect each other. B the result of Eercise 6, since P is the intersection AP CR point of two medians, Similarl, This RQ = PQ =.. implies that AP=PR=RC, so the diagonal has been trisected. Section 6. Dot Product of Vectors Eploration. u= -, 0- = -, v= -, 0- = -,. u # v=( -)(-)+( )( )= + + = +=0 Therefore, =90.. Answers will var. C Quick Review 6.. u = + (-) =. u = (-) + (-) =. u = cos + sin =. u =cos 7 + sin 7 =. AB= -( ), -0 =, 6. AB= -, -0 =, 7. AB= -, =, - 8. AB= -( ), - -0 =, - # k, l k, 6l 9. u= u # v = = ƒvƒ + = h, 6 i # k-, l k-, 9l 0. u= u # v = = ƒvƒ (-) + =h -, 9 i N B 0 Copright 0 Pearson Education, Inc.

5 6 Chapter 6 Applications of Trigonometr Section 6. Eercises. 60+=7. 0+6=. -= = 6. +8= = 7. +0= 8. 0+= 9. u = u # u = + = 0. u = u # u = 6 + = 7. u = u # u = 6 =. u = u # u = 9 =. u # v=-=, u = =, v = + = 6, = cos - a - 6 b L.6. u # v= = 0, u = 90. u # v= 6+=9, u = + 9 =, u = 9 + =, = cos - 9 a # b L u # v= 0-=, u = + = 9, ƒvƒ = 6 + = 7, = cos - - a 9 # 7 b L u # v= 6-6, ƒuƒ = = 8, v = + = 6 =, = cos a =6 8 # b 8. u # v=0, =90 p p 9. u has direction angle and v has direction angle a which is equivalent to - p, so the angle between the b p vectors is or. - a - p b = p p p 0. u has direction angle and v has direction angle, 6 p so the angle between the vectors is or p = p. u # v= +0=, ƒuƒ = 6 + = 89, ƒvƒ = =, u = cos - a - 89 b L u # v=-7= 69, ƒuƒ = = 7, ƒvƒ = + 8 = 8, u = cos a 7 # 8 b L.0. u # v=, # h = a, - i b + (-) = - = 0 Since u# v=0, u and v are orthogonal.. u # v=, #, = -() + (-)(-) = - + = 0 Since u# v=0, u and v are orthogonal. For # 8, first find proj v u. Then use the fact that u# v=0 when u and v are orthogonal.. proj v u= a 8-8, 9 # 8-6, -9 b8-6, = a b8-, -9 = 8-6, = - 8, 9 0 u= - 7 8, , proj v u= a 8, -79 # 8-, -69 b8-, = a 6 0 b8-, -69 = - 9 8, 9 u= - 9 8, , proj v u= a 88, 9 # 8-9, -9 b8-9, = a -8 8 b8-9, -9 = 89, u= 89, , proj v u= a 8-, 89 # 89, -9 b89, = a - 90 b89, -9 = 7 8-, 9 7 u= 8-, 9 + 8, 9 9. B(, ) 0 A(, 0) C(, ) CA # CB = -, 0- -, - =, 9 # 7, =+6=0, ƒ CA ƒ = + 8 = 8, ƒ CB ƒ = = 6, 0 C=cos a 8 # 6 b L 7.7 BC # BA = 7, # -( ), 0- = 7, #, =-0=, # Copright 0 Pearson Education, Inc.

6 Section 6. Dot Product of Vectors 7 ƒ BC ƒ = 6, ƒba ƒ = 0, B=cos a 6 # 0 b 7.7 A=80 -B-C =7. 0. CA # CB = -, -( 6) -, -( 6) =, 7 #, = 0+=, ƒ CA ƒ = + 9 = 7, ƒ CB ƒ = 6 + =, C=cos a 7 # b L 7.0 BA # BC = -, -( ) # -, 6-( ) = 9, #, =6-0=6, B=cos 6 a # b A=80 -B-C =.9 For # and, use the relationship u v= u v cos... A(, ) C(, 6) B(, ) ƒba ƒ = 8 + = 8, ƒ BC ƒ =, u # v = # 8 cos 0 L u # v = # 0 cos a p b = 0 For # 8, vectors are orthogonal if u # v = 0 and are parallel if u=kv for some constant k.. Parallel: h - 0 = 8, 9, - i = h 0, i. Neither: u # v = 0 and Z 0 v = h 0, ih, i Z u. Neither: u # v = -0 Z 0 and - v = , 9h, i Z u 6. Orthogonal: u # v = = 0 7. Orthogonal: u # v = = 0 8. Parallel: - v = - 8-, 9 = 8, -79 = u For #9 (b), first find the direction(s) of AP and then find the unit vectors. Then find P b adding the coordinates of A to the components of a unit vector. 9. (a) A is (, 0) and B is (0, ). # # (b) The line is parallel to AB= 0-, -0 =,, so the direction of AP is u=, or v=,. u AP = = 8,-9 or ƒ u ƒ = h, - i v 8-, 9 AP = =. ƒ v ƒ = h -, i So, P is (.6, 0.8) or (., 0.8). 0. (a) A is (, 0) and B is (0, ). (b) The line is parallel to AB= 0-( ), -0 =,, so the direction of AP is u=, or v=,. AP u 8-, 9 = = or ƒ u ƒ + = h - 9, 9 i v 8, -9 AP = =. ƒ v ƒ 9 = h 9, - 9 i So P is a - - (.7, 0.9) or 9, 9 b a - + (.6, 0.9). 9, - 9 b. (a) A is (7, 0) and B is (0, ). (b) The line is parallel to AB= 0-7, -0 = 7,, so the direction of AP is u=, 7 or v=, 7. u 8, -79 AP = = or ƒ u ƒ = h 8, i v 8-, 79 AP = =. ƒ v ƒ 8 = h - 8, 7 8 i So P is a7 + ( 7.9, 0.9) or 8, b a7 - (6.6, 0.9). 8, 7 8 b. (a) A is (6, 0) and B is (0, ). (b) The line is parallel to AB= 0-6, -0 = 6,, so the direction of AP is u=, 6 or v=, 6. u 8, 69 8, 69 AP = = = or ƒ u ƒ = h, i v 8-, , -69 AP = = = ƒ v ƒ = h -. So P is a6 +,, - i b (6., 0.89) or a6 - (., 0.89)., - b. v +v =0, v +v =7. Since v =- v, +v 9 a - =7, -v + v +v =7, v b v -v +8=0, v -60v +=0, 8 (v -)(v -8)=0, so v = or v =. Copright 0 Pearson Education, Inc.

7 8 Chapter 6 Applications of Trigonometr Therefore, v L 8-, 9 or v = h., 8 In this case F=60 cos, sin.8,.6. i L 8.07, 0.69 Since we onl want the force in the -direction, we immediatel find our answer of about.8 pounds.. v +v =, v +v =0. Since v = v +, 9. Since the car weighs 600 pounds, the force needed to lift a the car is 0, v =0, (v + )b W= F # AB = 0, 600 # 0,. =,00 foot-pounds v + 0v + 0. Since the potatoes weigh 00 pounds, the force needed to +v =0, lift the potatoes is 0, 00. 9, 9v +0v +8=0, v + 0 v + 8 W= F # AB = 0, 00 # 0, =00 foot-pounds = 0 8,9. F= # =, 8 ƒ8, 9ƒ (v +)(9v +8)=0, so v = or v =. 8 9 W= F # AB =, #, 0 = Therefore, or v = h - 9, i v = 8, -9.7 foot-pounds 8, 9.8,.79.. F= # =,, ƒ8, 9ƒ +. v=(cos 60 )i+(sin 60 )j= i+ j 0 F =proj v F=(F v)v= a -60 j # a v i + W= F # AB =, 0 = # jbb 80 W= F # AB =, #, 0 = = -80 a. i + jb = -0 i - 0 j 7.96 foot-pounds 8, 9 The magnitude of the force is. F= 0 # 0 =, =, ƒf ƒ = (-0) + (-0) ƒ8, 9ƒ = 9, pounds. Since we want to move feet along the line =,we 6. In this case, F= j and v remains the same as in solve for and b using the Pthagorean theorem: Eample 6. F =proj v F=(F# v)v= a v=6. (i+j). b + =, + a =9, =9, b The magnitude of the force is F = =, = pounds. 7. (a) v=(cos )i+(sin )j 6 AB = h F= 000j, i F =proj v F=(F# v)v =( 0, 000 # cos, sin ) cos, sin W = F #, 9 # h 6 =( 000 sin ) cos, sin., AB = 8 i Since cos, sin is a unit vector,the = = foot-pounds magnitude of the force being etended is A F =000 sin.8 pounds. 8, 9 0. F= 0 # 0 =, =, (b) We are looking for the gravitational force eerted ƒ8, 9ƒ + perpendicular to the street. A unit vector perpendicular to the street is w= cos ( 78 ), sin ( 78 ), =, we solve for and b using the Pthagorean Since we want to move the object feet along the line so F =proj w F=(F# w)w theorem: + =, + =, =, =( 000 sin ( 78 )) cos ( 78 ), sin ( 78 ) =., =.. Since cos ( 78 ), sin ( 78 ) is a unit vector, the magnitude of the force perpendicular to the street AB = 8.,.9. is 000 sin ( 78 ) 96.0 pounds. W=F # 0 AB =, # 8.,.9 = 6 8. We want to determine how much of the 60 pound A force is projected along the inclined plane.. foot-pounds F=60 cos, sin.88, 0.9 and v= cos 8, sin 8 0.9, 0.. W=F # AB = ƒfƒ ƒabƒ cos =00 cos 0 (8.88, 0.99 # 80.9, 0.9) 80.9, 0.9 proj v F= =00 # = 009 L 6. foot-pounds () , AB=, -, =,.7, The magnitude of W=F # AB = ƒfƒ ƒabƒ cos =79 cos 60 this force is ƒf ƒ = (.7) + (6.80) =79 # foot-pounds.8 pounds. Of note, it is also possible to evaluate = 79 L 0.9 this problem considering the -ais parallel to the inclined plane and the -ais perpendicular to the plane. Copright 0 Pearson Education, Inc.

8 7. (a) Let u= u, u, v= v, v, and w= w, w. 0 # u= 0, 0 # u, u =0 # u +0 # u =0 (b) u # (v+w)= u, u # ( v, v + w, w ) = u, u # v +w, v +w =u (v +w )+u (v +w ) =u v +u v +u w +u w = u, u # v, v + u, u # w, w =u# v+u# w (c) (u+v) # w=( u, u + v, v ) # w, w = u +v, u +v # w, w =(u +v ) w +(u +v )w =u w +u w +v w +v w = u, u # w, w + v, v # w, w =u# w+v# w (d) (cu) # v=(c u, u ) # v, v = cu, cu # v, v =cu v +cu v = u, u # cv, cv =u # (cv) =c(u v +u v )=c( u, u # v, v )=c(u# v) 8. (a) When we evaluate the projection of u onto v we are actuall tring to determine how much of u is going in the direction of v. Using Figure 6.9, imagine that v runs along our -ais, with the -ais perpendicular to it. Written in component form, u=( u cos, u sin ) and we see that the projection of u onto v is eactl u cos times v s unit vector v v. Thus, proj v u= u cos # ƒ v ƒ ƒ v ƒ = ƒ u ƒ a u # v b v = (u # v) v = a u # v ƒ u ƒƒvƒ ƒ v ƒ ƒ v ƒ ƒ v ƒ ƒ v ƒ b v (b) Recall Figure 6.9 and let w be PR=proj v u and w = RQ =u-proj v u. Then, (u-proj v u) # (proj v u)=w # w. Since w and w are perpendicular, w # w =0. 9. v u u+v u-v As the diagram indicates, the long diagonal of the parallelogram can be epressed as the vector u+v, while the short diagonal can be epressed as the vector u-v.the sum of the squares of the diagonals is u+v + u-v =(u+v) # (u+v) +(u-v) # (u-v)=u# u+u# v+v# u+v# v +u# u-u# v-u# v+v# v= u + v +u# v-u# v= u + v, which is the sum of the squares of the sides. 60. Let u= u, u. (u# i)i+(u# j)j =( u, u #, 0 )i+( u, u # 0, )j =(u )i+(u )j =u i+u j = u, u =u 6. False. If either u or v is the zero vector, then u# v=0 and so u and v are orthogonal, but the do not count as perpendicular. 6. True. u# u= ƒ u ƒ = () =. # Section 6. Dot Product of Vectors 9 6. u v=0, so the vectors are perpendicular. The answer is D. 6. u # v=, #, =( )+( )( ) = 8+ =7 The answer is C. 6. proj v u= a u # v ƒ v ƒ bv = a + 0, 0 b = a, 0 b = h, 0 i The answer is A. 66. The unit vector in the direction of u is,. The force is represented b times the unit vector. The answer is B. 67. (a) # 0+ # =0 and # + # 0=0 (b) AP= -0, 7- =, AB= -0, 0- =, w =proj AB = a 8, 9 # 8, -79 AB, + (-) b - 0 = a b, =, 9 9 w = AP-proj AB AP =, -, 9 = h - = 86, i 9 (c) w is a vector from a point on AB to point P. Since w is perpendicular to AB, w is the shortest distance from AB to P. w = B a 6 9 b + a 9 b 7,869 = B 9 = 9 9 (d) Consider Figure 6.9. To find the distance from a point P to a line L, we must first find u =proj v u.in this case, proj AB = a 8 0, 0-9 # 8, -9 AP b, A + (-) ) = a b, 9 = h , i and 9 9 AP-proj AB AP= 0, h , i 9 9 = ( 9 0, 9 ( 0 -) , ) = , Copright 0 Pearson Education, Inc.

9 0 Chapter 6 Applications of Trigonometr So, the distance is the magnitude of this vector. d= (e) In the general case, AP = ƒap - Section 6. Parametric Equations and Motion 9 ( ) + ( ) Eploration. = ( ) + ( ) 9 = 9( ) 9 = ƒ( )ƒ 9 = h 0, 0 - c, so proj AB APs b i a 0c a - (bc 0 - c ) b bh c a, - c b i proj AB c ` a ` + ` - c b ` AP ƒ AB = h c a, -c b i = h 0, 0 - c b i and -ab = N b 0 - ab 0 + ac 0 + a 0 - a c b a + b, a + b O -ab - N b 0 - ab 0 + ac 0 + a 0 - a c b a + b, a + b O = h a 0 + ab 0 - ac a + b, ab 0 + b 0 - bc a + b i The magnitude of this vector, ƒap - proj AB AP ƒ,is ƒa 0 + b 0 - cƒ the distance from point P to L:. a + b 68. (a) Yes, if v= 0, 0 or t=n, n=an integer. np (b) Yes, if u= 0, 0 or t=, n=odd integer. (c) Generall, no, because sin t Z cos t for most t. p Eceptions, however, would occur when t=, + np n=an integer, or if u= 0, 0 and/or v= 0, One possible answer: If au+bv=cu+dv au-cu+bv-dv=0 (a-c)u+(b-d)v=0 Since u and v are not parallel, the onl wa for this equalit to hold true for all vectors u and v is if (a-c)=0 and (b-d)=0, which indicates that a=c and b=d.. 0.(7)+.=0, so the point (7, 0) is on the graph, t= ( )+.= 0, so the point (, 0) is on the graph, t=.. =a=-t, t=-a, t = Alternativel, - a. b=-t, so t=-b.. Choose Tmin and Tma so that Tmin - and Tma Ú.. Eploration. It looks like the line in Figure 6... The graph is a vertical line segment that etends from (00, 0) to (00, 0).. For 9 and 0, the ball does not clear the fence, as shown below. 9 : 0 : [ 0, ] b [, ] [0, 0] b [0, 80] [0, 0] b [0, 80] For and, the ball clears the fence, as shown below. : [0, 0] b [0, 80] [0, 0] b [0, 80] Copright 0 Pearson Education, Inc.

10 Section 6. Parametric Equations and Motion Quick Review 6.. (a) OA=, (b) OB =, 6 (c) AB= -( ), 6-( ) = 7, 8. (a) OA=, (b) OB =, (c) AB= -( ), - =, 6. m = 6 - (-) - (-) = = 7 ( + ) or - 6 = 8 ( - ) 7. m = (-) = -6 -=. Graph = ; ( + ) or + = -6 ( - ) (b) a0, b (, ) p p 6. (a) t (b) (, ) a, b 8 (0, ) (, 0) (, 0) (0, ) [, 7] b [ 7, 7] 6. Graph = ; = 8. (+) +(-) = [ 7, ] b [ 7, 7] 600 rotations min 700 rotations min # min 60 sec # # min 60 sec # Section 6. Eercises. (b) [, ] b [, ]. (d) [, ] b [, ]. (a) [, ] b [, ]. (c) [ 0, 0] b [, 0] p rad = 0p rad/sec rotation p rad rotation = 70 p rad/sec. (a) t 0 0 p - p p undef. p p [ 0, 0] b [ 0, 0] [ 0, 0] b [ 0, 0] [ 0, 0] b [ 0, 0] [ 0, 0] b [ 0, 0]. =+, so =-: line through (0, ) and (, 0) Copright 0 Pearson Education, Inc.

11 Chapter 6 Applications of Trigonometr. t=-, so =-(-); = - : line + 7 through a0, 7 b and (7, 0). t= ; +, so = 9 - a + b = +, 7: line segment with endpoints (, ) and (7, ). t=-, so =-(-); = - : line segment with +, - 8 endpoints (8, ) and (, ). =(-) : parabola that opens to right with verte at (0, ) 6. = -: parabola that opens upward with verte at (0, ) 7. = -+: cubic polnomial 8. = -; parabola that opens to right with verte at (0, ) 9. =- ; parabola that opens to left with verte at (, 0) 0. t=, so =6 -: cubic, -. t=+,so =, on domain: -8, Z - +. t=-, so = -, Ú. + =, circle of radius centered at (0, 0). + =6, circle of radius centered at (0, 0). + =, three-fourths of a circle of radius centered at (0, 0) (not in Quadrant II) 6. + =9, semicircle of radius, Ú 0 onl 7. OA =,, OB,, OP =, OP - OA =ta OB - OA B +, - =t 6, +=6t Q =6t- -=t Q = t+ 8. OA,, OB =,, OP =, OP - OA =ta OB - OA B +, + =t 8, +=8t Q =8t- +=t Q =t- For #9, man answers are possible; one or two of the simplest are given. 9. Two possibilities are =t+, = - 7 t, 0 t, or = t +, = - 7t, 0 t Two possibilities are =-t, =- 7 t, 0 t 7, or =-7t, =-6t, 0 t.. One possibilit is =+ cos t, =+ sin t, 0 t p.. One possibilit is = + cos t, = + sin t, 0 t p.. In Quadrant I, we need >0 and >0, so - t >0 and t-0.>0. Then <t< and t>0., so 0.<t<. This is not changed b the additional requirement that - t.. In Quadrant II, we need <0 and >0, so - t <0 and t-0.>0. Then (t< or t>) and t>0., so t>. With the additional requirement that - t, this becomes 6 t.. In Quadrant III, we need <0 and <0, so - t <0 and t-0.<0. Then (t< or t>) and t<0., so t<. With the additional requirement that - t, this becomes - t In Quadrant IV, we need >0 and <0, so - t >0 and t-0.<0. Then <t< and t<0., so <t<0.. This is not changed b the additional requirement that t. 7. (a) One good window is [ 0, 00] b [, 0]. If our grapher allows, use Simultaneous rather than Sequential plotting. Note that 00 d is 00 ft. To show the whole race, use 0 t (upper limit ma var), since Ben finishes in.96 sec. Note that it is the process of graphing (during which one observes Ben passing Jerr and crossing the finish line first), not the final product (which is two horizontal lines) which is needed; for that reason, no graph is shown here. (b) After seconds, Jerr is at 0()=60 ft and Ben is at ()-0=6 ft. Ben is ahead b ft. 8. (a) If our grapher allows, use Simultaneous rather than Sequential plotting. To see the whole race, use 0 t. (upper limit ma var), since the faster runner reaches the flag after. sec. Note that it is the process of graphing, not the final product (which shows a horizontal line) which is needed; for that reason, no graph is shown here. (b) The faster runner (who is coming from the left in the simulation) arrives at t=. sec. At this instant, the slower runner is. ft awa from the flag; the slower runner doesn t reach the flag until t =. sec. This can be observed from the simulation, or b solving algebraicall =0 and =0. 9. (a) = 6t +v 0 t+s 0 = 6t +0t+000 = 6t +000 (b) Graph and trace: = and = 6t +000 with 0 t 6, on the window [0, ] b [0, 00]. Use something like 0. or less for Tstep. This graph will appear as a vertical line from (, ) to (, 000); it is not shown here because the simulation is accomplished b the tracing, not b the picture. (c) When t=, = 6() +000=7 ft; the food containers are 7 ft above the ground after sec. 0. (a) = 6t +v 0 t+s 0 = 6t +80t+ (b) Graph and trace: =6 and = 6t +80t+ with 0 t. (upper limit ma var) on [0, 7] b [0, 0]. This graph will appear as a vertical line from about (6, 0) to about (6, 0). Tracing shows how the ball begins at a height of ft, rises to over 00 ft, then falls back to the ground. Copright 0 Pearson Education, Inc.

12 Section 6. Parametric Equations and Motion (c) Graph =t and = 6t +80t+ with 0 t. (upper limit ma var). [0, 7] b [0, 0] (d) When t=, = 6() +80()+=69 ft. The ball is 69 ft above the ground after sec. (e) From the graph in (b), when t=. sec, the ball is at its maimum height of 0 ft.. Possible answers: (a) 0 6 t 6 p (t in radians) (b) 0 6 t 6 p p (c) 6 t 6 p. (a) Both pairs of equations can be changed to + =9 a circle centered at the origin with radius. Also, when one chooses a point on this circle and swaps the - and -coordinates, one obtains another point on the same circle. (b) The first begins at the right side (when t=0) and traces the circle counterclockwise. The second begins at the top (when t=0) and traces the circle clockwise.. (a) =00 when t L.80 about.80 sec. (b) When t L.80 sec, L 7.8 ft. (c) Reaching up, the outfielder s glove should be at or near the height of the ball as it approaches the wall. If hit at an angle of 0, the ball would strike the wall about 9.7 ft up (after.8 sec) the outfielder could not catch this.. (a) No: =(0 cos 0 )t; this equals 0 when t L.7. At this time, the ball is at a height of = 6t +(0 sin 0 ) t + L.9 ft. (b) The ball hits the wall about.9 ft up when t L.7 (see (a)) not catchable.. (a) Yes: =(+0 cos 0 )t; this equals 0 when t L.. At this time, the ball is at a height of = 6t +(0 sin 0 )t+ L.9 ft. (b) The ball clears the wall with about.9 ft to spare (when t L.). 6. For Linda s ball, =( cos )t and = 6t +( sin )t+. For Chris s ball, =78-( cos 9 )t and = 6t +( sin 9 )t+. Find (graphicall) the minimum of d(t)= ( - ) + ( - ). It occurs when t L. sec; the minimum distance is about 6.60 ft. 7. No: =(0 cos 70 )t and = 6t +(0 sin 70 )t +. The dart lands when =0, which happens when t L.86 sec. At this point, the dart is about 9. ft from Ton, just over 0 in. short of the target. 8. Yes: =( cos )t and = 6t +( sin )t +. The dart lands when =0, which happens when t L. sec. At this point, the dart is about 0.8 ft from Sue, inside the target. 9. The parametric equations for this motion are =(v+60 cos 0 ) t and = 6t +(60 sin 0 )t+, where v is the velocit of the wind (in ft/sec) it should be positive if the wind is in the direction of the hit, and negative if the wind is against the ball. To solve this algebraicall, eliminate the parameter t as follows: t =. So = -6a v + 60 cos 0 v + 60 cos 0 b + 60 sin 0 a. v + 60 cos 0 b + Substitute =00 and =0: 00 0 = -6a v + 60 cos 0 b sin 0 a. v + 60 cos 0 b + 00 Let u =, so the equation becomes v + 60 cos 0-6u +.7u - 6 = 0. Using the quadratic formula, we find that u = -.7 ;.7 - (-6)(-6) L ,.8. Solving 0.7 = and v + 60 cos =, v.0, v A wind v + 60 cos 0 speed of ft/sec (7.7 mph) is unrealistic, so we eliminate that solution. So the wind will be blowing against the ball in order for the ball to hit within a few inches of the top of the wall. To verif this graphicall, graph the equation 00 0 = -6a v + 60 cos 0 b sin 0 a, and find v + 60 cos 0 b + the zero. [, ] b [, 0] 0. Assuming the course is level, the ball hits the ground when = 6t +(80 sin )t equals 0, which 80 sin u happens when t = =. sin sec. At that 6 time, the ball has traveled =(80 cos )t =0(cos )(sin ) feet. The answers are therefore approimatel: (a) 06. ft. (b) 60.8 ft. (c) 77.6 ft. (d) ft. Copright 0 Pearson Education, Inc.

13 Chapter 6 Applications of Trigonometr. = cosa p 6 tb and = 0 + sina p 6 tb. t = = + +, = + a + b Since = (-) + and = = 6 = () + both (, ) and (, 6) are = 0 = 6, on the line.. (a) When t= (or, or, etc.), =. This corresponds to the highest points on the graph. (b) The -intercepts occur where =0, which happens when t=0,,, etc. The -coordinates at those times are (respectivel) 0,,, etc., so these are units apart.. (a) [.7,.7] b [.,.] (b) All s should be changed to s.. The particle begins at 0, moves right to ±. (at t=.), then changes direction and ends at. 6. The particle begins at, moves right to ± (at time t=), then changes direction and returns to. 7. The particle begins at, moves right to about (at time t 0.), changes direction and moves left to about 0.8 (at time t.), then changes direction and ends at The particle begins at 0, moves right to about (at t 0.6), changes direction and moves left to about 6.06 (at time t.9), then changes direction and ends at True. Eliminate t from the first set: t = + = ( + ) + = + Eliminate t from the second set: t = + = a + b = + Both sets correspond to the rectangular equation = True. =0 and = when t=, and = and = when t=. Eliminating t, t=+. =(+)-. =+, 0,. 6. = (-) - = -, = = - The answer is A. 6. The parametrization describes a circle of radius, centered at the origin and t represents the angle traveled counterclockwise from (, 0). The answer is A. 6. Set -6t + 80t + 7 equal to 9 and solve either graphicall or using the quadratic formula. The answer is D. 6. The equations are both linear, so the answer is either A, B, or C. Since t has a minimum value and no maimum value, the answer is C. 6. (a) (b) + =(a cos t) +(a sin t) =a cos t+a sin t =a The radius of the circles are a={,,, }, centered at (0, 0). (c) (d) -h=a cos t and -k=a sin t, so (-h) +(-k) =(a cos t) +(a sin t) =a cos t+a sin t =a The graph is the circle of radius a centered at (h, k). (e) If (+) +(-) =9, then a=, h=, and k=. As a result, = cos t- and = sin t (a) (b) [ 6, 6] b [, ] [ 6, 6] b [, ] [, 8] b [, 6] [, 8] b [, 6] (c) t = a - b a, = ca a - b a b + d = c ad - bc +, a Z 0. a a Copright 0 Pearson Education, Inc.

14 Section 6. Polar Coordinates c (d) Slope:, if a Z 0; 68. a -bc + ad -intercept: a0, b, if a Z 0; a bc - ad -intercept: a, 0b, if c Z 0. c (e) The line will be horizontal if c=0. The line will be vertical if a= (a) Jane is traveling in a circle of radius 0 feet and center (0, 0), which ields =0 cos (nt) and =0+0 sin (nt). Since the ferris wheel is making one revolution ( ) ever seconds, =n, so Thus, =0 cos a p and =0+0 sin a p in 6 tb 6 tb radian mode. (b) Since the ball was released at 7 ft in the positive -direction and gravit acts in the negative -direction at 6 ft/s, we have =at+7 and = 6t +bt, where a is the initial speed of the ball in the -direction and b is the initial speed of the ball in the -direction. The initial velocit vector of the ball is 60 cos 0, sin 0 = 0, 09, so a= 0 and b=0. As a result = 0t+7 and = 6t +(0 )t are the parametric equations for the ball. (c) [ 0, 00] b [ 0, 0] Our graph shows that Jane and the ball will be close to each other but not at the eact same point at t=. seconds. (d) d(t) = ( - ) + ( - ) = B (0 cos a p 6 t) + 0t - 7b (e) n = p = p 6. + a0 + 0 sin a p 6 tb + 6t - (0)tb Assuming that the bottom of the ferris wheel and the ball s initial position are at the same height, the position of Matthew is =7 cos a p = sin a p 0 tb and 0 tb. The ball s position is =90+(88 cos 00 )t and = 6t +(88 sin 00 )t. Find (graphicall) the minimum of d(t) = ( - ) + ( - ). It occurs when t L.9 sec; the minimum distance is about.7 ft. 69. Chang s position: =0 cos a p and =0 6 tb +0 sin a p Kuan s position: =+ 6 tb. cos a p and =+ sin a p. tb tb Find (graphicall) the minimum of d(t)= ( - ) + ( - ). It occurs when t.0 sec; the minimum distance is about. ft. 70. Chang s position: =0 cos a p and =0 6 tb +0 sin a p. Kuan s position: =+ sin a p 6 tb tb and =- cos a p Find (graphicall) the mini mum tb of d(t)= ( - ) + ( - ). It occurs when t. sec; the minimum distance is about 0.8 ft. 7. (a) (0)=0c+(-0)a=a and (0)=0d+(-0)b=b (b) ()=c+(-)a=c and ()=d+(-)b=d 7. (0.)=0.c+(-0.)a=0.(a+c)=(a+c)/, while (0.)=(b+d)/ the correct coordinates for the midpoint. 7. Since the relationship between and is linear and one unit of time (t=) separates the two points, t=, will divide the segment into three equal pieces. Similarl, t=,, will divide the segment into four equal pieces. Section 6. Polar Coordinates Eploration. a, p = (, ) b a -, p =(0, ) b [0, ] b [, ] The minimum distance occurs at t=., when d(t)=.6 feet. (, )=(, 0) a -, p =(0, ) b (, )=(, 0) Copright 0 Pearson Education, Inc.

15 6 Chapter 6 Applications of Trigonometr. (-, - ) = a -, p b (b) (0, )= a, p b (, 0)=(, 0) (, 0)=(, ) 6,, (0, )= a, p b Quick Review 6.. (a) Quadrant II (b) Quadrant III. (a) Quadrant I (b) Quadrant III. Possible answers: 7 /, 9 /. Possible answers: 7 /, /. Possible answers: 0, Possible answers: 0, (-) + = 8. +(+) =9 9. a = +0 -()(0)cos a. 0. a =9 +6 -(9)(6) cos 0 a.8 Section 6. Eercises. a -, b. (, ). (-, - ). a -, b p p p p. (a) 6 - r 0 0 (b), 6, (0, ) (0, ), π a, 7π b 0 O O a, π b a, π b π 6 6 (, 0 ) O 7π 0 0 (, 0 ) O O π a, b π (, 0 ) 0 O O 0 O (, ) p p p p 6. (a) 6 - r undefined undefined. a, b 6. a, b 7. (.70,.0) 8. (.6,.8) 9. (, 0) 0. (0, ) Copright 0 Pearson Education, Inc.

16 Section 6. Polar Coordinates 7. (0, ). (, 0). a, p and a -, p + (n + )pb, 6 + npb 6 n an integer. a, - p and a -, - p + (n + )pb, + npb n an integer. (., n ) and (-., n ), n an integer 6. (-., n ) and (., n ), n an integer 7. (a) a, p or a -, p b b (b) a, p or b p a -, - b (c) The answers from (a), and also a, 9p or b a -, p b 8. (a) (0, tan - ) L (0,.) or (- 0, tan - + p) L (- 0,.9) (b) (0, tan - ) L (0,.) or (- 0, tan - - p) L (- 0, -.89) (c) The answers from (a), and also (0, tan - + p) L (0, 7.) or (- 0, tan - + p) L (- 0, 0.67) 9. (a) (9, tan - (-.) + p) L (9,.9) or (- 9, tan - (-.) + p) L (- 9,.09) (b) (- 9, tan - (-.)) L (- 9, -.9) or (9, tan - (-.) + p) L (9,.9) (c) The answers from (a), plus (9, tan - (-.) + p) L (9, 8.) or (- 9, tan - (-.) + p) L (- 9,.8) 0. (a) (-, tan - ) L (-,.) or (, tan - + p) L (,.) (b) (-, tan - ) L (-,.) or (, tan - - p) L (, -.0) (c) The answers from (a), plus (-, tan - + p) L (-, 7.9) or (, tan - + p) L (, 0.). (b). (d). (c). (a). = a vertical line 6. = a horizontal line 7. r +r sin =0, or + +=0. Completing the square gives + a + a circle centered at b = 9 a0, - with radius b. 8. r +r cos =0, or + +=0. Completing the square gives (+) + = a circle centered at (, 0) with radius 9. r -r sin =0, or + -=0. Completing the square gives + a - a circle centered b = at a0, with radius. b 0. r -r cos =0, or + -=0. Completing the square gives a - a circle centered at b + = 9 a with radius, 0b.. r -r sin +r cos =0, or + -+=0. Completing the square gives (+) +(-) = a circle centered at (, ) with radius.. r -r cos +r sin =0, or + -+=0. Completing the square gives (-) +(+) =8 a circle centered at (, ) with radius.. r=/cos = sec a vertical line [, ] b [, ]. r=/cos = sec [0, 0] b [, ]. r= cos u - sin u [, ] b [, ] Copright 0 Pearson Education, Inc.

17 8 Chapter 6 Applications of Trigonometr 6. r= cos u + sin u [, ] b [, ] 7. r -6r cos =0, so r=6 cos [, 9] b [, ] 8. r -r sin =0, so r= sin [, ] b [, ] 9. r +6r cos +6r sin =0, so r= 6 cos -6 sin [, 6] b [ 9, ] 0. r -r cos +8r sin =0, so r= cos -8 sin [ 8, 0] b [ 0, ]. d= d= - 0 cos 0.6 mi. d= + - # # cos (70-0 ) + - # # cos ( - 7 ) = 0-6 cos 60 = = L. mi. Using the Pthagorean theorem, the center-to-verte a distance is The four vertices are then a a, p. b, a a a a and a a Other polar, 7p, p, p b, b, b. coordinates for these points are possible, of course.. The verte on the -ais has polar coordinates (a, 0). All other vertices must also be a units from the origin; their coordinates are aa, p b, aa, p b, aa, 6p b, and aa, 8p Other polar coordinates for these points b. are possible, of course.. False. Point (r, ) is the same as point (r, +n ) for an integer n. So each point has an infinite number of distinct polar coordinates. 6. True. For ( r, ) and ( r, + ) to represent the same point, ( r, + ) has to be the reflection across the origin of ( r, + ), and this is accomplished b setting r = -r. 7. For point (r, ), changing the sign on r and adding to constitutes a twofold reflection across the origin. The answer is C. 8. The rectangular coordinates are ( cos( /), sin( /))=(, ). The answer is C. 9. For point (r, ), changing the sign on r and subtracting 80 from constitutes a twofold reflection across the origin. The answer is A. 60. (, ) lies in Quadrant III, whereas (-, ) lies in Quadrant IV. The answer is E. 6. (a) If - is an odd integer multiple of, then the distance is ƒr + r ƒ. If - is an even integer multiple of, then the distance is ƒr - r ƒ. (b) Consider the triangle formed b O, P, and P (ensuring that the angle at the origin is less than 80 ), then b the law of cosines, P = OP P + OP - # OP # OP cos u, where is the angle between OP and OP. In polar coordinates, this formula translates ver nicel into d = r + r - r r cos (u - u ) (or cos ( - ) since cos ( - )=cos ( - )), so d = r + r - r r cos (u - u ). (c) Yes. If - is an odd integer multiple of, then cos ( - )= Q d = r + r + r r = ƒr + r ƒ. If - is an even integer multiple of, then cos (u - u ) = Q d = r + r - r r = ƒr - r ƒ. 6. (a) The right half of a circle centered at (0, ) of radius (b) Three quarters of the same circle, starting at (0, 0) and moving counterclockwise (c) The full circle (plus another half circle found through the TRACE function) (d) counterclockwise rotations of the same circle 6. d = + - ()() cos 0 L d = ()(6) cos L. 6. d = (-) + (-) - (-)(-) cos d = (6)(8) cos Since =r cos and =r sin, the parametric equation would be =f( ) cos ( ) and =f( ) sin ( ). Copright 0 Pearson Education, Inc.

18 Section 6. Graphs of Polar Equations = cos =(cos )(sin ) 69. =(cos )(sin ) = sin 70. =(cos )(sec )= =(sin )(sec )= tan 7. =(cos )(csc )= cot =(sin )(csc ) = Section 6. Graphs of Polar Equations Eploration Answers will var. Eploration. If r = cos( ), then r does not eist when cos( )<0. Since cos ( )<0 whenever is in the interval a p, n is an integer, the + np, p + npb domain of r does not include these intervals.. -rcos (u) draws the same graph, but in the opposite direction.. (r) - cos ( )=r - cos ( ) (since cos ( )=cos ( )). ( r) - cos ( )=r - cos ( ). ( r) - cos ( )=r - cos ( ) (b) a0,,,, 7 b 6 a, b (, 0) (, ) a, b. (a) 0 /6 / / / /6 r (b) a, b 6 6 b a0, 0,,, b 6 a, b. k= Quick Review 6. For #, use our grapher s TRACE function to solve.. Minimum: at = e p ; Maimum: at, p f = {0, p, p}. Minimum: at = ; Maimum: at = {0, p}. Minimum: 0 at = e p ; Maimum: at, p, p, 7p f = {0, p, p} p p. Minimum: 0 at = ; Maimum: 6 at =. (a) No (b) No (c) Yes 6. (a) No (b) Yes (c) No 7. sin ( - )=sin 8. cos ( - )= cos 9. cos ( + )=cos ( + )=cos =cos -sin 0. sin ( + )=sin ( + )=sin = sin cos Section 6. Eercises. (a) 0 / / / / / 7 / r [, ] b [, ]. k= [, ] b [, ]. k= [, ] b [, ] 6. k= [, ] b [, ] Copright 0 Pearson Education, Inc.

19 0 Chapter 6 Applications of Trigonometr 7. r is not shown (this is a -petal rose). r is not shown (this is a 6-petal rose), r is graph (b) cos sin =( cos u sin u) where u= ; this equals sin u= sin. r= sin is the equation for the 8-petal rose shown in graph (a). p 9. Graph (b) is r=- cos : Taking =0 and =, we get r= and r= from the first equation, and r=0 and r= from the second. No graph matches the first of these (r, ) pairs, but (b) matches the latter (and an others one might choose). 0. Graph (c) is r=+ cos : Taking =0, we get r= from the other equation, which matches nothing. An (r, ) pair from the first equation matches (c), however. p. Graph (a) is r=- sin where =, + cos =, but a, p is clearl not b p on graph (a); meanwhile - sin =0, and a0, p b (the origin) is part of graph (a). p. Graph (d) is r=-. sin where =, +. cos =, but a, p is clearl not b p on graph (d); meanwhile -. sin =0., and a0., p is part of graph (d). b. Smmetric about the -ais: Replacing (r, ) with (r, - ) gives the same equation, since sin( - )=sin.. Smmetric about the -ais: Replacing (r, ) with (r, ) gives the same equation, since cos( )=cos.. Smmetric about the -ais: Replacing (r, ) with (r, ) gives the same equation, since cos( )=cos. 6. Smmetric about the -ais: Replacing (r, ) with (r, - ) gives the same equation, since sin( - ) = sin. 7. All three smmetries. Polar ais: Replacing (r, ) with (r, ) gives the same equation, since cos ( )=cos. -ais: replacing (r, ) with (r, - ) gives the same equation, since cos [( - )]=cos ( - ) =cos ( )=cos. Pole: Replacing (r, ) with (r, + ) gives the same equation, since cos [( + )] =cos ( + )=cos. 8. Smmetric about the -ais: Replacing (r, ) with ( r, ) gives the same equation, since sin ( )= sin. 9. Smmetric about the -ais: Replacing (r, ) with (r, - ) gives the same equation, since sin ( - ) = sin. 0. Smmetric about the -ais: Replacing (r, ) with (r, ) gives the same equation, since cos( )=cos.. Maimum r is when =n for an integer n. p. Maimum r is (when r= ) when = +n for an integer n.. Maimum r is (when r= ; ) when =n / for an integer n.. Maimum r is (when r= ; ) when =n / for an odd integer n.. Domain: (- q, q) Range: r = Smmetric about the -ais, -ais, and origin Maimum r value: [ 6, 6] b [, ] 6. Domain: (- q, q) Range: r = Smmetric about the -ais, -ais, and origin Maimum r value: [.,.] b [, ] 7. Domain: = / Range: ( q, q) Smmetric about the origin Unbounded Maimum r value: none [.7,.7] b [.,.] 8. Domain: = / Range: ( q, q) Smmetric about the origin Unbounded Maimum r value: none [.7,.7] b [.,.] Copright 0 Pearson Education, Inc.

20 9. Domain: (- q, q). Domain: (- q, q) Range: [, ] Range: [0, 8] Smmetric about the -ais Smmetric about the -ais Maimum r value: Maimum r value: 8 Section 6. Graphs of Polar Equations [, ] b [, ] 0. Domain: (- q, q) Range: [, ] Smmetric about the -ais, -ais, and origin Maimum r value: [ 6, ] b [ 6, 6]. Domain: (- q, q) Range: [0, 0] Smmetric about the -ais Maimum r value: 0 [.7,.7] b [.,.]. Domain: (- q, q) Range: [, 9] Smmetric about the -ais Maimum r value: 9 [ 9, 9] b [ 0.,.]. Domain: (- q, q) Range: [, 7] Smmetric about the -ais Maimum r value: 7 [ 9, 9] b [., 9.]. Domain: (- q, q) Range: [, ] Smmetric about the -ais Maimum r value: [ 7, ] b [ 6, 6] 6. Domain: (- q, q) Range: [, ] Smmetric about the -ais Maimum r value: [ 6, 8] b [ 8, 8] [ 6, 6] b [ 6, ] Copright 0 Pearson Education, Inc.

21 Chapter 6 Applications of Trigonometr 7. Domain: (- q, q). Domain: (- q, q) Range: [, 7] Range: [0, q) Smmetric about the -ais No smmetr Unbounded Maimum r value: 7 Maimum r value: none Graph for Ú 0: [, 8] b [, ] 8. Domain: (- q, q) Range: [, 7] Smmetric about the -ais Maimum r value: 7 [, ] b [ 0, 0]. Domain: (- q, q) Range: [0, q) No smmetr Unbounded Maimum r value: none Graph for Ú 0: [ 7., 7.] b [ 8, ] 9. Domain: (- q, q) Range: [0, ] Smmetric about the -ais Maimum r value: [ 6, 6] b [, ]. Domain: c0, p ª cp, p d d Range: [0, ] Smmetric about the origin on each interval in domain Maimum r value: [,.] b [.,.] 0. Domain: (- q, q) Range: [, ] Smmetric about the -ais Maimum r value: [.,.] b [, ]. Domain: c0, p ª c p ª c 7p, p d d, p d Range: [0, ] Smmetric about the -ais, -ais, and origin on each interval in domain Maimum r value: [.7,.7] b [.,.] Copright 0 Pearson Education, Inc.

22 Section 6. Graphs of Polar Equations [.,.] b [.,.] For # 8, recall that the petal length is the maimum r value over the interval that creates the petal.. r= when = e p and r=6 when, 7p f = e p. There are four petals with lengths {6,, 6, }., p f 7. Answers will var but generall students should find that a controls the length of the rose petals and n controls both the number of rose petals and smmetr. If n is odd, n rose petals are formed, with the cosine curve smmetric about the polar -ais and sine curve smmetric about the -ais. If n is even, n rose petals are formed, with both the cosine and sine functions having smmetr about the polar -ais, -ais, and origin. 8. Smmetr about -ais: r- sin ( )=0 9. Q r- sin (( ))= r+ sin ( ) (since sin ( ) is odd, i.e., sin ( )= sin ( ))=r- sin ( )=0. Smmetr about the origin: r- sin ( )=0 Q r- sin ( + )=r- sin ( )=0. 6. r= when ={0, } and r=8 when = e p. There are four petals with lengths {, 8,, 8}., p f 7. r= when = e 0, p and, p, 6p, 8p f r= when = e p., p, p, 7p, 9p f There are ten petals with lengths {,,,,,,,,, }. 8. r=7 when = e p and 0, p, 9p 0, p 0, 7p 0 f r= when = e p. 0, 7p 0, p 0, p, 9p 0 f There are ten petals with lengths {7,, 7,, 7,, 7,, 7, }. 9. r and r produce identical graphs r begins at (, 0) and r begins at (, 0). 0. r and r produce identical graphs r begins at (, 0) and r begins at (, 0).. r and r produce identical graphs r begins at (, 0) and r begins at (, 0).. r and r produce identical graphs r begins at (, 0) and r begins at (, 0).. (a) A -petal rose curve with short petals of length and long petals of length. (b) Smmetric about the origin. (c) Maimum r value:.. (a) A -petal rose curve with petals of about length,., and units. (b) Smmetric about the -ais. (c) Maimum r value:.. (a) A 6-petal rose curve with three short petals of length and three long petals of length. (b) Smmetric about the -ais. (c) Maimum r value:. 6. (a) A 6-petal rose curve with three short petals of length and three long petals of length. (b) Smmetric about the -ais. (c) Maimum r value:. [0, ] b [0, 6] =- sin has minimum and maimum values of 0 and 6 on [0, ]. So the range of the polar function r=- sin is also [0, 6]. 60. [0, ] b [, ] =+ cos has minimum and maimum values of and on [0, ]. So the range of the polar function r=+ cos is also [, ]. In general, this works because an polar graph can also be plotted using rectangular coordinates. Here, we have representing r and representing on a rectangular coordinate graph. Since is eactl equal to r, the range of and range of r will be eactl the same. 6. False. The spiral r= is unbounded, since a point on the curve can be found at an arbitraril large distance from the origin b setting numericall equal to that distance. 6. True. If point (r, ) satisfies the equation r=+cos, then point (r, ) does also, since +cos ( )=+cos =r. 6. With r=a cos n, if n is even there are n petals. The answer is D. 6. The four petals lie along the - and -ais, because cos takes on its etreme values at multiples of /. The answer is D. 6. When cos =, r=. The answer is B. 66. With r=a sin n, if n is odd there are n petals. The answer is B. 67. (a) Smmetr about the polar -ais: r-a cos (n )=0 Q r-a cos ( n )=r-a cos (n ) (since cos ( ) is even, i.e., cos ( )=cos ( ) for all )=0. Copright 0 Pearson Education, Inc.

23 Chapter 6 Applications of Trigonometr (b) No smmetr about -ais: r-a cos (n )=0 Q r-a cos ( n )= r-a cos (n ) (since cos ( ) is even) Z r-a cos (n ) unless r=0. As a result, the equation is not smmetric about the -ais. (c) No smmetr about origin: r-a cos (n )=0 Q r-a cos (n ) Z r-a cos (n ) unless r=0. As a result, the equation is not smmetric about the origin. (d) Since cos (n ) for all, the maimum r value is a. (e) Domain: (- q, q) Range: [ a, a ] Smmetric about the -ais Maimum r value: a 68. (a) Smmetr about the -ais: r-a sin (n )=0 Q r-a sin ( n )= r+a sin (n ) (since sin ( ) is odd, sin ( )= sin ( ))= (r-a sin (n )) =( )(0)=0. (b) Not smmetric about polar -ais: r-a sin (n )=0 Q r-a sin ( n )=r+a sin (n ). The two functions are equal onl when sin (n )=sin (n )=0, or ={0, }, so r-a sin (n ) is not smmetric about the polar -ais. (c) Not smmetric about origin: r-a sin (n ) Q r-a sin (n )= (r+a sin (n )). The two functions are equal onl when r=0, so r-a sin (n )) is not smmetric about the origin. (d) Since sin (n ) for all, the maimum r value is a. (e) Domain: (- q, q) Range: [ a, a ] Smmetric about -ais Maimum r value: a 69. (a) For r : 0 (or an interval that is units long). For r : same answer. (b) r : 0 (overlapping) petals. r : (overlapping) petals. 70. Starting with the graph of r, if we rotate clockwise (centered at the origin) b / radians ( ), we get the graph of r ; rotating r clockwise b / radians ( ) gives the graph of r. (a) (b) (c) 7. Starting with the graph of r, if we rotate counterclockwise (centered at the origin) b / radians ( ), we get the graph of r ; rotating r counterclockwise b / radians (60 ) gives the graph of r. (a) (b) [, ] b [, ] [, ] b [, ] [, ] b [, ] [, ] b [, ] [, ] b [, ] (c) [, ] b [, ] [, ] b [, ] [, ] b [, ] Copright 0 Pearson Education, Inc.