Selected Answers and Solutions

Transcription

1 value for the w idth of the box w ould be negative or zero. 3. height =.9 in., length «. in., w idth ~. in., m axim um volum e ~. in 3 5. radius =.55 in., height = 3.9 in. m inim um surface area = 5.9 in 7. (6.67,.83), Sam ple answ er: A m inim um tim e o f.83 hours w ill be obtained if the racer takes a path that w ill have him her at the sidew alk 3.33 m iles before the pier. 9. m = ; Sam ple answ er: The slope of the tangent line is. The tangent line is a horizontal line. 9. vertex: ( - 8, - ) ; focus: ( 8, 3); directrix: = ; axis o f sm m etr: x = 8. Sam ple answ er: The focus is feet above the ground. CHAPTER 7 Conic Sections and Parametric Equations (3) a. - 8x + l = + = 8x Page Chapter 7 Get Read. x = ; (, - ) ; (, - 3 ) 3. x = - ; (, - 8 ) ; ( -, - ) 5. x = ; (, - ) ; (, - 6 ) 7.x = 5; (, 55); (5,53.75) vertical asm ptote at x = 5 7. vertical asm ptotes at x = 5 and x = ; horizontal asm ptote at = 9. vertical asm ptote at x =. Pages 8-3 Lesson 7-. vertex: (3, 7); focus: (3, ); directrix: = ; axis of sm m etr: x = 3 + l + 5 = 8x l + 5 = 8x - 5 ( + 5 ) = 8 (x - 3) b. The equation in standard form has as the squared term, w hich m eans that the parabola opens horizontall. Because p = 8, p = 5 and the graph opens to the right. The equation is in the form ( fc) = p(x h), so h = 3 and fc = 5. Since the stern is located at the vertex o f the parabola form ed, it is at the p oint (;, fc) or (3, 5). The sw im m er is at the focus, located at (h + p, fc), w hich is (3 + 5, 5) or (8, 5). The distance the sw im m er is from the stem o f the boat represents the length of rope needed to attach the sw im m er to the stern. U sing the distance form ula, the distance betw een these tw o points is ( 8-3 ) + ( ( 5)) or 5. Therefore, the length o f rope attaching the sw im m er to the stern o f the boat 3. vertex: (, ); focus: (3, ); directrix: x = 7; axis of sm m etr: = - ( - ) = (x+ ) is 5 feet. 5. x = 8( + 7); vertex: (, 7); focus: (, 5); directrix: = 9; axis o f sm m etr: x = 5. vertex: ( 8,3 ); focus: ( 8,5 ); directrix: = ; axis of sm m etr: x 7. x = ( + ); vertex: (, ); focus: (, 7); directrix: = 5; axis o f sm m etr: x = 7. vertex: (, 5); focus: (7, 5); directrix: x = 5; axis o f sm m etr: = 5 9. = (x - 3); vertex: (3, ); focus: (8, ); directrix: x = ; axis o f sm m etr: = R9 ; Selected Answers

2 . ( ) = (x ); vertex: (, ); focus: (.5, ); directrix: x =.5; axis of sm m etr: =. ( + 9 ) = (x + 6) 3. (x + 6) = 8( + 9); vertex: ( 6, 9); focus: ( 6,.5); directrix: = 3.5; axis of sm m etr: x = in. 7. ( + l ) = (x + ) 9. (x + 3 ) = 8( - ) i i 8 v ) i (x + 3) _ ) T f 8 + x = 6 J - o n o Y, k f 6- o * w ( + l ) = (x + ) 3. ( - ) = 3(x - 7) s r - - x >< CJ C CJ 33. (x - l ) = ( - 6) 35. (x - 8) = 6( + 7) * L S * ix ' T ( x - ) = ( 6) 37. (x - l ) = ( - 5) 39. (x + ) = - ( - ) 8: i, - 8 r-)z = (-- 5 ) - t t For Homework Help, go to H o tm a th.co m k -k x + = - -- )- A K = 8x 5 (! 7. = x + The graph opens verticall. D eterm ine the vertex and focus..5(x - 6) = - 9 (x - 6 ) = ( - 9) Because p =, p =. The vertex is (6,9 ) and the focus is (6,8 ). We need to determ ine d, the distance betw een the focus and the point of tangenc, C. This is one leg of the isosceles triangle. d = j{x - x l ) + ( - l ) = ( - 6 ) + (5-8 ) = 5 U se J to find A, the end point o f the other leg o f the isosceles triangle. Since p is negative, the parabola opens dow n and A w ill be above the focus. A = (6, 8 + 5) or (6, 3 ) Points A and C b o th lie on the line tangent to the parabola. To find an equation for this line, first use tw o points to find the slope m. Then use a p oint on the line and the slope to w rite an equation for the line. - i = m ( x - * i ) 5 = (x ) 5 = x + Point-slope formula (x t) = (,5 ) and m = - Simplif. = x + 5 Solve for. 5. opens dow nw ard 53. opens upw ard 55. (x 3 ) = ( 5) 57. ( l ) = 6(x + 5) 59a. Sam ple answ er: x = 8.7( + ) 65. (, 8) 67. (, - ) 69a. 7a. ( ) = 8(x + 3) or ( - 59b. about 3.35 m 63. ( + 5 ) = 8(x + 3) -6 "(5,3 h 5, 7) 6 8 (--, - ) 3 ) : 69b..53 ft 8(x + 3) R9

3 Sam ple answ er: To prove that the endpoints o f the latus rectum and W and the point o f intersection o f the axis and directrix D are the vertices of a right isosceles triangle A D W, w e need to show that ZDW is a right angle and that D = WD. Since F = FW, FD = FD, and Z F D = ZWFD, A F D = A FW D b SA S. Thus, D = WD. To prove that ZDW is a right angle, w e can first p rove that Z D F and ZFDW are b oth 5 angles. Since F = FD and ZFD is a right angle, A F D is an isosceles right triangle. Therefore, ZDF is 5. This also m eans that ZFDW is 5. Likew ise, ZDW is a right angle. Thus, ADW is an isosceles right triangle. 73. Sam ple answ er: x = ( ) 75. Sam ple answ er: = x 77a. i. unit; ii. units; iii. units 77c. A s the focus is m oved farther aw a from the vertex, the parabolas becom e wider. 77d. Sam ple answ er: (x + l ) = ( + 7) 77e. Sam ple answ er: The parabolas all have a vertex of (,- ) and open dow nw ard. T he first equation produces the narrow est parabola and the second equation p roduces the w idest parabola. (79) The w idth of the sector is 3 units, so =.5. A = x and the area o f the sector is. square units. Substitute the values into the equation and solve for x. - = x(.5). = x. = x Therefore, the vertex o f the parabola is (, ), and tw o other points on the parabola are (.,.5 ) and (.,.5). Set up and solve a sstem o f equations using as the independent variable. a + b + c = x a( ) + b{ ) + c = a(.5) + b(.5) + c =. a(.5) + b(.5) + c =..5 a +.5b =..5a -.5b =..5a =. c = 97. f i x ) = x 3 - x - x +. G 3. G Pages 38- Lesson x x = 9 A 9x + + 6x =.5 ^ +.5b =.. +.5b =..5b = b = The equation of the parabola is -j - 8. Q uadrants and ; the vertex is (, 5) and p =. Since the vertex is to the left of the -axis, and the parabola opens to the left, no points w ill be to the right o f the -axis, or in Q uadrants and. 5 : x or L = x m ax at (7,8.5 ) = 8.5, m in at (, ) = cot 6 = 7-, csc 6>= 6 6 R9 Selected Answers 7 (x 3) ( + 3) l a ( x - ) ( ) 36 5 (* + <>), ( - 3), «( * - ), ( - s ), a. 8.8 in. i' 3 b. J ^ + ^ = l 5.( } + = ; ellipse 3 ' 7 ( x - 7 ) ( + ) 7. ( + 9) = (x 6); parabola 9. n ellipse x x = x + x =

4 {x + x + 36) + 7( ) + 3 = (x + 6 ) + 7( + 6 ) = (x + 6) ( + 6) + = 7 The related conic is an ellipse becau se a ==b and the graph is of the form + ( 7 ^ =. a b 33. (x - 3) + = 35. (x + ) + ( + 3) = 36 53b. ** > * > 5 f f N JU 53c. 5 das 55. ( 7,- ), ( 3, - 5 ) 57. ( -,3 ), ( 6, - 5 ) 59. ^ ^ = i - H 3, ) - ^ K A S - t i r i r " n, * v - 6. (x - 6.5) + ( -.5 ) = (x - l ) + ( + 7) = 6; center: (, 7), radius: 65. (x + l ) + ( ) = 6; center: (, ), radius: Sam ple answ er: N o; if a = p and b = p + r, then c = r. f a = p + r and b = p, then c = fr and the foci are (, ± r ). 37. P F i + PF = a J(x - ) + ( - c) + ^(x - ) + ( - ( - c ) ) = a jx + ( - c) + Jx + ( + c) = a jx + ( - c) = a - Jx + ( + c) x + c + c = a - a ^ x + ( + c) + ajx ^J7-~tf = a + c ajx + ( + c) = a + c x + + c + c a(x + + c + c) = a + ac + c! ax + a + flc + ac = a + ac + c ^ <P~ + ax = a ac (a c) + ax = fl(a c) b + ax = ab - + ^ = a b 3 9. i i ^ + ( - 3) =. ^ + ^ = ) a. The length o f the m ajor axis is or The value o f a is or The distance 87 from the focus (the sun) to the vertex is or Therefore, the value o f c, the focus to the center, is or 7.. n an ellipse, c = a b. The value o f b is «c = ~ T he valu e o f b, the length o f the m inor axis, is about or about 7.35 m illion mi. c 7 b. The eccentricit e is equal to. e = ^ ^ :.3 5. center: (, 6); foci: (+ 5 3, 6); vertices: (±, 6 ) 7. center: (, ); foci: (, ± 7 ); vertices: (, ± 6 5 ) ( + ), (x ) 9. + = ' 6 53a. x + = 6, x The area o f the ellipse A = tt. Substitute into the area form ula to obtain an equation w ith a and b. tt = -nab = ab Substitute a = b + 5 into the equation and solve for b. = (b + 5 )b = b + 5b = b + 5b = (b + 8)(b - 3) Since b cannot b e negative, b = 3 and a = or 8. Therefore, an equation for the ellipse is x v x v Yes; sam ple answ er: f (x, ) is a point on the ellipse, then ( x, ) m u st also b e on the ellipse. x (- = a b ( - x ), ( - ) + - b = a h = b Thus, ( x, ) is also a p oint on the ellipse and the ellipse is sm m etric w ith respect to the origin. 73. Sam ple answ er: W hen a is m u ch greater than c, j is close to. Since e = the valu e o f e is close to zero and the foci are near the center o f the ellipse. So, the ellipse is nearl circular. 75. vertex: - j; focus: 7^; directrix: = av,'s o f sm m etr: M R eal Babies and 3 M First Babies 79. sin (» + f ) - cos ( + f j 7 ' = sin cos ^ + cos sin ~ cos cos ^ + sin sin = i sin + cos cos + j sin For Homework Help, go to H o tm a th.c o m $ connecte^ticgrav^iil^orj R 9 3

5 = j sin 8 + j sin 9 = sin 6 8. J>L 83. ^ 6 85.[ - 7, 5 ] 6 3 tu rn in g p o in ts;,, and tu rn in g p o in ts;,5, and Pages real zeros and real zeros and i 93. B 95. C Lesson x z + 3z - x + 6 = 8. j - -, k " 8 - (- l va '- -7, ^ s U M )* s J» i Ss L - 5 x x - 8 = 87 (8.89, ) t- ( - l ) 5 ( + 8) 6 s 8x a. 3 Y(. 9). m n t, 3)r <7 ik if. _ -(o, 9.3: 8-8 (x n 3. 5, < (* ) 9 = _ =. 33b. 9 ft h p erb o la 9. ellip se 5. h p erb o la 53. circle p arabo la ) a. T h e co m m o n d ifferen ce is 8 k ilo m eters and the ab so lu te v alu e o f the d ifferen ce o f the d istan ces from an p o in t on a h p erbo la to the fo ci is a, so a = 8, a = 9, an d a = 8. T h e tw o airp o rts are the fo ci o f the h p erb o la and are 7 k ilo m eters ap art, so c = 7, c = 36, and c = 96. c = a + bz, so b = 96 8 S) 3x 5) 5 7 or 5. A irp o rt B is d ue sou th o f airp ort A, so the tran sv erse axis is v ertical and n-term goes w ith the -term. T h e eq u atio n for the h p erbo la is t. 8 it a ( - ) ( - ) -^ JU -- t >f J (x + 7) 5 ( + 3) 7? - ( 9. [x + ) ( - C (x + 3) = 33 rfp T 9. 3 )- - 6 ( + ) 9 the x -term. T h e eq u atio n fo r the h p erbo la is v -K (x - 3) 7 x-co o rd in ates o f the v ertice s are d ifferent, so the tran sv erse axis is h o rizo n tal and the a -term go es w ith o 5. (, 3). T h erefore, h = and k = 3. T h e con ju g ate axis is o f le n g th, so b =, b = 6, and b = 36. The S - 5 8, Me,c )? 5, 8, ) i ( - 3 )-,r (x + l ) = 9 (3? T h e v e rtice s are u n its ap art, so a =, a =, and a =. T h e cen ter is the m id p o in t o f the v ertices, or b. ^ = B - ' 5 =. C ( 36)- Airpo r t be: f e, - - -* T h e p lan e is on the top b ran ch b e cau se it is clo ser to airp o rt A. x irport bj (, - 3 B) (x -5 ) _ 9 R9! Selected Answers (- ) 7 C. B ecau se th e h ig h w a is the tran sv erse axis and the p lan e is k ilo m eters from the highw a, x =. S u b stitu te fo r x in the equ ation.

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7 7. in tersectin g lin es 3x ± (3x) - ()(3x - ) = x and = x 9. p o in t at (, ) 5. b () = c -3x ± 96-3x G raph the conic. [ -, ] scl: b [ -, ] scl: 55b. A x - Bx - C + Dx + E + F = A '(x') + B 'x'' C '(') + D 'x' + E'' + F'; b su b tractin g o u t the x, x,, and F term s, the rem ain in g statem en t is true. A x - C = A '(x') C'('); A m u st eq u al A and B m u st equ al B' in o rd er fo r the statem en t to b e tru e. 55c. A x + Bx - C - Dx + E + F = A '(x') + B'x'' - C ( r) + D 'x ' + E'' - F'; b su btractin g ou t the x,, and F term s, the rem ain in g statem en t is true. A x + Bx + C = A ix ) + B'x'' - C '('); A m u st equal A', B m u st eq u al B', and C m u st eq u al C' in ord er for the statem en t to b e true. T h erefo re, B A C = (S ') (A'C); A (x cos 9 + sin 9) + B (x cos 9 - sin 9)( cos 9 x sin 9). 33. i 8(x') - 6 (-) = "s. ( v (57) For e ach eq u atio n, solv e fo r u sin g the qu ad ratic form u la. T h e n g rap h u sin g a g rap h in g calculator. ) 9 x - x = 5 - x() - (9 x ) = x ± J(x) (5)(9x ) x + 8-6x 39. Sx + bx - = - J [ - 6.6,.6 ] scl: b [ -, ] scl: Y AK x - x - - x - - = - ( x l) - (x x - ) = x + ± ( x l) ( )(x x ) x + +?x 6x , 7.5 8] scl: b [ - 5, 5] scl:. [ -.5 8,.5 8 ] scl: b [ -, 8] scl: [ -, ] scl: b [ -, ] scl: T h e g rap h s in te rsect at fo u r p o in ts: (.9,.), (.5,.5), (.9, -. ), (.9,.8). Equation Graph M inim um Angle of Rotation x - 5 x - 3 = parabola 36 6x = 5 x= 9 x + x - - x - fi = - ellipse 55a. A x + Bx - C + Dx + E + F = is eq u iv alen t to A'(x') + B'x'' + C '(') + D'x' + E'' - F' = b ro tatin g the A C con ic sectio n th ro u gh 9 su ch th at co t 9 = -. F is u n affected b th is ro tatio n, so F = F'. hperbola 8 59b. S am p le answ er: A p arab o la h as lin e o f s m m etr and the m in im u m an g le o f ro tatio n is a co m p lete circle. A n [ -, ] scl: b [ -, ] scl: R96 S e le c te d A n s w e r s ellip se and a h p erb o la h av e lin es o f s m m etr and the m in im u m an g le o f ro tatio n is a h a lf circle. 59c. 3

8 3. x = - f - r <r> _ i-t6 = [ (* + ( K ) = M + ( 63. 7^ -! CvJ 6. L et x = x' cos 9 + ' sin 9 and = x' sin 9 - ' cos 9. r = x + = (x ' cos 9 + ' sin 9) + ( x' sin 9 + ' cos 9) = ( x ^ c o s 9 - x''cos 9 sin 9 - ( ^ s in 9 + (x^ sin 9 - x''cos 9 sin 9 - ( ^ c o s 9 = [(xr) - ( ^cos 9 - [(x - ( ^ sin 9 = [ (* + ( ^ K c o s 9 - s in 9) t= ) ^ t = a ~ t = - cos 9 (x = x' cos 9 ' sin 9) 7. sin 9 ( = x' sin 9 - ' co s 9) x cos 9 - sin 9 x cos 9 - sin 9 x co s 9 - sin 9 x cos 9 - sin 9 8 = x' c o s 9 ' sin 9 cos 9 = x' s in 9 - v ' sin 9 cos 9 = x' c o s 9 - x' s in 9 = x' (cos 9 - s in 9) = x' = - F -5-6 ^ } p t h =7 - t--= 8 3 sin 9 (x = x' cos 9 ' sin 9) cos 9 ( = x' sin 9 - ' cos 9) x sin 9 = x' cos 9 sin 9 ' s in 9 cos 9 = x' cos 9 sin 9 - ' c o s 9 x sin 9 cos 9 = ' sin 9 ' c o s 9 9. =. 5 x -.5x -.5. = 5x + > x sin 9 cos 9 = ' (sin 9 - c o s 9) cos 9 x sin 9 = ' 65. Sam ple answ er: T he d iscrim in an t is d efin ed as B C, or in this in stan ce, (B ') AA'C'. S in ce a con ic th at is rotated has n o B' term, the d iscrim in an t red u ces to AA'C'. T hu s, onl the A' and C term s d eterm in e the tp e o f conic. T herefore, AA'C' < w o u ld b e an ellip se o r a circle, AA'C' = w ou ld be a parabo la, and AA'C' > w ou ld b e a h p erbola. For a circle or an ellip se, A' and C n e e d to share the sam e sign. For a p arab o la, eith er A! or C h as to b e equ al to. Fo r a h p erb o la, A' and C' n eed to h av e o p p o site signs. - t - f A M i f -r- A 5. = ^ i - 6 -f - -f 6x 8x i S 7 = o r 69. i > s J M a. s - d = 5,.35s -.5 d = b. sav in g s accou n t: $5; A -i - (x 3) 6 Pages ( 8 A 9. 9 = i =: certificate o f d ep o sit: $ es, es; (x ) (x - )(x - )(x - 3) ( 7) = r 5 7. = - x + ^ fx -l- 3 J = t 79. D 8. A Lesson 7-5. f =^ i x T +T J x = 3 cos e = 3 s in 9 = K = For Homework Help, go to H o t m a t h.c o m dl r R97 = b

9 x = tv co s 8 7.x = a n d = ^ 9 - ^ 5- x + 9 =. =.38 - v -^ 8x.9 = v, t = '- - = 9 - x - -t= - - A f= = J C 5a. ab o u t 7. ft s 5 b. ab o u t secon d 5c. abo u t.8 seco n d s 53. S am p le answ er: x = + f, = + 3f 55. S am p le answ er: T h e h o rizo n tal d istan ce is m o d eled b the cosin e fu n ctio n, w h ich is at 9. T h is w o u ld im p l th at the p ro jectile h as n o h o rizo n tal m o v em en t. T h e co rresp o n d in g p aram e tric eq u atio n w o u ld b e x =. 57. S am p le answ er: P aram etric equ atio n s sh ow both 3. x = f and = -f+ t+ 9. x = 5t and = - 5 f + f ou t= 6 * [ (x ' >u ll wl* + t--= 'i 6x ^ t= 5 ' h o rizo n tal and v e rtical p o sitio n s o f an o b ject o v er tim e, w h ile rectan g u lar e q u atio n s c a n o n l sh o w o n e or th e other. - - ( k s ** - CJ (x - 3) = tan x n 3 + n ; n - n 3;.97 = C t= 6 ( - ). 33. es (35) Solve the first eq u atio n for t. x = log t L6 69. h o riz o n ta l as m p to te at = ; v ertical as m p to te at x = 6; D = (x x ~ 6, x R) x = t u J Su bstitu te into second equation. k > = t+ 3 = * + 3 n x = lo g t, x m u st b e >, so the d o m ain restrictio n is x > = * = * 8, x ==.x = f + a n d = 7 + 3f 3. x = t and = 6 + f, < f < 5. b 7. a 7. v ertical as m p totes at x = 5; o b liq u e as m p to te at = x + 3; x -in tercep ts and 8; -in tercep t ; D = (, 5) U ( 5, ) (9) a. The p o sitio n equ atio n s are x = tv cos 9 and = fw sin 9 ~ gt + h. T h e in itial v elo cit z> is.75 and 9 is 5. T h e su rface o f the w ater is at =, so the v alu e o f the in itial h e ig h t h is.3. B su bstitu tion, the p o sitio n eq u atio n s are x = t.75 cos 5 and = t.75 sin 5.9 f +.3. b. Find the v alu e o f t for w h ich = in the v ertical b. ellip se 77. E 79a. x x = 79c. (x ') + 9 ( ') = 79d. ~.75 p o sitio n equation. = t.75 sin f +.3 = ~ t -.9 f +.3. The p o sitiv e zero in the grap h o f the eq u atio n is ab o u t.37, so t =.37 second s. S u b stitu te th is for t in x = t.75 cos 5, the h o rizo n tal position equ ation. x =.37(.75) ^ :.6. This m ean s th at w h en the frog reach es th e su rface o f the w ater, h e is o n l ab o u t.6 m from w here h e ju m p ed. T herefo re, h e is ab o u t.5.6 or.3 m eters from th e oth er bank. C. T h e. m eter d istan ce is a h o rizo n tal d istance, so su b stitu te the v alu es into the h o rizo n tal p o sitio n e q u atio n and solv e for u. R98 Selected Answers Pages con ic section Chapter 7 Stud Guide and Review 3. d irectrix 5. foci 7. cen ter 9. p aram etric. vertex: ( 3, ), focus: ( 3, ) ; axis o f sm m etr: x = 3; d irectrix: = * (-3, - )

10 3. vertex: (, ), focus: (, ); axis of smmetr: x = ; directrix: = (, - f ix t r. 3. -,] scl: b [-, ] scl: [-, ] scl: b [-, ] scl: (x - l) = 6(i - 5) ( 5-6 6x ( + ) = - ( x + 3). (x - 3) = (i+ 7) l- -3, - ' s M 8 C - [-5, 5] scl: b [-5, 5] scl: 7. (x - f3x'' + 3 (^ + (3 - )x' + (a3 + )' = ; parabola 9. 9(') + (x') = 36; ellipse ~ ~- f = - 3 l. = 3, t=-ia t= f = J - " f = t-.= n 53. = i (x - 3) ( ( + 6 ) _ 6 5 ( x - 5 ), ( - ) ^ ' l (x - l ) + ( ) = 3; circle 9. (x - ) = ( + 5); parabola 8 x - 57a. x + = 9 57b. 5 seconds 59a. 3x + 3x + + 8x 83 = 59b. * k! ix Pages Chapter 7 Connect to AP Calculus units units3 5. Sample answer: The approximation will get closer to the actual volume of the solid because as the clinders decrease in height, the will better fill the volume of the solid units3 v s r» i s u.*» ectors Page 8 Chapter 8 Get Read.3 ; ( ± ) 3. 9; ( - i - 8 ) ft. no solution 3. B «39, C as 5, c a 3. For Homework Help, go to Hotm ath.com connected.mcgraw-hill.com R 9 9