Chapter 4 Practice Exercises 303
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1 Chapter Practice Eercises y 5. y 55. y 56. y
2 30 Chapter Applicatios of Derivatives % 57. y 58. y y 60. y 6. lim lim Ä Ä a ac a a b bc b 6. lim b lim Ä Ä & 63. lim Ä ta ta! ta sec 6. lim lim Ä! si Ä! cos si si cos si cos taa b sec a b sec a b a sec a btaa b b sec! 65. lim lim lim lim Ä! Ä! Ä! Ä! siamb m cosamb m siab cosab 66. lim lim Ä! Ä! cosa b sia b ( ( ( ( 67. lim seca( bcosa b lim lim Ä Î c Ä Î c cosa b Ä Î c sia b a b È 68. lim È sec lim Ä! b Ä! b! cos! cos si! 69. lim acsc cot b lim lim Ä! Ä! si! Ä! cos
3 Chapter Practice Eercises lim ˆ lim Š lim a b lim a b lim Ä! % Ä! % Ä! % Ä! Ä! % È È È È 7. lim lim Ä_ Š È È Ä_ Š È È lim Ä _ È È È Notice that for! so this is equivalet to b lim lim Ä_ b b c Ä_ È È É É É É a b a b 7. lim Š lim lim lim lim Ä_ Ä_ a ba b Ä_ % Ä_ % Ä_ lim lim! Ä_ % Ä_ (l The limit leads to the idetermiate form : lim 0 lim l 0 Ä 0 Ä The limit leads to the idetermiate form 0: lim lim l 3 Ä 0 Ä 0 si 0 (l 33 si (l (cos 75. The limit leads to the idetermiate form 0: lim e lim l Ä 0 e Ä 0 0 c si c si (l ( cos 76. The limit leads to the idetermiate form 0: lim e lim e l Ä 0 Ä cos 5 si 5 cos 77. The limit leads to the idetermiate form 0: lim e lim lim 5 Ä 0 e e Ä 0 Ä 0 0 e e 78. The limit leads to the idetermiate form 0: lim e lim e e Ä 0 Ä 0 tl(t Š b 0 t 79. The limit leads to the idetermiate form 0: lim t lim t _ t Ä! t Ä! 0 si ( (si (cos 0 e 3 ec 80. The limit leads to the idetermiate form : lim Ä c lim Ä lim lim si( cos( e e Ä Ä c c 0 e t e t e t 8. The limit leads to the idetermiate form 0: lim Š t t lim Š t lim t Ä! t Ä! t Ä! 8. The limit leads to the idetermiate form _: lim e l y lim lim y Ä! y Ä! e y Ä! y lim 0 y Ä! e y c _ c Îy l y y y ey y c c ˆ c b k Îb bk bk 83. lim lim e Ä _ ˆ Ä Š _ Îb 8. lim 0 0 Ä _ ˆ 7
4 306 Chapter Applicatios of Derivatives 85. (a Maimize f( È È Î Î 36 (36 here 0 Ÿ Ÿ 36 Î Î È36 È È È36 Ê f ( (36 ( Ê derivative fails to eist at 0 ad 36; f(0 6, ad f(36 6 Ê the umbers are 0 ad 36 (b Maimize g( È È Î Î 36 (36 here 0 Ÿ Ÿ 36 Î Î È36 È È È36 Ê g ( (36 ( Ê critical poits at 0, 8 ad 36; g(0 6, g(8 È8 6È ad g(36 6 Ê the umbers are 8 ad 8 È Î Î Î 3 Î 86. (a Maimize f( (0 0 here 0 Ÿ Ÿ 0 Ê f ( Ê 0 ad 0 are critical poits; f(0 f(0 0 ad f ˆ 0 É 0 ˆ 0 0 È È0 3È3 Ê the umbers are ad. È Î È0 È0 (b Maimize g( 0 (0 here 0 Ÿ Ÿ 0 Ê g ( 0 Ê È Ê. The critical poits are ad 0. Sice g ˆ 79 8 ad g(0 0, 79 the umbers must be ad. 87. A( ( a7 bfor 0 Ÿ Ÿ È7 Ê A ( 3(3 (3 ad A ( 6. The critical poits are 3 ad 3, but 3 is ot i the domai. Sice A (3 8 0 ad A Š È7 0, the maimum occurs at 3 Ê the largest area is A(3 5 sq uits The volume is V h 3 Ê h. The surface area is S( ˆ 3 8, here 0 Ê S ( ( a 6b Ê the critical poits are 0 ad, but 0 is ot i the 56 domai. No S ( 0 Ê at there is a miimum. The dimesios ft by ft by ft miimize the surface area. 89. From the diagram e have ˆ h r Š È3 Ê h r. The volume of the cylider is h È 3 V r h Š h ah h b, here 0 Ÿ h Ÿ 3. The V (h ( h( h Ê the critical poits are ad, but is ot i the domai. At h there is a maimum sice V ( 3 0. The dimesios of the largest cylider are radius È ad height.
5 Chapter Practice Eercises From the diagram e have radius ad y height ad V( (, here 0 Ÿ Ÿ 6 Ê V ( ( ad V ( 8. The critical poits are 0 ad ; V(0 V(6 0 Ê gives the maimum. Thus the values of r ad h yield the largest volume for the smaller coe The profit P p py p p ˆ 0 0 5, here p is the profit o grade B tires ad 0 Ÿ Ÿ. Thus P p ( (5 a 0 0 b Ê the critical poits are Š 5 È5, 5, ad Š 5 È5, but oly Š 5 È5 is i the domai. No P ( 0 for 0 Š 5 È5 ad P ( 0 for Š 5 È5 Ê at Š 5 È5 there is a local maimum. Also P(0 8p, P Š 5 È5 p Š 5 È5 p, ad P( 8p Ê at Š 5 È5 there is a absolute maimum. The maimum occurs he Š 5 È5 ad y Š 5 È5, the uits are hudreds of tires, i.e., 76 tires ad y 553 tires. 9. (a The distace betee the particles is lfab t lhere fab t cos t cosˆ t. The, f ab t si t siˆ t. Solvig f ab! t graphically, e obtai t Þ(, t %Þ!, ad so o. % % Alteratively, f ab t may be solved aalytically as follos. f ab t si ˆ t si ˆ! t siˆ t cos cosˆ t si siˆ t cos cosˆ t si si cosˆ t so the critical poits occur he cosˆ t!, or t k. At each of these values, fab t cos!þ(& uits, so the maimum distace betee the particles is!þ(& uits. (b Solvig cos t cos ˆ t graphically, e obtai t Þ(%*, t &Þ*!, ad so o. % Alteratively, this problem ca be solved aalytically as follos. cos t cos ˆ t cos ˆ t cos ˆ t cosˆ t cos siˆ t si cosˆ t cos siˆ t si si ˆ t si! si ˆ t! %
6 308 Chapter Applicatios of Derivatives ( t k ( Þ(%* The particles collide he t. (plus multiples of if they keep goig. 93. The dimesios ill be i. by! i. by i., so Vab a! ba b % &! for! &. The V ab!%! % a ba! b, so the critical poit i the correct domai is. This critical poit correspods to the maimum possible volume because V a b! for! ad V a b! for &. The bo of largest volume has a height of i. ad a base measurig 6 i. by i., ad its volume is i. Graphical support: 9. The legth of the ladder is d d 8 sec 6 csc. We ish to maimize I( 8 sec 6 csc Ê I ( 8 sec ta 6 csc cot. The I ( 0 È 6 Ê 8 si 6 cos 0 Ê ta Ê d É È36 ad d È36 É È36 Ê the legth of the ladder is about Š È36 É È 36 Š È36 *Þ( ft. Î 95. g( 3 Ê g( 0 ad g(3 0 Ê g( 0 i the iterval [ß3] by the Itermediate 3 Value Theorem. The g ( 3 3 Ê b 33 ;! Ê. Ê.965, ad so forth to & % 96. g( 75 Ê g(3 0 ad g( 7 0 Ê g( 0 i the iterval [ ß% ] by the Itermediate Value Theorem. The g ( 3 Ê ; 3 Ê b Ê , ad so forth to & % 97. a 5 7 b 5 d 7 C % 75 3! 98. Š 8t t % dt % t 8t t t C t t t C 6 6 Î c 99. ˆ 3È t dt ˆ Î 3t t 3t t Î dt C t C t Š 3 t Î c Š dt ˆ Î % t 3t t 3t dt Œ C È % t C Èt t ( 3 t
7 Chapter Practice Eercises Let u r 5 Ê du dr dr du u ar 5b u ar 5b c u du C u C C 0. Let u r È Ê du dr c 6 dr 6 dr 6 du u 3 6 u du 6 Š C 3u C C Š r È Š r È u Š rè 03. Let u Ê du d Ê du d Î 3 È d Èu ˆ 3 du 3 Î 3 u Î Î u du Œ C u C a b C 3 0. Let u 7 Ê du d Ê du d Î Î u Î d ˆ du u du C u C 7 C Œ È È7 Èu % a % Î% Î% Î% Î% % Î% b ˆ u Œ a b 05. Let u Ê du d Ê du d d u du u du C u C C 06. Let u Ê du d Ê du d Î& ( Î& d Î& u ( du Î& u du u C 5 Î& u C 5 ( Î& C Š s 0 0 s s 0 a b Let u Ê du ds Ê 0 du ds sec ds sec u (0 du 0 sec u du 0 ta u C 0 ta C 08. Let u s Ê du ds Ê du ds a bˆ csc s ds csc u du csc u du cot u C cot s C 09. Let u È Ê du È d Ê du d È csc È cot È d (csc u cot u Š du ( csc u C csc È C 0. Let u Ê du d Ê 3 du d È È È sec ta d (sec u ta u(3 du 3 sec u C 3 sec C. Let u Ê du d Ê du d cos u si u si d asi u b( du ˆ du ( cos u du ˆ u C u si u C ˆ si ˆ C si C. Let u Ê du d Ê du d cos u si u cos d acos u b( du ˆ du ( cos u du u C si C 3. ˆ 3 d 3 lll C
8 30 Chapter Applicatios of Derivatives. ˆ 5 d ˆ 5 d 5 ta C ct 5. ˆ t t e e t e t t dt e C e e C s 6 6. a s 5 5 s b 5 s ds C l a b d c C 8. a r b dr b r C l È È 9. d d sec C d d 0. si ˆ C È6 d Ê6Š É 6 6. y d a bd C C; y he Ê C Ê C Ê y y ˆ d ˆ d a bd C C; y he Ê C Ê C Ê y 3. dr 3 dr dt Š 5Èt dt ˆ Î 5t Î 3t dt Î 0t Î È 6t C; dt 8 he t t Ê Î Î 0( 6( C 8 Ê dr Î Î C 8. Thus dt 0t 6t 8 Ê r ˆ Î Î 0t 6t 8 dt &Î Î t t 8t C; r 0 he t Ê &Î Î ( ( 8( C 0 Ê C 0. Therefore, &Î Î r t t 8t dr dt. cos t dt si t C; r 0 he t 0 Ê si 0 C 0 Ê C 0. Thus, si t dr dt Ê si t dt cos t C ; r 0 he t 0 Ê C 0 Ê C. The cos t Ê r (cos t dt si t t C ; r he t 0 Ê 0 0 C Ê C. Therefore, r si tt 5. Yes, si ad cos differ by the costat È 6. Yes, the derivatives of y cos C ad y cos ( C are both c da c c c 7. A y e Ê d e (( e e a b. Solvig d 0 Ê 0 Ê da ; 0 for da ad 0 for 0 Ê absolute maimum of Î e at dr dt dr dt da È d È d È È Èe Î È Èe uits log by y e uits high. 8. A y ˆ l l da l l da Ê d. Solvig d 0 Ê l 0 Ê e; da 0 for e ad da 0 for e absolute maimum of l e d d Ê e e at e uits log ad y e uits high.
9 Chapter Practice Eercises 3 9. y l Ê y ˆ l ( l ; Ê ˆ e e ˆ Ê solvig y 0 ; y 0 for ad y 0 for Ê relative miimum of at ; f ad f 0 absolute miimum is at ad the absolute maimum is 0 at e 30. y 0( l Ê y 0( l 0 ˆ 0 0 l 0 0( l ; solvig y 0 Ê e; y 0 for e ad y 0 for e Ê relative maimum at e of 0e; y! o Ð!ß e Ó ad y ae b 0e ( l e 0 Ê absolute miimum is 0 at e ad the absolute maimum is 0e at e ÎÈ Š È Š È 3 b ÎÈ ÎÈ 3 Š È Š È 3. fab e for all i a_, _ b; f ab e e ˆ ˆ a 3 Î b ÎÈ ÎÈ ÎÈ Ä_ Ä _ È ÎÈ e 0 Ê 0 Ê are critical poits. Cosider the behavior of f as Ä _; lim e lim e as suggested by the folloig table ( digit precisio, digits displayed: Î e _ 0 ã ã ã ã ã ã _ 0 Therefore, y is a horizotal asymptote i both directios. Check the critical poits for absolute etreme values: ÈÎ ÈÎ ÈÎ fab e 0.93, fab e.08 Ê the absolute miimum value of the fuctio is e at, È Î ad the absolute maimum value is e at. È 3 3. fab e ; The domai of g is all such that 3 0. The parabola y 3 is cocave do ith -itercepts at 3 ad, therefore 3 0 if 3 Ÿ Ÿ, ad this iterval is the domai of g; È3 0 È3 g a b e 0 Ê 0 Ê is a critical poit; g a 3 b g a b e,
10 3 Chapter Applicatios of Derivatives gab e Ê the absolute miimum value of the fuctio is at 3 ad, ad the aboslute maimum value is e at. l l l È È Î È 33. (a y Ê y 3 &Î &Î &Î Ê y ( l ˆ 3 l ; solvig y 0 Ê l Ê e ; y 0 for e ad ad y 0 for e Ê a maimum of ; y 0 8 Î Ê l 3 Ê e ; the curve is cocave do o ˆ Î 0 e ad cocave up o ˆ Î ß e ß_ ; so there is a iflectio poit at ˆ Î e ß. e %Î (b y e Ê y e Ê y e e a be ; solvig y 0 Ê 0; y 0 for 0 ad y 0 for 0 Ê a maimum at 0 of! e ; there are poits of iflectio at È ; the curve is cocave do for ad cocave up otherise. e È È c c c c (c y ( e Ê y e ( e e Ê y ec e c ( e c ; solvig y 0 c Ê e 0 Ê 0; y 0 for 0 ad y 0 for 0 Ê a maimum at 0 of ( 0 e! ; there is a poit of iflectio at ad the curve is cocave up for ad cocave do for. 3. y l Ê y l ˆ l ; solvig y 0 Ê l 0 Ê l Ê e ; y 0 for e ad y 0 for e Ê a miimum of e l e at e. This miimum is a absolute miimum e sice y is positive for all!. 35 I the iterval the fuctio si 0 si Ê (si is ot defied for all values i that iterval or its traslatio by.
11 Chapter Additioal ad Advaced Eercises v l ˆ dv (l l l Ê l ˆ dv ( l ; solvig 0 d d cî dv cî dv cî d d c Î r h Ê Î È Ê l 0 Ê l Ê e ; 0 for e ad 0 for e Ê a relative maimum at e ; ad r h e e.65 cm CHAPTER ADDITIONAL AND ADVANCED EXERCISES. If M ad m are the maimum ad miimum values, respectively, the m Ÿ f( Ÿ M for all I. If m M the f is costat o I. 3 6, Ÿ 0. No, the fuctio f( has a absolute miimum value of 0 at ad a absolute 9, 0 Ÿ Ÿ maimum value of 9 at 0, but it is discotiuous at O a ope iterval the etreme values of a cotiuous fuctio (if ay must occur at a iterior critical poit. O a half-ope iterval the etreme values of a cotiuous fuctio may be at a critical poit or at the closed edpoit. Etreme values occur oly here f 0, f does ot eist, or at the edpoits of the iterval. Thus the etreme poits ill ot be at the eds of a ope iterval.. The patter f ± ± ± ± idicates a local maimum at ad a local % miimum at (a If y 6( (, the y 0 for ad y 0 for. The sig patter is f ± ± Ê f has a local miimum at. Also y 6( ( ( 6( (3 Ê y 0 for 0 or, hile y 0 for 0. Therefore f has poits of iflectio at 0 ad. There is o local maimum. (b If y 6( (, the y 0 for ad 0 ; y 0 for 0 ad. The sig sig patter is y ± ± ±. Therefore f has a local maimum at 0 ad! È7 È7 local miima at ad. Also, y Š Š, so y 0 for È7 È7 È7 ad y 0 for all other Ê f has poits of iflectio at. f(6 f(0 6. The Mea Value Theorem idicates that 6 0 f (c Ÿ for some c i (0ß6. The f(6 f(0 Ÿ idicates the most that f ca icrease is. 7. If f is cotiuous o [aßc ad f ( Ÿ 0 o [aßc, the by the Mea Value Theorem for all [aßc e have f(c f( c Ÿ 0 Ê f(c f( Ÿ 0 Ê f( f(c. Also if f is cotiuous o (cß b] ad f ( 0 o (cßb], the for f( f(c c all (cß b] e have 0 Ê f( f(c 0 Ê f( f(c. Therefore f( f(c for all [aßb]. 8. (a For all, ( Ÿ 0 Ÿ ( Ê a b Ÿ Ÿ a b Ê Ÿ Ÿ. c f(b f(a f(b f(a c (b There eists c (aßb such that Ê ¹ ¹ Ÿ, from part (a Ê kf(b f(a k Ÿ kb a k. c ba ba c 9. No. Corollary requires that f ( 0 for all i some iterval I, ot f ( 0 at a sigle poit i I. 0. (a h( f(g( Ê h ( f (g( f(g ( hich chages sigs at a sice f (, g ( 0 he a, f (, g ( 0 he a ad f(, g( 0 for all. Therefore h( does have a local maimum at a.
12 3 Chapter Applicatios of Derivatives (b No, let f( g( hich have poits of iflectio at 0, but h( has o poit of iflectio (it has a local miimum at 0. a. From (ii, f( bc 0 Ê a ; from (iii, either lim f( or lim f(. I either case, Ä_ Ä_ lim f( lim lim b 0 ad c. For if b, the Ä _ Ä _ b Ê c Ä _ b c lim! ad if c 0, the lim lim _. Thus a, b 0, ad c. Ä _ c Ä _ b Ä _ È k k 36 d 6. 3 k 3 0 Ê Ê has oly oe value he k 36 0 Ê k 9 or k The area of the? ABC is A( ( È a b Î, here 0 Ÿ Ÿ. Thus A ( Ê 0 ad are È critical poits. Also A a b 0 so A(0 is the maimum. Whe 0 the? ABC is isosceles sice AC BC È. f(ch f(c. lim f (c for f (c 0 there eists a 0 such that 0 h h Ä 0 h Ê % k k k k f(ch f(c f(ch h h Ê ¹ f (c ¹ kf (c k. The f (c 0 Ê kf (c k f (c kf (c k f(ch h 3 f(ch f(ch 3 h h Ê f (c kf (c k f (c kf (c k. If f (c 0, the kf (c k f (c Ê f (c f (c 0; likeise if f (c 0, the 0 f (c f (c. (a If f (c 0, the h 0 Ê f (c h 0 ad 0 h Ê f (c h 0. Therefore, f(c is a local maimum. (b If f (c 0, the h 0 Ê f (c h 0 ad 0 h Ê f (c h 0. Therefore, f(c is a local miimum. 5. The time it ould take the ater to hit the groud from height y is É y g, here g is the acceleratio of gravity. The product of time ad eit velocity (rate yields the distace the ater travels: y Î Î h g g g D(y É È6(h y 8 É ahy y b, 0 Ÿ y Ÿ h Ê D (y É ahy y b (h y Ê 0, ad h are critical poits. No D(0 0, D ˆ h 8h h ad D(h 0 Ê the best place to drill the hole is at y. Èg b a ta ta a h ta ta h b a a ta h ta a bh 6. From the figure i the tet, ta ( ; ta ( ; ad ta. These equatios h give h a ta h a ta. Solvig for ta gives ta h a(b a or h ah a(b a bta bh. Differetiatig both sides ith respect to h gives d d bh dh dh h a(b a È h ta ah a(b a bsec b. The 0 Ê h ta b Ê h Š b Ê bh bh ab(b a Ê h a(b a Ê h a(a b.
13 7. The surface area of the cylider is S r rh. From the diagram e have r H h h RH rh R H Ê R ad S(r r(r h r ˆ H r H r R ˆ H r Hr, here 0 Ÿ r Ÿ R. R Chapter Additioal ad Advaced Eercises 35 Case : H R Ê S(r is a quadratic equatio cotaiig the origi ad cocave upard Ê S(r is maimum at r R. Case : H R Ê S(r is a liear equatio cotaiig the origi ith a positive slope Ê S(r is maimum at r R. Case 3: H R Ê S(r is a quadratic equatio cotaiig the origi ad cocave doard. The ds ˆ H r H ad ds 0 Ê ˆ H r H 0 Ê r RH. For simplificatio dr R dr R (H R e let r RH (H R. RH Ê Ê (H R (a If R H R, the 0 H R H (H R R hich is impossible. R (b If H R, the r R R Ê S(r is maimum at r R. H RH (c If H R, the R H H Ê H (H R Ê Ê R Ê r R. Therefore, RH (H R S(r is a maimum at r r. (H R (H R Coclusio: If H (0ßR] or H R, the the maimum surface area is at r R. If H (RßR, the r R RH hich is ot possible. If H (R ß_, the the maimum is at r r. (H R 8. f( m Ê f ( m ad f ( 0 he 0. The f ( 0 Ê yields a Èm miimum. If f Š 0, the Èm Èm Èm 0 Ê m. Thus the smallest acceptable value for m is. Èm sia& b sia& b! sia& b!! 9. (a lim lim lim Ä! Ä! a& b Ä! a& b & sia& bcosa b sia& bsia b& cosa& bcosa b (b lim sia& bcota b lim lim Ä! Ä! sia b Ä! cosa b (c lim csc È lim lim lim lim Ä! Ä! È siè siè cosè Ä! Ä! siš È È Ä! lim Ä! cosš È & È cosš È È si (d lim asec ta b lim lim Ä / Ä / cos!þ Ä / si si cos cos cos si (e lim lim lim lim lim li Ä! ta Ä! sec Ä! ta Ä! ta Ä! ta sec m Ä! cos lim Ä! cos sia b cosa b a bsia b cosa b (f lim lim lim Ä! si Ä! cos si Ä! si cos sec sec ta sec ta sec! (g lim lim lim Ä! Ä! Ä! a ba % b % %%% (h lim lim lim Ä % Ä a ba b Ä % si si cos È b & & È & È É 0. (a lim lim lim Ä_ È & Ä_ È Ä_ b& & È È (É (b lim lim lim Ä_ ( È Ä_ Ä_! b( È
14 36 Chapter Applicatios of Derivatives. (a The profit fuctio is Pab ac eb aa bb e ac bb a. P ab e c b!. (a cb c b e e Ê. P ab e! if e! so that the profit fuctio is maimized at. (b The price therefore that correspods to a productio level yeildig a maimum profit is p¹ c e ˆ cb cb dollars. c b c e e e e % e (c The eekly profit at this productio level is Pab eˆ cb ac bbˆ cb ac bb a a. (d The ta icreases cost to the e profit fuctio is Fab ac eb aa b tb e ac b tb a. tbc cbt No F ab e c b t! he e e. Sice F ab e! if e!, F is maimized cbt he uits per eek. Thus the price per uit is p c e ˆ cbt cbt dollars. Thus, such a ta e e cbt cb t icreases the cost per uit by dollars if uits are priced to maimize profit. f a b f a b c The -itercept occurs he!ê Ê. (b By Netos method,. Here f a b. So Š a b. q q q q f a b a q a aq ba q a f a! b q q q q q!!!! a q q! q q!!!!! 3.!!! so that is a eighted average of qc qc qc Š qc Š! ad ith eights m ad m.! c a q a q a a q a!! q q q q q q q q I the case here qc e have a ad Š Š Š.! c c c c!!!! d y d d d yd d d d y y d d y yd d d d d d d. We have that a hb ay hb r ad so a hb ay h b! ad ay h b! hold. d Thus y h h, by the former. Solvig for h, e obtai h. Substitutig this ito the secod equatio yields y Œ!. Dividig by results i y Œ!. ds ds kt 5. dt ks Ê s k dt Ê l s kt C Ê s s! e Ê the th cetury model of free fall as epoetial; ote that the motio starts too sloly at first ad the becomes too fast after about 7 secods
15 6. To vies of the graph of y 000 (.99 are sho belo. Chapter Additioal ad Advaced Eercises 37 (a At about there is a miimum (b There is o maimum; hoever, the curve is asymptotic to y 000. The curve is ear 000 he 63. kt kt!! Ê!!!!! È!! k k ab! È!! k È!! k k k 7. (a aab t s ab t k ak! bês ab t kt C, here s a! bêc Êsab t kt. So sab t t C here sa b C so sab t t. No sab t he kt t. Solvig for t e obtai t. At such t e at s t, thus k Š! or k Š!. I either case e obtai!! k!!! so that k Þ( ft/sec. kt %% (b The iitial coditio that s a! b%% ft/sec implies that s ab t kt %% ad sab t %% t here k is as above. The car is stopped at a time t such that s ab t kt %%!Êt k. At this time the car has traveled a distace s ˆ %% ˆ %% k ˆ %% %% * ˆ!! k k %% k k k * & feet. Thus halvig the iitial velocity quarters stoppig distace. 8. hab f ab g ab h ab fabf ab gabg ab Ê fabf abgabg ab fabgabgabafabb!!. Thus hab c, a costat. Sice h a! b &, ha b & for all i the domai of h. Thus h a! b &. d 9. Yes. The curve y satisfies all three coditios sice everyhere, he!, y!, ad! everyhere. 30. y for all Êy C here C ÊC %Êy %. d y d t! % % t t ab a b ab a b Î % Î a Cb Î Î Î Î %Î % a b a b Ê a b ˆ C Ê a b ˆ C b % Ê Î% Î% a% bb a% bb È Î% Ê C. Thus v! s a! b b. 3. s ab t a t Ê v s ab t C. We seek v s a! b C. We ko that sat b b for some t ad s is at a maimum for this t. Sice s t Ct k ad s!! e have that s t Ct ad also s t! so that t C. So C C b C C b C b C Î Î Î Î % Î Î % % &Î % Î % % % &Î % Î % % ab & a! b & ab & & 3. (a s ab t t t Ê vab t s ab t t t k here va! b k Ê vab t t t Þ (b s t t t t k here s k. Thus s t t t t. 33. (a L k ˆ a b cot b csc dl b csc b csc cot dl Ê k Š ; solvig 0 R% r% d R% r% d % % % % Ê r b csc br csc cot 0 Ê (b csc ar csc R cot b 0; but b csc Á 0 sice % % r R % % Á Ê r csc R cot 0 Ê cos Ê cos Š, the critical value of % 6 (b cos ˆ 5 cos ( r R % %
16 38 Chapter Applicatios of Derivatives ad f ab f a b 3 f a b 3. (a If 3 0, the 0 Ê. (b fab Ê 3 a3 b 35. The graph of fab a b c ith a! is a parabola opeig upards. Thus fa b! for all if fa b! for at most b Éa bb % ac a oe real value of. The solutios to fa b! are, by the quadratic equatio. Thus e require a bb % ac Ÿ! Ê b ac Ÿ!. 36. (a Clearly fab aa b b ÞÞÞaa b b! for all. Epadig e see f a b aa ab b bþþþaa ab bb aa a ÞÞÞab aab ab ÞÞÞ abbab b ÞÞÞ b b!. ÞÞÞ Ÿ ÞÞÞ ÞÞÞ Thus aa b a b ÞÞÞ a b b aa a ÞÞÞa bab b ÞÞÞb b Ÿ! by Eercise 35. Thus aab ab abb aa a abab b b b. (b Referrig to Eercise 35: It is clear that fa b! for some real Íb % ac!, by quadratic formula. No otice that this implies that f a b aabb ÞÞÞaa bb aa a ÞÞÞab aab ab ÞÞÞ abbab b ÞÞÞbb! Íaab ab ÞÞÞ abb aa a ÞÞÞabab b ÞÞÞbb! Íaab ab ÞÞÞ abb aa a ÞÞÞabab b ÞÞÞbb But o fab!íab! for all i ßßÞÞÞßÍ a b! for all i ßßÞÞÞß. i i i i
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