274 Chapter 4 Applications of Derivatives

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1 274 Chapter 4 Applicatios of Derivatives Ê % a%) Ê * %) Ê %) Ê. We discard as a etraeous solutio, leavig. Sice D a for % ad D a for %, the critical poit correspods to the miimum distace. The miimum distace is ÈD a. Geometry Method: The semicircle is cetered at the origi ad has radius %. The distace from the origi to Š ß È is Ê Š È. The shortest distace from the poit to the semicircle is the distace alog the radius () cotaiig the poit Š ß È. That distace is %. The miimum distace is from the poit Š ß È to the poit Š ß È o the graph of y È, ad this occurs at the value here Da, the distace squared, has its miimum value. a 6. (a) The ase radius of the coe is r ad so the height is h Èa r a É a ˆ. Therefore, V a rh ˆ a Éa ˆ a. () To simplify the calculatios, e shall cosider the volume as a fuctio of r: volume fa r r Èa r, here r a. f r r a r r d r a r a Š È a Š È r ra a r a r dr Èa r Èa r ar r ra a r a aè Èa r Èa r. The critical poit occurs he r, hich gives r a É. The h Èa r Éa a Éa aè aè aè. Usig r ad h, e may o fid the values of r ad h for the give values of a. % È % È & È & È Whe a % : r, h à Whe a & : r, h à Whe a : r È, h Èà Whe a ) : r, h à ) È ) È aè aè r h È (c) Sice r ad h, the relatioship is. 62. (a) Let represet the fied value of at the poit P, so that P has the coordiates a ß a, ad let m f a e the slope of the lie RT. The the equatio of the lie RT is y ma a. The y-itercept of this lie is m a ma a a m, ad the -itercept is the solutio of ma a, or. Let O desigate the origi. The m

2 (Area of triagle RST) (Area of triagle ORT) (-itercept of lie RT)(y-itercept of lie RT) ˆ m a m aam mˆ m a ˆ m a m m a m a m mˆ m ˆ m Sustitutig for, f a for m, ad fa for a, e have Aa f a. Sectio 4.5 Applied Optimizatio Prolems 275 f a f a () The domai is the ope iterval a. To graph, let y fa É ß & &, y f a NDERay, ad y y y Aa y Š. The graph of the area fuctio y Aa is sho elo. The vertical asymptotes at ad correspod to horizotal or vertical taget lies, hich do ot form triagles. (c) Usig our epressio for the y-itercept of the taget lie, the height of the triagle is a m fa f a & È & È 2 È 2 È We may use graphig methods or the aalytic method i part (d) to fid that the miimum value of Aa occurs at )Þ. Sustitutig this value ito the epressio aove, the height of the triagle is 5. This is times the y-coordiate of the ceter of the ellipse. (d) Part (a) remais uchaged. Assumig C B, the domai is aß C. To graph, ote that fa B BÉ B B ÈC B B ad f a a. Therefore e have C f a C B C ÈC CÈC B B ÈC B Š BC B È C Š È C A a fa f a CÈ C Œ C c B CÈ È C B C C c B Š BC BÈC Š ÈC B BCÈC BaC BCÈC BCÈC BCŠ C C BCŠ C ÈC È BCÈC ÈC ÈC a C ÈC C ÈC ÈC a ac Aa BC Š Š Š c Š Š c È C c ÈC c BCŠ C ÈC ac Š CÈC Š ÈC È BCŠ C ÈC ac C C CÈC ac ÈC BCŠ C È C BCŠ C ÈC ac ÈC ac Î C Š CÈ C C C CaC C ÈC BC C C Š È ac Î Š C CÈC

3 276 Chapter 4 Applicatios of Derivatives To fid the critical poits for C, e solve: C CÈC % % % Ê % % C C C C % Ê% C Ê a% C. The miimum value of Aa for C occurs at the critical poit CÈ C %, or. The correspodig triagle height is am fa f a BÈ C B CÉC C B C B C % C % C % BÈ C BŠ CÉC B C B Cˆ B B B BC % C B This shos that the traigle has miimum arrea he its height is B. 4.6 INDETERMINATE FORMS AND LHOPITALS ^ RULE. lhopital: ^ lim or lim lim lim Ä 2 ¹ Ä 2 Ä 2 a a Ä lhopital: ^ si 5 5 cos 5 si 5 si 5 lim 5 or lim 5 lim 5 5 Ä 0 ¹ Ä Ä lhopital: ^ lim lim lim or lim lim Ä_ Ä_ Ä_ Ä_ Ä_ 4. lhopital: ^ lim lim or lim lim Ä Ä Ä Ä a lim Ä a a a a a lhopital: ^ cos si cos cos acos cos lim lim lim or lim lim Ä Ä 2 Ä 2 Ä ˆ Ä co si si si lim lim Ä 2 a cos ˆ ˆ ˆ Ä cos 2 s 6. lhopital: ^ % % lim lim lim or lim lim Ä_ Ä_ Ä_ Ä_ Ä_ lim lim Ä 2 Ä lim lim 0 Ä5 5 Ä5 t 4t5 t 4 ( ) 4 2 t t 2 2t 2( ) 7 9. lim lim t Ä t Ä t t 4t t 2t 0. lim lim t Ä t Ä lim lim lim Ä_ Ä_ Ä_

4 lim lim lim Ä_ Ä_ Ä_ si t acos t (2t). lim lim 0 t Ä 0 t t Ä 0 si 5t 5 cos 5t 4. lim lim 5 t Ä 0 t t Ä lim lim lim 6 Ä 0 cos Ä 0 si Ä 0 cos si cos si cos lim lim lim lim Ä 0 Ä 0 Ä 0 Ä 0 2) lim lim 2 ) Ä Î2 cos(2 ) ) ) Ä Î2 si(2 ) ) si ˆ ) 8. lim lim ) Ä Î si ˆ ) ) ÄÎ cos ˆ ) si ) cos ) si ) cos 2 2 si 2 4 cos 2 ( 4)( ) 4 9. lim lim lim ) Ä Î2 ) ) Ä Î2 ) ) Ä Î2 ) l si ( ) cos ( ) 20. lim lim Ä Ä lim lim lim lim 2 Ä 0 l (sec ) Ä 0 ˆ sec ta Ä 0 ta Ä 0 sec sec l (csc ) ˆ csc cot csc 22. lim lim lim lim Ä Î2 ˆ ˆ Ä Î2 2ˆ ˆ Ä Î2 2ˆ ˆ Ä Î2 t( cos t) ( cos t) t(si t) si t (si t t cos t) t si t cos t si t 2. lim lim lim t Ä 0 t Ä 0 t Ä 0 cos tcos tcos tt si t 0 lim t Ä 0 cos t t si t si t t cos t cos t (cos t t si t) ( 0) Sectio 4.6 Idetermiate Forms ad LHopitals ^ Rule 277 cot csc 24. lim lim lim 2 t Ä 0 cos t t Ä 0 si t t Ä 0 cos t ˆ 25. lim ˆ sec lim lim ˆ ÄÐÎ2Ñc ÄÐÎ2Ñc cos ÄÐÎ2Ñc si ˆ 26. lim ˆ ta lim lim ˆ lim si ÄÐÎ2Ñc ÄÐÎ2Ñc cot ÄÐÎ2Ñc csc ÄÐÎ2Ñc si si (l )(cos )) a (l )() ) ) 27. lim lim l ) Ä 0 ) ) Ä 0 ) ) ˆ ˆ lˆ ˆ 28. lim lim l l l 2 l 2 ) Ä 0 ) ) Ä 0 ˆ 2 () a2 ()(l 2) a (l 2) 2 (l 2) 2 l 29. lim lim Ä 0 Ä 0 a l l l 2 2 l 2 2 l 2 l 0. lim lim Ä 0 Ä 0 l () l () ˆ. lim lim (l 2) lim (l 2) lim (l 2) lim l 2 Ä_ log 2 Ä_ ˆ l Ä_ ˆ Ä_ Ä_ l

5 278 Chapter 4 Applicatios of Derivatives ˆ l log 2 l l l l l 2. lim lim lim lim Ä_ log ( ) ˆ Ä_ l ( ) l Ä_ l ( ) ˆ l ˆ Ä_ ˆ l lim Ä_ ˆ lim Ä_ l l l l Š l ˆ 2 2 l a 2 Š lim lim 2 lim lim lim Ä l Ä ˆ Ä 2 Ä 2 2 Ä e l ae Š e e e 0 e c 4. lim lim lim lim Ä l Ä ˆ Ä e Ä e È5y 25 5 Î (5y 25) 5 ˆ cî (5y 25) (5) 5 5. lim lim lim lim y Ä 0 y y Ä 0 y y Ä 0 y Ä 0 2È 5y 25 Èay a a aay a a ˆ aay a (a) a 6. lim lim Î lim lim, a 0 y Ä 0 y y Ä 0 y y Ä 0 y Ä 0 2È ay a 7. lim [l 2 l ( )] lim l ˆ l lim l lim l 2 Ä_ Ä_ Š Š Ä_ Ä_ 8. lim (l l si ) lim l ˆ l lim l lim l 0 Ä Ä si Š Š Ä si Ä cos siaa hsi a cosaa h cos a 9. lim lim h Ä 0 h h Ä lim ˆ lim Š lim Ä Ä Ä lim Ä Š cos cos ( )( si ) ()(0) 6 cos cos si 0 ( )(si ) si ( )(cos ) si si si cos 4. lim ˆ lim lim lim Ä Ä Š Ä Š l ( ) Ä Š l ( )(l ) (l ) ( ) ˆ ( l ) lim Ä Š (l ) (0 ) 42. lim (csc cot cos ) lim ˆ cos cos ( cos ) (si )(cos ) lim Š Ä Ä si si Ä si si cos si 00 lim Ä Š cos cos ) si ) cos ) 4. lim ) lim ) lim ) ) Ä 0 e ) ) Ä 0 e ) Ä 0 e e (h) e e 44. lim lim lim h Ä 0 h h Ä 0 h h Ä 0 t t t t e t e 2t e 2 e 45. lim lim lim lim t Ä_ et t Ä_ et t Ä_ et t Ä_ et lim e lim lim lim 0 Ä_ Ä_ e Ä_ e Ä_ e _ ÎÐcÑ ÎÐcÑ l 47. The limit leads to the idetermiate form. Let f() Ê l f() l a. No ÎÐcÑ lim l f() lim lim. Therefore lim lim f() lim e e Ä Ä Ä Ä Ä Ä l ˆ l fðñ e

6 Sectio 4.6 Idetermiate Forms ad LHopitals ^ Rule 279 _ ÎÐcÑ ÎÐcÑ l 48. The limit leads to the idetermiate form. Let f() Ê l f() l a. No l ˆ ÎÐcÑ lim l f() lim lim. Therefore lim lim f() lim e e e Ä Ä Ä Ä Ä Ä Î Î l (l ) l fðñ 49. The limit leads to the idetermiate form _. Let f() (l ) Ê l f() l (l ). No l (l ) ˆ lim l f() lim Ä_ Ä_ lim Ä_ l 0. Therefore lim (l ) lim f() Ä_ Ä_ lim e e Ä_ l fðñ _ ÎÐ c Ñ 50. The limit leads to the idetermiate form. Let f() (l ) Ê l f() e lim l f() Ä e l (l ) ˆ e l (l ) l fðñ Îe l ÎÐceÑ lim lim. Therefore (l ) lim f() lim e e e e Ä e Ä e Ä e Ä e 5. The limit leads to the idetermiate form 0. Let f() Ê l f() l. Therefore cîl l fðñ lim lim f() lim e e e Ä Ä Ä c Î l l l l Îl Î 52. The limit leads to the idetermiate form _. Let f() Ê l f() l. Therefore lim Ä_ fðñ lim f() lim e e e Ä_ Ä_ ÎÐ Ñ l ( 2) 2l 5. The limit leads to the idetermiate form _. Let f() ( 2) Ê l f() 2 l l ( 2) Ê lim l f() lim lim lim. Therefore lim ( 2) Ä_ Ä_ 2 l Ä_ 2 Ä_ Ä_ l fðñ Î lim f() lim e e Ä_ Ä_ 54. The limit leads to the idetermiate form. Let f() ae Ê l f() ÎÐ2lÑ _ Î lae l ae e Ä 0 Ä 0 Ä 0 e a Ä 0 Ä 0 l fðñ lim e e Ä 0 Ê lim l f() lim lim 2. Therefore lim e lim f() 55. The limit leads to the idetermiate form 0. Let f() Ê l f() l Ê l f() l ˆ lim l f() lim lim lim ( ) 0. Therefore lim lim f() Ä Ä ˆ Ä Š Ä Ä Ä l fðñ lim e e Ä c 56. The limit leads to the idetermiate form _. Let f() ˆ l a Ê l f() c Ê lim l f() Ä Š c c lim c lim lim 0. Therefore lim ˆ c c lim f() Ä Ä Ä Ä Ä l fðñ lim e e Ä È lim lim lim 9 Ä_ È É Ä_ É Ä_ È l ˆ È 58. lim Ä È si Ê si É lim Ä sec cos 59. lim lim lim Ä Î2c ta ˆ ˆ Ä Î2c cos si Ä Î2c si

7 280 Chapter 4 Applicatios of Derivatives cot ˆ cos 60. lim lim lim cos Ä csc si Ä ˆ Ä si 0 _ 6. Part () is correct ecause part (a) is either i the or form ad so lhopitals ^ rule may ot e used. 62. Part () is correct; the step lim lim i part (a) is false ecause lim is Ä 0 2 cos Ä 0 si Ä 0 2 cos ot a idetermiate quotiet form. 0 _ Part (d) is correct, the other parts are idetermiate forms ad caot e calculated y the icorrect arithmetic d d f () f() d d g () g() 64. (a) For Á 0, f () ( 2) ad g () ( ). Therefore, lim, hile lim Ä 0 Ä () This does ot cotradict lhopitals ^ rule ecause either f or g is differetiale at 0 (as evideced y the fact that either is cotiuous at 0), so lhopitals ^ rule does ot apply. 9 si 9 9 cos If f() is to e cotiuous at 0, the lim f() f(0) Ê c f(0) lim lim Ä 0 Ä 0 5 Ä 0 27 si 8 cos 27 lim lim. Ä 0 0 Ä (a) () The limit leads to the idetermiate form : È È È È lim Š È lim Š È a Š lim Š lim Ä_ Ä_ Ä_ Ä_ lim Ä_ È É 67. The graph idicates a limit ear. The limit leads to the idetermiate form 0 2 ( ) È 2 0 : lim Ä 9 Î Î Î lim lim Ä Ä cî _ 68. (a) The limit leads to the idetermiate form. Let f() ˆ Ê l f() l ˆ Ê lim l f() Ä_ l ˆ l a Š c lim lim lim c lim Ä_ ˆ c c Ä_ Ä_ Ä_ ˆ 0 Ê lim lim f() lim e e e Ä_ ˆ l fðñ Ä_ Ä_ c c

8 Sectio 4.6 Idetermiate Forms ad LHopitals ^ Rule 28 () ˆ , , Both fuctios have limits as approaches ifiity. The fuctio f has a maimum ut o miimum hile g has o etrema. The limit _ of f() leads to the idetermiate form. (c) Let f() ˆ Ê l f() l a 2 l a Š c c Ê lim l f() lim Ä_ Ä_ c a a 6 c lim Ä_ c lim Ä_ lim Ä_ lim Ä_ 0. Therefore lim ˆ l fðñ lim f() lim e e Ä_ Ä_ Ä_ rk k l a rk l a rk Š r k k k k rk 69. Let f(k) ˆ r c c rk Ê l f(k) Ê lim lim c lim k Ä_ k Ä_ k Ä_ rk r r k lim lim r. Therefore lim ˆ l fðkñ r lim f(k) lim e e. k Ä_ k r k Ä_ k Ä_ k k Ä_ k Ä_ c c c c c c y ˆ () l y 70. (a) y Î l Ê l y Ê Ê y ˆ l Î a. The sig patter is y Îe ±±hich idicates a maimum value of y e he e e Î l y ˆ a 2 l 2 l y () y Ê l y Ê Ê y ˆ Î a. The sig patter is % y Î2e ±± hich idicates a maimum of y e he Èe Èe Î l ˆ a (l ) ˆ c c ( l ) Î (c) y Ê l y 2 Ê y 2. The sig patter is Îe y ±± hich idicates a maimum of y e he È e È e Î l Î l ÐlÑÎ (d) lim lim ˆ e lim e ep Š lim ep lim e Ä_ Ä_ Ä_ Ä_ Š ˆ Ä_ 7. (a) We should assig the value to f() (si ) to make it cotiuous at 0. l (si ) l (si ) ˆ si (cos ) Š () l f() l (si ) ˆ Ê lim l f() lim ˆ lim Ä Ä Ä 2 lim lim 0 Ê lim f() e Ä 0 ta Ä 0 sec Ä 0 (c) The maimum value of f() is close to ear the poit.55 (see the graph i part (a)).

9 282 Chapter 4 Applicatios of Derivatives (d) The root i questio is ear (a) Whe si 0 there are gaps i the sketch. The ih of each gap is. () Let f() (si ) ta Ê l f() (ta ) l (si ) Ê lim l f() lim Ä Î2c Ä Î2 ˆ si (cos ) c l (si ) cot cos lim lim 0 Ä Î2c csc Ä Î2c ( si ) Ê lim f() e. Similarly, Ä Î2 c lim f() e. Therefore, lim f(). Ä Î2 Ä Î2 (c) From the graph i part () e have a miimum of aout at 0.47 ad the maimum is aout.49 at cos 7. Graphig fa o th ido Òß Ó y ÒÞ&ß Ó it appears that lim fa. Hoever, e see that if e let Ä 0 u, the lim f a lim cos u lim si u lim cos u. Ä 0 u Ä 0 u u Ä 0 u u Ä 0 f a c fa fa ga c ga ga % c 74. (a) We seek c i aß so that. Sice f a c ad g a c c e have that Ê c. fa c f a fa a a ga c g aga a a c a () We seek c i aaß so that. Sice f a c ad g a c c e have that Ê c a. f ac (c) We seek c i aß so that fa fa. Sice f a c c % ad g a c c e have that gac ga ga * c % È( È( c Ê c Ê c.

10 PA CE 75. (a) By similar triagles, here E is the poit o AB such that CE ¼ AB : AB EB Sectio 4.7 Netos Method 28 cos ) acos ) ) ) si ) ) si ) ) acos ) ) si ) cos ) ) cos ) si ) si ) ) cos ) si ) a ) Ä 0 ) Ä 0 ) si ) ) Ä 0 cos ) ) Ä 0 si ) ) Ä 0 si ) ) asi ) cos ) cos ) ) si ) cos ) lim lim ) Ä 0 cos ) ) Ä 0 cos ) Thus, sice the coordiates of C are acos ) ß si ). Hece,. () lim lim lim lim lim ) acos ) (c) We have that lim cos ) a a ) lim acos ) lim acos ) ) Ä_ ) Ä_ ) si ) ) Ä_ ) si ) As ) Ä _, a cos ) oscillates etee ad, ad so it is ouded. Sice lim ˆ ), ) Ä_ ) si ) ) lim a cos ). Geometrically, this meas that as ) Ä_, the distace etee poits P ad D ) Ä_ ) si ) approaches Throughout this prolem ote that r y, r y ad that oth r Ä_ ad y Ä_ as Ä Þ (a) lim r y lim ) Ä Î2 ) Ä Î2 ry () lim r y lim ) Ä Î2 ) Ä Î2 r ry y y y y y y y r y r r r y ) Ä Î2 r ) ) Ä Î2 ) Ä Î2 (c) We have that r y ar yar ry y y. Sice lim y lim si y e have that lim r y. ) 4Þ7 NEWTONS METHOD y Ê y 2 Ê ; Ê 9 Ê Ê.6905; Ê 2 Ê y Ê y Ê ; 0 Ê 0 7 Ê % % y Ê y 4 Ê ; Ê Ê.6542; Ê Ê

11 284 Chapter 4 Applicatios of Derivatives ; y 2 Ê y 2 2 Ê ; 0 Ê 0 Ê Ê Ê Oe ovious root is 0. Graphig e ad 2 shos that 0 is the oly root. Takig a aive approach e ca use Netos Method to estimate the root as follos: Let fa e 2,, ad ec 2 0 ec 2. Performig iteratios o a calculator, spreadsheet, or CAS gives 0.594, , You may get differet results depedig upo hat you select for fa ad, ad hat calculator or computer you may use Graphig ta a ad 2 shos that there is oly oe root ad it is etee 0. ad 0.4. Let f a tac a 2 f a 2 2 fa ta a2, 0., ad. Performig iteratios o a calculator, spreadsheet, or CAS gives , 0.729, You may get differet results depedig upo hat you select for fa ad, ad hat calculator or computer you may use. fa 7. f( ) 0 ad f () Á Ê gives Ê f a Ê for all 0. That is, all of the approimatios i Netos method ill e the root of f() It does matter. If you start too far aay from, the calculated values may approach some other root. Startig ith 0.5, for istace, leads to as the root, ot. f a f a 9. If h 0 Ê h È f( ) f() f(h) f(h) h h hš Èh Š 2Èh h; Š È h if h 0 Ê h È f( ) f( h) f() f( h) h h hš Èh Š 2Èh h. c Š È 2 h Î 0. f() Ê f () ˆ Î Ê ˆ Î cî 2 ; Ê 2, 4, 8, ad % 6 ad so forth. Sice kk 2l cl e may coclude that Ä _ Ê k k Ä _.. i) is equivalet to solvig. ii) is equivalet to solvig. iii) is equivalet to solvig. iv) is equivalet to solvig. All four equatios are equivalet. 0.5 si 2. f() 0.5 si Ê f () 0.5 cos Ê 0.5 cos ; if.5, the.49870

12 Sectio 4.7 Netos Method 285. For Þ, the procedure coverges to the root Þ)&&ÞÞÞÞ (a) () (c) (d) Values for ill vary. Oe possile choice is Þ. (e) Values for ill vary. 4. (a) f() Ê f () Ê Ê the to egative zeros are.5209 ad () The estimated solutios of 0 are.5209, 0.470, (c) The estimated -values here % g() has horizotal tagets are the roots of g (), ad these are.5209, 0.470, ta a 2 5. f() ta 2 Ê f () sec 2 Ê sec a ; Ê Ê Ê % 6. f() Ê f () Ê ; if 0.5, the % ; if 2.5, the % %

13 286 Chapter 4 Applicatios of Derivatives 7. (a) The graph of f() si 0.99 i the ido 2 Ÿ Ÿ 2, 2 Ÿ y Ÿ suggests three roots. Hoever, he you zoom i o the -ais ear.2, you ca see that the graph lies aove the ais there. There are oly to roots, oe ear, the other ear 0.4. () f() si 0.99 Ê f () cos 2 Ê ad the solutios si ( ) 0.99 cos ( ) 2 are approimately ad (a) Yes, three times as idicted y the graphs () f() cos Ê f () si Ê cos a si a ; at approimately , , ad e have cos % 9. f() 2 4 Ê f () 8 8 Ê ; if 2, the ; if % , the ; the roots are approimately ad ecause f() is a eve fuctio. ta a 20. f() ta Ê f () sec Ê sec a ; Ê.97 Ê.459 ad e approimate to e Graphig e ad shos that there are to places here the curves itersect, oe at 0 ad the other c f e etee 0.5 ad 0.6. Let fa e, 0.5, ad f 0 a. 2 2 e c2 Performig iteratios o a calculator, spreadsheet, or CAS gives , , , (You may get differet results depedig upo hat you select for fa ad, ad hat calculator or 4 0 computer you may use.) Therefore, the to curves itersect at 0 ad Graphig la ad shos that there are to places here the curves itersect, oe etee ad 0.9, ad the other etee 0.5 ad 0.6. Let fa la, ad a 2 f a f a. Performig iteratios o a calculator, spreadsheet, or CAS ith 0.5 gives , lˆ 2 c2 c , , ad ith 0.9 gives , , , (You may get differet results depedig upo hat you select for fa ad, ad 4 0 hat calculator or computer you may use.) Therefore, the to curves itersect at ad If f() 2 4, the f() 0 ad f(2) 8 0 Ê y the Itermediate Value Theorem the equatio has a solutio etee ad 2. Cosequetly, f () 2 ad The Ê.2 Ê.7975 Ê Ê % Ê the root is approimately.795.

14 % % 24. We ish to solve Let f() 8 4 9, the f () Ê. approimatio of correspodig root % Sectio 4.8 Atiderivatives 287 % i i i 25. f() 4 4 Ê f () 6 8 Ê. Iteratios are performed usig the f a f a i i i i % procedure i prolem i this sectio. (a) For or Þ), i Ä as i gets large. () For Þ& or Þ&, i Ä as i gets large. (c) For Þ) or, i Ä as i gets large. (d) (If your calculator has a CAS, put it i eact mode, otherise approimate the radicals ith a decimal value.) È2 È2 7 7 i È2 È2 7 7 For or, Netos method does ot coverge. The values of alterate etee or as i icreases. 26. (a) The distace ca e represeted y D() É( 2) ˆ, here 0. The distace D() is miimized he ˆ f() ( 2) ˆ is miimized. If f() ( 2), the f () 4 a ad f () 4 a 0. No f () 0 Ê 0 Ê a Ê. 2 a () Let g() a Ê g () a (2) Ê ; Ê to five decimal places. Œ Î Ñ c2 ÏŠ c Ò % i % * 27. f() ( ) Ê f () 40( ) Ê. With 2, our computer % a 40 a * 9 40 gave )( )) )* â.05, comig ithi 0.05 of the root. 28. f() Ê f () 7.2 Ê ; 2 Ê Ê Ê Ê % hich is 2.45 to to decimal places. Recall that 0 % ch O d Ê ch O d () a0 % (2.45) a0 % ANTIDERIVATIVES. (a) () (c) 8 8 ) ) 2. (a) () (c) 8 c c. (a) () (c)

15 288 Chapter 4 Applicatios of Derivatives 4 2 c c 4. (a) () (c) (a) () (c) 2 % (a) () (c) 7. (a) È () È 2 (c) È 2È %Î Î %Î Î 4 8. (a) () (c) Î Î Î 9. (a) () (c) Î Î Î 0. (a) () (c). (a) l l l () 7 l l l (c) 5 l ll (a) l l l () l l l (c) l ll cos ( ). (a) cos ( ) () cos (c) cos () 4. (a) si ( ) () si ˆ (c) ˆ 2 si ˆ si 5. (a) ta () 2 ta ˆ 2 (c) ta ˆ 6. (a) cot () cot ˆ (c) 4 cot (2) 7. (a) csc () csc (5) (c) 2 csc ˆ (a) sec () sec () (c) sec ˆ /2 9. (a) e () e (c) 2e 2 4/ / (a) e () e (c) 5e 5 2. (a) () 2 (c) ˆ l l 2 la5/ È È2 È È2 22. (a) () (c) 2. (a) 2 si () ta (c) ta a l a/2 l 2 l (a) ˆ () 2 (c) l ll 25. ( ) d C 26. (5 6) d 5 C 27. ˆ t t % C 28. Š 4t t t t t t C 4 6

16 % 5 & Sectio 4.8 Atiderivatives a2 5 7 d 7 C 0. a d C c. ˆ d ˆ d C C c 2. ˆ 2 2 d ˆ d Š C C Î Î &Î% 4 2 4È 4 Î cî%. d C C 4. d C C Î %Î 5. ˆ È È d ˆ Î Î 2 Î %Î d C C 4 4 È 2 È Î Î 6. Š d ˆ Î Î 2 Î Î d Š 2 Š C 4 C Î% 2 7. Š 8y dy ˆ Î% 8y 2y 8y y 8 Î% dy 2 Š C 4y y C y Î% 4 cî% 8. Š dy ˆ &Î% y y y 4 dy y Š C C 7 &Î% y c 9. 2 a d a2 2 d 2 Š C C c c 40. ( ) d a d Š C C È È y Î% t t t Î Î Î cî t t Î Î t t 2 4. Š ˆ t t Š C 2Èt C t t t È 4 t Î c cî 4 t &Î t t Š ˆ 4t t 4 Š Š C Î C t t t t 4. 2 cos t 2 si t C si t 5 cos t C ) ) si d) 2 cos C 46. cos 5 ) d ) si 5) C sec ta 47. csc d cot C 48. d C csc ) cot ) d ) csc ) C 50. sec ) ta ) d ) sec ) C e ae 5e d 5e C 52. a2e e d 2e e C 4 a. l 4 l a. 5. ae 4 d e C 54. a. d C Èt t 55. a 4 sec ta 2 sec d 4 sec 2 ta C a csc csc cot d cot csc C 57. asi 2 csc d cos 2 cot C 58. (2 cos 2 si ) d si 2 cos C

17 290 Chapter 4 Applicatios of Derivatives cos 4t si 4t t si 4t 59. ˆ cos 4t t ˆ C C cos 6t si 6t t si 6t 60. ˆ cos 6t t ˆ C C ˆ /4 d l ll 5 ta C 62. Š dy 2 si y y C È y2 y/4 Š È È 6. d C 64. Š È2c È2 d C È È2 65. a ta ) d) sec ) d) ta ) C 66. a2 ta ) d) a ta ) d ) a sec ) d) ) ta ) C 67. cot d acsc d cot C 68. a cot d a acsc d a2 csc d 2 cot C 69. cos )(ta ) sec )) d ) (si ) ) d) cos ) ) C ˆ ˆ csc ) csc ) si ) csc ) si ) csc ) si ) si ) si ) cos ) 70. d ) d ) d ) d) sec ) d) ta ) C d (7 % 2) 4(7 2) (7) d Š C (7 2) d ( 5) ( 5) () d c c 72. Š C Š ( 5) d 7. ˆ ta (5 ) C asec (5 ) (5) sec (5 ) d 5 5 d 74. ˆ cot ˆ C ˆ csc ˆ ˆ csc ˆ d d 75. ˆ C d ( )( ) () ( )( )( ) 76. ˆ C d ( ) d ( ) ( ) d d d d 77. ala C 78. a e e e a e e e 79. ˆ ta ˆ ˆ d d d a a a ˆ 2 d a a Š a a a ˆ si ˆ ˆ d d d a É d a ˆ 2 aéˆ 2 Èa 2 2 a a ta Š ta 8. If y l l a c C, the dy d ta a ata a ta a a c c c Š d d d, hich verifies the formula c

18 Sectio 4.8 Atiderivatives If y asi 2 2È si C, the c 2 asi 2 dy asi 2 si 2È Š d asi d, hich verifies the formula È È È d 2 8. (a) Wrog: d Š si C si cos si cos Á si d d d d () Wrog: ( cos C) cos si Á si (c) Right: ( cos si C) cos si cos si d sec ) sec ) d) 84. (a) Wrog: Š C (sec ) ta )) sec ) ta ) Á ta ) sec ) d () Right: ˆ ta C d) ) (2 ta )) sec ) ta ) sec ) d (c) Right: ˆ sec C d) ) (2 sec )) sec ) ta ) ta ) sec ) d (2 ) (2 ) (2) d 85. (a) Wrog: Š C 2(2 ) Á (2 ) d d a d d a () Wrog: (2 ) C (2 ) (2) 6(2 ) Á (2 ) (c) Right: (2 ) C 6(2 ) d Î Î (a) Wrog: a C a C (2 ) Á È2 È d 2 C d Î Î 2 () Wrog: Š a C a (2 ) Á È2 È d 2 d d (c) Right: Œ Š È2 C ˆ Î (2 ) C Î (2 ) (2) È2 d d 6 dy 87. Graph (), ecause d 2 B Ê y C. The y() 4 Ê C. dy d 88. Graph (), ecause B Ê y C. The y( ) Ê C. dy d Ê y 7 C; at 2 ad y 0 e have 0 2 7(2) C Ê C 0 Ê y 7 0 dy d Ê y 0 C; at 0 ad y e have 0(0) C Ê C Ê y 0 dy d 2 9. Ê y C; at 2 ad y e have 2 C Ê C Ê y or y dy d Ê y 2 5 C; at ad y 0 e have 0 ( ) 2( ) 5( ) C Ê C 0 Ê y dy d Î Î Î Î 9. Ê y C * ; at 9 C; at ad y & e have &* ( ) C Ê C % Î Ê y 9 % dy d È Î Î Î Î 94. Ê y C; at 4 ad y 0 e have 0 4 C Ê C 2 Ê y 2 ds 95. cos t Ê s t si t C; at t 0 ad s 4 e have 4 0 si 0 C Ê C 4 Ê s t si t 4

19 292 Chapter 4 Applicatios of Derivatives ds 96. cos t si t Ê s si t cos t C; at t ad s e have si cos C Ê C 0 Ê s si t cos t dr d) 97. si ) Ê r cos ( ) ) C; at r 0 ad ) 0 e have 0 cos ( 0) C Ê C Ê r cos ( ) ) dr d) 98. cos ) Ê r si( ) ) C; at r ad ) 0 e have si ( 0) C Ê C Ê r si ( ) ) dv 99. sec t ta t Ê v sec t C; at v ad t 0 e have sec (0) C Ê C Ê v sec t dv 00. 8t csc t Ê v 4t cot t C; at v 7 ad t e have 7 4 ˆ cot ˆ C Ê C 7 Ê v 4t cot t 7 dv tèt 0., t Ê v sec t C; at t 2 ad v 0 e have 0 sec 2 C Ê C Ê v sec t dv 8 t sec t Ê v 8 ta t ta t C; at t 0 ad v e have 8 ta a0taa0c Ê C Ê v 8 ta t ta t d y dy dy d d d dy d Ê 2 C ; at 4 ad 0 e have 4 2(0) (0) C Ê C 4 Ê 2 4 Ê y 4 C ; at y ad 0 e have 0 0 4(0) C Ê C Ê y 4 d y dy dy dy d d d d Ê C ; at 2 ad 0 e have C 2 Ê 2 Ê y 2 C ; at y 0 ad 0 e have 0 2(0) C Ê C 0 Ê y 2 dr 2 dr dr dr t Ê r t 2t C ; at r ad t e have 2() C Ê C 2 Ê r t 2t 2 or r t 2t t Ê t C ; at ad t e have () C Ê C 2 Ê t 2 d s t ds t ds (4) ds t t t s 4 ad t 4 e have 4 6 C Ê C 0 Ê s Ê C ; at ad t 4 e have C Ê C 0 Ê Ê s C ; at dy dy dy dy d d d d dy dy dy d d d Ê 6 C ; at 8 ad 0 e have 8 6(0) C Ê C 8 Ê 6 8 Ê 8 C ; at 0 ad 0 e have 0 (0) 8(0) C Ê C 0 Ê 8 Ê y 4 C ; at y 5 ad 0 e have 5 0 4(0) C Ê C 5 Ê y 4 5 d ) d ) d ) d ) d) d) Ê C ; at 2 ad t 0 e have 2 Ê 2t C ; at ad t 0 e d) have 2(0) C Ê C Ê 2t Ê ) t t C ; at ) È2 ad t 0 e have È 2 0 (0) C C È Ê 2 Ê t tè2 ) Ð%Ñ 09. y si t cos t Ê y cos t si t C ; at y 7 ad t 0 e have 7 cos (0) si (0) C Ê C 6 Ê y cos t si t 6 Ê y si t cos t 6t C ; at y ad t 0 e have si (0) cos (0) 6(0) C Ê C 0 Ê y si t cos t 6t Ê y cos t si t t C ; at y ad t 0 e have cos (0) si (0) (0) C Ê C 0 Ê y cos t si t t Ê y si t cos t t C %; at y 0 ad t 0 e have 0 si (0) cos (0) 0 C % Ê C% Ê y si t cos t t

20 Ð%Ñ Sectio 4.8 Atiderivatives y cos 8 si (2) Ê y si 4 cos (2) C ; at y 0 ad 0 e have 0 si (0) % cos (2(0)) C Ê C 4 Ê y si 4 cos (2) 4 Ê y cos 2 si (2) 4 C ; at y ad 0 e have cos (0) 2 si (2(0)) 4(0) C Ê C 0 Ê y cos 2 si (2) 4 Ê y si cos (2) 2 C ; at y ad 0 e have si (0) cos (2(0)) 2(0) C Ê C 0 2 Ê y si cos (2) 2 Ê y cos si (2) C ; at y ad 0 e have % 2 2 % % cos (0) si (2(0)) (0) C Ê C 4 Ê y cos si (2) 4. m y È Î Î Î Î Ê y 2 C; at (*ß4) e have 4 2(9) C Ê C 50 Ê y 2 50 dy dy dy d d d Ê y C ; at y ad 0 e have C Ê y 2. (a) 6 Ê C ; at y 0 ad 0 e have 0 (0) C Ê C 0 Ê () Oe, ecause ay other possile fuctio ould differ from of the iitial coditios y a costat that must e zero ecause dy 4 Î 4. Ê y ˆ Î %Î %Î d C; at (ß0.5) o the curve e have 0.5 C d Ê C 0.5 Ê y %Î dy ( ) d Ê C Ê y 4. Ê y ( ) d C; at ( ß) o the curve e have ( ) C dy 5. si cos Ê y (si cos ) d cos si C; at ( ß) o the curve e have d cos ( ) si ( ) C Ê C 2 Ê y cos si 2 dy 6. si Î si Ê y Î si d Î È ˆ cos C; at ( ß ) o the d Î curve e have 2 cos () C Ê C 0 Ê y È cos ds 7. (a) 9.8t Ê s 4.9t t C; (i) at s 5 ad t 0 e have C 5 Ê s 4.9t t 5; displacemet s() s() ((4.9)(9) 9 5) (4.9 5).2 uits; (ii) at s 2 ad t 0 e have C 2 Ê s 4.9t t 2; displacemet s() s() ((4.9)(9) 9 2) (4.9 2).2 uits; (iii) at s s ad t 0 e have C s Ê s 4.9t t s ; displacemet s() s() ((4.9)(9) 9 s ) (4.9 s ).2 uits () True. Give a atiderivative f(t) of the velocity fuctio, e ko that the odys positio fuctio is s f(t) C for some costat C. Therefore, the displacemet from t a to t is (f() C) (f(a) C) f() f(a). Thus e ca fid the displacemet from ay atiderivative f as the umerical differece f() f(a) ithout koig the eact values of C ad s. 8. a(t) v (t) 20 Ê v(t) 20t C; at (0 ß ) e have C 0 Ê v(t) 20t. Whe t 60, the v(60) 20(60) 200 m/sec. d s ds ds ds 9. Step : k Ê kt C ; at 88 ad t 0 e have C 88 Ê kt 88 Ê t kt s kš 88t C ; at s 0 ad t 0 e have C 0 Ê s 88t ds 88 k k ˆ 88 k ˆ 88 (88) (88) (88) k 2k k 2k Step 2: 0 Ê 0 kt 88 Ê t Step : Ê 242 Ê 242 Ê k 6

21 294 Chapter 4 Applicatios of Derivatives d s ds ds ds kt k(0) Ê kt ds k s 44t. The 0 Ê kt 44 0 Ê t ad s ˆ 44 k k 44 k 44 ˆ k ft Ê k k 45 Ê k 45 Ê k sec. 20. k Ê k kt C; at 44 he t 0 e have 44 k(0) C Ê C 44 Ê kt 44 Ê s 44t C ; at s 0 he t 0 e have 0 44(0) C Ê C 0 ˆ Î Î Î Î ds Î Î Î 2. (a) v a 5t t 0t 6t C; () 4 Ê 4 0() 6() C Ê C 0 Î Ê v 0t 6t ˆ Î Î &Î Î &Î Î () s v 0t 6t 4t 4t C; s() 0 Ê 0 4() 4() C Ê C 0 &Î Ê s 4t 4t Î d s ds ds ds Ê 5.2t C ; at 0 ad t 0 e have C 0 Ê 5.2t Ê s 2.6t C ; at s 4 ad t 0 e have C 4 Ê s 2.6t 4. The s 0 Ê 0 2.6t 4 Ê t.24 sec, sice t 0 É d s ds ds ds at a(0) at he t 0 Ê s v (0) C Ê C s Ê s v t s 2. a Ê a at C; v he t 0 Ê C v Ê at v Ê s v t C ; s s 24. The appropriate iitial value prolem is: Differetial Equatio: g ith Iitial Coditios: v ad ds ds s s he t 0. Thus, g gt C ; (0) v Ê v ( g)(0) C Ê C v ds Ê gt v. Thus s a gt v gt v t C ; s(0) s (g)(0) v (0) C Ê C s. Thus s gt v t s 25. (a) f() d È C È C () g() d 2 C C ds (c) f() d ˆ È C È C (d) g() d ( 2) C C (e) [f() g()] d ˆ È ( 2) C È C (f) [f() g()] d ˆ È ( 2) C È C ds 26. Yes. If F() ad G() oth solve the iitial value prolem o a iterval I the they oth have the same first derivative. Therefore, y Corollary 2 of the Mea Value Theorem there is a costat C such that F() G() C for all. I particular, F( ) G( ) C, so C F( ) G( ) 0. Hece F() G() for all Eample CAS commads: Maple: ith(studet): f := -> cos()^2 + si(); ic := [=Pi,y=]; F := uapply( it( f(), ) + C, ); eq := eval( y=f(), ic ); solc := solve( eq, {C} ); Y := uapply( eval( F(), solc ), ); DEplot( diff(y(),) = f(), y(), =0..2*Pi, [[y(pi)=]], color=lack, liecolor=lack, stepsize=0.05, title=sectio ); Mathematica: (fuctios ad values may vary) The folloig commads use the defiite itegral ad the Fudametal Theorem of calculus to costruct the solutio of the iitial value prolems for eercises 27-0.

22 Chapter 4 Practice Eercises 295 Clear[, y, yprime] yprime[_] = Cos[] 2 Si[]; iitvalue = ; iityvalue = ; y[_] = Itegrate[yprime[t], {t, iitvalue, }] iityvalue If the solutio satisfies the differetial equatio ad iitial coditio, the folloig yield True yprime[]==d[y[], ] //Simplify y[iitvalue]==iityvalue Sice eercise 06 is a secod order differetial equatio, to itegratios ill e required. Clear[, y, yprime] y2prime[_] = Ep[/2] ; iitval = 0; iityval = 4; iityprimeval = ; yprime[_] = Itegrate[y2prime[t],{t, iitval, }] iityprimeval y[_] = Itegrate[yprime[t], {t, iitval, }] iityval Verify that y[] solves the differetial equatio ad iitial coditio ad plot the solutio (red) ad its derivative (lue). y2prime[]==d[y[], {, 2}]//Simplify y[iitval]==iityval yprime[iitval]==iityprimeval Plot[{y[], yprime[]}, {, iitval, iitval }, PlotStyle Ä {RGBColor[,0,0], RGBColor[0,0,]}] CHAPTER 4 PRACTICE EXERCISES. No, sice f() 2 ta Ê f () 2 sec 0 Ê f() is alays icreasig o its domai cos 2 2. No, sice g() csc 2 cot Ê g () csc cot 2 csc si si si (cos 2) 0 Ê g() is alays decreasig o its domai. No asolute miimum ecause lim (7 )( ) _. Net f () Ä_ Î Î ( ) (7 ) 4( ) ( ) (7 )( ) Ê ad are critical poits. Î ( ) Î ( ) Sice f 0 if ad f 0 if, f() 6 is the asolute maimum. Î a a a 2(a ) aa 2 a a a % a 8 a a a 4. f() Ê f () ; f () 0 Ê (* a a) Ê& a. We require also that f(). Thus Ê a ). Solvig oth equatios yields a 6 ad 0. No, f () so that f ± ± ± ±. Thus f chages sig at from / positive to egative so there is a local maimum at hich has a value f(). 5. ga e Ê g a e Ê g ± Ê the graph is decreasig o a_, 0, icreasig o a0, _ ; 0 a asolute miimum value is at 0; 0 is the oly critical poit of g; there is o asolute maimum value 2e ˆ 2 2e 2e 2 2 2e a 6. fa 2 Ê f a 2 2 Ê f ± Ê the graph is icreasig o a_, _ ; a2 a2 is the oly critical poit of f; there are o asolute maimum values or asolute miimum values fa 2 l o Ÿ Ÿ Ê f a Ê f ± ± ± Ê the graph is decreasig o a, 2, 2 icreasig o a2, ; a asolute miimum value is 2 2 l 2 at 2; a asolute maimum value is at.

23 296 Chapter 4 Applicatios of Derivatives fa l o ŸŸ 4 Êf a 2 2 Êf ±±± Êthe graph is decreasig o 2 4 a, 2, icreasig o a2, 4 ; a asolute miimum value is 2 l 4 at 2; a asolute maimum value is 4 at. 9. Yes, ecause at each poit of [ß Ñ ecept 0, the fuctios value is a local miimum value as ell as a local maimum value. At 0 the fuctios value, 0, is ot a local miimum value ecause each ope iterval aroud 0 o the -ais cotais poits to the left of 0 here f equals. 0. (a) The first derivative of the fuctio f() is zero at 0 eve though f has o local etreme value at 0. () Theorem 2 says oly that if f is differetiale ad f has a local etreme at c the f (c) 0. It does ot assert the (false) reverse implicatio f (c) 0 Ê f has a local etreme at c.. No, ecause the iterval 0 fails to e closed. The Etreme Value Theorem says that if the fuctio is cotiuous throughout a fiite closed iterval a Ÿ Ÿ the the eistece of asolute etrema is guarateed o that iterval. 2. The asolute maimum is kk ad the asolute miimum is k0k 0. This is ot icosistet ith the Etreme Value Theorem for cotiuous fuctios, hich says a cotiuous fuctio o a closed iterval attais its etreme values o that iterval. The theorem says othig aout the ehavior of a cotiuous fuctio o a iterval hich is half ope ad half closed, such as Òß Ñ, so there is othig to cotradict.. (a) There appear to e local miima at.75 ad.8. Poits of iflectio are idicated at approimately 0 ad. ( & % () f () 5 5 a a 5. The patter y ± ± ± ± È È& È idicates a local maimum at È 5 ad local miima at È. (c)

24 Chapter 4 Practice Eercises (a) The graph does ot idicate ay local etremum. Poits of iflectio are idicated at approimately ad. % ( % 0 ( () f () 2 5 a 2a 5. The patter f )( ± ± idicates ( È & È a local maimum at (È 5 ad a local miimum at È 2. (c) 5. (a) g(t) si t t Ê g (t) 2 si t cos t si (2t) Ê g 0 Ê g(t) is alays fallig ad hece must decrease o every iterval i its domai. () Oe, sice si t t 5 0 ad si t t 5 have the same solutios: f(t) si t t 5 has the same derivative as g(t) i part (a) ad is alays decreasig ith f( ) 0 ad f(0) 0. The Itermediate Value Theorem guaratees the cotiuous fuctio f has a root i [ ß 0]. dy d) 6. (a) y ta ) Ê sec ) 0 Ê y ta ) is alays risig o its domai Ê y ta ) icreases o every iterval i its domai () The iterval 4 ß is ot i the tagets domai ecause ta ) is udefied at ). Thus the taget eed ot icrease o this iterval. % 7. (a) f() 2 2 Ê f () 4 4. Sice f(0) 2 0, f() 0 ad f () 0 for 0 Ÿ Ÿ, e may coclude from the Itermediate Value Theorem that f() has eactly oe solutio he 0 Ÿ Ÿ. 2 È48 () 0 Ê È ad 0 Ê È () 8. (a) y Ê y 0, for all i the domai of Ê y is icreasig i every iterval i its domai () y 2 Ê y 2 0 for all Ê the graph of y 2 is alays icreasig ad ca ever have a local maimum or miimum 9. Let V(t) represet the volume of the ater i the reservoir at time t, i miutes, let V(0) a e the iitial amout ad V(440) a (400)(4,560)(7.48) gallos e the amout of ater cotaied i the reservoir after the rai, here 24 hr 440 mi. Assume that V(t) is cotiuous o [ß440] ad differetiale o (ß 440). The Mea Value Theorem says that for some t i (ß 440) e have V (t ) a (400)(4,560)(7.48) a 456,60,20 gal mi V(440) V(0) ,778 gal/mi. Therefore at t the reservoirs volume as icreasig at a rate i ecess of 225,000 gal/mi.