180 Chapter 3 Applications of Derivatives

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1 80 Chapter Applicatios of Derivatives. Let represet the legth of the rectagle i meters (! % ) The the idth is % ad the area is Aab a%b %. Sice A ab %, the critical poit occurs at. Sice, A a b! for! ad A a b! for %, this critical poit correspods to the maimum area. The rectagle ith the largest area measures m by % m, so it is a square. Graphical Support:. (a) The lie cotaiig poit P also cotais the poits (!ß ) ad (ß!) Ê the lie cotaiig P is y Ê a geeral poit o that lie is (ß ). (b) The area A() ( ), here 0 Ÿ Ÿ. (c) Whe A(), the A () 0 Ê 0 Ê. Sice A(0) 0 ad A() 0, e coclude that A ˆ sq uits is the largest area. The dimesios are uit by uit.. The area of the rectagle is A y a b, È here 0 Ÿ Ÿ. Solvig A () 0 Ê 6 0 Ê or. No is ot i the domai, ad sice A(0) 0 ad A Š È 0, e coclude that A() square uits is the maimum area. The dimesios are uits by 8 uits. 5. The volume of the bo is V() (5 )(8 ) 0 6, here 0 Ÿ Ÿ. Solvig V () 0 5 Ê 0 9 (6 )(5 ) 0 Ê or 6, but 6 is ot i the domai. Sice V(0) V() 0, V ˆ 5 %&! ( * i must be the maimum volume of 5 5 the bo ith dimesios iches. b È da È b db È00 b 6. The area of the triagle is A ba 00 b, here 0 Ÿ b Ÿ 0. The 00 b 00 b 0 Ê the iterior critical poit is b 0È. È00 b Whe b 0 or 0, the area is zero Ê A Š 0È is the maimum area. Whe a b 00 ad b 0 È, the value of a is also 0È Ê the maimum area occurs he a b.

2 Sectio.6 Applied Optimizatio 8 7. The area is A() (800 ), here 0 Ÿ Ÿ 00. Solvig A () Ê 00. With A(0) A(00) 0, the maimum area is A(00) 80,000 m. The dimesios are 00 m by 00 m. 8. The area is y 6 Ê y 08. The amout of fece eeded is P y, here! ; dp d 0 Ê 8 0 Ê the critical poits are 0 ad 9, but 0 ad 9 are ot i the domai. The P (9) 0 Ê at 9 there is a miimum Ê the dimesios of the outer rectagle are 8 m by m Ê 7 meters of fece ill be eeded. 9. (a) We miimize the eight ts here S is the surface area, ad t is the thickess of the steel alls of the tak. The surace area is S % y here is the legth of a side of the square base of the tak, ad y is its depth. The &!!!!! volume of the tak must be &!! ft Ê y. Therefore, the eight of the tak is ab tˆ. Treatig the thickess as a costat gives ab tˆ!!! for Þ!. The critical value is at!. Sice %!!! a! b t ˆ!, there is a miimum at!. Therefore, the optimum dimesios of the tak are! ft o! the base edges ad & ft deep. (b) Miimizig the surface area of the tak miimizes its eight for a give all thickess. The thickess of the steel alls ould likely be determied by other cosideratios such as structural requiremets. & 0. (a) The volume of the tak beig & ft, e have that y & Ê y. The cost of buildig the tak is cab ˆ & (&! &!, here!. The c ab!!ê the critical poits are! ad &, but! is ot a i the domai. Thus, c & b! Ê at & e have a miimum. The values of & ft ad y & ft ill miimize the cost. (b) The cost fuctio c & a % yb! y, ca be separated ito to items: () the cost of the materials ad labor to fabricate the tak, ad () the cost for the ecavatio. Sice the area of the sides ad bottom of the taks is a % y b, it ca be deduced that the uit cost to fabricate the taks is &/ft. Normally, ecavatio costs are per uit volume of ecavated material. Cosequetly, the total ecavatio cost ca be take as! y ˆ! a y b. This suggests that the!î ft uit cost of ecavatio is here is the legth of a side of the square base of the tak i feet. For the least!îft!þ'( ) epesive tak, the uit cost for the ecavatio is & ft ft yd. The total cost of the least epesive tak is (&, hich is the sum of '& for fabricatio ad (&! for the ecavatio.. The area of the pritig is (y )( 8) 50. Cosequetly, y ˆ The area of the paper is A() ˆ 50, here 8. The 8 A () ˆ ( 8) 00 Š 0 8 ( 8) ( 8) Ê the critical poits are ad 8, but is ot i the domai. Thus A (8) 0 Ê at 8 e have a miimum. Therefore the dimesios 8 by 9 iches miimize the amout of paper.. The volume of the coe is V r h, here r 9 y È ad h y (from the figure i the tet). Thus, V(y) 9 y (y ) 7 9y y y Ê V (y) a b a b a9 6y y b ( y)( y). The critical poits are ad, but is ot i the domai. Thus V () ( '6()) 0 Ê at y e have a maimum volume of V() (8)() cubic uits.

3 8 Chapter Applicatios of Derivatives ab si ). The area of the triagle is A( )), here 0 ). ab cos ) Solvig A ()) 0 Ê 0 Ê ). Sice A ()) ab si ) ˆ ) Ê A 0, there is a maimum at.. A volume V r h 000 Ê h. The amout of 000 r material is the surface area give by the sides ad bottom of 000 the ca Ê S rh r r, 0 r. The r ds 000 r 000 dr r r 0 are 0 ad È r! Ê 0. The critical poits, but 0 is ot i the domai. Sice d S 000 dr r r È È 0, e have a miimum surface area he r cm ad h cm. Comparig this result to the result foud i Eample, if e iclude both eds of the ca, the e have a miimum surface area he the ca is shorter-specifically, he the height of the ca is the same as its diameter. 5. With a volume of 000 cm ad V r h, the h. The amout of alumium used per ca is 000 r r 000 r r r r A 8r rh 8r. The A (r) 6r 0 Ê 0 Ê the critical poits are 0 ad 5, but r 0 results i o ca. Sice A (r) 6 0 e have a miimum at r 5 Ê h ad h:r 8:. & a! ba& b 6. (a) The base measures! i. by i., so the volume formula is Vab & (&. (b) We require!,!, ad &. Combiig these requiremets, the domai is the iterval a!ß & b. (c) The maimum volume is approimately 66.0 i. he Þ*' i. &! Éa&! b % a' ba(& b &! È(!! ' a b (d) V ab ' &! (&. The critical poit occurs he V a b!, at & & È( ', that is, Þ*' or 'Þ(. We discard the larger value because it is ot i the domai. Sice V ab &!, hich is egative he Þ*', the critical poit correspods to the maimum volume. The maimum volume occurs he && È( ' Þ*', hich comfimrs the result i (c).

4 Sectio.6 Applied Optimizatio 8 7. Let the radius of the cylider be r cm,! r!. The the height is È!! r ad the volume is Vab r r È!! r cm. The, V ab r r Š a rb Š È!! r a rb r % ra!! r b ra!! r b!! r!! r È!! r!! È. The critical poit for!r! occurs at r!. Sice V r! for È É É ab!r! É ad V ab r! for! É r!, the critical poit correspods to the maimum volume. The! %!!! È È dimesios are r! É )Þ' cm ad h Þ&& cm, ad the volume is %)Þ%! cm. 8. From the diagram the perimeter is P r h r, here r is the radius of the semicircle ad h is the height of the rectagle. The amout of light trasmitted proportioal to A rh r r(p r r) r da rp r r. The dr P r r 0 P P P ( )P Ê r Ê h P. Therefore, r 8 h dr da most light sice 0. gives the proportios that admit the V r r 9. The fied volume is V r h r Ê h, here h is the height of the cylider ad r is the radius of the hemisphere. To miimize the cost e must miimize surface area of the cylider added to tice the surface area of the hemisphere. Thus, e miimize C rh r r ˆ V r V 8 r r. r r The r 0 Ê V r Î Êr ˆ. From the volume equatio, h dc V 6 8 V V r dr r 8 r Î Î Î Î Î Î Î V V V V V Î ˆ dc V 6 Î. Sice Î Î Î dr r 0, these dimesios do miimize the cost. 0. The volume of the trough is maimized he the area of the cross sectio is maimized. From the diagram the area of the cross sectio is A( )) cos ) si ) cos ), 0 ). The A ()) si ) cos ) si ) a si ) si ) b ( si ) )(si ) ) so A ()) 0 Ê si ) or si ) Ê ) 6 because si ) Á he 0 ). Also, A ()) 0 for 0 ) 6 ad A ()) 0 for 6 ). Therefore, at ) 6 there is a maimum.. Note that h r ad so r È h. The the volume is give by V r h ah bh h h for!h È dv dv, ad so dh r ar b. The critical poit (for h! ) occurs at h. Sice dh! for dv!h, ad! for h È dh, the critical poit correspods to the maimum volume. The coe of greatest volume has radius È m, height m, ad volume m. a. (a) f() Ê f () a a b, so that f () 0 he implies a 6 a (b) f() Ê f () a a b, so that f () 0 he implies a. (a) sab t ' t *' t Ê vab t s ab t t *'. At t!, the velocity is v a! b *' ft/sec. (b) The maimum height ocurs he vab t!, he t. The maimum height is s a b&' ft ad it occurs at t sec. (c) Note that satb ' t *' t ' at bat ( b, so s! at t or t (. Choosig the positive value of t, the velocity he s! is v a( b ) ft/sec.

5 8 Chapter Applicatios of Derivatives. Let be the distace from the poit o the shorelie earest Jae's boat to the poit here she lads her boat. The she eeds to ro È % mi at mph ad alk ' mi at 5 mph. The total amout of time to reach the village is È% ' & È% & È% & fa b hours (! Ÿ Ÿ '). The f ab a b. Solvig f a b!, e have: % Ê & È% Ê & % a% b Ê ' Ê. We discard the egative È% & È value of because it is ot i the domai. Checkig the edpoits ad critical poit, e have f a! b Þ, % % È È f Š Þ, ad f a' b Þ'. Jae should lad her boat!þ)( miles do the shorelie from the poit earest her boat. ) h ' 5. Ê h ) ad Lab Éh a ( b ( Ɉ ' ) a ( b he!. Note that La b is miimized he fab ˆ ' ) a ( b is miimized. If f a b!, the ˆ ' ) ˆ ' a( b! () Ê a ( bˆ! Ê ( (ot acceptable sice distace is ever egative or. The L a b È*( %'Þ)( ftþ 6. (a) s s Ê si t si ˆ t Ê si t si t cos si cos t Ê si t si t cos t Ê ta t È Ê t or È (b) The distace betee the particles is s(t) ks s k si t si ˆ t ¹ si t È cos t¹ Š si t È cos t Š cos t È si t ¹ si t È cos t¹ Ê s (t) sice k k Ê critical times ad edpoits È d d kk 5 5 are 0,, 6,, 6, ; the s(0), s ˆ 0, s ˆ 6, s ˆ 0, s ˆ 6, s( ) Ê the greatest distace betee the particles is. Š si t cos t Š cos t È si t ¹ si t È cos t¹ (c) Sice s (t) e ca coclude that at t ad, s (t) has cusps ad the distace betee the particles is chagig the fastest ear these poits. 7. (a) s 0 cos ( t) Ê v 0 si ( t) Ê speed k0 si ( t) k 0ksi ( t) k Ê the maimum speed is 0. cm/sec sice the maimum value of ksi ( t) k is ; the cart is movig the fastest at t 0.5 sec,.5 sec,.5 sec ad.5 sec he ksi ( t) k is. At these times the distace is s 0 cos ˆ 0 cm ad a 0 cos ( t) Ê kk a 0 kcos ( t) kê kk a 0 cm/sec (b) kk a 0 kcos ( t) kis greatest at t 0.0 sec,.0 sec,.0 sec,.0 sec ad.0 sec, ad at these times the magitude of the cart's positio is kk s 0 cm from the rest positio ad the speed is 0 cm/sec. 8. (a) si t si t Ê si t si t cos t 0 Ê ( si t)( cos t) 0 Ê t k here k is a positive iteger È

6 Sectio.6 Applied Optimizatio 85 (b) The vertical distace betee the masses is s(t) ks s k ˆ as s b a(si t si t) b Î Î Î (cos t cos t)(si t si t) ksi t si tk ( cos t )(cos t )(si t)(cos t ) ksi t si tk critical times at 0,,,, ; the s(0) 0, ˆ ˆ È ˆ 8 ˆ È È the greatest distace is at t ad Ê s (t) ˆ a(si t si t) b ()(si t si t)( cos t cos t) Ê s ˆ si si, s( ) 0, s ˆ si si, s( ) 0 Ê 9. From the diagram above e have! ) Ê )!. From the left triagle e have cot! Ê cot ˆ 50 50!, ad from the triagle o the right side e have cot Ê cot ˆ Ê ) cot ˆ cot ˆ 50 d) Ê 0 Œ Œ 60 0 d ˆ 60 ˆ 50 c d ) 60 0 d Š ˆ Š ˆ 50 c 0 a50 b a50 c b a50 b 0 Ê 60Š 0 a50 b 0a60 b Ê Ê 0Š 5 È7. Sice 0 Ÿ Ÿ 50 Ê 0Š 5 È7 7.5 Þ a0b ) a50 0b ad a 0 b. 0Š 5È7 50 0Š 5È7 ) a50 0b Ê the maimum agle is ) cot 60 cot rad or 6.55 he the solar statio is 7.5 m est of the left ( 60 m) buildig. 0. The distace OT TB is miimized he OB is a straight lie. Hece! Ê ) ).. The profit is p c ( c) ca( c) b(00 ) d( c) a b(00 )( c) c a (bc 00b) 00bc b. The p () bc 00b b ad p () b. Solvig p () 0 Ê 50. c At 50 there is a maimum profit sice p () b 0 for all.. Let represet the umber of people over 50. The profit is p() (50 )(00 ) (50 ) The p () 68 ad p. Solvig p () 0 Ê 7. At 7 there is a maimum sice p (7) 0. It ould take 67 people to maimize the profit.

7 86 Chapter Applicatios of Derivatives h h hq km q. (a) A(q) kmq cm q, here q 0 Ê A (q) kmq ad A (q) kmq. The km km km critical poits are É, 0, ad É, but oly É km is i the domai. The A Š É 0 Ê at h h h h q É km h there is a miimum average eekly cost. (kbq)m q h h km É h (b) A(q) cm q kmq bm cm q, here q 0 Ê A (q) 0 at q as i (a). Also A (q) kmq 0 so the most ecoomical quatity to order is still q É km hich miimizes the average eekly cost. c a b c a b d c abcab c a b. We start ith ca b the cost of producig items,!, ad the average cost of producig items, assumed to be differetiable. If the average cost ca be miimized, it ill be at a productio level at hich d Š! Ê! (by the quotiet rule) Ê c ab ca b!(multiply both sides by ) Ê c a b here c a b is the margial cost. This cocludes the proof. (Note: The theorem does ot assure a productio level that ill give a miimum cost, but rather, it idicates here to look to see if there is oe. Fid the productio levels here the average cost equals the margial cost, the check to see if ay of them give a mimimum.) dr d R C d R C 5. We have dm CM M. Solvig dm C M! Ê M. Also, dm! Ê at M there is a maimum.!!!!! r! r! r! r! 6. (a) If v cr r cr, the v cr r cr cr ar r bad v cr 6cr c ar r b. The solutio of v 0 is r 0 or, but 0 is ot i the domai. Also, v 0 for r ad v 0 for r Ê at r there is a maimum. (b) The graph cofirms the fidigs i (a). h 7. If 0, the ( ) 0 Ê Ê. I particular if a, b, c ad d are positive itegers, a b c d a b c d the Š Š Š Š 6. Î cî aa b aa b a a Èa aa b aa b aa b 8. (a) f() Ê f () 0 Ê f() is a icreasig fuctio of d Èb (d) (b) g() Ê g () ab (d) b(d) b Î ab (d) b ab (d) b Î ab (d) b (d) ab (d) b b (d) 0 Ê g() is a decreasig fuctio of Î dt (c) Sice c, c 0, the derivative d is a icreasig fuctio of (from part (a)) mius a decreasig dt d t fuctio of (from part (b)): d c f() c g() Ê d c f () c g () 0 sice f () 0 ad g () 0 Ê dt is a icreasig fuctio of. d cî Î

8 Sectio.6 Applied Optimizatio At c, the tagets to the curves are parallel. Justificatio: The vertical distace betee the curves is D() f() g(), so D () f () g (). The maimum value of D ill occur at a poit c here D 0. At such a poit, f (c) g (c) 0, or f (c) g (c). 0. (a) f() cos cos is a periodic fuctio ith period (b) No, f() cos cos cos a cos b a cos cos b ( cos ) 0 Ê f() is ever egative. (a) If y cot È csc here 0, the y (csc ) Š È cot csc. Solvig y 0 Ê cos (b) È Ê. For 0 e have y 0, ad y 0 he. Therefore, at there is a maimum value of y. The graph cofirms the fidigs i (a). È È. (a) If y ta cot here 0, the y sec csc. Solvig y 0 Ê ta (b) Ê, but is ot i the domai. Also, y sec ta csc cot 0 for all 0. Therefore at there is a miimum value of y. The graph cofirms the fidigs i (a). %. (a) The square of the distace is Dab ˆ ˆ È *!, so D ab ad the critical poit occurs at. Sice D a b! for ad D a b! for, the critical poit correspods to the miimum distace. The miimum distace is ÈD a b È &.

9 88 Chapter Applicatios of Derivatives (b) The miimum distace is from the poit ˆ ß! to the poit aß bo the graph of y È, ad this occurs at the value here Da b, the distace squared, has its miimum value.. (a) Calculus Method: The square of the distace from the poit Š ß È to Š ß È' is give by D a b a b Š È' È ' È%)! È%). ' The D ab a' b. Solvig D a b! e have: ' È%) È%) È%) Ê ' % a%) b Ê * %) Ê %) Ê. We discard as a etraeous solutio, leavig. Sice D a b! for % ad D a b! for %, the critical poit correspods to the miimum distace. The miimum distace is ÈD a b. 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 (b) 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 b, the distace squared, has its miimum value.

10 .7 INDETERMINATE FORMS AND L'HOPITAL'S ^ RULE Sectio.7 Idetermiate Forms ad L'Hopital's ^ Rule 89. l'hopital: ^ lim or lim lim lim Ä ¹ Ä Ä a ba b Ä. l'hopital: ^ si 5 5 cos 5 si 5 si 5 lim 5 or lim 5 lim 5 5 Ä 0 ¹ Ä 0 5! 5 Ä l'hopital: ^ lim lim lim or lim lim Ä_ Ä_ Ä_ Ä_ Ä_. l'hopital: ^ lim lim or lim lim Ä Ä Ä Ä a b lim Ä a + + b a ba b a ba + + b 5. l'hopital: ^ cos si cos cos acos b cos lim lim lim or lim lim Ä! Ä! Ä! Ä! ˆ Ä! co si si si lim lim Ä! a cos b ˆ ˆ ˆ Ä! cos s 6. l'hopital: ^ % %! lim lim lim or lim lim Ä_ Ä_ Ä_ '! Ä_ Ä_! 7. lim lim Ä Ä 5 8. lim lim 0 Ä5 5 Ä5 t t5 t ( ) t t t ( ) 7 9. lim lim t Ä t Ä t t t t t 0. lim lim t Ä t Ä lim lim lim lim Ä_ Ä_ Ä_ Ä_ lim lim lim Ä_ Ä_ Ä_ si t acos t b(t). lim lim 0 t Ä 0 t t Ä 0 si 5t 5 cos 5t 5 t. lim lim t Ä 0 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 ) 7. lim lim ) Ä Î cos( ) ) ) Ä Î si( ) ) si ˆ ) 8. lim lim ) Ä Î si ˆ ) ) ÄÎ cos ˆ )

11 90 Chapter Applicatios of Derivatives si ) cos ) si ) cos si cos ( )( ) 9. lim lim lim ) Ä Î ) ) Ä Î ) ) Ä Î ) l si ( ) cos ( ) 0. lim lim Ä Ä. lim lim lim lim Ä 0 l (sec ) Ä 0 ˆ sec ta Ä 0 ta Ä 0 sec sec l (csc ) ˆ csc cot cot csc csc. lim lim lim lim Ä Î ˆ ˆ Ä Î ˆ ˆ Ä Î ˆ ˆ Ä Î t( cos t) ( cos t) t(si t) si t (si t t cos t) t si t cos t si t. 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). lim lim lim t Ä 0 cos t t Ä 0 si t t Ä 0 cos t ˆ 5. lim ˆ sec lim lim ˆ ÄÐÎÑc ÄÐÎÑc cos ÄÐÎÑc si ˆ 6. lim ˆ ta lim lim ˆ lim si ÄÐÎÑc ÄÐÎÑc cot ÄÐÎÑc csc ÄÐÎÑc si si (l )(cos ))! a b(l )() ) ) 7. lim lim l ) Ä 0 ) ) Ä 0 ) ) ˆ ˆ lˆ ˆ 8. lim lim l l l l ) Ä 0 ) ) Ä 0 ˆ () a b ()(l ) a b 0 (l ) (l ) l! 9. lim lim Ä 0 Ä 0 a b! l l l l l l! 0. lim lim! Ä 0 Ä 0 l () l () ˆ b. lim lim (l ) lim (l ) lim (l ) lim l Ä_ log Ä_ ˆ l Ä_ ˆ Ä_ Ä_ l ˆ l log l l l l l. lim lim lim lim Ä_ log ( ) ˆ Ä_ l ( ) l Ä_ l ( ) ˆ l ˆ b Ä_ ˆ l lim Ä_ ˆ lim Ä_ l l l l Š l ˆ b b l a b Š. lim lim b lim lim lim Ä! b l Ä! b ˆ Ä! b Ä! b Ä! b e l ae b Š c e e e 0 e. lim lim lim lim Ä! b l Ä! b ˆ Ä! b e Ä! b e È5y 5 5 Î (5y 5) 5 ˆ cî (5y 5) (5) 5 5. lim lim lim lim y Ä 0 y y Ä 0 y y Ä 0 y Ä 0 È 5y 5 Èay a a aay a b a ˆ aay a b (a) a 6. lim lim Î lim lim, a 0 y Ä 0 y y Ä 0 y y Ä 0 y Ä 0 È ay a

12 Sectio.7 Idetermiate Forms ad L'Hopital's ^ Rule 9 7. lim [l l ( )] lim l ˆ l lim l lim l Ä_ Ä_ Š Š Ä_ Ä_ 8. lim (l l si ) lim l ˆ l lim l lim l 0 Ä! b Ä! b si Š Š Ä! b si Ä! b cos siaa hbsi a cosaa h b 0 9. lim lim cos a h Ä 0 h h Ä 0 0. lim ˆ lim Š lim Ä! b Ä! b Ä! b lim Ä! b Š cos cos ( )( si ) ()(0) 6 cos cos si 0 ( )(si ) si ( )(cos ) si si si cos. lim ˆ lim lim lim Ä b Ä b Š Ä b Š l ( ) Ä b Š l ( )(l ) (l ) ( ) ˆ ( l ) lim Ä b Š (l ) (0 ). lim (csc cot cos ) lim ˆ cos cos ( cos ) (si )(cos ) lim Š Ä! b Ä! b si si Ä! b si si cos si 00 lim Ä! b Š cos cos ) si ) cos ). lim ) lim ) lim ) ) Ä 0 e ) ) Ä 0 e ) Ä 0 e e (h) e e. lim lim lim h Ä 0 h h Ä 0 h h Ä 0 t t t t e t e t e e 5. lim lim lim lim t Ä_ et t Ä_ et t Ä_ et t Ä_ et 6. lim e lim lim lim 0 Ä_ Ä_ e Ä_ e Ä_ e _ ÎÐcÑ ÎÐcÑ l 7. The limit leads to the idetermiate form. Let f() Ê l f() l a b. No ÎÐcÑ lim l f() lim lim. Therefore lim lim f() lim e e Ä b Ä b Ä b Ä b Ä b Ä b l ˆ l fðñ e _ ÎÐcÑ ÎÐcÑ l 8. The limit leads to the idetermiate form. Let f() Ê l f() l a b. No l ˆ ÎÐcÑ lim l f() lim lim. Therefore lim lim f() lim e e e Ä b Ä b Ä b Ä b Ä b Ä b! Î Î l (l ) l fðñ 9. 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() b Ä e l (l ) ˆ e l (l ) l fðñ Îe l ÎÐceÑ lim lim. Therefore (l ) lim f() lim e e b e b e b b 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 b b b e Ä! Ä! Ä! c Î l l

13 9 Chapter Applicatios of Derivatives! l l Îl Î 5. The limit leads to the idetermiate form _. Let f() Ê l f() l. Therefore lim Ä_ lim f() lim e e e Ä_ Ä_ fðñ! ÎÐ Ñ l ( ) l 5. The limit leads to the idetermiate form _. Let f() ( ) Ê l f() l l ( ) Ê lim l f() lim lim lim. Therefore lim ( ) Ä_ Ä_ l Ä_ Ä_ Ä_ lim f() lim e e Ä_ Ä_ l fðñ Î 5. The limit leads to the idetermiate form. Let f() ae b Ê l f() ÎÐlÑ _ Î lae b l ae b e Ä 0 Ä 0 Ä 0 e a b Ä 0 Ä 0 l fðñ lim e e Ä 0 Ê lim l f() lim lim. 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! b Ä! b ˆ Ä! b Š Ä! b Ä! b Ä! b l fðñ! lim e e Ä! b! c 56. The limit leads to the idetermiate form _. Let f() ˆ l a b Ê l f() c Ê lim l f() Ä! b c c Š b lim c lim lim 0. Therefore lim ˆ c c lim f() Ä! b Ä! b Ä! b Ä! b Ä! b l fðñ! lim e e Ä! b È lim lim lim 9 Ä_ È É Ä_ É Ä_ È l ˆ È 58. lim Ä! b È si Ê si É lim Ä! b sec cos 59. lim lim lim Ä Îc ta ˆ ˆ Ä Îc cos si Ä Îc si cot ˆ cos 60. lim lim lim cos Ä! b csc si Ä! b ˆ Ä! b si 0 _ 6. Part (b) is correct because part (a) is either i the or form ad so l'hopital's ^ rule may ot be used. 6. Part (b) is correct; the step lim lim i part (a) is false because lim is Ä 0 cos Ä 0 si Ä 0 cos ot a idetermiate quotiet form. 0 _ 6. Part (d) is correct, the other parts are idetermiate forms ad caot be calculated by the icorrect arithmetic

14 Sectio.7 Idetermiate Forms ad L'Hopital's ^ Rule 9 6. (a) (b) The limit leads to the idetermiate form : È È È È lim Š È lim Š È a b Š lim Š lim Ä_ Ä_ Ä_ Ä_ lim Ä_ È É! 65. The graph idicates a limit ear. The limit leads to the idetermiate form 0 ( ) È 0 : lim Ä 9 Î Î Î lim lim Ä Ä 9 5 cî _ 66. (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 b Š c b 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_ (b) ˆ , , Both fuctios have limits as approaches ifiity. The fuctio f has a maimum but o miimum hile g has o etrema. The limit _ of f() leads to the idetermiate form. c c (c) Let f() ˆ Ê l f() l a b l a b Š c c c b Ê lim l f() lim lim c lim c c lim lim 0. Ä_ Ä_ Ä_ Ä_ a b Ä_ a b Ä_ 6 Therefore lim ˆ l fðñ! lim f() lim e e Ä_ Ä_ Ä_ rk k l a rk b l a rk b Š r k k k k rk 67. Let f(k) ˆ r c c b 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

15 9 Chapter Applicatios of Derivatives y ˆ () l y 68. (a) y Î Ê l l y Ê Ê y ˆ l Î a b. The sig patter is y Îe ±±hich idicates a maimum value of y e he e! e Î l y ˆ a b l l y (b) y Ê l y Ê Ê y ˆ Î a b. The sig patter is % y Îe ±± hich idicates a maimum of y e he Èe! Èe Î l ˆ a b (l ) ˆ c c ( l ) Î (c) y Ê l y Ê y. 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_ Þ8 NEWTON'S METHOD b! 6 9 9! 5. y Ê y Ê ; Ê 9 Ê Ê.6905; Ê Ê.66667!. y Ê y Ê ; 0 Ê 0 b 7 Ê % % 6 b! ! y Ê y Ê ; Ê 65 5 Ê.65; Ê Ê 00 b! ;! y Ê y Ê ; 0 Ê 0 Ê Ê Ê Oe obvious root is 0. Graphig e ad shos that 0 is the oly root. Takig a aive approach e ca use Neto's Method to estimate the root as follos: Let fab e,, ad ec 0 b ec. Performig iteratios o a calculator, spreadsheet, or CAS gives 0.59, , You may get differet results depedig upo hat you select for fa b ad, ad hat calculator or computer you may use Graphig ta a bad shos that there is oly oe root ad it is betee 0. ad 0.. Let f a b tac ab b f a b b fab ta ab, 0., ad. Performig iteratios o a calculator, spreadsheet, or CAS gives 0.705, 0.79, You may get differet results depedig upo hat you select for fa b ad, ad hat calculator or computer you may use. fa b 7. f(!) 0 ad f (!) Á! Ê b gives! Ê f ab! Ê! for all 0. That is, all of the approimatios i Neto's method ill be the root of f() 0. f a b f ab

16 Sectio.8 Neto's Method 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.! 9. If h 0 Ê h!! È f( ) f()! f(h) f(h) h h hš Èh Š Èh h; Š È h! if h 0 Ê h!! È f( ) f( h) f()! f( h) h h hš Èh Š Èh h. c Š È h Î 0. f() Ê f () ˆ Î Ê b ˆ Î cî ;! Ê,, 8, ad % 6 ad so forth. Sice kk l 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.. f() 0.5 si Ê f () 0.5 cos Ê ; if.5, the.9870 b ta a b 0.5 si 0.5 cos!. f() ta Ê f () sec Ê b sec a b ;! Ê 905 Ê Ê (a) Yes, three times as idicted by the graphs (b) f() cos Ê f () si Ê cos ab si a b b ; at approimately , , ad e have cos 5. (a) The graph of f() si 0.99 i the ido Ÿ Ÿ, Ÿ y Ÿ suggests three roots. Hoever, he you zoom i o the -ais ear., you ca see that the graph lies above the ais there. There are oly to roots, oe ear, the other ear 0.. (b) f() si 0.99 Ê f () cos Ê ad the solutios b si ( ) 0.99 cos ( ) are approimately ad

17 96 Chapter Applicatios of Derivatives % 6. f() Ê f () 6 Ê ; if! 0.5, the % ; if!.5, the % b % 6 7. Graphig e ad shos that there are to places here the curves itersect, oe at 0 ad the other c f e betee 0.5 ad 0.6. Let fab e, 0.5, ad f 0 b a. b e c Performig iteratios o a calculator, spreadsheet, or CAS gives , , 0.585, (You may get differet results depedig upo hat you select for fa b ad, ad hat calculator or 0 computer you may use.) Therefore, the to curves itersect at 0 ad Graphig la bad shos that there are to places here the curves itersect, oe betee ad 0.9, ad the other betee 0.5 ad 0.6. Let fab la b, ad a b f a b b f ab. Performig iteratios o a calculator, spreadsheet, or CAS ith 0.5 gives , lˆ c c , , ad ith gives 0.987, 0.97, 0.99, (You may get differet results depedig upo hat you select for f ad, ad a b 0 hat calculator or computer you may use.) Therefore, the to curves itersect at 0.99 ad % i i i 9. f() Ê f () 6 8 Ê. Iteratios are performed usig the f a b f a b ib i i i % procedure i problem i this sectio. (a) For! or!!þ), i Ä as i gets large. (b) 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.) È È For or, Neto's method does ot coverge. The values of alterate betee! 7! 7 i È È! 7! 7 or as i icreases. 0. (a) The distace ca be represeted by D() É( ) ˆ, here 0. The distace D() is miimized he ˆ f() ( ) ˆ is miimized. If f() ( ), the f () a b ad f () a b 0. No f () 0 Ê 0 Ê a b Ê. i

18 a b (b) Let g() a b Ê g () a b () Ê ; Ê 0.68 to five decimal places. b Œ b Î Ñ c ÏŠ b c Ò! % Sectio.9 Hyperbolic Fuctios 97.9 HYPERBOLIC FUNCTIONS ˆ 6 6 cosh ˆ tah cosh 5 si. sih Ê cosh È sih sih É ˆ É É, tah, coth, sech, ad csch ˆ 9 9 cosh ˆ 5 5 tah, ad csch cosh 5 sih. sih Ê cosh È sih sih 5 É É, tah, coth, sech 7. cosh, 0 Ê sih Ècosh Ɉ sih É É, tah ˆ cosh ˆ 7 5 8, coth 7, sech 5, ad csch 5 7 tah 8 cosh 7 sih 8 69 sih ˆ. cosh, 0 Ê sih Ècosh 5 É É, tah, cosh ˆ tah cosh sih coth, sech, ad csch l l e ec l el 5. cosh (l ) Š e l l l l e ec c Š % e e 6. sih ( l ) e e 5 e e 5 5 c c 7. cosh 5 sih 5 e e e e e c c 8. cosh sih e e ec e ec % % % 9. (sih cosh ) ˆ ae b e 0. l (cosh sih ) l (cosh sih ) l acosh sih b l 0. (a) sih sih ( ) sih cosh cosh sih sih cosh (b) cosh cosh ( ) cosh cosh sih si cosh sih c c e! a bae b ae b (). cosh sih ˆ e e ˆ e e cae e bae e bdcae e bae e bd dy. y 6 sih Ê 6 ˆ cosh ˆ cosh d dy d. y sih ( ) Ê [cosh ( )]() cosh ( )

19 98 Chapter Applicatios of Derivatives 5. y Èt tah Èt t Î tah t Î dy Ê sech ˆ t Î ˆ t Î ˆ t Î ˆ tah t Î ˆ t Î sech È t tah È t Èt dt dy t dt t t 6. y t tah t tah t Ê csech at bdat bat b(t) atah t b sech t tah dy cosh z dy sih z dz sih z dz cosh z 7. y l (sih z) Ê coth z 8. y l (cosh z) Ê tah z dy 9. y (sech ))( l sech )) Ê ˆ sech ) tah ) (sech )) ( sech ) tah ))( l sech )) d) sech ) sech ) tah ) (sech ) tah ))( l sech )) (sech ) tah ))[ ( l sech ))] (sech ) tah ))(l sech )) dy 0. y (csch ))( l csch )) Ê (csch )) ˆ csch ) coth ) ( l csch ))( csch ) coth )) d) csch ) csch ) coth ) ( l csch ))(csch ) coth )) (csch ) coth ))( l csch )) (csch ) coth ))(l csch )) dy dv cosh v sih v. y l cosh v tah v Ê ˆ ( tah v) asech vb tah v (tah v) asech vb (tah v) a sech v b (tah v) atah vb tah v dy dv sih v cosh v. y l sih v coth v Ê ˆ ( coth v) acsch vb coth v (coth v) acsch vb (coth v) a csch v b (coth v) acoth vb coth v. y a b sech (l ) a bˆ a bˆ a bˆ dy Ê c c el e l d. y a b csch (l ) a bˆ a bš a bˆ c c dy d Ê el e l () cî Š É aî b ÈÈ È( ) 5. y sih È sih ˆ Î dy Ê d cî () Š ( ) Éc( ) Î d ÈÈ È 7 6. y cosh È cosh ˆ Î ( ) dy Ê d dy 7. y ( ) ) tah ) Ê ( ) ) ˆ ( ) tah ) tah ) d) ) ) dy d) ( ) ) 8. y a) ) b tah () ) Ê a) ) b () ) tah () ) ) ) ) ) () ) tah () ) () ) tah () ) 9. y ( t) coth È t ( t) coth ˆ Î t dy Î Ê ( t) ( ) coth ˆ t coth È dt t Î Š t cî at b Èt dy 0. y a t b coth t Ê a t bˆ ( t) coth t t coth t dt t dy d È È È È. y cos sech Ê Š () sech sech sech

20 . y l È Î sech l a b sech Ê dy d Sectio.9 Hyperbolic Fuctios 99 Î Î È È È a b Š ˆ a b ( ) sech sech sech. y csch ˆ ) dy l () l () Ê d) l Š Š l Š Ë Š ÊŠ ÊŠ ) ) ) ) ) ) dy (l ) d) ) É ) a b ). y csch Ê l È ) dy sec sec sec ksec k ksec k d È (ta ) Èsec ksec k ksec k 5. y sih (ta ) Ê ksec k dy (sec )(ta ) (sec )(ta ) (sec )(ta ) d Èsec Èta kta k 6. y cosh (sec ) Ê sec, 0 7. sih ˆ l l ˆ 8. cosh ˆ 5 Š 5 5 É l Š É l 9 9. tah ˆ l (/) l 0. coth ˆ 5 Š (9/) l Š l 9 l (/) (/) È. sech ˆ (9/5) l Š l. csch Š l È / l Š È 5 (/5) È e ec e ˆ e 0 È Š / È. (a) lim tah lim lim lim lim Ä_ Ä_ e Ä_ e Ä_ e e e e Ä_ 0 c e ˆ e e e e e e e e e c e ˆ e e e 0 e ec e e e e 0 e ˆ e e e e c e ˆ e e e e c ˆ e e c e e 0 eec e 0 e e e e ec e ˆ e 0 (b) lim tah lim lim lim lim Ä_ Ä_ Ä_ Ä_ Ä_ (c) lim sih lim lim lim 0 Ä_ Ä_ Ä_ Ä _ (d) lim sih lim lim 0 Ä_ Ä_ Ä _ (e) lim sech lim lim lim Ä_ Ä_ Ä_ Ä_! (f) lim coth lim lim lim lim Ä_ Ä_ e Ä_ e Ä_ e e e e Ä_ 0 c e ˆ e e e e e e e c e e (g) lim coth lim lim lim Ä 0b Ä 0b e e e Ä 0b e Ä 0b e _ c e e e e e c e e e (h) lim coth lim lim lim Ä 0c Ä 0c eec Ä 0c e Ä 0c e _ e e (i) lim csch lim lim lim Ä_ Ä_ Ä_ e e Ä_ 0 e e e e 0! c e y e ecy y ey y y y y. y sih Ê sih y Ê Ê e Ê e e Ê e e 0 È e y y Ê e Ê e È y Ê sih y l Š È. Sice e 0, e caot choose e y È because È 0. e mg gk mg gk gk gk k m dt k m m m 5. (a) v É tahœé dv t Ê É sech ŒÉ t ŒÉ g sech ŒÉ t. gk gk dt m m dv Thus m mg sech ŒÉ t mgœtah ŒÉ t mg kv. Also, sice tah! he!, v! he t!. mg kg mg kg mg mg k m k m k k (b) lim v lim É tah t lim tah t () t Ä_ t Ä_ ŒÉ É É É t Ä_ ŒÉ

21 00 Chapter Applicatios of Derivatives 60 60,000 (c) É É 00 80È ft/sec È5 ds d s dt dt 6. (a) s(t) a cos kt b si kt Ê ak si kt bk cos kt Ê ak cos kt bk si kt k (a cos kt b si kt) k s(t) Ê acceleratio is proportioal to s. The egative costat k implies that the acceleratio is directed toard the origi. ds d s dt dt (b) s(t) a cosh kt b sih kt Ê ak sih kt bk cosh kt Ê ak cosh kt bk sih kt k (a cosh kt b sih kt) k s(t) Ê acceleratio is proportioal to s. The positive costat k implies that the acceleratio is directed aay from the origi. H 7. (a) y cosh ˆ dy H Ê ta 9 ˆ sih ˆ sih ˆ H d H H H (b) The tesio at P is give by T cos 9 H Ê T H sec 9 HÈ ta 9 HÉ ˆ sih H cosh ˆ ˆ H cosh ˆ y H H È a a a a 8. s sih a Ê sih a as Ê a sih as Ê sih as; y cosh a cosh a Èsih a Èa s É s a a a H CHAPTER PRACTICE AND ADDITIONAL EXERCISES. No, sice f() ta Ê f () sec 0 Ê f() is alays icreasig o its domai cos. No, sice g() csc cot Ê g () csc cot csc si si si (cos ) 0 Ê g() is alays decreasig o its domai. No absolute miimum because lim (7 )( ) _. Net f () Ä_ Î Î ( ) (7 ) ( ) ( ) (7 )( ) Ê ad are critical poits. Î ( ) Î ( ) Sice f 0 if ad f 0 if, f() 6 is the absolute maimum. Î a b a a b(a b) aa b ab a b a b '% ab 8 a ba b a b. f() Ê f () ; f () 0 Ê (* a ' b a)!ê& a b!. We require also that f(). Thus Ê a b ). Solvig both equatios yields a 6 ad b 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. gab e Ê g ab e Ê g ± Ê the graph is decreasig o a_, 0 b, icreasig o a0, _ b; 0 a absolute miimum value is at 0; 0 is the oly critical poit of g; there is o absolute maimum value e ˆ e e e a b 6. fab Ê f ab Ê f ± Ê the graph is icreasig o a_, _ b; ab ab is the oly critical poit of f; there are o absolute maimum values or absolute miimum values. 7. fab l o Ÿ Ÿ Ê f ab Ê f ± ± ± Ê the graph is decreasig o a, b, icreasig o a, b; a absolute miimum value is l at ; a absolute maimum value is at.

22 Chapter Practice ad Additioal Eercises 0 8. fab l o ŸŸ Êf ab Êf ±±± Êthe graph is decreasig o a, b, icreasig o a, b; a absolute miimum value is l at ; a absolute maimum value is at. 9. Yes, because at each poit of [!ß Ñ ecept 0, the fuctio's value is a local miimum value as ell as a local maimum value. At 0 the fuctio's value, 0, is ot a local miimum value because 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. (b) Theorem says oly that if f is differetiable 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, because the iterval 0 fails to be closed. The Etreme Value Theorem says that if the fuctio is cotiuous throughout a fiite closed iterval a Ÿ Ÿ b the the eistece of absolute etrema is guarateed o that iterval.. The absolute maimum is kk ad the absolute 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 about the behavior 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) g(t) si t t Ê g (t) si t cos t si (t) Ê g 0 Ê g(t) is alays fallig ad hece must decrease o every iterval i its domai. (b) 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). (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 (b) The iterval ß is ot i the taget's domai because ta ) is udefied at ). Thus the taget eed ot icrease o this iterval. % 5. (a) f() Ê f (). Sice f(0) 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. È8 (b) 0 Ê È ad 0 Ê È () 6. (a) y Ê y 0, for all i the domai of Ê y is icreasig i every iterval i its domai (b) y Ê y 0 for all Ê the graph of y is alays icreasig ad ca ever have a local maimum or miimum 7. Let V(t) represet the volume of the ater i the reservoir at time t, i miutes, let V(0) a! be the iitial amout ad V(0) a! (00)(,560)(7.8) gallos be the amout of ater cotaied i the reservoir after the rai, here hr 0 mi. Assume that V(t) is cotiuous o [!ß0] ad differetiable o!! (!ß 0). The Mea Value Theorem says that for some t i (!ß 0) e have V (t ) a (00)(,560)(7.8) a 56,60,0 gal 0 0 mi V(0) V(0) 0 0!! 6,778 gal/mi. Therefore at t the reservoir's volume as icreasig at a rate i ecess of 5,000 gal/mi.!

23 0 Chapter Applicatios of Derivatives 8. Yes, all differetiable fuctios g() havig as a derivative differ by oly a costat. Cosequetly, the d differece g() is a costat K because g () (). Thus g() K, the same form as F(). d ˆ ( ) () d ˆ d ( ) ( ) d. 9. No, Ê differs from by the costat. Both fuctios have the same derivative a b 0. f () g () Ê f() g() C for some costat C Ê the graphs differ by a vertical shift. d. The global miimum value of occurs at.. (a) The fuctio is icreasig o the itervals Òß Ó ad Òß Ó. (b) The fuctio is decreasig o the itervals Òß!Ñ ad Ð!ß Ó. (c) The local maimum values occur oly at, ad at ; local miimum values occur at ad at provided f is cotiuous at!

24 Chapter Practice ad Additioal Eercises

25 0 Chapter Applicatios of Derivatives 9. (a) y 6 Ê y ± ± Ê the curve is risig o ( %ß %), fallig o ( _ß ) ad (%ß _) % % Ê a local maimum at ad a local miimum at ; y Ê y ± Ê the curve (b) is cocave up o ( _ß!), cocave do o (!ß _) Ê a poit of iflectio at 0! 0. (a) y 6 ( )( ) Ê y ± ± Ê the curve is risig o ( _ß ) ad ( ß _), fallig o ( ß ) Ê local maimum at ad a local miimum at ; y Ê y ± Ê cocave up o ˆ ß _, cocave do o ˆ _ß Ê a poit of iflectio at Î (b). (a) y 6( )( ) 6 6 Ê y ± ± ± Ê the graph is risig o ( ß!)! ad (ß _), fallig o ( _ß ) ad (!ß ) Ê a local maimum at 0, local miima at ad (b) È7 È7 ; y 8 6 a b 6 Š Š Ê y ± ± Ê the curve is cocave up o Š _ß ad Š ß _, cocave do È( È( È 7 È 7 poits of iflectio at È 7 ß Ê o Š È7 È7. (a) y (6 ) 6 Ê y ± ± Ê the curve is risig o ˆ _ß, fallig o ˆ ß _! Î Ê a local maimum at ; y ( ) Ê y ± ± Ê cocave up o! (!ß ), cocave do o ( _ß!) ad (ß _) Ê poits of iflectio at 0 ad

26 Chapter Practice ad Additioal Eercises 05 (b). (a) y % a b Ê y ± ± ± Ê the curve is risig o Š _ß È ad È! È Š È ß_, fallig o Š ÈßÈ Ê a local maimum at È ad a local miimum at È ; y ( )( ) Ê y ± ± ± Ê cocave up o ( ß 0) ad (ß _),! cocave do o ( _ß ) ad (0ß ) Ê poits of iflectio at 0 ad (b) %. (a) y a b Ê y ± ± ± Ê the curve is risig o ( ß 0) ad (0ß ),! fallig o ( _ß ) ad (ß _) Ê a local maimum at, a local miimum at ; y 8 a b Ê y ± ± ± Ê cocave up o Š _ß È ad Š 0ß È, cocave È! È do o Š Èß0 ad Š È ß_ Ê poits of iflectio at 0 ad È (b)

27 06 Chapter Applicatios of Derivatives y 6. y 7. y 8. y 9. y 50. y % 5. lim lim Ä Ä a ac a a b bc b 5. lim b lim Ä Ä & 5. lim Ä ta ta!

28 Chapter Practice ad Additioal Eercises 07 ta sec 5. 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! 55. lim lim lim lim Ä! Ä! Ä! Ä! siamb m cosamb m siab cosab 56. lim lim Ä! Ä! cosa b sia b ( ( ( ( 57. lim seca( bcosa b lim lim Ä Î c Ä Î c cosa b Ä Î c sia b a b È 58. lim È sec lim Ä! b Ä! b! cos! cos si! 59. lim acsc cot b lim lim Ä! Ä! si! Ä! cos È È È È 60. lim Š È lim Ä_ È Š È Ä_ 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 The limit leads to the idetermiate form : lim 0 lim l 0 Ä 0 Ä 0 (l 0)0 si 0 si (l )(cos ) 6. The limit leads to the idetermiate form 0: lim e lim l Ä 0 e Ä cos 5 si 5 cos 6. The limit leads to the idetermiate form 0: lim e lim lim 5 Ä 0 e e Ä 0 Ä 0 0 e e 6. The limit leads to the idetermiate form 0: lim e lim e e Ä 0 Ä 0 tl(t) Š b 0 t 65. The limit leads to the idetermiate form 0: lim t lim t _ t Ä! t Ä! 0 e t e t e t 66. The limit leads to the idetermiate form 0: lim Š t t lim Š t lim t Ä! t Ä! t Ä! b k Îb bk bk 67. lim lim e Ä _ ˆ Ä Š _ Îb 68. lim 0 0 Ä _ ˆ (a) Maimize f() È È Î Î Î Î 6 (6 ) here 0 Ÿ Ÿ 6 Ê f () (6 ) ( ) È6 È È È6 Ê derivative fails to eist at 0 ad 6; f(0) 6, ad f(6) 6 Ê the umbers are 0 ad 6

29 08 Chapter Applicatios of Derivatives (b) Maimize g() È È Î Î 6 (6 ) here 0 Ÿ Ÿ 6 Î Î È6 È È È6 Ê g () (6 ) ( ) Ê critical poits at 0, 8 ad 6; g(0) 6, g(8) È8 6È ad g(6) 6 Ê the umbers are 8 ad 8 È Î Î Î Î 70. (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 0È0 È Ê 0 0 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. 7. A() () a7 bfor 0 Ÿ Ÿ È7 Ê A () ( )( ) ad A () 6. The critical poits are ad, but is ot i the domai. Sice A () 8 0 ad A Š È7 0, the maimum occurs at Ê the largest area is A() 5 sq uits. 7. From the diagram e have ˆ h r Š È Ê h r. The volume of the cylider is h È V r h Š h ah h b, here 0 Ÿ h Ÿ. 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 () 0. The dimesios of the largest cylider are radius È ad height. 7. 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:

30 Chapter Practice ad Additioal Eercises 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 È6 ad d È6 É È6 Ê the legth of the ladder is about Š È6 É È 6 Š È6 *Þ( ft. Î 75. g() Ê g() 0 ad g() 0 Ê g() 0 i the iterval [ß] by the Itermediate Value Theorem. The g () Ê b ;! Ê. Ê.965, ad so forth to & % 76. g() 75 Ê g() 0 ad g() 7 0 Ê g() 0 i the iterval [ ß% ] by the Itermediate Value Theorem. The g () Ê ; Ê.5959 b Ê.9050, ad so forth to & c da c c c % 75! da 77. 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 È d È d È È Èe Î È Èe uits log by y e uits high. 78. A y ˆ l l da l l da da Ê d. Solvig d 0 Ê l 0 Ê e; d 0 for da l e e ad 0 for e Ê absolute maimum of at e uits log ad y uits high. d e e e 79. y l Ê y ˆ l () l ; solvig y 0 Ê ; y 0 for ad y 0 for Ê relative miimum of at ; f ˆ e ad f ˆ 0 Ê absolute miimum is at ad e e the absolute maimum is 0 at e 80. 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

31 0 Chapter Applicatios of Derivatives ÎÈ Š È Š È b ÎÈ ÎÈ Š È Š È 8. fab e for all i a_, _ b; f ab e e ˆ ˆ a Î b ÎÈ ÎÈ ÎÈ Ä Ä 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): È ÎÈ 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 0.9, 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 8. fab e ; The domai of g is all such that 0. The parabola y is cocave do ith -itercepts at ad, therefore 0 if Ÿ Ÿ, ad this iterval is the domai of g; È È 0 g a b e 0 Ê 0 Ê is a critical poit; g a b g a b e, gab e 7.89 Ê the absolute miimum value of the fuctio is at ad, ad the aboslute maimum value is e at. l l l È È Î È 8. (a) y Ê y &Î &Î &Î Ê y ( l ) ˆ 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 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

32 Chapter Practice ad Additioal Eercises 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. 8. 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!. dy dy/dt sec t ta t ta t dy È d d/dt sec t sec t d¹ tî È ta ad y È sec y dy dy/dt cos t dy ; d d/dt cos t sec t d 85. ta t, y sec t Ê si t Ê si ; t Ê Ê Ê cos ˆ tî dy dy/dt Š dy 5 t t d d/dt d Š t t dy dy/dt ˆ dy d d/dt d ¹ Š t t t 86., y Ê t Ê ¹ () ; t Ê ad y Ê y ; t Ê () y sih Ê y sih cosh 6sih cosh Ê y sih 6 dy sec dy dy d tah y sechy d d 88. ta tah y Ê sec tah y sech y Ê Ê sec coth y cosh y dy dy dy cosh y coth y È d sechytahyè d d È 89. si sech y Ê sech y tah y Ê Ê dy dy sec ta dy d d cosh y d 90. sih y sec Ê cosh y sec ta Ê Ê sec ta sech y dy dy + y y d d 9. ta y tah Ê Ê 9. y tahal b Ê y sech al b Ê y sech al b dy sihal yb dy y dy dy y d y d sihal yb d d 9. coshal yb Ê sihal yb Ê Ê Ê y cschal yb aeb 9. y sihata e b Êy coshata e b e Êy e coshˆ tac e e 6

33 Chapter Applicatios of Derivatives dy sec sec dy 95. y sih ata b Ê sec Èsec Ê lsec l d É È ta ta a b Èsec d dy dy d d dy dy dy d d a bd dy coshysihcosh coshysihcosh d ysihy ysihy 96. y cosh y sih 50 Ê y sih y cosh y sih cosh 0 Ê y sihy coshysihcoshê ysihy coshysihcosh