130 Chapter 3 Differentiation

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1 0 Capter Differentiation 20. (a) (b) 2. C position, A velocity, an B acceleration. Neiter A nor C can be te erivative of B because B's erivative is constant. Grap C cannot be te erivative of A eiter, because A as some negative slopes ile C as only positive values. So, C (being te erivative of neiter A nor B) must be te grap of position. Curve C as bot positive an negative slopes, so its erivative, te velocity, must be A an not B. Tat leaves B for acceleration. 22. C position, B velocity, an A acceleration. Curve C cannot be te erivative of eiter A or B because C as only negative values ile bot A an B ave some positive slopes. So, C represents position. Curve C as no positive slopes, so its erivative, te velocity, must be B. Tat leaves A for acceleration. Inee, A is negative ere B as negative slopes an positive ere B as positive slopes. 2. (a) c(00),000 Ê c 0 av, (b) c() Ê c () Marginal cost c () Ê te marginal cost of proucing 00 macines is c (00) 80 st 20 (c) Te cost of proucing te 0 macine is c(0) c(00) (a) r() ˆ Ê r (), ic is marginal revenue. a (b) r!! b Þ (c) lim r () lim Ä_ Ä_ 0. Te increase in revenue as te number of items increases itout boun ill approac zero. ' % % 25. b(t) 0 0 t 0 t Ê b (t) 0 (2) a0 tb 0 (0 2t) % (a) b (0) 0 bacteria/r (b) b (5) 0 bacteria/r % (c) b (0) 0 bacteria/r 26. Q(t) 200(0 t) 200 a900 60t t b Ê Q (t) 200( 60 2t) Ê Q (0) 8,000 gallons/min is te rate te ater is running at te en of 0 min. Ten Q(0) Q(0) 0 0 0,000 gallons/min is te average rate te ater flos uring te first 0 min. Te negative signs inicate ater is leaving te tank.

2 27. (a) y 6 ˆ t t t t 6 Š Ê 6 2 Section. Te Derivative as a Rate of Cange (b) Te largest value of is 0 m/ en t 2 an te flui level is falling te sloest at tat time. Te smallest value of is m/, en t 0, an te flui level is falling te fastest at tat time. (c) In tis situation, Ÿ 0 Ê te grap of y is alays ecreasing. As increases in value, te slope of te grap of y increases from to 0 over te interval 0 Ÿ t Ÿ 2. V V r r r=2 V 28. (a) V r Ê r Ê (2) 6 ft /ft (b) Wen r 2, r 6 so tat en r canges by unit, e epect V to cange by approimately 6. Terefore en r canges by 0.2 units V canges by approimately (6)(0.2) ft. Note tat V(2.2) V(2).09 ft km/r 55 m/sec m/sec, an D t Ê V t. Tus V Ê t Ê t 25 sec. Wen t 25, D (25) m v! v! v! v!!! 2! 2 6! È È 80È9 ft sec 60 sec min 60 min mi r 5280 ft 0. s v t 6t Ê v v 2t; v 0 Ê t ; 900 v t 6t so tat t Ê 900. Ê v (6)(900) 80 9 ft/sec an, finally, 28 mp. (a) v 0 en t 6.25 sec (b) v 0 en 0 Ÿ t 6.25 Ê bo moves up; v 0 en 6.25 t Ÿ 2.5 Ê bo moves on (c) bo canges irection at t 6.25 sec () bo spees up on (6.25ß2.5] an slos on on [0ß6.25) (e) Te bo is moving fastest at te enpoints t 0 an t 2.5 en it is traveling 200 ft/sec. It's moving sloest at t 6.25 en te spee is 0. (f) Wen t 6.25 te bo is s 625 m from te origin an fartest aay.

3 2 Capter Differentiation 2.. (a) v 0 en t sec (b) v 0 en 0 Ÿ t.5 Ê bo moves on; v 0 en.5 t Ÿ 5 Ê bo moves up (c) bo canges irection at t sec () bo spees up on ˆ an slos on on ß&!ß (e) bo is moving fastest at t 5 en te spee kv(5) k 7 units/sec; it is moving sloest at t en te spee is 0 (f) Wen t 5 te bo is s 2 units from te origin an fartest aay. (a) v 0 en t 6 5 sec È 6È5 6È5 6È5 6È5 (b) v 0 en t Ê bo moves left; v 0 en 0 Ÿ t or t Ÿ Ê bo moves rigt (c) bo canges irection at t sec 6 È5 6 È Š È È È ß ß% ß ß () bo spees up on Š an slos on on 0 Š. (e) Te bo is moving fastest at t 0 an t en it is moving 7 units/sec an sloest at t 6 5 sec 6 È 5 (f) Wen t te bo is at position s 6.0 units an fartest from te origin. È

4 Section. Derivatives of Trigonometric Functions. È 6È5 6È5 6È5 6È5 t Ê bo is moving rigt 6 È5 6 È5 6 Š È an slos on on È È ß ß%!ß ß (a) v 0 en t 6 5 (b) v 0 en 0 Ÿ t or t Ÿ Ê bo is moving left; v 0 en (c) bo canges irection at t sec () bo spees up on Š Š (e) Te bo is moving fastest at 7 units/sec en t 0 an t ; it is moving sloest an stationary at t 6 È5 6 È 5 (f) Wen t te position is s 0.0 units an te bo is fartest from te origin. 5. (a) It takes 5 secons.? F &! & (b) Average spee? t (! (!Þ!') furlongs/sec.? F % (c) Using a symmetric ifference quotient, te orse's spee is approimately? t &* '!Þ!(( furlongs/sec. () Te orse is running te fastest uring te last furlong (beteen te 9t an 0t furlong markers). Tis furlong takes only secons to run, ic is te least amount of time for a furlong. (e) Te orse accelerates te fastest uring te first furlong (beteen markers 0 an ).. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS. y 0 cos Ê 0 (cos ) 0 sin 2. y 5 sin Ê 5 (sin ) 5 cos. y csc È 7 Ê csc cot 0 csc cot 2 È È y cot Ê (cot ) cot a b csc (cot )(2) csc 2 cot 5. y (sec tan )(sec tan ) Ê (sec tan ) (sec tan ) (sec tan ) (sec tan ) (sec tan ) asec tan sec b(sec tan ) asec tan sec b asec tan sec tan sec sec tan basec tan sec tan sec tan sec b 0. Š Note also tat y sec tan atan btan Ê 0.

5 Capter Differentiation (sin cos ) sin (sin cos )(sec tan ) (sec )(cos sin ) cos sin cos sin cos cos sin cos cos 6. y (sin cos ) sec Ê (sin cos ) (sec ) sec (sin cos ) sec Š Note also tat y sin sec cos sec tan Ê sec. cos sin cos cot ( cot ) (cot ) (cot ) ( cot ) ( cot ) acsc b(cot ) acsc b cot ( cot ) ( cot ) 7. y Ê csc csc cot csc cot csc ( cot ) ( cot ) cos ( sin ) (cos ) (cos ) ( sin ) ( sin ) asin b(cos ) acos b sin ( sin ) ( sin ) sin sin cos sin ( sin ) ( sin ) ( sin ) ( sin ) sin 8. y Ê cos tan 9. y sec cot Ê sec tan csc cos ( sin ) (cos )() (cos )() ( sin ) sin cos cos sin cos cos cos 0. y Ê. y sin 2 cos 2 sin Ê a cos (sin )(2) ba(2)( sin ) (cos )(2) b2 cos cos 2 sin 2 sin 2 cos 2 cos cos 2. y cos 2 sin 2 cos Ê a ( sin ) (cos )(2) ba2 cos (sin )(2) b2( sin ) sin 2 cos 2 cos 2 sin 2 sin sin t s t. s tan t e Ê sec t e t s t. s t sec t 5e Ê 2t sec t tan t 5e csc t s ( csc t)( csc t cot t) (csc t)(csc t cot t) csc t (csc t) 5. s Ê csc t cot t csc t cot t csc t cot t csc t cot t 2 csc t cot t ( csc t) ( csc t) sin t s ( cos t)(cos t) (sin t)(sin t) cos t cos t sin t cos t cos t ( cos t) ( cos t) ( cos t) cos t 6. s Ê cos t r 7. r ) sin ) Ê ˆ ) (sin )) (sin ))(2)) a ) cos ) 2 ) sin ) b ) () cos ) sin )) ) ) r ) 8. r ) sin ) cos ) Ê () cos ) (sin ))()) sin ) ) cos ) r 9. r sec ) csc ) Ê ) (sec ))( csc ) cot )) (csc ))(sec ) tan )) ˆ ˆ ˆ cos ) ˆ ˆ ˆ sin ) sec ) csc ) cos ) sin ) sin ) sin ) cos ) cos ) sin ) cos ) r ) 20. r ( sec )) sin ) Ê ( sec )) cos ) (sin ))(sec ) tan )) (cos ) ) tan ) cos ) sec ) p cot q q 2. p & 5tan q Ê sec q p q 22. p ( csc q) cos q Ê ( csc q)( sin q) (cos q)( csc q cot q) ( sin q ) cot q sin q csc q

6 sin q cos q p (cos q)(cos q sin q) (sin q cos q)( sin q) cos q q cos q 2. p Ê cos q cos q sin q sin q cos q sin q cos q cos q sec q Section. Derivatives of Trigonometric Functions 5 tan q p ( tan q) asec q b(tan q) asec qb sec q tan q sec q tan q sec q sec q tan q q ( tan q) ( tan q) ( tan q) 2. p Ê 25. (a) y csc Ê y csc cot Ê y a(csc ) acsc b(cot )( csc cot ) b csc csc cot (csc ) acsc cot b (csc ) acsc csc b 2 csc csc (b) y sec Ê y sec tan Ê y (sec ) asec b(tan )(sec tan ) sec sec tan (sec ) asec tan b (sec ) asec sec b 2 sec sec Ð%Ñ 26. (a) y 2 sin Ê y 2 cos Ê y 2( sin ) 2 sin Ê y 2 cos Ê y 2 sin Ð%Ñ (b) y 9 cos Ê y 9 sin Ê y 9 cos Ê y 9( sin ) 9 sin Ê y 9 cos 27. y sin Ê y cos Ê slope of tangent at is y ( ) cos ( ) ; slope of tangent at 0 is y (0) cos (0) ; an slope of tangent at is y ˆ cos 0. Te tangent at ( ß! ) is y 0 ( ), or y ; te tangent at (0ß0) is y 0 ( 0), or y ; an te tangent at ˆ ß is y. 28. y tan Ê y sec Ê slope of tangent at is sec ˆ ; slope of tangent at 0 is sec (0) ; an slope of tangent at is sec ˆ. Te tangent at ˆ tanˆ ß Š ßÈ is y È ˆ ; te tangent at (0ß0) is y ; an te tangent at ˆ ßtan ˆ Š ßÈ is yè ˆ. 29. y sec Ê y sec tan Ê slope of tangent at is sec ˆ tan ˆ 2È ; slope of tangent at is sec ˆ tan ˆ È 2. Te tangent at te point ˆ sec ˆ ˆ ß ß is y 2 È ˆ ; te tangent at te point ˆ sec ˆ ß Š ßÈ2 is y È2 È2ˆ. 0. y cos Ê y sin Ê slope of tangent at È is sin ˆ ; slope of tangent at is sin ˆ. Te tangent at te point ˆ cos ˆ ß ˆ ß is y È ˆ ; te tangent at te point ˆ cos ˆ ˆ ß ß is y

7 6 Capter Differentiation. Yes, y sin Ê y cos ; orizontal tangent occurs ere cos 0 Ê cos Ê 2. No, y 2 sin Ê y 2 cos ; orizontal tangent occurs ere 2 cos 0 Ê cos. But tere are no -values for ic cos.. No, y cot Ê y csc ; orizontal tangent occurs ere csc 0 Ê csc. But tere are no -values for ic csc.. Yes, y 2 cos Ê y 2 sin ; orizontal tangent occurs ere 2 sin 0 Ê 2 sin 5 Ê sin Ê or We ant all points on te curve ere te tangent line as slope 2. Tus, y tan Ê y sec so tat y 2 Ê sec 2 Ê sec È 2 Ê. Ten te tangent line at ˆ ß as equation y 2 ˆ ; te tangent line at ˆ as equation y 2 ˆ ß. 6. We ant all points on te curve y cot ere te tangent line as slope. Tus y cot Ê y csc so tat y Ê csc Ê csc Ê csc Ê. Te tangent line at ˆ ß! is y y cot 2 csc Ê y csc 2 csc cot ˆ ˆ 2 cos sin (a) Wen, ten y ; te tangent line is y 2. (b) To fin te location of te orizontal tangent set y 0 Ê 2 cos 0 Ê raians. Wen, ten y % È is te orizontal tangent. sin 8. y È 2 csc cot Ê y È 2 csc cot csc ˆ Š (a) If, ten y ; te tangent line is y. sin È 2 cos sin È (b) To fin te location of te orizontal tangent set y 0 Ê 2 cos 0 Ê raians. Wen, ten y 2 is te orizontal tangent. 9. lim sin ˆ sin ˆ sin 0 0 Ä 2 0. lim È cos ( csc ) É cos ˆ csc ˆ È cos a 2 b È2 Ä 6 a b 6. lim sec e tan ˆ sec tan ˆ sec tan ˆ sec Ä! sec sec 0

8 2. lim sin ˆ tan sin ˆ tan 0 sin ˆ Ä! tan 2 sec tan 0 2 sec 0 Section. Derivatives of Trigonometric Functions 7. lim tan ˆ sin t sin t tan Š lim tan ( ) 0 t Ä! t t Ä! t. lim cos ˆ ) ) cos lim cos cos ˆ Š Œ ) Ä! sin ) ) Ä! sin ) lim )Ä! sin ) ) s v a 5. s sin t Ê v 2 cos t Ê a 2 sin t Ê j 2 cos t. Terefore, velocity v ˆ È2 m/sec; spee v ˆ È2 m/sec; acceleration a ˆ È 2 m/sec ; jerk j ˆ È 2 m/sec. s v a 6. s sin t cos t Ê v cos t sin t Ê a sin t cos t Ê j cos t sin t. Terefore velocity v ˆ 0 m/sec; spee v ˆ 0 m/sec; acceleration a ˆ È 2 m/sec ; jerk j ˆ 0 m/sec. 7. lim f() sin lim lim 9 ˆ sin ˆ sin 9 so tat f is continuous at 0 Ê lim f() f(0) Ä! Ä! Ä! Ä! Ê 9 c. 8. lim g() lim ( b) b an lim g() lim cos so tat g is continuous at 0 lim g() Ä! c Ä! c Ä! b Ä! b Ê Ä! c lim g() Ê b. No g is not ifferentiable at 0: At 0, te left-an erivative is Ä! b ( b), but te rigt-an erivative is (cos ) sin 0 0. Te left- an rigt-an =0 =0 erivatives can never agree at 0, so g is not ifferentiable at 0 for any value of b (incluing b ). *** % 9. (cos ) sin because (cos ) cos Ê te erivative of cos any number of times tat is a *** % *** multiple of is cos. Tus, iviing 999 by gives Ê (cos ) %* % %*% (cos ) (cos ) sin. *** (cos )(0) ()( sin ) sin 50. (a) y sec Ê ˆ ˆ sin sec tan 5. cos (cos ) cos cos cos Ê (sec ) sec tan (sin )(0) ()(cos ) cos (b) y csc Ê ˆ ˆ cos csc cot Ê sin (sin ) sin sin sin (csc ) csc cot cos (sin )( sin ) (cos )(cos ) sin cos sin (sin ) sin sin (c) y cot Ê csc Ê (cot ) csc sin ( ) sin sin ( ) sin As takes on te values of, 0.5, 0. an 0. te corresponing ase curves of y get closer an closer to te black curve y cos because (sin ) lim cos. Te same Ä! is true as takes on te values of, 0.5, 0. an 0..

9 8 Capter Differentiation 52. cos ( ) cos cos ( ) cos As takes on te values of, 0.5, 0., an 0. te corresponing ase curves of y get closer an closer to te black curve y sin because (cos ) lim sin. Te Ä! same is true as takes on te values of, 0.5, 0., an (a) (b) sina bsina b Te ase curves of y are closer to te black curve y cos tan te corresponing ase curves in Eercise 5 illustrating tat te centere ifference quotient is a better approimation of te erivative of tis function. cosa bcosa b Te ase curves of y are closer to te black curve y sin tan te corresponing ase curves in Eercise 52 illustrating tat te centere ifference quotient is a better approimation of te erivative of tis function. k0kk0k kkkk 5. lim lim lim 0 0 te limits of te centere ifference quotient eists even Ä! 2 Ä! 2 Ê Ä! toug te erivative of f() k koes not eist at y tan Ê y sec, so te smallest value y sec takes on is y en 0; y as no maimum value since sec as no largest value on ˆ ß ; y is never negative since sec.

10 Section. Derivatives of Trigonometric Functions y cot Ê y csc so y as no smallest value since csc as no minimum value on (!ß ); te largest value of y is, en ; te slope is never positive since te largest value y 2 csc takes on is. sin 57. y appears to cross te y-ais at y, since sin sin 2 lim ; y appears to cross te y-ais Ä! sin 2 sin at y 2, since lim 2; y appears to Ä! sin cross te y-ais at y, since lim. Ä! Hoever, none of tese graps actually cross te y-ais since 0 is not in te omain of te functions. Also, sin 5 sin ( ) sin k lim 5, lim, an lim Ä! Ä! Ä! sin 5 sin ( ) k Ê te graps of y, y, an sin k y approac 5,, an k, respectively, as Ä 0. Hoever, te graps o not actually cross te y-ais. sin 58. (a) ˆ sin ˆ sin sin ˆ sin ˆ sin ) lim lim lim lim Ä! Ä! Ä! 80 ) Ä! ) 80 () 80) (converting to raians) cos (b) cos lim Ä! 0, eter is measure in egrees or raians. (c) In egrees, (sin ) lim ) sin lim cos sin ) sin Ä! Ä! lim ˆ cos sin lim ˆ sin cos (sin ) lim ˆ cos (cos ) lim ˆ sin Ä! Ä! Ä! Ä! (sin )(0) (cos ) ˆ cos () In egrees, (cos ) lim ) cos lim sin sin ) cos Ä! Ä! (cos )(cos ) sin sin lim lim ˆ cos sin cos lim sin Ä! ˆ Ä! Ä! (cos ) lim ˆ cos (sin ) lim ˆ sin (cos )(0) (sin ) ˆ sin Ä! Ä! (e) (sin ) ˆ cos ˆ sin ; (sin ) ˆ Š sin ˆ cos ; (cos ) ˆ sin ˆ cos ; (cos ) ˆ Š cos ˆ sin

11 0 Capter Differentiation.5 THE CHAIN RULE AND PARAMETRIC EQUATIONS %. f(u) 6u 9 Ê f (u) 6 Ê f (g()) 6; g() Ê g () 2 ; terefore f (g())g () f(u) 2u Ê f (u) 6u Ê f (g()) 6(8 ) ; g() 8 Ê g () 8; terefore f (g())g () 6(8 ) 8 8(8 ). f(u) sin u Ê f (u) cos u Ê f (g()) cos ( ); g() Ê g () ; terefore f (g())g () (cos ( ))() cos ( ). f(u) cos u Ê f (u) sin u Ê f (g()) sin ˆ ; g() Ê g () ; terefore f (g())g () sin ˆ ˆ sin ˆ 5. f(u) cos u Ê f (u) sin u Ê f (g()) sin (sin ); g() sin Ê g () cos ; terefore f (g())g () (sin (sin )) cos 6. f(u) sin u Ê f (u) cos u Ê f (g()) cos ( cos ); g() cos Ê g () sin ; terefore f (g())g () (cos ( cos ))( sin ) 7. f(u) tan u Ê f (u) sec u Ê f (g()) sec (0 5); g() 0 5 Ê g () 0; terefore f (g())g () asec (0 5) b(0) 0 sec (0 5) 8. f(u) sec u Ê f (u) sec u tan u Ê f (g()) sec a 7btan a 7 b; g() 7 Ê g () 2 7; terefore f (g())g () (2 7) sec a 7 btan a 7b & u % % u 9. Wit u (2 ), y u : 5u 2 0(2 ) * u ) ) u 0. Wit u ( ), y u : 9u ( ) 27( ). Wit u ˆ u, y? : 7u ˆ ˆ ( ) ) 7 u Wit u ˆ u, y? : 0u ˆ 5 ˆ! u 8 u 8 % u. Wit u Š, y? : u ˆ Š ˆ. Wit u ˆ u, y? : 5u ˆ ˆ ˆ & % % 5 5 u u u 5. Wit u tan, y sec u: (sec u tan u) asec b (sec (tan ) tan (tan )) sec u 6. Wit u, y cot u: acsc u bˆ csc ˆ u u u 7. Wit u sin, y u : u cos asin b(cos ) u % u & & 8. Wit u cos, y 5u : a20u b( sin ) 20 acos b(sin )

12 y e Ê y e ( 5) Ê y 5e 20. y e Ê y e ˆ 2 2 Ê y e 2Î 2Î 2Î y e Ê y e (5 7) Ê y 7e 22. y e Ê y e ˆ È 2 Ê y Š 2 e ˆ È ˆ È ˆ È È Section.5 Te Cain Rule an Parametric Equations È Î p Î Î 2È t 2. p t ( t) Ê ( t) ( t) ( t) 2. q È2r r Î q Î Î a2r r b Ê a2r r b a2r r b a2r r b (2 2r) r r r È 2r r s s sin t cos 5t Ê cos t (t) ( sin 5t) (5t) cos t sin 5t (cos t sin 5t) 26. s sin ˆ t cos ˆ t s Ê cos ˆ t ˆ t sin ˆ t ˆ t cos ˆ t sin ˆ t 2 2 ˆ cos t sin t 2 r csc ) cot ) csc ) csc )(cot ) csc )) ) ) (csc ) cot )) (csc ) cot )) 27. r (csc ) cot )) Ê (csc ) cot )) (csc ) cot )) csc ) csc ) cot ) r sec ) tan ) sec ) sec )(tan ) sec )) ) ) (sec ) tan )) (sec ) tan )) 28. r (sec ) tan )) Ê (sec ) tan )) (sec ) tan )) sec ) sec ) tan ) % % % 29. y sin cos Ê asin bsin a b acos bcos () ˆ sin (sin ) % 2 sin ˆ 2 cos (cos ) cos % a sin cos b2 sin aa2 cos b ( sin ) bcos % sin cos 2 sin 2 sin cos cos & & & 0. y sin cos Ê y asin bsin ˆ acos bcos ˆ ' & 5 sin cos sin a b a bˆ aa cos b( sin ) bacos bˆ 5 ' & sin cos sin cos sin cos (. y ( 2) ˆ 7 ' Ê ( 2) ( 2) ( ) ˆ 2 2 ˆ 7 ' ' ( 2) ( ) ˆ ˆ ( 2) 2 Š % y (5 2) ˆ 2 % Ê (5 2) ( 2) ˆ 2 % ˆ 2 2 6(5 2) ˆ ˆ 6 (5 2) % 2 Š % % %. y ( ) ( ) Ê ( ) ( )( ) ( ) ( ) ()( ) ( ) ( ) % ( )( ) % () ( ) ()( ) () ( ) % ( ) % 6( ) ( ) ( ) ( ) 6( ) ( ) ( c 7) ( ) ( ) % %

13 2 Capter Differentiation ' & '. y (2 5) a 5 b Ê (2 5) (6) a 5 b (2 5) a 5 b ( )(2 5) (2) & 2a 5b (2 5) 6a 5b ' 5. y e e Êy e ab ab e e a be e y a 2be Êy a 2b e a2b a2b e e 7. y a 2 2be Êy a 2 2b e ˆ 5 a2 2b e ˆ 5 e 2 5Î2 2 5Î2 5Î2 2 5Î y a9 6 2be Êy a9 6 2b e a ba8 6b e a be 9. () tan ˆ 2È 7 Ê () ˆ tan ˆ Î 2 tan ˆ Î 2 () 0 sec ˆ Î 2 Î Î ˆ 2 tan ˆ 2 sec ˆ 2È tan ˆ 2È È sec ˆ 2È tan ˆ 2È 0. k() sec ˆ Ê k () ˆ sec sec ˆ a b sec ˆ tan ˆ ˆ 2 sec ˆ sec ˆ tan ˆ ˆ 2 sec ˆ 2 sec ˆ sec ˆ tan ˆ È ( cos ))(cos )) (sin ))( sin )) cos ) cos ) ) cos ) cos ) (cos )). f( )) ˆ sin ) Ê f ()) 2 ˆ sin ) ˆ sin ) 2 sin ) (2 sin )) acos ) cos ) sin ) b (2 sin )) (cos ) ) 2 sin ) (cos )) (cos )) (cos )) (sin t)( sin t) (cos t)(cos t) sin t sin t sin t ( cos t) (sin t) 2. g(t) ˆ cos t Ê g (t) ˆ cos t ˆ cos t sin t asin t cos t cos tb ( cos t) cos t r. r sin a) b cos (2)) Ê ) sin a) b( sin 2 )) ) (2)) cos (2)) acos a) bb ) a) b sin a) b( sin 2 ))(2) (cos 2 )) acos a) bb(2 )) 2 sin a) bsin ()) 2 ) cos (2)) cos a) b. r Š sec È tan ˆ r Ê Š sec È ˆ sec ˆ tan ˆ ) ) Š sec È ) tan È ) Š ) ) ) ) ) È) sec sec tan È ) ˆ ˆ sec È) tan È) Š sec È ) ) ) ) ) È) ) tan ) tan sec ˆ È) ) t q t t t Èt Èt Èt Èt 5. q sin Š Ê cos Š Š cos Š Èt() t ˆ Èt ˆ Èt t È t t È b t 2(t ) t t 2 t Èt t Èt Î 2(t ) Î 2(t ) Èt 2 t cos Š cos Š Š Š cos Š 6. q cot ˆ sin t q Ê csc ˆ sin t ˆ sin t ˆ csc ˆ sin t ˆ t cos t sin t t t t t t ) ) ) ) ) ) ) ) ) ) 7. y cos Š e Ê sin Š e Š e Š sin Š e Š e a) b 2) e sin Š e ) ) ) ) 8. y ) e cos 5 ) Ê a) bˆ e cos 5 ) a) cos 5) be ( 2 )) 5(sin 5 )) ˆ ) e ) ) e ( cos 5) 2 ) cos 5) 5 ) sin 5 )) ) ) 9. y sin ( t 2) Ê 2 sin ( t 2) sin ( t 2) 2 sin ( t 2) cos ( t 2) ( t 2) 2 sin( t2) cos( t2)

14 Section.5 Te Cain Rule an Parametric Equations 50. y sec t Ê (2 sec t) (sec t) (2 sec t)(sec t tan t) ( t) 2 sec t tan t % & & 8 sin 2t ( cos 2t) 5. y ( cos 2t) Ê ( cos 2t) ( cos 2t) ( cos 2t) ( sin 2t) (2t) & 52. y ˆ cot ˆ t Ê 2 ˆ cot ˆ t ˆ cot ˆ t 2 ˆ cot ˆ t ˆ csc ˆ t ˆ t csc ˆ t ˆ cot ˆ t 2 cos at b 2 cos at b 2 cos at b 5. y e Ê e 2cosat b asinat bb 2sinat bcosat be sinatî2b sinatî2b 2 sinatî2b sinatî2b 2sinatÎ2b sinatî2b y ˆ e Ê ˆ e e cosˆ t cosˆ t e e cosˆ t e 55. y sin acos (2t 5) b Ê cos (cos (2t 5)) cos (2t 5) cos (cos (2t 5)) ( sin (2t 5)) (2t 5) 2 cos (cos (2t 5))(sin (2t 5)) 56. y cos ˆ 5 sin ˆ t Ê sin ˆ 5 sin ˆ t ˆ 5 sin ˆ t sin ˆ 5 sin ˆ t ˆ 5 cos ˆ t ˆ t 5 sin ˆ 5 sin ˆ t ˆ cos ˆ t % ˆ t ˆ t ˆ t % ˆ t ˆ t ˆ t 57. y % tan ˆ t Ê % tan ˆ t % tan ˆ t % tan ˆ t tan ˆ t tan ˆ t 2 tan tan sec tan tan sec y c cos (7t) Ê c cos (7t) 2 cos (7t)( sin (7t))(7) 7 c cos (7t) (cos (7t) sin (7t)) Î Î Î Î t sin at b È cosat b 59. y a cos at bb Ê a cos at bb a cos at bb a cos at bb ˆ sin at b at b a cos at bb asin at b b 2t 60. y sin ŒÉ È t Ê cos ŒÉ Èt ŒÉ Èt cos ŒÉ È t ˆ Èt 2 cos ŒÉ Èt cos ŒÉ Èt ÉÈt 2Èt ÉtÈt ÉÈt ˆ ˆ 2ˆ ˆ ˆ 6 ˆ 6 ˆ 6 ˆ 6 ˆ ˆ % 6 ˆ ˆ 2 6. y ˆ Ê y ˆ ˆ ˆ Ê y ˆ ˆ ˆ ˆ Î Î Î Î Î ˆ È ˆ ˆ È ˆ 62. y ˆ È Ê y ˆ È ˆ ˆ È Ê y ( 2) Î ˆ È ˆ È ˆ È Î ˆ È È È ˆ È Š ˆ È Š 6. y cot ( ) Ê y csc ( )() csc ( ) Ê y ˆ 2 (csc ( ) csc ( )) csc ( )( csc ( ) cot ( ) ( )) 2 csc ( ) cot ( )

15 Capter Differentiation 6. y 9 tan ˆ Ê y 9 ˆ sec ˆ ˆ sec ˆ Ê y 2 sec ˆ ˆ sec ˆ tan ˆ ˆ 2 sec ˆ tan ˆ y e 5 Êy 2e 5 Êy 2 e a2b2e a 2be y sina e b Ê y cosa e b a e 2e b a 2be cosa e b Ê (Use triple prouct rule: Dafgb f g fg fg ) y a2 2be cosa e ba 2be cosa e ba 2be a sina e b a e 2e bb a 2be cosa e be a bsina e b È & % È 67. g() Ê g () Ê g() an g () ; f(u) u Ê f (u) 5u Ê f (g()) f () 5; 5 terefore, (f g) () f (g()) g () 5 ( ) u ˆ u 68. g() ( ) Ê g () ( ) ( ) Ê g( ) an g ( ) ; f(u) Ê f (u) Ê f (g( )) f ; terefore, (f g) ( ) f (g( ))g ( ) 69. g() 5È 5 5 Ê g () Ê g() 5 an g () ; f(u) cot ˆ u Ê f (u) csc ˆ u ˆ È u csc ˆ Ê f (g()) f (5) csc ˆ ; terefore, (f g) () f (g())g () = 70. g() Ê g () Ê g ˆ an g ˆ ; f(u) u sec u Ê f (u) 2 sec u sec u tan u 2 sec u tan u Ê f ˆ g ˆ f ˆ 2 sec tan 5; terefore, (f g) ˆ f ˆ g ˆ g ˆ 5 2u u 7. g() 0 Ê g () 20 Ê g(0) an g (0) ; f(u) Ê f (u) 2u 2 u a b Ê f (g(0)) f () 0; terefore, (f g) (0) f (g(0))g (0) 0 0 au b(2) (2u)(2u) u a b u u u u u (u )() (u )() 2(u )(2) (u ) u (u ) (u ) (u ) g() Ê g () Ê g( ) 0 an g ( ) 2; f(u) ˆ u u u Ê f (u) 2 ˆ ˆ 2 ˆ Ê f (g( )) f (0) ; terefore, (f g) ( ) f (g( ))g ( ) ( )(2) 8 7. (a) y 2f() Ê 2f () Ê ¹ 2f (2) 2 ˆ 2 =2 = (b) y f() g() Ê f () g () Ê ¹ f () g () 2 5 = f() g()f () f()g () g(2)f (2) f(2)g (2) (2) ˆ (8)( ) 7 g() [g()] ¹ [g(2)] 6 =2 (c) y f() g() Ê f()g () g()f () Ê ¹ f()g () g()f () 5 ( )(2 ) 5 8 () y Ê Ê =2 Î Î f () f (2) ˆ È 2 Èf() ¹ Èf(2) È8 6È8 È2 2 =2 (e) y f(g()) Ê f (g())g () Ê ¹ f (g(2))g (2) f (2)( ) ( ) (f) y (f()) Ê (f()) f () Ê 5 = Î Î a b a b a b Î Î =2 (g) y (g()) Ê 2(g()) g () Ê ¹ 2(g()) g () 2( ) 5 () y (f()) (g()) Ê (f()) (g()) 2f() f () 2g() g () Ê ¹ a (f(2)) (g(2)) b a 2f(2)f (2) 2g(2)g (2) b a 8 2 b ˆ ( ) 5 È7

16 Section.5 Te Cain Rule an Parametric Equations (a) y 5f() g() Ê 5f () g () Ê ¹ 5f () g () 5 ˆ ˆ = =0 ˆ (b) y f()(g()) Ê f() a(g()) g () b (g()) f () Ê ¹ f(0)(g(0)) g (0) (g(0)) f (0) ()() () (5) 6 f() (g() )f () f() g () (g() )f () f()g () g() (g() ) = (g() ) ( ) ˆ () ˆ 8 ( ) (c) y Ê Ê ¹ () y f(g()) Ê f (g())g () Ê ¹ f (g(0))g (0) f () ˆ ˆ ˆ 9 =0 (e) y g(f()) Ê g (f())f () Ê ¹ g (f(0))f (0) g ()(5) ˆ (5) 8 0 =0! ¹ = ˆ ˆ 2 ˆ 2 (f) y a f() b Ê 2 a f() b a f () b Ê 2( f()) a f () b 2( ) (g) y f( g()) Ê f ( g()) a g () b Ê ¹ f (0 g(0)) a g (0) b f () ˆ =0 9 ˆ ˆ 75. s s ): s cos s sin s sin so tat s s ) ) Ê ) Ê ˆ 5 5 ) ) ) ) )= 2 = 76. : y 7 5 Ê 2 7 Ê ¹ 9 so tat 9 u u u 5 u 5 u 5 u u u u u ( ) u u () ( ), again as epecte ( ) ( ) c ( ) a b 77. Wit y, e soul get for bot (a) an (b): (a) y 7 Ê ; u 5 5 Ê 5; terefore, 5, as epecte (b) y Ê ; u ( ) Ê ( ) () ; terefore Î Î u u ˆ u È u È È 78. Wit y, e soul get for bot (a) an (b): (a) y u Ê u ; u È Ê ; terefore, u È È, as epecte. È u u Î u Èu u Èu È (b) y u Ê ; u Ê ; terefore,, again as epecte. 79. y 2 tan ˆ Ê ˆ 2 sec ˆ sec (a) ¹ sec ˆ slope of tangent is 2; tus, y() 2 tan ˆ Ê 2 an y () Ê tangent line is = given by y 2 ( ) Ê y 2 (b) y sec ˆ an te smallest value te secant function can ave in 2 is Ê te minimum value of y is an tat occurs en sec ˆ Ê sec ˆ Ê sec ˆ Ê (a) y sin 2 Ê y 2 cos 2 Ê y (0) 2 cos (0) 2 Ê tangent to y sin 2 at te origin is y 2; y sin ˆ Ê y cos ˆ Ê y (0) cos 0 Ê tangent to y sin ˆ at te origin is y. Te tangents are perpenicular to eac oter at te origin since te prouct of teir slopes is. (b) y sin (m) Ê y m cos (m) Ê y (0) m cos 0 m; y sin ˆ Ê y cos ˆ m m m Ê y (0) cos (0). Since m ˆ, te tangent lines are perpenicular at te origin. m m m

17 6 Capter Differentiation (c) y sin (m) Ê y m cos (m). Te largest value cos (m) can attain is at 0 Ê te largest value y can attain is km kbecause ky k km cos (m) k kmkkcos mk kmk km k. Also, y sin ˆ Ÿ m Ê y cos ˆ Ê ky k cos ˆ Ÿ cos ˆ Ÿ Ê te largest value y can attain is. m m m m m m kmk m () y sin (m) Ê y m cos (m) Ê y (0) m Ê slope of curve at te origin is m. Also, sin (m) completes m perios on [0ß2]. Terefore te slope of te curve y sin (m) at te origin is te same as te number of perios it completes on [0ß 2]. In particular, for large m, e can tink of compressing te grap of y sin orizontally ic gives more perios complete on [0ß2], but also increases te slope of te grap at te origin. 8. cos 2t, y sin 2t, 0 Ÿ t Ÿ 82. cos ( t), y sin ( t), 0 Ÿ t Ÿ Ê cos 2t sin 2t Ê y Ê cos ( t) sin ( t) Ê y, y! 8. cos t, y 2 sin t, 0 Ÿ t Ÿ 2 8. sin t, y 5 cos t, 0 Ÿ t Ÿ 2 6 cos t sin t y 6 sin t 25 cos t y Ê Ê Ê Ê 85. t, y 9t, _ t _ Ê y 86. Èt, y t, t 0 Ê Èy or y, Ÿ 0

Chapter 2. Limits and Continuity 16( ) 16( 9) = = 001. Section 2.1 Rates of Change and Limits (pp ) Quick Review 2.1

Chapter 2. Limits and Continuity 16( ) 16( 9) = = 001. Section 2.1 Rates of Change and Limits (pp ) Quick Review 2.1 Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 969) Quick Review..... f ( ) ( ) ( ) 0 ( ) f ( ) f ( ) sin π sin π 0 f ( ). < < < 6. < c c < < c 7. < < < < < 8. 9. 0. c < d d < c