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# 4 - SECTION 3.4 DERIVATIVES 0F TRIGONOMETRIC FUNCTIONS I...

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Unformatted text preview: SECTION 3.4 DERIVATIVES 0F TRIGONOMETRIC FUNCTIONS I: 173 PV PV 1 33' PV ‘nRT ‘> T * nR _ (10)(0.0821) 0.821 dT_ 1 , _ 1 ~1 ~ 67?: _0—821 [P(t)V 't()+V(1)P(1)]_m[(8)(—0.15)+(10)(0.10)]~ 0.2436 K/mln. 34. (a) If dP/dt : 0. the population is stable (it is constant). (b)dp_0 _> ﬁP_ro(1 P>P .7» 2‘1 P ? 13—1—6 => P=PC(1—£>. —.(PV) Using the Product Rule we have dt PC To PC PC 7.0 7'0 If PC : 10000.1"0 = 5% : 0.05, and 0 = 4% : 0.04, then P = 10,000(1 — g) : 2000. (c) Ifﬂ : 0.05. then P : 10.000(1 ~ g) z 0. There is no stable population. 35. (a) If the populations are stable. then the growth rates are neither positive nor negative; that is. dC dW — d — ~ —0 dt 0 an dt (b)‘ The caribou go extinct” means that the population is zero. or mathematically, C = 0. (c) We have the equations §_ ¥ aC— bCW and 310% : —cW + dCW. Let dC/dt : dW/dt : 0, a = 0.05. b = 0.001, c : 0.05. and d = 0.0001 to Obtain 0.050 * 0.001C’W 2 0 (l) and —0.05W + 0.0001CW : O (2). Adding 10 times (2) to (l) eliminates the CW—terms and gives us 0.050 — 0.5W : 0 => C : 10W. Substituting C : 10W into (1) results in 0.05(10W) , 0.001(10W)W : 0 e» 0.5W — 0.01W2 : 0 41) 50W — W2 : 0 4:» W(50 — W) : 0 4:) W = 0 or 50. Since C = 10W. C = 0 or 500. Thus. the population pairs (C, W) that lead to stable populations are (07 0) and (500, 50). So it is possible for the two species to live in harmony. 3.4 Derivatives of Trigonometric Functions \ .f(;r:)::c~3sina: => f'(;c):1—3cos:c .f(:c):xsinx :> f'(\$):m~cosm+(sin.1:)~12xcosar+sinx .y:sinz+10tan\$ => y’:cosm+108ec2m 1 2 3 4.y:2csc.7:+5tzos:c :> y’:~2cscxcotx758inx 5. g(t) : t3 cost => g’(t) = t3(~sint) + (cost) - 3t2 : 3t2 cost — t3 sint or t2(3cost — tsint) 6. g(t) = 4sect+tant :> g'(t) = 4secttant+sec2t 7 . [1(6) : 0500 + eecotQ :> h'(9) = — csc0 cot 6 + e6 (7 csc2 9) + (cot mes = — cschot «9 + 69 (cot 19 — cscr’z 0) 8. y: eu(cosu + cu) :> y _ —e “(—sinu +0) + (00521 +cu)e“ e"(cosu ~ sinu + cu+ c) 9. y: as :> y' _ _—(cos:c)(1) (xx—sinus) : cosm+zcsinx cos :0 (cos 90)? cos2 x 10.y:1+smx : x+cosx y, _ (ac +c0sx)(cos:1:) i (1 +sin at)(1 ~ sinx) xcosx +cos2ac — (1 — sin2 :17) (:0 + cos 1:)2 _ (a: + cos x)2 _ xcosnv + 0032 00 ~ ((3052 x) accosx (a: + cos x)2 _ (:0 + cos :0)? 174 I: CHAPTER3 DIFFERENTIATION RULES sec6 11. 6 = —— f< ) 1+sec6 f'(6) , (1 + see 6)(sec6tan 6) 7 (sec 6)(sec6tan 6) (sec6tan 6)[(1 —I— see 6) 7 sec 6] _ sec6tan6 (1 + sec6)2 (1+ sec 6)2 _ (1 + sec6)2 12.y:Ei‘ﬂ;l:> seem dy , secwsecga: i (tang: —1)secwtanm # Secw (seczx ~ tan2m + tanm) _ 1 + tang: dac sec2 ac _ sec2 :3 _ seem Another method: Simplify y ﬁrst: y = sinx — cos ac => 1/ : c059: —I— sinac. since , 3:2 cosac — (sinzc)(2:1:) 10(95 cosa: — 23in w) wcosm — 2sina: y * (\$2)2 \$4 : CE3 13. y: m2 14. y = csc6(6 + cot6) :> y’ = csc6 (1 - C862 6) + (6 + C0t6)(— csc6c0t6) : csc6 (1 * csc2 6 — 6cot6 — cot2 6) : csc6 (— cot2 6 # 6cot 6 * cot2 6) [1+ cot2 6 = csc2 6] = csc6 (#6c0t6 — 2c0t2 6) : —csc6cot6(6 + 2c0t6) 15. y : sec6 tan6 => 3/ : sec6 (sec2 6) + tan6 (sec6tan 6) : sec6 (sec2 6 + tan2 6) Using the identity 1 —I— tan2 6 2 sec2 6. we can write alternative forms of the answer as sec6 (1 + 2tan2 6) or sec6 (2 sec2 6 i 1) 16. Recall that ify : fgh. then 3/ : f’gh + fg’h + fgh’. y : msinarcosac => d . . . . E: = smmcosx + xcossccoscc +scsm:13(—sm;c) : smaccoscc + xcos2x — aicsin2 a: d d 1 ' ! 1 — 17. ‘ (cscsc) : ._ . _ (S111 :c)(0.) 2 (cos x) _ icqsm _ '1 . cosx : cscmcotw dac dac smz‘ 3111 x 3m x smw smart 18. i (secx) : i 1 1 (cosx)(0) — 1(—sinx) # sinm _ 1. 'sincc :secmtanm dw da: cos :6 cos2 cc cos2 ac cos :3 cos cc 19. _d_ (cotzc) : i (cosm) _ (sinm)(isin:c) ; (cosx)(cosx) _ *sin2 ac.+2coszx : _ . 12 : —CSC2 ac dm dzc smm sm x 8111 m sm so 20. f(2:) : cosw => f'(m) : lim f(a: + h) , f(9:) : lim cos(zc + h) # coszc _ lim coszvcosh — sinmsinh , cosm h—m) h h—>0 h h-»0 h. i 1, cosh i 1 'n Sinh 1 os lim cosh ; 1 _ sinm lim sinh ! 11131) coszc h SI :c h ; C m 1180 h h—>0 h : (coszc)(0) ! (sinx)(1) : —sinx 21. y : tanm :> y’ : sec2 at :> the slope of the tangent line at (g 1) is sec2 % : (\/§)2 : 2 and an equation ofthetangentlineisy— 1 :2(ac— %) ory :2\$+1 — g. 22. : ex cosy: :> y’ : 693(7 sinm) + (cos x)ew : e": (cosx - sin 1:) => the slope of the tangent line at (0.1) is 60(COSO 4 sinO) : 1(1 — 0) : 1 and an equation is y - 1 : 1(1: — 0) ory : ac + 1. 23. y : a: + coszz: : y' :1— since At (0,1).y' : Land an equation of the tangent line is y — 1 2 1(9: 7 0). or : cc + 1. 1 cosx—smx 1#0 :71.and __ ,:_ R' 1R1.ma1J:———— sings + 00531: :> y (sincc + cos cc)2 l ecrproca u e] ( ) y (0 -I— 1)2 an equation of the tangent line is y — 1 : —1(:c — 0). or y : —a: + 1. 24. y: SECTION 3.4 DERIVATIVES 0F TRIGGNOMETRlC FUNCTIONS C 175 25. (a) y = azcosz => 3/ : \$(— sinw) + cosx(1) = cosa: — xsinac. (b) 1 So the slope of the tangent at the point (7r. —7r) is '1 gm 5 cos7r — 7rsin7r = 71— 7r(0) : —1. and an equation is y+7r:i(w—7r)ory=ix. 26. (a)y:sec:c—2cosx => y'=secmtanz+28inz : the slope of the tangent line at (g, 1) is sec—tan—+2sin—=2 ﬁ+2~£ 23:3x/gandanequationis y~1:3\/§(\$—§)ory=3\/§z+li7r\/§. H‘ZIV'IE" 27. (a) f(m) 22x+cotx => f’(a:) :2—csc2\$ (b) 6 Notice that f'(m) : 0 when f has a horizontal tangent. f’ is positive when f is increasing and f’ is negative when f is decreasing. Also, f’(:c) is large negative when the graph A of f is steep. sin .1: 28. (a) f(a:) : ﬁsinm => f’(x) : ﬁcosw+(sinm)(%x_l/2) = ﬁcosm—l— 2%; (b) 3 Notice that f’(\$) = 0 when f has a horizontal tangent. f’ is positive when f is increasing and f’ is negative when f is decreasing. 0 A 1 2 -3 v w 29. f(\$)2 : z + 25mm has a horizontal tangent when f'(x) — 0 <:> 1+ Zeosw = 0 <:> cosa: : —§ <=> 9:- ¥ —+ 27771 or4~ 7r+ 27m. where n is an integer. Note that“3 "and23 — 7rare :l:— 7‘ units from 7r. This allows us to write the solutions in 3the more compact equivalent form (271 + 1)7r ig 71 an integer. cosac 30. : “ => y 2 + sinm , (2+sinac)(—sin\$)—Cosmcoszc —2sinz—sin2x—c052x —2sin.7;—1 y 2 , 2 . 2 — 3 : 0 when (2 + s1n m) (2 + Sin :6) (2 + 8111 1:)2 —2sinx — 1 : 0 4:) sinm : —% (:r x : 11—" +27TTLOF32— — 7—6? +27m.nan integer. Soy = f 0r __ 111r 771' 1 y— —\/— and the points on the curve with horizontal tangents are: (6 + 27m «5) ( 6 + 27m, J93) 71 an integer. \ 176 3 CHAPTEB3 DIFFERENTIATION RULES 31. (a) m(t) = 555th => v(t) = \$’(t) : Boost (b) The mass at time t : 2—375 has position mfg”) : 8 sin 2?” : 8(1?) : 4 \/§ and velocity 242?") : 8cos 2?" : 8(—%) = —4. Since 11(23’5) < 0. the particle is moving to the left. 32. (a)s(t):2cost+351nt :> v(t):—2sint+3cost (b) (c) 3 z 0 :> t2 :3 2.55. So the mass passes through the equilibrium position for the ﬁrst time when t m 2.55 s. (d) v : 0 :> t1 m 0.98, 3(t1) z 3.61 cm. So the mass travels a maximum of about 3.6 cm (upward and downward) from its equilibrium position. (e) The speed |v| is greatest when 3 = 0; that is. when t = 252 + Wt, 71 a positive integer. 33. From the diagram we can see that sin6 : 93/10 4:» x : 10 sin 9. We want to ﬁnd the rate of change of a: with respect to 6; that is. day/dd. Taking the derivative of the above expression. dcc/dG : 10(Cos 0). So when 6 : g dz/dO = 10cos§ : 10%) : 5 ft/rad 10 r. _ ’ . i - __6 34- (a) F: MW _> dF (,usmH—tcosGXO) uWUtcosQ sm0) # ,uW(s1n0 pcos ) usin6+cosﬁ E _ (Msin6+cos0)2 (,usint94rc056)2 (b)%%:0 :> uW(sin9~,ucos€):0 : sin6:peos€ :> tan0:p => 0:tan_1,u 0.6(50) : ——’- < 6 < 1. th t (c) 30 From the graph of F 0.65in6 + 6080 forO ; _ we see a dF . . . It? 2 0 => 6 z 0.54. Checking this With part (b) and pt 2 0.6. we calculate 6 = tan‘1 0.6 m 0.54. So the value from the graph is consistent with the value in part (b). 0 1 25 . sin 3:8 , 3 sin 3:1: . . 35. 11m : 11m [multiply numerator and denominator by 3] x—>0 {E 160 31‘ : 3 lim 5‘“ 3x [as m a 0. 3x —> 0] Bar—>0 3.21) . sin 6 : 3 11m [let 0 : 3m] 0—»0 : 3(1) [Equation 2] : 3 sin 4:0 sin 4w :0 . 4 sin 45!; , 6m ‘ : ' - : l - l , 36' i113) sin 62: 0161:?) < a: sin 6x) sol—IR) 4m 221310 6 Sin 6w Sin 4316 1 65C 1 2 _ ‘ - ' #41~—1 :— 4il—»mo 4:5 6212113) sin6m () 6() 3 38. 39. 41. 42. 45. SECTION 3.4 DERIVATIVES 0F TRIGONOMETRIC FUNCTIONS tan 6t . sin 6t 1 t . 68in 6t . 1 , 2t - 2 . __ . _ : 1 . l - 11m , I135 sinZ PIE) ( t cos 6t sin 2t) \$136 615 9—135 cos 6t t—»0 25m 2t sin 6t 1 1 . 2t 1 1 : . “ ~-l =61~—-—1:3 6:11—13) 6t igcosm 2t1—Iftllsin2t () 1 2() c050—1 lim 0050—1 , 0080 — 1 . 9 9—.0 0 0 _ : , _ — O (515% sin0 \$135 sini9 1. sm0 1 1m 0 6—»0 0 . 0 sin<lim cos 6) 'nl lim sm(cos ) : 9_,0 : 51 : sinl 9—40 secﬁ lim sec0 6H0 . sin2 3t . sin 3t sin 3t . sin 3t , sin 3t 11m 11m ~ ﬂ 11m - 11m t——>0 t2 we t t H0 t t—>0 t . 2 . 2 Sln 3t . Sin 3t 2 : ' I : . 1 = 9 (12% i ) (321:2, 3i > <3 ) 1’ wt 23: i 1’ ‘COS 2msinm — lim cos 2m “(sin me — lim COS 2x ﬁiﬂﬁsm me] \$13?) cscx _ 233:) sin 21' ﬂ x—so (sin 2x)/:c — 1—»0 2 hit}) [(sin 2x)/2:z:] 1 1 2 1 - —— : — 2 - 1 2 I' sing; — cosm _ l‘m sinm — cosx ﬂ lim sinx — cosx \$3,554 COS 21L" _ szr/4 cos2 x ~ sin2 a: _ z—n‘r/4 (cosm + sin :L')(cosac — sinx) . *1 —1 —1 2 11m = _ 2—»7r/4 cos x + sinx cosg + sin f ﬂ :l 177 . Divide numerator and denominator by 6. (sin 6 also works.) sin0 l' sin0 l‘m Sing lim ‘9 9135 9 1 1 l .— 2 S 2 . : _ 9—.00+tan0 e—io sm6 1 , smﬂ , 1 1+1-1 2 1 — . 1 + 11m — 11m 0 cos6 9—»0 6’ 9—0 cost? . sin(z ~ 1) . sin(\$ — 1) . 1 , sin(m — 1) 1 1 l §:1 ﬁ—l l ——-1—- zl—vrnlxz—f—x—2 xl—>Inl(x+2)(:1;~1) zLIrllm+2zl—»rnl \$~1 3 3 (a) itanx : i sina: :> 56022: _ coszcosx ~ sina:(—sinx) _ cos2ac +sin2x‘ So sec2x : 1 ‘ d3: dzc cos ac c052 m c052 2: cos2 cc d d 1 — 1 — ' ' (b) —secx:— => secwtanx2w.805ecxtanm2 smxi d2: dx cos x cos2 x 0052 cc . d 1 t (C) ﬂ (smut: +Cos cc) : ﬁ‘:s:: a: 3 . CSC\$(—CSC2.’E) —(1+COt£L’)(—CSC;ECOt.’E) csca: [— CSC2\$+(1+COI§.Z‘) cotx] cosx — smac— one2 a: csczac _ —csczm+cot2x+cotx _ —1+cot;c cscm ‘ cscm , cota: ~ 1 Socosx —smz = *. cscx 178 46. 47. j CHAPTER3 DIFFERENTIATION RULES Let |PR| : ac. Then we get the following formulas for r and h in terms of 6 and an: sing—Z => r—ccsingandcosg—E => h— 6 A 2 T ac _ 2 2 _ 9: _ “OS 2' P I Q Now A(6) 2 %7rr2 and 3(9) : 5mm : m. So - A(9) . l7”; . r . xsin(6/2) x 1 —— : 1 2 = A _ 2 ; ___ 9361+ B 6) 69—13(1)le Th ”6133+ h 2W6E151+ xcos(6/2) a : %7r 6133+ tan(6/2) : 0. R By the deﬁnition of radian measure. 3 : r6, where r is the radius of the circle. 6 d 6 By drawing the bisector of the angle 6. we can see that sin ~2- = 7/72 2 d = 2r sin 5. So lim f = lim ——7ﬁ—— : lim 2 ' (0/2) : lim —6—/2—— = 1. [This is just the reciprocal of the limit 6—.0+ d 9—.0+ 2r sin(6/2) 960+ 28in(6/2) 6—»0 sin(6/2) sinz lim z—>0 : 1 combined with the fact that as 6 —+ 0. g —> 0 also] 3.5 The Chain Rule 1. 10. 11. Letu : g(:I:) = 4m and y : ﬂu) : sinu. Then gal—c : %§% : (cosu)(4) : 4cos4x. d d du 3 3 — — z : :1/2 _y:l_:l’1/23:__:._———. Letu7g(:c)—4+3mandy f(u) ﬂ u .Then dw dudm 2U () 2W 2m d .Letu : 9(13) : 1 ~ 902 andy = ﬂu) = um. Then % : gig—1;: : (10u9)(—2:c) : —20:c(1 - x2)9. Letu — (a3) — since and : f(u) : tanu Then 13—, : ﬂd—u : (see2 u) (coszc) : sec2(sinac) ~ cossc or —g — y ’ dsc du dm ’ equivalently, [sec(sin 16)]2 cos x. .Letu 2 9(30) 2 ﬁandy : f(u) : e“. ﬁ dy-d£@_ u 1 71/2 _ ﬁ._1___ e The“ dm * dudm '(6 )(295 )’6 zﬁ ’ 2ﬁ Letu = (x) = 6”” andy : f(u) = sinu. Then (—13% = d—yd—u = (cosu)(em) 2 ex 0036’”. 9 ‘ da: du dw . F(:c) : (x3 + 4:07 :> F/(x) : 7(;.:3 + 4m)6(3w2 + 4) [or 735%»? + 4)6(3x2 + 4)] pm : (x2 _ a: +1)3 => F'(;I:) : 3(z2 — a: + 1)2 (2x # 1) F(:1:) : \4/1 +2zc+sc3 : (1 +2\$+\$3)1/4 => I ,1 34/511 2 3: 1 -2+3x2 F(m)_4(1+2x+w) dw< "I" 554—9”) 4(1+2m+m3)3/4 ( I 2+3m2 _ 2+3x2 4(1 + 206 + 933W“ 4 {/(1 + 2x + m3)3 rm , (1 H4)“ 2 f’(m) = 2(1 +m“)‘”3<4m3) = —§””:— ‘ 1 3 3m g(t) : —-i— : (t4 + 1)‘3 => g'(t) : —3(t4 + 1)‘4(4t3) = #12t3(t4 + 1Y4 : _%12_t3_4 (t4+1)3 (t +1) ...
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