ma1012_hw13_201012

# ma1012_hw13_201012 - m is c4 when x œ 0 and y œ 1(c We...

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Unformatted text preview: m is c4 when x œ 0 and y œ 1. # # (c) We want the slope and x0 may vary) Ê y œ 8 Ê 3x c 4 œ 8 Mathematica: (functionsof the curve to be 8(see section 2.5 re. RealOnly ):Ê 3x œ 12 Ê x œ 4 Ê x œ „ 2. When c2 dr 2 c\$ 5 c# 5 dr 2 5 c" c% c\$ Ê ds œx<Miscellaneous`RealOnly` line athematics IcHomework or 3. œ erivatives Part Ic2, œ s œ 1 and the b 2s M ds equation 5s œ c s <c 3 2, y b # s œ 3stangent Ê has œ 2s cy : 1 œ s8(x c 2) 1 y D 8x c 15; when x œ x] \$ Section y œ (c m, c 4( h] 3.4 Derivatives of Trigonometric Functions S b 141 y œ 8x b 17. 2) or 133 Clear[f, 2) x, y, c2) b 1 œ 1, and the tangent line has equation y c 1 œ 8(xection 3.2 Differentiation Rules 0 c 3, x0 b 3}] dr c12 12 4 Differentiation d c\$ c% c# c% c& c& Section 3.2 Ê d)r œ 24)c\$ c Rulesb 20133 b ) Ê0= )1œ c12) b 12) c 4) œ ) b ) c ) 48) ) c' x d /4; b h] c f[x])/h \$ Thomas’ Calculus 11th Edition, Chapter 3 (a) fy œ x c 3x c 2 Ê yw œ 5,# w (1) For the tangent to be horizontal, we need m œ yw œ 0 Ê 0 œ 3x# c 3 0 42. 40. u(1) œ 2, uw (1) œ 0, v(1) œ 3x vc 3. œ c1 2 q[x, h], h Ä[x_]:=x Cos[x] 1: 35, 37, 41, 43 0] d Section ) œ 5, vw (1) (œ c3x(uv)¸3 Ê x œ „ 1.b v(1)ux œ œ 1, †y(c1) b 5 † 0 œ c2 line has equation y œ 0. The line Ê 1 # œ x0œ u(1)vw (1) 3}] When w (1) c 2 œ 0 Ê the tangent a) m[x0] (x c Plot[f[x], {x,x=1 c 3, : 5, 7, 15, 19, 43, 47a x0) dx Section 2x0 b w wperpendicular to this line at (c"ß !) is x œ c1. When x œ 1, y œ c4 Ê the tangent line has equation (1) b v(1)u(sin œb † ‰cx) b 5 † ")cc 2 5 †(c1) [x_,Êˆ u ( œ1) v(1)ub œdu(1)v f[x])/h )#basecx0 xd b(1)x0h_]:=(f[x3 c x#0((v(1)) a2 \$(1) œ b†0c2b ax\$ c25 b 1b † d a3 c x# b x œ2 b x\$ c c (3,b dxb v w ¸ b 1dy b x n},{x, (1) dx qb)c2x(c1)cos x=1a œ h] c † dx x c x 1(5) 3}] dx )cu(1)v 5†0œ †c4. Section 4:1, 5, 9, 15, this line at ("ß c%) is x œ 1. 2 y [x_]:=Limit[q[x, h], h Ä 0] to 17, 23, 25 The line perpendicular in x b cos x) sin x cos v œ b a(5) dcc sin x25b (cu(1)v (")cv(1)u%(1) 12x†#c1)c5†0c 3 1 # m x\$ xˆ xœ 1 2 ( c 2x (1)) b cos x‰¸ œ 2x) œ c5x bœ b a3x c 1b (c) b w cos x (u(1)) c3, (2) The dx u Total: 8+9+7=24 problems œ c 2 is )cv(1)u (1)(b) y†(c1)smallest value of yc x0) # and this occurs when x œ 0 and y œ c2. The tangent to the curve at (!ß c2) c tan:=f[x0] x=1 % D \$ # c23x b 5†0 œ b 1œ c2 4xifferentiation(7v cywm[x0] (x 7v (1) c 2x c 3 † (c1) c 2 b (1)) c x œ (d) (2) d 3 Ê 3 2u)¸ c5xline w12xc 2uw (1) œ 7 the tangent † 0 (œ c2) has slope " Ê y b 2 œ " (x c 0) or œ has slope c Ê x=1 x0 c perpendicular to the at !ß 7 dx 3 3 Section 3.2 Differentiation Rules 131 ytan},{x, 3, x0 b 3}] ]},{x, c cwPlot[{f[x], c 3, x0 7vwdy œ s2u#(1)‹ "b †3}] 1) c 2equation 7 the perpendicular line. (1) x0ec y œœ 7 c 2 is an † 0 œ c of (Answers d x.dy œx(sin x b cos x) (sec x) b sec x d (sin x b cos x) s) x)dx b x b mb dx 3 yw œ (x c 1)(2x dx 1) b ax# b x b dxb (") œ 3x# sec ax# x Ê [x0 c 1]//N 1Ê b 1 3 y [x0 \$ c Chapter 3.1 ( " b cos the x cos œ 2\$ b 1. c x )(sec \$41. mbœœ(sec 4x 3xc%x Note # # œ Ê) x d y œ 0 c 12xc&1œsinx12c 4(2) b 1 c # RULESdy œxxc1]//NÊ ywœ \$sinthat x #ß(sin is on x) sincurve: x c cx x ax x)(cos œ b IONx bxxtan 1dx b b # b )hat (#ß "Ê x) the\$curve: 0 œc c x) b 1 dx w x # b cos x b on c35. 1 ≠ 1 3x4(2) 1 cos tcot x) accsc) isPa) Slope ofxbthe atangentc (4x)(2x) b 3}]œ 4 c 8xc 4 4 acxslope of the tangent at (#ß ") is yw (2) œ 3(2)# c 4 œ 8. Thus 2c x b b(4) at (xß y) is 4x b 3x 4x x lot[{f[x], dy ac 43. y œ (c (cot œ# "m[x]},{x, 1x0bc b3, x0 œ y axbb 1b œ Ê b bb"b . When x œ 0, y œ 0 and yw œ 4(0 1 1) x b cos x c cosbxxsin x x) Êcsc œ b sec# x 1 x (1 b cot wx) 13x cosdx œ / 2 ≠a2 of the tangent at (#ß ") is yw ax œ 3(2)# c 4"œ 8. Thus t at x cß"y) is y œthe slope of thecline perpendicular tor the tangent at1(\$ ") 2 c 5 Ê the equation of the line perpendicular to ccx Ê5 slope c2 4 (2) c#ß is cos (x 37. 1 # d d 5 c dy s 0 œ c#Bœ d 2 w s œ c2 yd cb5ˆb d a5 b#,dssbÊ (3)yœ\$ ab#to sbb0œ 3sˆxb "2s bdy"is theˆline y %b 4x.† When b œb1, y œ 2 Ê yw œ 0, so the tangent to the 2x b œ % dr c 3 œ c 1 # xdx Êx " ‰o the y tangent#b 1#ß†"dx is c 5 Ê d the equation of"5s dxœxs x x1s8 5 x b x 2s œ cx b œ c at Ê Ê ‰ b œ b 5 c ‰ d a # c dxpœ sin x sec x todx tangent 2x the curve #B (!ß !)dx œœ c2x.‹ œ xthe line perpendicular to y erpendicular b thedx41. x atRULES ") 8is y dxx 1 dssec# " (x c 2) or y œ c b . tocos x sec œ (tan x(b 1 Ê c œ #c 8 the tangent line at) #ß 3.2 DIFFERENTIATION 8 " c # c # b b a (b 5 "isxc" line œ x# x "ß 2) b a1 c)xisurve atxœ c b(xtheb2) oryyœa2. c bb . c xc# b b a2x# b 10x b 2b œ 3x4 b 10x b 2 c x 1 t (#ß " d y y c%1 Thedrslopec (2x) curve at8%x1is 5mc& 3x cc 4 and the4 smallestr value for m is cc&when xc' 0 and y œ 1. (b)€ of the# d œ cc 4 œ 8 4œ 12 12 d c\$ cdyx)œ\$2x œc# dÊ Ê) xœ b 1 d # b 10xœ 2 c " 1sin4)cdxbxbc1 bx)#acos 2x œ 12)c by12) # c 4) œ x#Ê csin cot x) #3x œ 4 wand the smallest value for c is xb 4b cx Êand yœ 24) c 48) b 20) b) œ " 0 x) y c Êb dx b b (cos ) dxat (1 b (c) (cot d x) dx rve ba2xisœ5x)x€Wec cb(cotdy3xslopecx8(2x) (1 b cot x) amcsc8)cc (cot)wx) ac8)œx0Ê d#d y œ œ 2 Ê 3x# œ 12 Ê x# œ 4 Ê x œ „ 2. When when cscÊ dy 1. xyb mc dxxb 3 want the (1 d cot x) # b b d x(3) œ c2x b 0 œ c#B b 3x c œ 1. 8 4c Ê œ dx a of the curve to be Ê y œ È15œ (1 6 b È15 8 dx c b 20 b sin x  t(1Ÿ cot wx)wdx ax b 4b(0) c #œ œleft;16x . bWhen dx 44. b (1 cot x) 2, œ 4dxÊ w œ c16(2) œ c " ) ocurve of the r y œ to b 4 8Ê2,yyyœœbody is 4moving œx8bvhas3xwhen x œc 1yœ 8(x c 2) or y„ b8xWhen #when x œ c2, be œ Ê Ê 1 8 aÊtheb tangent a Ê 0 # œ 12 yÊ x# œ 1 and yx œ a2œ2. b c 15; , so the tangent x 4 x ccsc4x œ and b 3x c 4 line 4b equation 3 3 cot b d x ds xc csc 1% \$ Ê d s œ d a15t# # c d a % % œ s % \$ 43. dy 5t#tangent‰cot to œdy(1œ &(dt œ \$ (2bßc )1has15t1bœd1, or cy60t\$ #cc d"linec b œor y c 60tcb 2. 8(x b 2) or y œ 8x b 17. 15t x dt 3t cot x)at y c 15t the equation d ac tangent (x has œ c2, c"b œ dt 1 cot œ dt a5tline hasaequation c 4(cœ b Ê dt 30t y œ15t œ 15; ya15t 2), 30t œ c x 1 œ bb e x)x odylineb c the curve2)15t " dt2) 8(x c 2) and the8x b cddt when x equation y # c dt Ê b .# y is moving b 8 Ê w œ 2x b 1 b 0 œ 2x b 1 Ê #\$ œ # d d 2 œ x# dt x right # # "x dx 1b 3 c 1 b 11,\$ c #bbtangentc" œ hasa1 c # b y \$2x x 1 1 b b 2 y œ 8x † 17. x c a xÈ15c bb b c x b ) b" bx†œ6a„bandxthe 1ax Ê xyline a3 c x bx†cdx axcc b b cbb 2) orc x b 1bb dx a3 c x b equation œ 1 œ 8(x x adxx # d % at t " b a3x3sindx)1dx (cosax)\$ \$c x#x) dx1(1 (c2x) œwcb sin ba12x# bc (cos x) acos xb csc # x b# dy œ c #45. (1y"dw c wsec œ xxbb 1bdcb2b Ê yd œ!ß !126zcsin œ cthe tangent b c Ê c œ 0; y we ax# bm œ yw œ 0 Ê 0 œ 3x) c 3 b 42z Êœ (a) 21zœcc21z 3x 42z sin 2 w 3x ) x) 3. cx42z2x b b(0) to be horizontal, œ need bx passes through ("ß # # b Ê dz dz bœ (1 b &c 42z " through 5x Ê& b # b 42 1zxœc42. Ê\$ #yb126z (cospassesb 42xx) œ ((1œ # c È 0Fora(0) c 3 œ y 'bx c b b Ê dz ax wb 2x c c dx x (1 È& 3. sxœ 5tb c15 & sin x) ds œ x d a5t\$ b c d a3t& b œ15 sin x) 15t% È15 d s œ d a15t# b c d a15t% b œ 30t c 60t\$ 15 w \$ # 6 # È 3t Ê w % x b 4xrc x œ Forb 3 wtangentœxc5x„b ’!ß 6 c 2x‹ r3Š#ß 6œ 3w Ê œ dtat the # c dt equation 1 at x The c dt ß ß “ b Ê Ê be 1. When 15t c Š c 3x%b; # sin x) y dt down on 12x# c x need y b Ê dt the dt œœ œ 0 has 3 yx #‹3xsin 2 c3 a bc(1and slowsb c"and since the curve ism œ y œ‹ythe x0 œ 3xorigin, its slope is y œ 0. œ 0 line cx1 cos œ Ê c 3. Ê 3x y œ 32ax to b horizontal, 3we œ c1,tangent to0 Ê tangent line 3 œx sin c x) xœ (1 b sin x) x 1 sin x x) cos b x œ c# (1 bxsinwcosp1bc 1,xy œsin0 Êbtheœ 2a(0) b b) has equationWhen0. œ 1, lineÊ uv œ 1.the tangent line œ b equation c œ 0 so dyhe „ 1. units/sec when 2xx œ0andt v 1line at (2 "ß ! uisœ b œ 1. vThen aÊTheœ vu c 4 Ê œœquotient y Ê erpendicular toÊœœ 3x c movingÊx œand w y œ x b wœ y œ c a Whenœ d yœ t8x x xu œœ Êx1rule: œwhencos 05and this 4;tangent Ê wslowest 1. œ 3 b y at 2 In summary a has œ 1 and c line cand stationary test at 7 c it is t b 2 4x v dxc 1) ax# b x b 1b Ê yw œ # c 1)(2x b 1) b ax# b x b 1b (") œ w # 4. atthe "ß 4 c cdxycbx# b19line perpendicular4 Ê b line at ("ßdc3x is126z&1. 42z b 42 w(c curve is 15œ 21zWhen x œœ 21z' c 21z# the tangentdz %) has œ c œ c !( 6x œ œ 4. 1. c (xÊ3dw 1, y œ c to this 42z Ê line x equation 3z ) y c \$Chapter .2 7z c Thex. œ isbline œ 6x x 5)(3) dz ifferentiation is x œ dy w (3x œ 4 sec#x a (3x c 1)bsin x\$ bœc 2)sec is b c # cos # c 1) ax bb) c œ b# x)(2x)b bb(2x)(The linedxc (cos x)(2)tanw x2 c3, x x erpendicularxcotthisÊ x) aty(1 4Ê%)yx œyœisc cscand this occurs when x œ 0 and y œ c2. The tangent to the curve at (!ß c2) to x" smallest value ofx 3x 1. "ß c ( œ dy bx# b 0 b 4 Ê 1 ddy œ d 2x b bx b 0 œd b 1 sitionœc1 ¸cx œ 2xxÊ dxand 2x b(1 0 and y 2xxcdc(1) the cos x) w cosyx.isÊsœ andc as slope dxwhen xthe 1x)) b sec œ dx (sin totangent to theÊ !ß c2) (!ß c2) œ"x Êx# . b 2 œ " (x c 0) or 5 y 1b 3 x this passes œ Ê (secline Ê y0 œ 8x The thea tangent c# b sec46. y 3, (2)œ\$(2xxxoccurs 3 the#body is farthest from c b Êx c x b 1 ( the curve is y x (sin units c x) 4x œ œ 2. of dx axiscœdy4 10.3031)(2x) costhroughc"ß !Êperpendicular x1origin. œ 1b atcurve at has slope 3 c y For this curve, h dx b 3 dz c # x)(1) c 2xd 2 c dyˆ dx csin x)b (cos x)(1) 2x (cos c 4x c x(csindx 2x c 2xsin2x c cos 2 x( c " c 5. œ d w # cc œ ‰ ˆ x)bx c x " œ xcx #x x cos x" w x#Ê 1dx yx œx5c 1b2x andtheœ 1axx batw bœx cœxhas5slope‰x"b Êcb andb yxb x†xsin xc ax or b have common tangents at x œ 0, œ aÊ b b dx linebperpendicular Ê x tangent yb (!ß dx 2) Since xb b x ˆ sin 5 1 œ‰ " (x ax b 1 c x)(cos y 2bsin c 11 (sin b the perpendicular line. cos 3 dx a 1 by x " he b"tan œb w(secxœ #tox c 6cc\$ anx)18cosd† w x1. cosœ sin 1y xœ3coscxxa18b b 2 œ x# b 0) b b y c2 ec dw x x) d c\$ b œ 2 cos ax) c% x is " equation of x c% c c\$ cos x " # b z œ c6zc#œ 18z 3 œ 2z c"œ Ê c z œ 18z # b zc x must b have b c b 1 Ê sinc b axx xbc (sin zœ(2x) dz x# x) c 2z c#œ z 2xw# œ 2x b cosdz b ay œ dz #ab x " b 7 sec dyx)(2) œslope c x)(2x)b x 2x a bœ 5 cos x6. cosofœxx perpendicular #x also x# b xacsin Êbd1yc x1. b b ab 0zb 2x b a 2Ê c1 œ 2 † 1 b a c x a œ c3 cy xb 4 quation1xcthe œ bcos b.bzb line. , z b c 2(b "c 11 at x œ 2x Thus y œ 10xb 1 b œ 3x b 10x 2 Ê b1 x# 3 # x 4 Ê dx dx œ x& cos " # \$ # c ! #œ & ¸ !Þ!') furlongs/sec.# x(\$sin x c 2xsin x 4x " dy xywc cosb 1v(4) last 2 c x passes 8x and a "ß ! 1 we have œ cos x (w 5x Ê# se cos x b (\$ ybrule:cu15.œ b.x Sincebbb (sin x)(2x)b4x ua(2x)(csin4x) xbb(cos x)(2)b0c 2vu c uv b b Ê b œ 4(0 b 1) summary, œ and 10x(4x)(2x) œ b 0.5 b w œ 2x gw (x) c ! Ê2 œ Ê dx xbthe quotient 5xb x 3x bdya œ43x 4 xthiscxb curveÊ b 4 cthrough vc œ ),"b ÊWhen xœ 10,cy3œ 0 and yw œ 2. In 8 ds sec xc# œ db 43. s y cos tx \$ b x œ tan ax b bs Ê b dy œ\$ sec# x.cb 1bc # 4 a# b 1b . œv c8 c1 Ê2 dx %œ xc d x 24 1 # c œ?F ax ‹# % c œ 24 sin x œ dt t œc b œ 2 and c œ 1 is the b dx are 1 1 # c1sindt("(4x)(2x)c3,thebhorse'sxcct4œcosapproximatelyby24t x% œ 3x œ 2x and!Þbœ xfurlongs/sec. Ê a 2 sin x4 2 œ 24t 2cxso dt tcurves cx a Êt(4)bcquotient,cx4xb sec8x speedÊ bxbœœ c4t #x cœ "c t b wt ¸4(0 !(( c xc.\$ 8txbb c t x b xab 4t " c\$x cos bw y c% rence c œ2x ax c x b b 4 x ) 2x b # c 4 4 1) dw 6 œ &* c \$\$ d #' ccos " c 7. w œ 3z b 0.5) z œ œb 0.5) c6z. When xœ ?zt y œ 0 Ê dz œ 18z c 2z œ 18 c z2 w Ê(x dz œ bz b œ1 œ œc z ) ax b 1 b " (x%, ax b 1 œ theso theb tangent xto the curve 9th !ßœ)0, the line y œ 4x. When x œ 1, y œ 2 Ê y œ 0, so the tangent to the at ( and 10th furlong y ! is z and markers). This furlong takes a b 1b "‰ ˆ x c x b 1‰ last19. furlong (between the x tastest during d dc \$( c b " dy y ! xc (cot10 Ê the x) y œ 2. x) a c% he curve(a) dx!ßc)x)is#b"ßx) bisybœ#"4x.12 cb0cotWhen 10xœ 121,yyœ"12xandsow the 2 cÊœthe tangent line to the curve at (c"ß !) is œ 2 yw c#47. cot at (curvec thexline(1 ywcot\$ 3x# (1 1. c c# 1, x b c c dÊaccscœ b0, y œ tangent to the (1\$ cœ x) dœ œ 10 c 102) 10xy line When x10 ccscy œ (cot c 30 0 c 0 c 30x % (cot atc dydx 10xsin x b the1cos x amount œ axds(œœ bcfurlong.bœ 2 Êb8dxa2x cos x b (sin x)(2)bc122(c30 x) Ê 12x 12x wbcx t c" a btcx b 43. c xc b, 1 12xb (cos x)(2x)b 1b # œ cs" 2 "c\$ for a# c 2xhichbaisbdxb2 least " Êbcaxof timecsinatc #c" xc atbb "xba2x x) cx)cxcxtd b x c c cx4 24 sin x at " † 1 œ 8. s œ c2t b cot x) x dx Ê dxwœ œœ a x) c 8t30x b cot t Ê " # % (1 b bt 4t " Ê f (t) œ b ba b c\$ (1 "bc 2t œab yatœ #2.t c "b œœthe, first or y œ 2x"b 2. 2 at b 2b2 œ t œt a dt œ ct4t 2b2 b 24t œ t b t b y" cos ybœ sin bb csin b œ cx# sin x 2(x wÁ1)2x x dt 2 # csc x bax"duringt b 2 xxc 2 furlong (between markers 0 and 1).t b 2b2 e fastest 2x csc 2x x b xcœ x cÊ 47a. 1 x x x b bcos x c cot c ) (1 b cot x) Ê Derivatives of œ c16x . When x œ 2, y œ 143 44. ySection 3.4yw œ ax ba4b(0)4b 8(2x)Trigonometric Functions 1 and yw œ ac16(2) œ c " , so the tangent œx84 b # x ax b 4b c\$ 2 b 4b dy b c# 4b(0) c" y œ 6x#c16x c 8(2x) c16(2) " c 10 w cbc 2t b 2t œ 12x2t c " b 10 Ê d y œ 12 c 0 c 30xc% œ 12 c 30 10 t t)(2t) c y 1 ct t # 9. œ t) to 10x c 5x t x c )"dxhas When b œ x x ONOMETRICc 41dv. œ ad1#b2x (b"ßœ.4 œ t12xequationa10x4 1 œ c c , soœ 3 tangent c x c 2. FUNCTIONS Ê c y œ tdsb4thedquotientbq)the curvetat (2b2,b(1 c# 1 and c" œ b2t b œ and (x the or w dx y ux) acos xb " t axc (sin q œcosbt axsin rule:Chapter 5t and v the 3x c 2 1 Ê bcw bœ œca1#bvwbc 2), Ê yyœœ vu buv b œ b (1 dlineÊ bdt u œ ca1 3 )tan œ œ b dt tan c c œ sec ; use b sin x) dxq)( x) c (cos x) dx (1 b sin x) 2# 1 q) dt #v (1 (cos (1 b sin x) acsin a b c (cos x œ œ (1 x b 2. 2)cos(2xthe equation sinc 1 œ c (x 19 2), or y œ cb sin x) ) c q b 5)(3) œ (1 b y cx) c 15 œ" c c has 6x c 4 6x # # x 2) csin# c (3x b (3x" 2) cd# c sin (1 cc x c œ 10dsx q 1 d (cos x)#) x) 10 c 3 sin x # œw c sec c 3 dxœ ax 5)(2)œ t) œ cpasses through 17 1 Ê bb 7)(1) c b b1x) c2x 17 c sec q œ 45. x) 2t c b sin sin bx wœ (1(2xsinœy œ (x(1 b(sec.œ b c2t c 2xxc 10t tan t c(!ß !) Ê 0 œ a(0) b b(0) b c Ê c œ 0; y œ ax b bx passes through ("ß #) œ to the tangent line at (#ß ") is y c 1 œ c 8 (x c 2) or y œ c 8 b 4 . evalf[4](< convert(Xvals,Matrix) , c12 c\$ c& us`RealOnly`c# b 3xc% œ c# b 3 Ê d y œconvert(Yvals,Matrix) >); Ê (dy œ b) plot( L, x=x0-5..x0+3, color=black, 0 # c 4 andœ 3.1 #62(f)" ); for at m the dx The slope of the curve x x is dx œ 3x c 12x title="Section xsmallest value y, h] w # s through (aˆxÈÊ b (2)œ0 5. 1)(2x)wb‰ b(0)cbcb4xÊ 2x œ 0;theœ2x c#2 b bx 2passesbybœ x at("ß #) !ß !) 2 a œ a(0) œ 2ax b c and c Ê b 0 y ax b; 1 since y curve is tangent to through the origin, its slope is 1 at x œ 0 c nÊ dz cos t c "s b "‰ Šc (2x b ˆÈs c œ Š2x ‹ 2 t c Ès ˆÈs b œ c2x c œ c 1 a ctan ‹ " œ c ax b x 1 " 3 q) w (s)dœ cos t)secœ wt cdycc tq csc t)(csc t ax t)c 1b sec "‰ c ˆ a c 1‰ anbœ and csc t)(ccscxbcot st)b ("is c tanœsecscot œ œ œ #2a(0) b x c 1bœb slope is 1 at a b b0œ 2 Ê a œ 1. In summary a œ b œ 1 and c œ 0 so sec c c 1 1 cos q curve q0Ê dsb a5 (1 qb since the œ bq sec b tangent to y 1 x atqthe origin, its œ 1. ax c 1b x œ fœdx(1b(sinÊ œ c3Èœ54cosxxx tan qx œ cscÈxˆÈ ‰ Ê È ˆÈ Then x sec xdx cot x) Ê x ˆ swhen 4b sec c (1 b tan q) s b 1 b œ x y (1dx csc tan q) 2 s s s b 1‰ c 9bt)‰ dt (1 b 1 . # 143 t t c csc Ê Ê 1 2a(0) b bt cotis y œ 2xcscb x. a b b œ 2 Ê a œ 1. In summary a œ b œ 1 and c œ 0 so t cot c cothe curve t b c 1. Then t Section 3.4 Derivatives of Trigonometric Functions s‰(1œ tcsc"csc tfrom Exampleœ in c csccot t 2.1 2 (1 Section s y4 dyuse t) csin x) c (cos x)(1) . (cos x)(1)t) 4 csin x) œ x x2sin x c cos x u\$ œ 2x and x w œ 1 Ê gw (x) œ vu c uv 15 #u œ xccsc and v œ 4 ;cww#Èthexquotient rule:b 0 œ # c x( x cot x cc b 0.5 Ê w cos x b x sin v # x( c . œ cœ c a(cscxx) accsc xb b (cot x)(ccsc x cot x)b œ csc x b csc x cot x 0.5 y œcsc cot c È b v b Èx x x dp # (cos q)(cos q c sin q) c (sin q b coscos csin q) #x q)( x dx x cos x 2x)dq)œc46.())b 2# sinx)bxœ c)()x cos b 4 ("ß sin )) 0 œ c(1) c 1 Ê c œ 1 Ê the curve is y œ x c x# . For this curve, c c ax cos " y œ ) b x# passes through # !) Ê a # 4b œ 2x c c b 4 œ \$ b x ) b cx cos q b xc (1 x b c (sin0.5)œ ds acsc c cos t)(cos b c 17. 2 cos t c cos t x " xx) 0.5) ˆ2Èx‰ (5)csc t)x(x b1t)(sin t) csc(x b 0.5)cscc sin t cos t c " b wc (5x 1) b wwœq b !) sin œ cos t) Š c # dt dusin ("ß cos qÊ qcœ #c"2xxœ 1 x5x qc œ c cosw t) the1. Since yt)œ x c x# . cos t ythisx# b ax b b have common tangents at x œ 0, y(1 0 œ dy d and œœ 21 (1 1 y # œ c curve is 1 c(1) ‹ œ Ê c 1 Ê Ê # œ \$ c cos œ c # c and œ curve, ughd # x)(sec x tan x) œ sec (1x b sec x tan1 2 For sec4x x b (sin x)(2x)b b a(2x)(csin x) b (cos x)(2)b c 2 cos x x q .x xw osq w (cot x) b Ê dx† b (tan # s y x œ 2 sin x sec x b œ a cos dx# œ c (sec x) a4x # cos 23ax # b b x cot (cot x)(2x) b x dx x œ cx csc x have Ê Ê dxosa#) cbaybœb#x bbax b c must alsoœ x#t b#axbb abbat xb œ 1.bThus y œ 2x" bxaœ 0, c1 œ 2 † 1 b a Ê a œ c3 y "x " ) c œ x 1. Since ty œ œ b secandw y sec a slope c1 have common tangents at "b x 2 x x) œ t c "bt sec œ c 25t. 2 cos \$ œ(t) cos b ba" c t "ba" œ t # c t c " œ sec n2 x ac 2xbsin tx b 2 cos1x, cÁ " Êxxfwcx#œ x x at b 2b2 at #ba Ê b y œt b# c 3x b b. Since this last curve passes through ("ß !), we have 0 œ 1 c 3 b b Ê b œ 2. In summary, tx 2 c c" 2 at a 2Ê a œ tc3b2 b b2 a b2 lso have tan q) asec2 1b at (tanœ asec Thus y œ b tan q sec Ê tan1 sec 2 † 1 b sec q slope c x 1. p q 2x b a q c c q œ q c x x œ (1 b x Š1 c xq‹ c ˆ1 bq) c 4Èqb‰ œ sec x c " œ q Ê this last ˆ)b tan q) œ)) through 21 so), web tan)# 0 œ ) b 2)b(1 b)bb bc)()y In summary, )) (1 tan a curve passes and œ (È !)‰ œ c a q) are 1 œ x# sin3x œ and œ ) b ## . c Ê q) inceww wd))œ c œ #cd) dy sin # (sin "ß the have ) tyv drœ c2(csind3,x(b 2 2 b x cœ))(2)œ 2(1curves cosyy%Ñc 3 c2bsin x 2 œ 2. cos x c x sin an œ Ê) www Ð d (c" a1 b t x) x (cos t)(2t) Êc a ccos x in1xb tÊc"dy xœ1c#x)Êœdvsin(x)sin (sec) x (1 ccos x)œb c" c2xœ b 2t b (sin x)(2)b c 2(csin x) c # b# œ Êt dxœb axœ c Ê b y b c c tan x b (sect x c tan x) d c 2t c x b tan x) 2 cos x)(2x) # 2t tanœ sec )dx csct y œ dttan b and (sec ) aso the curves1are ) xwww x# c 3x dx 2a1 b t by œ x c x . Ð%Ñb t b œ t (sec " 1 ww x) cb y œ c9 cos cotÊ Ê xww c9(ca(csc x)#accsc xxb b (cot x)(ccsc xdx x)b œ csc\$ x b csc x cot# x sin x) œ # sin # x ) w x c 2x cos x c 2 y œb 2 sin x œ cx9 sin # Ê ya1 œ 9 cos x a1 b t b cos œ cdr b (a)(secxx x\$ tanxx) ac yw x tan x c 1. When x œ c1, y œ 0 cot yw œ 2 Ê the tangent line to the curve at (c"ß !) is sin y Ê y sec#cscb x y œ c c œ sec œ 3x b sec xb x xc x and s ) Ê 47. œ () cos ) b (sin ))(1)) c #sin ) œ ) cos ) \$ sc#3x# c\$d) #(2xbcœy(csc2(xxbb#a# 2x#and2xw c 102 sec# xctantangent line c tan x sec# at (œ "ß !) is x bw cot c sec# œtancab 1)œ bcœc yxbc b Ê 2 17 # xxb sec\$ x to the curve xb c 0. x 7)(1)x œx)5)(2)y or0 csc 2x x 2. œ the csc x y x tan œ 1 c csc c csc x x c w 1. œÊtan )œx When) c (x"b b tan secœ 7cos ) b sec ) œ (2x c )w dsec ) (cos b # 1, œ ) (2x c 7) ) tangentsecœ c 1(2x csecyww c 1 œ tanasec# xb b (tanc 7) at x œ œ(tan t) tan x Ê t œ # œ 7) y (sec x) # t dy x)(sec x tan x) œ sec\$ x b sec x tan# x 2x dr œ sec È ‰ tan#b 2. # ) x bcsc c tan )x b (csc ‹ œ È Ê dt)x œ (atan# ˆ)(cb1b Š) cotc ˆÈs c1 ‰ Š#))(sec )ˆ0.‹ )) \$ˆÈ ‰ ) œ 1 Ê dx tan ‰ lope db tan xb œ (sec x) asec# x b sec xs c 1b œ s2 sec x c sec x ‹ sc1 b" c " s c# x of w (s) œ s " " Ê œ " " c " sin # ‰ and ) ‰fb ˆ sin d ‰ ˆ cos ) ‰ ˆˆcoss) ‰‰œ sin ") b cos ) œÈs ˆÈ)b 1‰csc#œ Ès ˆÈs b 1‰ ˆ sin È b1 ) 2 sec s c ; 1 cos ) ds ) (sec t) œ 2t c sec t tan t ) Ê dt œ 2t c dt 1 " ˆÈ 3‰ cos s# w œ #c2 cos x Example 2 in Section 2.12 sin x Ê ywww œ 2 cos x Ê yÐ%Ñ œ c2 sin x Ê y œ Ès from Ê yww œ c2(csin x) œ dr sin(11(x b cotœt (" b sec ww q cos ) c cot# q )(sec ) tanc)csc# x) ) b ") b tan# ) Ðœ cos ) b sec# ) c œc %Ñ )( cw Ê 1), x c ( y c 1) b (sin Ê )csc t)(d 9 sin œ Ê b ))œ c t cos ) www œ q ) œ (cos œ cy cscqc)cscq)cot t) (c"sincsc t)(csc9 cot t) x Êœycsin c9(csin q œ 9 sin x Ê y œ 9 cos x (1 c csc t) ˆ2Èx‰ (5) c (5x b 1) Š ‹ t t c csc t cot t c csc t cot t csc du œ c2 c tx cot t 5x c 1 dp cÊ tanœ (1 gent 5 cosat Ê slope ofsec# q csc t) œ 4x b œcsc t)dxx q Ê dq œ4xtangent at dt (2x c 7) dt (2x c 7) (2x c 7) [email protected] x x ) œ cos (c1) œ c"; slope of c t) c (sin 2 c cos cos t c œ y(1(0)cos t)(cos cos t)1œ t)(sin t) ˆ1 bcos t(1 cxcost c sin tx c " (1 c cos " œ c 1 c" t œ 4w c sÈx qÊ vwdp (0)(1cb x ‹ c q)(cc 4È ‰ bt)(cos q)(ccsc qt)cot q) œ (cossin q c 1) c cot# q œ csin q c csc# q œ cos x and x sin q) œ 2È œ cos Ê (1dq œ Š 1;csc c œ 1 ...
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