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Unformatted text preview: 188 CHAPTER 3 DIFFERENTIATION RULES 3.6 Implicit Differentiation d d 1. (a)&—$(:cy+2cc+3m2)=(—i;(4) :> (a:~y’+y-1)+2+6ac:0 é my'z—y~2—6zc => , —y—2—61‘ , y+2 =——-—- 2—6— . y ac ory an 4—2 — 2 4 (b)my+2m+3x2:4 fi myz4—2w—3x2 => y=~—Z—3x—=:—2-3x,soy'——F—3 (c)Frompan(a),y'=—y——Z—6—m: (4/:c 2 3:0) 2 6x:_4_/$_£:__4§_3. a: as m m d 2 2 _ d I_ ,__8_x_#4.m 2. (mam +9y)—d$(36) => 8$+18y y—O => y— 1831— 9y (b)4x2+9y2-36 _> 9312—36 49:2 > y2 3(9 932) —> y—iéJQ—flwo _ 2m 'Zi2.lg_2 ”2 —2:c= ____ y 3 2( m) ( ) $3M 4:0 4:0 29: I:~_:________: ——-—. (C)F‘°mpa“(a)’y 9y mam) $3M 3'(a)d%(5+%>:3%(1) => #EIE—y—liy’—O —> 12 ’—% —> y’——; (b)%+;=1=> %:1_%:x;1 => y=xi1 '—(x_1(:(1_)1—)2(m)(1) (33:11)2 (C) y, : {lo—Z : ALE/(mac; 1)]2 _ ”362(52— 1)2 _ _(:c—11)2 4-<a>%(fi+\/z7)=§;(4) : 2—3/7+§1—fiy’=0 ;» 212i; , 4 (b)\/g:4—\/i => y=(4—¢E)2:16—8fi+x => 3121—5“ fl 4—fi —i+1 (C)y’=—fi— fi fi (1) => 2z+2yy’=0 => Mil/=43” :> y, y I 9. d3: I tEli-2 (1) => 2x—2yy'20 => 2m=2yy' => y 2 7. Ed— (103 + $2?! + 4y?) : di (6) :> 3m2 + (xgy’ +y - 2x) + 8113/ = 0 => 1323/ + 8yy’ = —3m — 2mg 93 x 31:2 +2262; _ M => (w2+8y)y':'3m2_2$9 : y,:_ $2+8y x2+8y 2/ 8.-d—(2-2:cy+y3)=i(c) => 2m—2(my'+y~1)+3y2y’=0 => 2w—2y=2$y'*3yy => drz: dw 2m—2y I 2 ’_ __ 2w—2y:y(2xfi3y) :> y *2m—3y2 SECTION 3.6 IMPLICIT DIFFERENTIATION — 139 d 2 2 d 2 I 2 I 2 1 3 9.E(my+$y)»a(3x) => (my+y~2m)+($~ yy+y-)— => 2/ I 2 l 2 2 _3 2 k 2 I_3_2$y“y2 wy+2xyy:3—2a:y—y :> y($+$y)— — my y 2 y_fix2+2zcy d 5 23_d x2 4/ 2 2/ 3. _ .32. $2.: 10. dm<y +xy)~dm 1+ye => Syy +(cc -3yy +y 2x)—0+y e 2x+e y => 4 2 2 2 2 3 2273/0312 _y2) I z — I I“ y(5y +3my—8 )—2$y€ —2$3/ : 34—W d . 11.;(x2y2+zsiny)—d$ (4) => m2~2yy'+y2-2w+mcosy~y'+sm?l'1:0 :> m 2 I I 2 . 2 I —2$y2_8iny 2m yy +Lccosy'y = —2zy “5111?! => (233 y+mcosy)y : E 2 - /_ -2my —smy : y _2z2y+$cosy 12.6110”): 1 {E 33 [sin(:cy2)] 2 1 : [cos(acy2)](x - 2yy’ + y2 - 1) => 1 = 2223/ cos(a:y2)y’ + y2 cos(acy2) 1 ~ g2 cos($y2) =>1— 2cossc2=2$ (:0st ' => '=%’ y ( y ) y < y )y y 2xycos(my2) d 1 I _ . : 3 day (4 coszsm y) 3(1) => 4[cosac-cosy‘yl+siny- (—sinx)]: 0 .—_> yl(4COS$COSy)=4sinxsiny => y’:w = ta $t 4cosaccosy n any day [m sin(y2)] => ycos(w2) . 2m + sin(ac2) - y' = mcos(y2) - 2yy' + sin(y2) - 1 => ‘- 2 2 y’ [sin(m2) — 2mg cos(y2)] = sin(y2) — 2mg cos(x2) => 3/ = W sm(9:2) — 2mg cos(y2) 15.1(ez2y)=—(ac+y) => €12y($2'+ ~2w)—1+ ’ 2ez2y 2 $211 ’ dx dx 3/ y — y => I y+xye :1+y :> 2 $82M ,_1—2xyezy y —y — 1 — 2333161” 2'” => y'($2ez2y — 1) = 1 — Qxyezzy :> y _ {[26221’ ~1 16- diww ): d:(1+$2y2) => %($+y)‘1/2(1+y’)=x2-2yy'+y2-2m => / y _ 2 l 2 2‘/‘w+y+2‘/‘x+y“2$yy+2xy => 1+y’=4962yx/964ryy’+4903!2 $+y => 2 _ yl—42: ydm—ky y'=4$y2‘/x+y—1 :> y'(1—4:L'2y‘/a:+y)=4my2vx—i—y—1 :> y'— 4my2 w/m—i—y —1 1 — 4x2y‘m: +y 17.,/:cy=1+:1:2y : %(my)‘l/2(my'+y‘l)20+:c2y'+y-2:zr => a: y'+ y =$2'+2x 2‘/:cy 2,/.7;y y y 2 : y,(2w _$2):2$y* y i, y, r-2xx/fi :4mym—y : y,:4xy¢x‘y—y m 2m 2m 2m m—2z2m 190 CHAPTER 3 DIFFERENTIATION RULES y 1-I-sc2 (1+$2)SGC2($ — y) ' (1 - y') + tan(93 — y) - 2x = 3/ => 18- tan(m - y) = :> (1 + 9:2) tan(m — y) : y 2} (1+$2)sec2(gc — y) — (1 + $2) sec2(m — y) . y’ + 2xtan(m — y) : y' 2 (1 + $2) sec2(a: — y) + 2x tan(a: — y) 2 [1+ (1+ :3) sec2(x — y)] - y’ :> I WW 1 + (1 + x2)secz(:c — y) 19. my 2 c0t(xy) : y + my' : —cscz(:cy)(y + wy’) => (y + xy') [1 + csc2(a:y)] 2 0 => y + my' 2 0 [since 1 + csc2(33y) > 0] => 3/ : —y/m 20.5inx—I—cosyzsinsccosy :> cosx—siny-y'2sinx (—siny-y')+cosycosm => (SinxSiny_Siny)yl:COSLUCOSy—COSQL' ___.> y,zcesm(cesy-1) s1ny(smx — 1) 21.%{1+f(x)+m2[f(x)]3}=%(0) : f'<x)+x2.3tf<m>12-f’<x)+{f<z>13~2x=0.In:Lwehave f'(1)+12-3If(1)12'f'(1)+If(1)I3~2(1)=0 => f’(1)+1-3~22~f’(1)+23-2=0 => f’(1)+12f’(1)=—16 :> 13f’(1):—16 => f/(1):_%_ d . d , 22- " [9(93) + msmg(ac)] 2 aw“ => 9 (3:) + xcosg(x) -g’(:c) + sing(a:) - 1 = 2m.1f;c = 1. we have dm g'(1)+1cosg(1)-g'(1)+sing(1)22(1) => g'(1)+cosO~g’(1)—I—sin0=2 => g’(1)+g'(1)=2 :> 2g'(1) = 2 => g'(1) = 1. d 23.y4+ac2y2+ya:4:y+1 ;> 4213+($2.2y+y2~2m8§)+(y-4x3%§+w4.1)=1 :~ 9 dm dac dac 1 —4y3 —2:L'2y—x4 2m2—+4x3——:1—43;22—4 :> -2/ y dy ydy y as y a: dy 2$y2+4$3y dm dm dam aw2 — 4y (:52 + 3/2) 24. 2 22: 2 :22 22—— 2 = — 2 —=———” (cc +74) amy (m +y) xdy+ y 2ayzv dy+ax : dy 4m(a:2+y2)—2aa;y 25.x2+wy+y2:3 => 2m+xy'+y-1+2yy’=0 => xy'+2yy’:—2a:‘y :> _2 _ ——$——£.Whenx=1andy:1,wehavey’= m+2y an equation of the tangent line is y — 1 = —1($ — 1) cry 2 —av + 2. y'(m+2y)=—2m-y => y’= 26.m2+2$y—y2+m:2 : 23+2(my’+y~1)~2yy'+1=0 => 2xy’—2yy':#2sc—2y—1 => -2:I:—2y—1 y’(2:c#2y) : —2:c#2y— 1 2> y' = .Whena: = 1andy:2,wehave 2m—2y I#_2#4_1_#7 . . . 7 y — fl. _2 ,so an equatlon of the tangent km [S y — 2 2 5(93 — 1) cry 2 5:1: — 5. [OI-\l 27. 28. 29. 30. 31. SECTION 3.6 IMPLICIT DIFFERENTIATION 191 ,Wfi NIH x2 + 3/2 = (2x2 + 2y2 2 an)? => 235 + gyy’ : 2(2x2 + 2y2 — m)(4:c + 4yy’ — 1). When w 2 0 and y 2 have 0 + y' 2 2(%)(2y’ 2 1) 2> y' 2 2y' — 1 2 y' 2 1. so an equation of the tangent line is y—%21(x20)ory2x+%. 1 7/, / \3/33 2/3 2/3— 2 ‘1/3 2 ’1/3 ’2 — ‘ 2 2> 2——.Whenz2—3\/§ :6 +3; 2 4 2 3:6 + 3y y 0 => \3/5 + W 0 y \3/5 2/3 1 (—3x/E) 3 1 . . . : ’ = 2 _ — —. so ane uation of the tangent lme lS andy 1.we havey (73%)1/3 ,3\/§ 3%? \/§ q y212%(x+3x/§) 0ry2fix+4 2(562 + 312)2 2 25(302 — y2) 2 4(th2 + y2)(2m + 2yy’) 2 25(23: — 2yy') 2> 4(cc + yy')(a:2 + yz) 2 25(50 — yy') 2 Aim/($2 + yz) + 25yy' 2 252: 2 4:5(202 + 3/2) 2> 25m—4m(z2 +112) . ’2g.Whn 23and 21.wehave ’2M22fl223,soane uationofthe 25y+4y($2+y2) e 33 y y 25+40 65 13 q tangent line isy — 1 : —%(m _ 3) ory : _%$+ ‘11—‘33. 3/ y2(y2 — 4) 2 x2(a:2 2 5) 2> y4 — 4y2 2 2:4 — 5m2 2 4y3y' 2 8yy' 2 40:3 2 1056. When a: 2 0 and y 2 —2, we have —32y' + 16y' 2 0 2> —16y' 2 0 2> y' 2 0. so an equation of the tangent line is y+220(;c—0)0ry2 —2. 1 3 — (a) y2 2 5x4 2 m2 2> Zyy’ 2 5(4273) 2 23: 2> y' 2 M. (b) y 3 _ So at the point (1, 2) we have 3/ 2 % 2 g and an equation of the tangent line is y — 2 2 %(ar — 1) or y 2 Ex 2 g. , 3 2 6 32. (a) y2 2 m3 + 3x2 2 2yy 2 3202 + 3(2w) 2> y' 2 12:54. So at the point (1, —2) we have ,_3(1)2+6(1)_ 9 . . . 9 y 2 W — —Z, and an equation ofthe tangent line 18 y + 2 2 23(31 — 1) ory 2 —%:r + i. (b) The curve has a horizontal tangent where y’ 2 0 <2 (0) 32724—63220 4:» 3m($+2) :0 <2 m2Oor$2 —2. But note that at £1: 2 O. y 2 0 also. so the derivative does not exist. At :15 : —2, 3,2 : (—2)3 + 3(—2)2 : —8 +12 2 4. so y = i2. So the two points at which the curve has a horizontal tangent are (22. —2) and (—2, 2). 192 CHAPTER 3 DIFFERENTIATION RULES 3m2~6$+2 b ’_ a I,_ ()y 2(2y373y2—y—E1) y lat (0,1) and y' = g at (0, 2). 33. (a) Equations of the tangent lines are y = 7w + 1 and y 2 %x + 2. (c)y'_0 _> 3:02 6z+2—0 —> m 2 1i éx/Z’: There are eight points with horizontal tangents: four at at a“ 1.57735 and four at a: % 0.42265. (d) By multiplying the right side of the equation by ac — 3, we obtain the first graph. By modifying the equation in other ways. we can generate the other graphs. 4 5 O *2 I 6 i5 —3 Mf—®@*m : m(x ! 1)(x 7 2) 3 4 4 '3 III -' =3 A A ‘4 4 3 $(y + 1)(y2 -1)(y — 2) My2 + 1)(y - 2) W; + 1)(y2 - 2) :y(:8—1)(x—2) :x(x2—1)(x—2) SECTION3.6 IMPLICITDIFFERENTIATION :l 193 34. (a) 3 (b) There are 9 points with horizontal tangents: 3 at x— — 0 3 at m- — g, and 3 at :c = 1. The three horizontal tangents along the top of the ' wagon are hard to find, but by limiting the yrange of the graph (to :‘4 [1.6, 1.7], for example) they are distinguishable. 35. From Exercise 29, a tangent to the lemniscate will be horizontal if y’ = 0 => 2556 — 427(122 + y2) = 0 => my x[25 — 4(202 + y2)] : 0 ¢ 11:2 + y2 2 ¥ (1). (Note that when m is 0, y is also 0, and there is no horizontal tangent at the origin.) Substituting % for $2 + y2 in the equation of the lemniscate, 2(932 + y2)2 2 25(112 ~ yz), we get :32 — y2 : % (2). Solving (l) and (2), we have :52 2 % and y2 : 1—56, so the four points are (iii, i— 2.) :3 2x 2 132$ . 36. E + 35—2 : 1 => E + % : 0 :> y, : _a_2y => an equation of the tangent line at ($0,310) is 412930 . 1 . yo . gay :13 580:1: m2 . y * yo : agyo (:1: — am). Multiplying both Sides by E gives 11—2 — b—2 = — a2 + (1—2. Since ($0,110) lies on y2 310?! _$0 the ellipse, we havex F +— b2 _ — 2+ (2—3 = 1, 2 2 / 2 a: y 2m 2 b :L' . 37. (1—2 — b—2 — 1 —> (1—2 733a a 0 —> y’ — 3 ¢ an equation of the tangent line at ($0,310) is (72.730 guy 2 2 _ y (Boil: 130 y yo _ (12340 b2 b3 = —a2 — a—g Since ($0,310) lies on the 2 2 x0 99 L0?! _ 530 yo _ hyperbola, we have $— — b—2 _ (1—2 ._ E _ 1_ 38 _' x/37 . $5 + V37 = x/5 :> 2—‘/— + 2f =0 => 342 TE => an equation of the tangent line at ($0, yo) is _ _yV V310 y — yo — __ x261: — 220) Now ac — 0 —> y — yo ——$0 ( 1:0) — yo + \/m—0\/y_, so the y-intercept is y0+‘/xo\/3E.Andy:0 3 —y0: MCI—$0) z) w~$02y0~$0 x/I—o m z = x0 + «mo ‘ /y , so the x—intercept is 200 + «we , /yo. The sum of the intercepts is (310+x/x—0x/y—0)+($0+x/$—0x/%) =zo+2mm+yo : («z—Hm)? : («6f = 39. If the circle has radius 7', its equation is m2 + y2 = r2 => 22: + 2343/ : O 2 y’ = —£, so the slope of the y r , LC _ tangent line at P(mo, yo) IS ——0. The negative reciprocal of that slope is “I 2 31—0 310 —$o/yo 180 OP, so the tangent line at P is perpendicular to the radius OP. , which is the slope of 194 40. 41. 42. 43. 45. 47. 48. 49. I: CHAPTER3 DIFFERENTIATION RULES yq : $12 :> qu—ly/ :pmpil : y = z = — : _m(P/€l)—1 _ 7 I 1 d 1 1 y~tan 1 :c => 2—.._ _ 1 —1/2 _ f y 1+(x/é?)2 defi) 1+ (296 >'2 w flaw) y 2 tan—1x : (tan—lacy” : y' = %(tan_1m)‘1/2> i(tan‘1 :17) = 4—. 1 = —1__— 61$ 2Vtan‘la: 14—552 2Vtan‘1x(1+cc2) y = sin*1(2m + 1) :> I 1 d y=————. (2x+1)=———-—1—-—.2:_..2__:_1_ 1 — (256+ 1)2 d3? V1 — (4m2 +4m+ 1) \/-4zc2 —4a: \/——:c2 —$ h(:c) = \/1 —:c2 arcsincl: => 1 . h/(x :x/l—m2.__ +arcsin5c[l 1—x2 ~1/2 _2 ]: lfi warcsmx ) 1/1—552 2( I ( 33) ”m3 H65) 2 (1 + $2) arctanm => H'(ac) :2 (1 + 232) + (arctan :c)(2;1:) = 1 + 2x arctanm 1+:c2 .y:tan‘1(x—\/m2+l) => y _ 1 <1 2: > 1 (x/sc2+1—:c) 1+($_ /——$2+1)2 x2+1 1+z2g2m/m2+1+m2+1 x/x2—I—1 \/$2+1,$ _ x/mz—I—lflsc 2(1+x2—x\/x2+1) x2+1 2[\/x2+1(1+$2)—x(m2+1)] \/$2+1—m 1 ( w2+1—x)] 2(1+$2) h(t) = cot—1(t + cot"(1/t) : h’(t) 1 1 .d1 1 t2 i __ 1 + 1 _0 ‘ 1+2:2 1+(1/t)2 dtt 1+t2 t2+1 t2 1+t2 t2+1” Note that this makes sense because h(t) : g fort > 0 and h(t) :- _% fort < 0. 71 I —1 '73 :xcos :c-x/l—ar:2 => :cos m—————+——=c0s as y y x/lflm2 x/l—a? 1 d 262a: : COS_1 2x _> I _ - 62m ’ —‘——‘ y (6 ) y 1 _ (62m)? d]: ( ) fl -— 64:” I 1 . _ sin0 50. y — arctan(cos 0) i y f 1 + (cos 0)2 (_ 5m 9) * 1 + C032 6 51. f(m) : ex ‘ m2 arctancc => f'(ac) : e” — [93(1 +1332) + (arctanx)(2$)] :62 L 4’ : eac i 1 + $2 7 2xarctan$ _2 A 3 This is reasonable because the graphs show that f is increasing when f ’ is positive, and f ' is zero when f has a minimum. SECTION 3.6 IMPLICIT DIFFERENTIATION 195 mg“, f'(m) : .7; ——2m__ +arcsin (1 — 9:2) ~1 V 1 — (1 — m2)2 2 2 2 arcsin (1 — $2) — 4 2202 — 9:4 52. flan) 2 :carcsin(1 — $2) 2> This is reasonable because the graphs show that f is increasing when f ’ is positive, and that f has an inflection point when f’ changes from increasing to decreasing , d 53. Lety2c0s‘1x.Thencosy2acand0§y§7r 2> —sinyd—:2 1 2> dy — 1 ——;2—;. (Notethatsiny20f0r0§y£7r.) E: Sinyh x/l—COSZy s/1—232 54. (a) Let y 2 sec‘1 2:. Then secy 2 3c and y e (0, g] U (7r, Sill Differentiate with respect to m: c tan d_y — 1 2> fl2—1—2—1- Note that se y y d9: — d9” secytany secy «sec? y— :mx/xz’ — 1 tan2y23ec2y—1 => tany: t/secgy—1sincetany>0when0<y< g 0r7r<y< 3?" _1 dy dy 1 (b)y2sec cc 2> secy2x 2> secytanyZ-x 21 2> E2W.Now tan2y2sec2y—12w2 —1sotany2i\/ac2— LForyE [0, )zc>1sosecy2x2|a:|and dy 1 tan >0 => —=h:%.F0r E 1,7r.:cS—1,so:L-:—xand y¥ dz‘ 3; $2_1 iml‘/$2~1 y (2 i ii d 1 1 1 1 tany2~ 19—1 :> _y_ 2 _‘: dx secytany $(_ 352 _ 1) (_$) «($2 _1 W m 55. 2$2 +y2 2 3andsc 2 3/2intersectwhen21:2 +x—3 2 0 <2 (2$+3)(x— 1) 2 0 (2) a: 2 —g or 1,but ~g is extraneous since a: 2 y2 is nonnegative. When a: 2 1, 1 2 y2 2> y 2 i1, so there are two points of intersection: (1,:t1). 2x2 + 3/2 2 3 => 4x + 2yy' 2 0 2> y’ 2 —2:1:/y, ands: 2 y2 2> 1 2 2yy' 2> y' 2 1/(2y). At (1, 1), the slopes are m1 2 22(1)/12 —2 and m2 2 1/(2 ~ 1) 2 5, so the curves are orthogonal (since m1 and Wm are negative reciprocals of each other). By symmetry, the curves are also orthogonal at (1, —1). 56- 372 * 92 2 5 and 4:52 + 9y2 2 72 intersect when 4302 + 9(m2 — 5) 2 72 (is 13m2 : 117 (is x = i3, so there are four points ofintersection: (i3, i2). $2 — y2 2 5 2 2:1: — 2yy' 2 0 2 y’ 2 22/31, and 4552 + 9.712 2 72 2> 8.7: + 18yy' = 0 <=> y' 2 —4x/9y. At (3, 2), the slopes are m1 2 g and m2 2 ~ .30 who the curves are orthogonal. By symmetry. the curves are also orthogonal at (3, ~2), (—3, 2) and (—3, —2). 196 El CHAPTEB3 DIFFERENTIATIONRULES 57. A 58. The orthogonal family represents the direction of the wind. §\\ 59. x2 + y2 = 7‘2 is a circle with center 0 and am + by : 0 is a line through 0. y x2 + y2 = r2 => 22: + 2yy’ = 0 fi 3/ : far/y, so the slope of the tangent line at P0 ($0,310) is ewe/yo. The slope of the line 0P0 is yo/mo, which is the negative reciprocal of —mo /y0. Hence, the curves are orthogonal, and the families x of curves are orthogonal trajectories of each other. 60. The circles x2 + y2 2 am and m2 + 3/2 = by intersect at the origin where the Y tangents are vertical and horizontal. If (9:0, yo) is the other point of intersection, then :53 + 3/8 2 auto (1) and 23% + yg : byo (2). Now :62 + y2 2 am :> _ 2 x 2m+2yy’ 2a => 3/: a 211$ andm2+y2 =by => 29: + 2yy' = by' => 3/ = b 2132 . Thus, the curves are orthogonal at — y —2' b—2 (9307310) © a2y£0=~ nyo (it 2am0—4m324y8—2by0 (I) o 0 (m, + bye 2 2(x3 + ya). which is true by (1) and (2). 6Ly=cw2 => y’:2cwandm2+2y2:k z) 2$+4yy':0 => ny’ : —;z: => 3/ = ;_x_ = — m = #—,sothecurvesare 2(y) 2(czt2) 20m orthogonal. 62.y:a:c3 :> y’:3am2andm2+3y2=b => 256+6ny=0 Z) ’ ’ as a: — — 1 so the curves are 3yy — —m —> y — , [email protected] _ 3(asc3) 30,502 orthogonal. SECTION3.6 IMPLlClTDIFFERENTIATlON C 197 63. To find the points at which the ellipse 2:2 — my + y2 : 3 crosses the m—axis. let y : 0 and solve for w. y = 0 => 932 1 33(0) + 02 : 3 4:) a: = ix/3. So the graph of the ellipse crosses the m—axis at the points (ix/3. 0). Using implicit differentiation to find 31', we get 2m — my' — y + 2343/ = 0 :> y’(2y — x) : y _ 2g; yi2m / . 0—2\/§ , . 0+2\/§_ ¢> ' : .So at 3,0 is — = 2 and at —\/3.0 is -——— — 2. Thus, the tangent y 21,750 g (f) 2(0)—\/§ y ( ) 2(0)+\/?_> lines at these points are parallel. . . . i . . , y — 2m . . 64. (a) We use impliCit differentiation to find y : 2y _ an as in Exercrse 63. (b) . 1 — 2 —1 3 The slope ofthe tangent line at (*1,1) is m : WM 2 E :1. 1 . . . so the slope of the normal line is —— = —1. and its equation IS in y — 1 : —1(zc +1) 42> y : —m. Substituting —x fory in the equation of the ellipse, we get :02 — z(—w) + (—902 = 3 :> 3102 = 3 4:) a: : :l:1. So the normal line must intersect the ellipse again at m : 1, and since the equation of the line is y 2 ~22, the other point of intersection must be (1, —1). 65.$2y2+my:2 :> x2-2yy’+y2-2$+x-y'+y~120 4:) y’(2m2y+m)=—2xy2—y (it 2 2 2 y’~ wy +3130 2mg +y 2w2y+x 2x2y+w y(2$y+1)—:c(2acy+1):0 <=> (2my+1)(y—m):0 4:) xy:—%ory=;v.Butzy=*% :> ——1 4:) 2my2+y:2m2y+x 4:} y(2my+1):m(2xy+1) 4:) m2y2+scy=i—%#2,sowemusthavew:y.Then m2y2+zy=2 => x4+m2=2 {i} 3:4 +322 — 2 = 0 4:) ($2 + 2) ($2 — 1) = 0. So 302 = —2, which is impossible, or $2 :1 (it I : :1. Since :5 2 y. the points on the curve where the tangent line has a slope of —1 are (~1, —1) and (1., 1). 66. 2:2 + 43/2 = 36 :> 29v + 8yy' = 0 => y' : —%. Let (a, b) be a point on m2 + 43/2 = 36 whose tangent line passes through (12. 3). The tangent line is then y — 3 : —i (:z — 12), so b — 3 = T421) (a — 12). Multiplying bothsidesby4bgives4b2—12b=~a2+12a.so4b2+a2:12(a+b).But4b2+a2236,8036212(a+b) => a+b=3 :> b:3—a.Substituting3—aforbintoa2+4b2=36givesa2+4(3—a)2:36 4:» a2+36—24a+4a2:36 c) 5a2—24a20 «I» a(5a~24)=0.soa:00ra=154.1fa20, b23—023.andifa=%.b:3—%:~§.Sothetwopointsontheellipseare(0,3)and(%.—§). Using 3; — 3 : —4%(sc — 12) with (a, b) : (O. 3) gives us the tangent liney — 3 : Dory 2 3. With (a. b) : (%, —9). we have 4 y—32—M—Z_S%(z—12) 41> y—3:§(x—12) 4:» y 2 gay — 5‘ A graph of the ellipse and the tangent lines confirms our results. 198 C CHAPTERS DIFFERENTIATION RULES 67. (a) If y : f _1(m), then f (y) : m. Differentiating implicitly with respect to a: and remembering that y is a function , dz; d 1 7' ofx.wegetf(y)E=1.SOEZ:-m => (f1)($):m (b) f<4> : 5 : f"1<5> = 4- Bypama» (f’1)’<5> = 1/f’(f’1(5))= 1/f’(4) : 1/<%) = 68. (a) f(rc) : 25c + cossc => f’(x) : 2 — since > 0 for all ac. Thus, f is increasing for all a: and is therefore one-to-one. (b) Since f is one—to—one, f_1(1) : k 4:} f(k) = 1. By inspection, we see that f(0) = 2(0) + c030 : 1, so k : f’1(1) : 0. (c) (f’1)'(1)= 1/f’(f’1(1)) : 1/f’(0) : 1/(2 — sine) =% 69. x2 + 4:112 2 5 => 2x + 4(2yy') = 0 => 3/ = —4£y. Now let h be the height of the lamp. and let (a, b) be the point of tangency of the line passing through the points (3, h) and (—5, 0). This line has slope (h , 0)/[3 * (—5)] = éh. But the slope of the tangent line through the point ((1,1)) can be expressed as y’ : — b — 0 _ b . . a b a ; (—5) — a + 5 [smce the line passes through (*5, 0) and (a, b)], so _4—b = a + 5 4b2 = 7(12 ~ 5a 4:} a2 + 4b2 : —5a. But a2 + 4b2 : 5 [since (a,b) is on the ellipse], so 5 = #511 4:) or as a = —1. Then 4b2 — (12 5a : 1 5( 1) ‘ 4 -> b — 1. since the point is on the top half of the ellipse. h b 1 8 , So e h i 2. So the lamp is located 2 units above the w—axis. 3.7 Higher Derivatives ”——’————————’———— 1. a : f, b = f’. c = f". We can see this because where a has a horizontal tangent. b : 0, and where b has a horizontal tangent, c : 0. We can immediately see that c can be neither f nor f’ . since at the points where c has a horizontal tangent, neither (1 nor b is equal to 0. 2. Where d has horizontal tangents, only c is 0, so (1' : c. c has negative tangents for :c < 0 and b is the only graph that is negative for x < 0, so c' = b. b has positive tangents on IR (except at a: 2 0), and the only graph that is positive on the same domain is a, so 17' : a. We conclude that d : f, c : f’, b : f", and a : f’”. 3. We can immediately see that a is the graph of the acceleration function, since at the points where a has a horizontal tangent, neither c nor b is equal to 0. Next, we note that a : O at the point where b has a horizontal tangent, so b must be the graph of the velocity function, and hence, b’ : a. We conclude that c is the graph of the position function. 4. a must be the jerk since none of the graphs are 0 at its high and low points. a is 0 where b has a maximum. so b’ = a. b is 0 where c has a maximum, so 0’ = b. We conclude that d is the position function, 0 is the velocity, b is the acceleration, and a is the jerk. 5. f(;c) : x5 + 6302 7 7:16 :> f’(:£) : 5x4 +1256 — 7 => f”($) : 209:3 + 12 6. f(t) : t8 — 71:6 + 22:4 :> f’(t) : 8t7 7 425’ + 85’ => f”(t) : 56t6 7 210:4L + 241:2 7. y = 00320 => y' : *2sin29 :> y” : —4cos26 8. y : 6sin0 => y' : 0c0s6 +sin0 => y" : 6(~sin0) +cos§ - 1 +cos0 : 2cos€ ~ 65in6 ...
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