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File0047 - Section 3.6 Implicit Differentiation...

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Unformatted text preview: Section 3.6 Implicit Differentiation 46_x:,/5_\fl:>3—f=%(5_\/t)’1/2(_%r1/2)=——m;y(tv1)—\fl=>y+(t—1)‘% —%t‘”2 12y1 _ d _23Z.’Y_ 1—2y 1, 9X_ . fi:1—2Y\/ 4‘__\/V51‘ 3 (t 1)%1 =7 3’ => m— (1~1) —W’thusdx”€€= T W 1% 4\/1S \/1 2(1—2M1 5- _ 173/? t;t=4.>x_ [_fi-t_4 y(3)2 fl=2 1—22) 4 if therefore, % — L—(jjg— 2 74 t=4 47. x+2x3/2212+t 2 d—X+3x1/2d—’[‘22t+1:>(l+3x1/2)0d—x=21+12> %= l—i‘fiflw (n+1 1+2: y24 git/1+1 1+y- ,2)_(t+1)1/2+2\/_+2t(2y‘1 ’29,!) T %1\/t+1+7L+2\/37+(7‘-y%1) :0 , . d _ 2v d _ (WLlfiH—Zx/(i)_ -y\/'- 4N!“ fl $( t ' 1 ' ()7) 3% 271+] 2\/y if (m+71§)_2\/—(t+1)+2t;71+1’ 2! Ge‘m/H—l 9x _ dy/dt _— zV/flii lHZt 111 1 dx _ dx/dt _ 21+] I+3x112 thus 3:0 :> x+2xm=o :> x(1+2x1/2):0 2 X20;t20 24\/324(4)3Zo+. " 4(0+ l)+2(0)\/0+1 ‘V 2 yx/0+1+2(0)\/§24 2> y :4; therefore gfipo 2 - _ d___x dx ~ dx, # ggul—xcost. 48. xs1nt+2x-t => d1 smt+xcost+2 212(smt+2)d[—1 xcost > at“ Sim” , tsint—2t2y => sint+tcost—2= did;thus%2%;t27r => xsin7r+2x=7r sin1+2 E. El 7 sin7r+7rcos7r—2 A —477—8 _ _ 2x22,theref0redx‘[=w— 1777mm — 2+” —— 4 sin7r+2 qu-Fy 49. x2 + ny — 3y2 2 0 2 2x + ny’ + 2y — 6yy’ 2 0 2> y’(2x — 6y): —2x — 2y 2> y’: 2> the slope of the 21 2> the equation of the normal line at (1, 1)1s y — 1 2 —1(x —- 1) (1.1) 2> y 2 —x + 2. To find where the normal line intersects the curve we substitute into its equation: x2+2x(2—x)——3(2—x)220 2> x2+4x—2x2—3(4—4x+x2) 20 2> —4x2+16x-— 1220 2> x2 — 4x + 3 2 0 => (x — 3)(x - 1) O > x 3 andy x + 2 — 1. Therefore, the normal to the curve at (17 1) intersects the curve at the point (3, 71). Note that it also intersects the curve at (1, 1). tangent line In 2 y’|(1 1) 2 fl 50. xy+2x—y20 2 x3 —§—+y+2 1:0 => 33V;2%;theslopeoftheline2x+y=015—2. lnordertobe parallel, the normal linesd must also have slope of —2. Since a normal is perpendicular to a tangent, the slope of the tangent is % Therefore, ¥:—: 2 % 2 2y + 4 2 1 — x 2 x 2 -3 — 2y. Substituting in the original equation, y(—3—2y)+2(—3—2y)~y20 2> y2+4y+320 2 y2—3ory2—1.Ify2—3,thenX23and y+32—2(x—3) => y2—2x+3. Ify2—1,thenx:—1andy+12—2(x+1) 2 y2—2x—3. 51. y2 2 x 2 §§ 2 217 . Ifa normal is drawn from (a,0) to (x1,y1) on the curve its slope satisfies XELTJ 2 —2y1 > y1 2y1(x1 a) or a 2 x1 1 %' Since x1 2 0 on the curve, we must have that a 2 % . By symmetry, the two points on the parabola are (m. fl?) and (x1, — (/71) . For the normal to be perpendicular, xl—a 11— x1 2 )2—12> W212 x12 (a~x1)2 2>x12(x1+%—x1)2> xlziandylz :t%. Therefore, (27 :1: 5) anda: %. 143 ...
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