Chapter 3 41

# Chapter 3 41 - SECTION 3.6 IMPLlClT DIFFERENTIATION 191 1...

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Unformatted text preview: SECTION 3.6 IMPLlClT DIFFERENTIATION 191 1 27. m2 + y2 2 (211:2 + 2y2 — a:)2 => 230 + 2yy' 2 2(2332 -I— 2y2 — x)(43: + 4yyl — 1). When a: 2 0 and y 2 5. we have 0 + y’ — 2(l)(2y’ — 1) 2 y/ 2 2y’ 2 1 2> y’ 2 1, so an equation ofthe tangent line is — 2 y—%21(at—0)ory2a:+%. M 3 28.x2/3+y2/324 :> 2x‘1/3+§y‘1/3y’=0 => i+f§2=0 => y’=~ﬂ-Whenx=-3x/§ 3 {75 t/z7 t/E 2/3 1 (~3\/§) 3 1 . . . d 2 1. ha ’ : ———— 2 —— 2 —— 2 —. so an equation of the tangent line is an y we veg (23%)”3 —3\/§ 3\/§ x/i y212%(\$+3\/§) ory2¢i§\$+4 29. 2(x2 + y2)2 2 25(272 — yz) 2 4(ac2 + y2)(2m + 2yy’) 2 25(23: — 2313/) 2 4(m + yy’)(w2 + y2) 2 25(1) — yy’) 2> 4yy’(a:2 + 312) + 25yy' 2 253: 2 4m(a:2 + 312) 2> I _ 25w — 433(m2 + y2) 2 .Whena: 2 3 andy 2 1, we have y' 2 ‘75‘120 2 7g 2 23, so an equation ofthe y 25y+4y(m2+y2) 25+40 65 13 tangent line isy — 1 : 21—93(93 2 3) ory : _%\$+ 411—3 30. y2(y2 — 4) 2 332(332 2 5) 2> y4 — 4y2 2 2:4 — 53:2 2> 4y3y’ ~ 8nyI 2 4:53 ~ 102:. When as 2 0 and y 2 —2, we have —32y’ + 16yI 2 0 2> —16yl 2 0 2> y/ 2 0, so an equation of the tangent line is y+220(:1:—0)0ry2 —2. 10 3 — 31. (a) y2 2 55’ ~ :32 2> 2313/2 5(4LE3) — 2:13 2> y' = # (b) 1 1 321 9 So at the point (1,2) we have y’ 2 —L)2— 2 5, and an equation of the tangent line is y — 2 2 gm — 1) or y 2 3;): 2 [\JIU‘ 3 2 32. (a) y2 2 m3 + 3x2 2 2yyI 2 3m2 + 3(213) 2 y/ 2 L633. So at the point (1, —2) we have 21/ I _ 3(1)2 +60) _ 9 - - - 9 9 1 y 2 T2) 2 —4, and an equatlon ofthe tangent line 18 y + 2 2 21(zc — 1) org 2 ‘23: + Z (b) The curve has a horizontal tangent where y’ 2 0 (2 (c) 33324—63520 <2» 32:(m—I—2) :0 <2 m200r\$2 22. But note that at x 2 0. y 2 0 also. so the derivative does not exist. At m 2 22,312 = (—2)3 + 3(—2)2 : —8 +12 2 4, so y = :2. So the two points at which the curve has a horizontal tangent are (227 —2) and (—2, 2). ...
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