Chapter 3 43

# Chapter 3 43 - SECTION3.6 D 193 34(a(b There are 9 points...

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Unformatted text preview: SECTION3.6 IMPLICITDIFFERENTIATION D 193 34. (a) (b) There are 9 points with horizontal tangents: 3 at x— — O 3 at a:- — NIH and 3 at x = 1. The three horizontal tangents along the top of the wagon are hard to ﬁnd, but by limiting the y-range of the graph (to [1.6, 1.7], for example) they are distinguishable. 35. From Exercise 29, a tangent to the lemniscate will be horizontal if y’ = 0 => 25x — 4:c(;v2 + 312) : 0 :> m[25 — 4(ac2 + y2)] : 0 => :32 + y2 2 ¥ (1). (Note that when a: is 0, y is also 0, and there is no horizontal tangent at the origin.) Substituting 27.5 for 3:2 + y2 in the equation of the lemniscate, 2(932 + y2)2 = 25 (m2 — yZ), we get m2 — y2 = % (2). Solving (1) and (2), we have m2 = % and y2 = 1—56, so the four points are (dz—343, 21:3.) 2 2 l 2 cc 2% 2 has . . . 36. E + Eli—2 : 1 :> E + % : 0 z; y' : _a—2y => an equat10n of the tangent hnc at (2:0, yo) 1s ~ ‘4?“ — Mlt‘l' bth'd b yo ' yoy y3_ W” ”:2 - y yo - (12310 (x me). u 1p y1ng 0 s1 es y b_2 g1ves b—2 — 11—2 — — (12 + a—g. Since (\$0,140) hes on y2 :32 the ellipse, we have 3:1—2 + gig—2y} - _0 + 13—3 = 1_ 2 2 / 2 a: 2x 2 b x . . 37. 11—2 — Eli—2 — 1 —> E % — 0 y' — aTy : an equat10n of the tangent l1ne at (1:0,y0) is y — yo — b2“ ) Multiplying both sides b @ ives M — y—g — “—3: — a: 5'1 h azyo . y b2 g b2 b2 — a2 a2. 1nce (\$0,310) 1es ont e 2 hyperbola we have if — Log 2 ﬂ ~ y—0 = 1. a2 b—2 a2 b2 y/ 38.¢E+\/37:\/E => +—=0 => ’:~ﬂ => aneuat' ftht t' ' 2—\/_ 2f y \/_ q 1on0 e angen l1ne at (3:0,y0)1s CU —g(ar—\$o).Nowx—O —> y—yo 9‘90: ( \$0) — yo + «mo \/y—, so the y-intercept is as y0+‘/.’170\/y_0.Andy:0 i —y0: mtg—\$0) :> (If—(Eozyo'mo Vito m as 2 x0 + M220 , /y , so the m-intercept is \$0 + t/wo , /yo. The sum of the intercepts is (yowx—wy—o) + (amt/awe) mammal—owe = («w—HM? : (x/EV =0 39. If the circle has radius r, its equation is 3:2 + y2 = r2 => 2:1: + 2yy’ : 0 => 31' = —£, so the slope of the y . . a: — tangent hne at P(ato, yo) 1s ——0. The negative reciprocal of that slope is “I 2 y—0 210 —\$0/ yo 130 OP, so the tangent line at P is perpendicular to the radius OP. , which 15 the slope of ...
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