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Chapter 3 106

# Chapter 3 106 - 256 U CHAPTER3 PROBLEMSPLUS tangent to y(m...

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Unformatted text preview: 256 U CHAPTER3 PROBLEMSPLUS tangent to y : (m — 2)2 at (1.1) is 1712 = 2(1 # 2): —2. Therefore. tana = % : 1_:—2_2(—:22_) = g and soa : tan—1(3) Z 53° (0r1270). (ii) :32 —y2 =321ndx2—4.72+3;2+3=0intersectwhencr:2 —4:B+(\$2 —3) +3 2 0 4:) 21‘(;L'—2) : 0 => 1: = 0 0r 2. but 0 is extraneous. Ifzv : 2. then y = i1. If 2:2 — y2 : 3 then 25c 7 2yy’ : 0 => 2 — at 3/ m1: 2 and TTL2 = 0. so tana : 113% = —2 => (1 % 117°. At (2. —1) the slopes are m1 = —2 and y'zz/yandw2—4\$+y2+3=0 => 2m—4+2yy’:0 2 y'= .At(2.1)theslopesare m2:0.sotana=T3_—‘(£;—)2()O—):2 => am630(0r1170). 18. y2 4pm S 2313/ 4p 5 y’ 2p/y > slope of tangent at P(m1,y1) is m1 : 2p/y1. The slope of FP is m2 : a: y1 17‘ so by the formula from Problem 15(a). 1 _ tam : yl/(ml —p) — 2p/y1 . y1(w1 ep) _ y? — 217021 *p) 1 + (2p/yl)[y1/(w1 110)] 111 (931 —p) y1(\$1 —p) + 212211 _ 4pm — 219% + 2P2 2P(P “ 931) 7 2p _ = \$10 e of tan ent at P : tan m1y1 — 10311 + 211341 11109 -- \$1) 311 p g ﬂ Since 0 S a. [3 g g. this proves thata : B. 17. Since ZROQ = ZOQP = 0, the triangle QOR is isosceles. so |QR| : |RO| : w. By the Law of Cosines. x2 : \$2 + r2 — 27% cos 9. Hence. 2 2r\$cos0 : r2. sow : 2r:030 : 2cosE)‘ Note that asy —> 0+. 6 —+ 0+ (since sin 9 : y/r). and hence a: —> 2 c; 0 : Z. Thus. as P is taken closer and closer s to the m—axis. the point R approaches the midpoint of the radius A0. ms) — 1(0) 11m ﬂan) — f(0) , g 7 0 _ 9H — 0 18.11111 NC) #11111 “3“") 0 _ lim “33) ﬂ ) _ lim 9” 0 _ 0 m 0 ———f,() 140 9(1) WO 9(96) e 0 160 9(90) e 9(0) HO 9(1) , 9(0) lim 9(06) * 9(0) 9 0) a: e 0 EH0 (E # 0 19' lim sin(a + 2m) * 2 si2n(a, + ac) + sin a 3—“) :17 1_ Sinacos2a: + cosasin2m # 2sinacosm ~ 2cosasina: + sina — @135 x2 , sina(cosQw -2cosw+ 1) +cosa(sin2m — 2sinm) — 11m 2 2—14) I) . sinu(2cos2 :c — 1 ~ 2005]: —|— 1) + cosa(2sinmc0sm _ 2sinw) # 11m 2 \$44) \$ 1, sina(2cosm)(c03\$ # 1) +cosa(2 sinm)(cos;l: — 1) - rel—>Ino \$2 , 2(c0sm # 1)[sinac0sa; + cosasinm](cosm + 1) i 11m m—vO w2(cosac + 1) . #2 Sin2 m [sin(a + 23)] , sinan 2 sin(a + m) 2 sin(a + 0) . _ e i - : 2 1 - — sma 33%) m2(cosa:+1) 21111210 1: cosac+1 ( ) cos0+1 ...
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