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Chapter 3 104

Chapter 3 104 - 254 El CHAPTER3 PROBLEMSPLUS 10 The...

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Unformatted text preview: 254 El CHAPTER3 PROBLEMSPLUS 10. The equation of T1 is y — at? = 2331(3) — \$1) = 23:17; — 2:5? or y 2 2mm — as? The equation of T2 is y 2 2mm — 93%. Solving for the point ofintersection. we get 2:1:(x1 ~ :52) = 13% -— x3 => 9: 2 am + 202). Therefore. the coordinates of P are @(m + 11:2),111111‘2). So if the point of contact ofT is (a. a2). then Q1 is (ﬂu + 331).,am1) and Q2 is @(a + m2),am2). Therefore. |PQ1|2 : an # \$2)2 + 33%(11 — 932)2 : (a i 372)2 (i + as?) and IPQII2 _ (a , 962)2 |PP1l2 : i031 — I2)2 + maxi — m2)2 : (\$1 7 “)2 (i- + ﬂ). So 2 — 2,and similarly IPP1| (\$1 7 332) P 2 — 2 P P — ~ lel2 2 (901 a)2.Finally. | Q1|+ | Q2| = a \$2 + an a _ 1' iPPZl (131*962) IPP1| IPP2l Iii—x2 3317172 11. Consider the statement that d— (6” sin bcc) : rne” sin(b\$ + 719). For n : 1. in” d 0.11) ‘ 0.513 ‘ an? a (e sm bar) 2 ae Sln bIL’ + be cos bat. and am - an: - . am a . b 're sm(bm + 6) : re [sm bx cos 0 + cos b3: sm 6] : re — Sln bat + ; cos baa r : ac” sin bm + be” cos bx b b since tanB : -— => sin0 : — and cost9 : 2. a r 7’ So the statement is true for n : 1. Assume it is true for n = 19. Then k+1 5ka (en sin bar) 2 dis: [rkeuz sin(b:n + k6)] : rkae” sin(bw + k6) + rke‘nbcoswm + k0) : rkeazla sin(ba: + k0) + bcos(bm + k6)] But sin[ba: + (k + 1)0] : sin[(b:r + k0) + 0] : sin(ba: + k6) cos0 + sin 9 cos(bm + k6) : % sin(bac + k0) + 2c0s(bm + k0) Hence. a sin(bm + k0) + bcos(ba: + k9) = r sin[b:v + (k + 1)9]. So dk+1 d\$k+1 (em sin bar) : rke”[a sin(b:z: + k0) + bcos(bx + [96)] = rke‘nlr sin(b\$ + (k + 1)9)] : rk+1em[sin(b\$ + (k + 1)l9)] Therefore. the statement is true for all n by mathematical induction, 12. We recognize this limit as the deﬁnition of the derivative of the function f (m) = 651‘” at m : 7r, since it is of the form lim M. Therefore. the limit is equal to f’(7r) : (cos ﬂesm" : —1 - e0 : —1. 17"" m 4‘ 7T ...
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