Solutions Calculus II | MATH 1592, Assignments of Calculus

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10. Solutions . Section 9.4: 4. 2 = 0cos(—77/6) = 0,y = Osin (—77/6) = 0.(z,y) = (0,0). x z 0) © . Section 9.4: 6. 2 = ~scoe(-157) = = —0.0024, y = —3sin (—1.57) © 3.(z,y) = —0.0024, 3). , Cooomysyd s* . Section 9.4: 14. r = +V16 i tan? = =2 = —1,0 = —0.464. So Gi ty u” (r,0) = (2V5, 0.464) or (—2 we 64). x of 7 (4, -2) . Section 9.4: 22. 2? +y? —2ar = 0. x = rcos6,y = rsin@ => (rcos@)* + (rsin 0)? — 2a(r cos 9) =0— r? — 2ar cos = 0 => r(r — 2acos0) =0 =} r= 2a cos 0. x > Oo . Section 9.4: 32. r= 5cos0 => r? = 5rcosO => a? + y? = br => 2? — Bet 3 B44? =% = (r-3)' + =(8 ; . Section 9.4: 54. r = 2(1—sind) => 4 = =Beosdsinttteosh-sind at (2,0), 0 = 0. eos 0 cos 0—2sin O(1—sin 0) * So @ = —1. At (3,77/6), 0 = 77/6. So # is undefined. At (4,37/2), 0 = 3n/2. So#@= dr . Section 9.5: 2. (a) A = 7(3/2)? = 9n/4. (b) A = 2(1/2) a Geen tia = 9 Jo"? cos? dd = 3 fg?(1 + 2cos26)d0 = 3 [0+ saa)" =9n/4. 2 t -2 . Section 9.5 : 14. r = 3(1+sin@),r = 3(1 — sin@). Solving simultancously, 3(1 + sin®) = 3(1 — sind) => sind = 0 => 0 = 0,7. Also both curves pass through the pole: (0,37/2), and (0,7/2), respectively. So points of intersection: (3, 0), (3, 7), (0;0). . Section 9.5 : 42. r = 2acos@ => 1’ = —2asind => s = J, (2acos 6)? + (—2asin edd = J7"?, 2ad = 2rra. Section 9.5 : 52. r= acos@ => r' = —asind => S = In fe” acos6 cos 0,/(2acos 8)? + (—2a sin )2d0 = 2na? fy"? cos? Odd = ma? fi/(1 + 2008 28)d0 = ma? [9 + S22)" = 2a2/2.