Question

1. Find the points of horizontal and vertical tangency (if any) to the polar curve r = 2 – 2 cos(0)

Answer 1

r = 2-2 Cane x=rCaro = (2-2 Caro) Gaxo 2 Ceen o - 2 Cax 20 y = -r Sino = (2-2 600) Sino 2 = 2 Sino 2 Sino Gno da = - 2 Sino - 2 (2 Come) (- Sto) de da do = - 2 Sino + 4 Sind Care dy do = 2 Caso -2 (- Sin? + A + Cox?o dy de = 2 Como – 2 Cox?o + 2 Sim’e Slope m= dy da = dylde da / do 2 Coxo - 2 Cox² + 2 Sin² ma - 2 Sino + 4 Sing Cart m = CORO - Cop²o + sin²o Sino + 2 Sino GAO for Marizontal tangency dy m0 do =0R da do 70 2Coro -60x² +28in2 -28ing t4 Sind Cano #0 ::2610-2600²0 +25m²=0 Coxo - Cax²0 + 1 - conto =0 2 Cas²o - Cano-1=0 (cosa?o - Como DH com0-1) = 0 Cono (10-1) + (0-1)(610 +1) = 0 (Con 0-1) ( 2 Como +1) = 0 Cano el = 0, 268 +1 -0 = Coro -- 1 © = 21 4TT 21 A AT =0 2 O3

at o, o dx do - Sin 0 + 4 Smo Cono dx do o, is not a solution at 0, - 211 3 da do 2 sm 21 3 +4 Som 2 11 Coming dx do - 253 0 = is a solution at o 03 4 IT بی da do - 28m411 + 4 sm Sm Al Canal = 2.13 0 = , is a solution Now 2 at 0 - 271 3 2 = 2-2 Pax 211 = 2-21-^)=3 3 - ? paint ( 3, 211 3 at Oz - 41T 3 = 2 Con Al ( 3,40 2-2/-Ź) = 3 point 5,47) for vertical da -0²-28mo + 4 SmO GAD=0 tangency do dy do 0 2 Cono - 260120 + 2 fm²00 - 28mb + 4 smo con D=0 2 8m8 (2Cn0-1)=0 Sind=o o 04 = = 0, 0g = CanB =

at Og = 0 - 26x²o - 26x²0 + 2 sm²0 dy do dy do =o o il net a solution at Os - TT dy = 2 lant - 2 Cax²n + 2 fm² do dy --4 → © do is a solution at Op = 7 3 dy do = 2 can I 3 - 2 Cox ² 7 23 + 2 8m² Thi dy = 20 Og is a solution - 3 at o 511 dy do = 2.com 9. - 2 ray? 59 dy 70 = Oz is a solution de + 2 sm? s T How ☆ Nuts at Os - TT 5 - 2 - 2001 = 4 point 14,7) at TT - w/7 Of = 2-2 Com a = 1 point 11,” - 577 at Oz 5T q=2-2 Car 50 วร = 1 3 paint (1,51)