Question

please don't write in Cursive as i cant read it. And please write it in a clear handwritting. NO CURSIVE please

8. Given the polar equation r = 1 + cos (30) (a) Sketch the curve. (b) Find the area enclosed by the curve. (c) Set up the integral for the arc length of the curve with 0 <0s (d) Find the equation for the line tangent to the curve at 0 = 2 Show all work. (5 points) (10 points) (5 points) (5 points)

Answer 1

Solution:

Given that: Given the polar Equation, Y = 1+cob (30) forca): sketch the curve Graph: -2 2 2 forch): find the area E Enclosed by the curve NOW, . Hcos 30) =0 → C05030)=-1 so cos (1) 30 TT TT -> 0 Area - s! T 2 do 7113 ** CI+ cos 30) do on

It cos (60) T13 * S cit200530 + cos²30) do 7/3 [1+2C05 30+ 7713 į (otsin sin (30) +ã o + sin(60 501) => Ž (5) j) do 2 => T - 1 .: Area प 4 for cc):- Are Length d=5db oxoles db 12+() do dr -3 bin3o do Therefore, db=1 cit cos 30j++9 sint 30 do ,21113 (Vi+260530 +605?30 +9 5in? 30 ) do -271/3 2+2605 30 +8 sinf 30 do dapat -> -> 4.83096

for cd): dr do Sino+YcOSO Tangent equation Noul, slope: dy doc dr cobo-rsino do => -3 sin (30) sinot C14 COS C30)) c03 30 -3 bin (30) COSO -(1+COS (30)) sino * At o=25 osobe => -3 sin (3[?]] sin gu + [++ cos [3 [39]]] -3 sin (3[***]).cos PT - {1+cos cos [3 (37)]] sin => -300) [v] + CI+)(-) -360) (d)-C1+1)() cosa aper cupen -117 co-ordinates:- OC - Y Cobo => CI+ Cob 30) cobo => cobo+cos3o Coso

=> CO3 cos[?]+cos[3[?]]cos (31) - Q => -1 YsYbino => C17 c0530 sino >> Bino +cos30 sino QIT 3 3 => sin 3 +61) (3) => 253 २ => 3 Tangent Equations 9- - n (8-501) 9.13 = (x^(-1)) Y-13 Cox+1) ( z tv3] you +