Find the maximum value of the y-coordinate of the points on the limacon r=1+2cos(theta)

Question

anonymous · Answer

R u there?

dumbcow · Answer

Let theta =A
y= rsin(A)
y = sinA(1+2cosA) = sinA +2sinAcosA

set dy/dA = 0
dy/dA = cosA+2cos^2A-2sin^2A = 0
**substitute sin^2 = 1-cos^2 
cosA +2cos^2A-2(1-cos^2A) = 0
4cos^2A +cosA -2 = 0
using quadratic formula
cosA = -.843 or .593
max at .593 since it would lead to greater r
A = 53.625 degrees

r = 1+2(.593) = 2.186
max_y = 2.186*sin(53.625) = 1.76

zarkon · Answer

if you want an exact answer it is 

$$\frac{\sqrt{6(11\sqrt{33}+69)}}{16}$$

anonymous · Answer

ok thanks :D

anonymous · Answer

Thanks dumbcow :D

anonymous · Answer

See only once someone helps me out it becomes so clear. I dont know y i cldnt figure this one out on my own