## anonymous 4 years ago Find the maximum value of the y-coordinate of the points on the limacon r=1+2cos(theta)

1. anonymous

R u there?

2. anonymous

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

3. Zarkon

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

4. anonymous

ok thanks :D

5. anonymous

Thanks dumbcow :D

6. anonymous

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