
Hilary W. answered 10/23/14
Tutor
New to Wyzant
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i would start here:
r = cos(theta) + sin(theta) ; multiply both sides by r
r^2 = r[cos(theta) + sin(theta)] = r*cos(theta) + r*sin(theta)
we know r^2 = x^2 + y^2; r*cos(theta) = x; r*sin(theta) = y; substitute
x^2 + y^2 = x+y from here we derivitize
D[x^2 + y^2] = D[x+y]
2x + 2y*y' = 1 + y' solve for y'
2y*y' - y' = 1-2x
y'*(2y-1) = 1-2x
y' = (1-2x)/(2y-1)
let y' = 0 for horizontal tangent
1-2x = 0
x = 1/2
let y' = undefined at vertical tangent
2y-1 = 0
y = 1/2
using these values, you should be able to get back into polar coordinates (keep in mind the question is only looking for solutions where theta is between 0 and 2pi)