Hilary W. answered • 10/23/14

Tutor

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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)