well, you know it is a limaçon with a vertical axis, so it has a horizontal tangent at θ=π/2
To confirm that, recall that
y = rsinθ
x = rcosθ
dy/dθ = r' sinθ + rcosθ
dx/dθ = r'cosθ - rsinθ
r' = 2cosθ
so, at θ = π/2, r=3, and we have
dy/dθ = 0*1 + 3*0 = 0
dx/dθ = 0*0 - 3*1 = -3
dy/dx = (dy/dθ) / (dx/dθ) = 0/-3 = 0
Find the slope dy/dx of the polar curve r = 1 + 2 sin θ at θ = pi over 2
a) Undefined
b) 1
c) -1
d) 0
1 answer