Find the slope dy dx of the polar curve r = 1 + 2 sin θ at θ = pi over 2

Question

Find the slope  of the polar curve r = 1 + 2 sin θ at θ =  

A	Undefined
B	1
C	-1
D	0

Answer 1

x = rcos(θ)

y = rsin(θ)

dy = r'sin(θ) + rcos(θ)

dx = r'cos(θ) -rsin(θ)

dy/dx = (r'sin(θ) + rcos(θ))/( r'cos(θ) -rsin(θ))

r = 1+2sin(θ)

r' = 2cos(θ)

dy/dx = (2cos(θ)sin(θ)+(1+2sin(θ))cos(θ))/(2cos(θ)cos(θ) -(1+2sin(θ))sin(θ))

at θ=π/2

y'(π/2) = (0+0)/(0-(3)(1)) = 0 Answer D

If you plot 1-2sin(θ), at θ= π/2, you have two y-intercepts at y=1 and y=3 which are tangent lines with slope = 0.