Question

You are given the polar curve r=1+cos(θ).

Answer 1

For the answer to the question above,
 r = 1 + cos θ 

x = r cos θ 
x = ( 1 + cos θ) cos θ 
x = cos θ + cos^2 θ 
dx/dθ = -sin θ + 2 cos θ (-sin θ) 
dx/dθ = -sin θ - 2 cos θ sin θ 

y = r sin θ 
y = (1 + cos θ) sin θ 
y = sin θ + cos θ sin θ 
dy/dθ = cos θ - sin^2 θ + cos^2 θ 

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

For horizontal tangent line, dy/dθ = 0 

cos θ - sin^2 θ + cos^2 θ = 0 
cos θ - (1-cos^2 θ) + cos^2 θ = 0 
cos θ -1 + 2 cos^2 θ = 0 
2 cos^2 θ + cos θ -1 = 0 
Let y = cos θ 

2y^2+y-1=0 
2y^2+2y-y-1=0 
2y(y+1)-1(y+1)=0 
(y+1)(2y-1)=0 
y=-1 
y=1/2 

cos θ =-1 
θ = π 
cos θ =1/2 
θ = π/3 , 5π/3 

θ = π/3 , π, 5π/3 
when θ = π/3, r = 3/2 
when θ = π, r = 0 
when θ = 5π/3 , r = 3/2 
(3/2, π/3) and (3/2, 5π/3) give horizontal tangent lines 
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For horizontal tangent line, dx/dθ = 0 

-sin θ - 2 cos θ sin θ = 0 
-sin θ (1+ 2 cos θ ) = 0 
sin θ = 0 
θ = 0, π 

(1+ 2 cos θ ) =0 
cos θ =-1/2 
θ = 2π/3 
θ = 4π/3 

θ = 0, 2π/3 ,π, 4π/3 
when θ = 0, r=2 
when θ = 2π/3, r=1/2 
when θ = π, r=0 
when θ = 4π/3 , r=1/2 

(2,0) , (1/2, 2π/3) , (0, π), (1/2, 4π/3) 
At (2,0) there is a vertical tangent line