The graph of the polar curve r 1 2cos

Asked by maham237 @ 20/02/2021 in Mathematics viewed by 240 People

You are given the polar curve r=1+cos(θ).
(a) List all of the points (r,θ) where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of r and limit yourself to r≥0 and 0≤θ<2π. If two or more points share the same value of r, list those starting with the smallest value of θ.
(b) List all of the points (r,θ) where the tangent line is vertical. In entering your answer, list the points starting with the smallest value of r and limit yourself to r≥0 and 0≤θ<2π. If two or more points share the same value of r, list those starting with the smallest value of θ..

Answered by maham237 @ 20/02/2021

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
---------------------------------------------------------------------------------
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

The graph of the polar curve r 1 2cos

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