Q.138. Horizontal and vertical tangents to a curve, cardioid, defined by polar coordinates.

Question 138.
Find the horizontal and vertical tangents to the cardioid r = 1 - cos(theta), [0, 2pi].

Answer 138.
If r = 1 - cosθ, then
x = (1 - cosθ) * cosθ = cosθ - cos^2 θ and
y = (1 - cosθ) * sinθ = sinθ - (1/2)sin2θ

=> dx/dθ = - sinθ + 2cosθsinθ
and dy/dθ = cosθ - cos2θ  

=> dy/dx
= (dy/dθ) / (dx/dθ)
= (cosθ - cos2θ) / (sin2θ - sinθ)
= 2sin(3θ/2)sin(θ/2) / 2cos(3θ/2)sin(θ/2)
= tan(3θ/2)

For horizontal tangent, tan(3θ/2) = 0
=> 3θ/2 = kπ
=> θ = 2kπ/3
=> θ = {0, 2π/3, 4π/3, 2π}

Corresponding y-coordinates are given by y = sinθ - (1/2)sin2θ
θ = 0 and 2π => y = 0
θ = 2π/3 => y = √3/2 + √3/4 = 3√3/4
θ = 4π/3 => y = - √3/2 - √3/4 = - 3√3/4
=> Horizontal tangents are
y = 0 and y = ± 3√3/4

For vertical tangents, 3θ/2 = kπ + π/2 => θ = 2kπ/3 + π/3
=> θ = {π/3, π, 5π/3}
Corresponding x-coordinates are given by x = cosθ - cos^2θ
θ = π => x = - 2
θ = π/3 and 5π/3 => x = 1/4
=> vertical tangents are
x = - 2 and x = 1/4.

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