Question

Find the length of the parabolic segment r = 6/(1+cos theta), 0 less than or equal to theta less than or equal to pi/2. Please show a graph. If you could, please write legibly and indicate why each step was taken. It is a great help!

Answer 1

solution - we have r = 6/(1+cos theta)

The arc length = ∫(θ = 0 to π/2) √(r^2 + (dr/dθ)^2) dθ

= ∫(θ = 0 to π/2) √[(6/(1 + cos θ))^2 + (6 sin θ/(1 + cos θ)^2)^2] dθ

= ∫(θ = 0 to π/2) √[(6/(1 + cos θ)^2)^2 * ((1 + cos θ)^2 + (sin θ)^2)] dθ

= ∫(θ = 0 to π/2) (6/(1 + cos θ)^2) * √(2 + 2 cos θ) dθ

= ∫(θ = 0 to π/2) (6/(1 + cos θ)^2) * √(2(1 + cos θ)) dθ

= ∫(θ = 0 to π/2) (6/(2 cos^2(θ/2))^2) * √(2 * 2 cos^2(θ/2)) dθ, via half angle identity

= ∫(θ = 0 to π/2) (3/cos^4(θ/2)) * 2 cos(θ/2) dθ

= ∫(θ = 0 to π/2) 3 sec^3(θ/2) dθ

Now, letting w = θ/2, dw = (1/2)dθ yields

∫(w = 0 to π/4) 3 sec^3(w) * 2 dw

= 6 * (1/2) [sec w tan w + ln |sec w + tan w|] {for w = 0 to π/4}

= 3(√2 + ln(√2 + 1)