Problem 11 - Entrance Test

The arc length L of a polar curve r = f(θ) from θ = α to θ = β is given by the formula: L = ∫ from α to β √(r² + (dr/dθ)²) dθ. Given r = 2(1 + cos(θ)): dr/dθ = d/dθ [2 + 2cos(θ)] = -2sin(θ). Now, calculate r² + (dr/dθ)²: r² = [2(1 + cos(θ))]² = 4(1 + 2cos(θ) + cos²(θ)). (dr/dθ)² = (-2sin(θ))² = 4sin²(θ). r² + (dr/dθ)² = 4 + 8cos(θ) + 4cos²(θ) + 4sin²(θ). Using the identity cos²(θ) + sin²(θ) = 1: r² + (dr/dθ)² = 4 + 8cos(θ) + 4(1) = 8 + 8cos(θ) = 8(1 + cos(θ)). Use the half-angle identity 1 + cos(θ) = 2cos²(θ/2): r² + (dr/dθ)² = 8(2cos²(θ/2)) = 16cos²(θ/2). Take the square root: √(16cos²(θ/2)) = 4|cos(θ/2)|. The integral for the arc length is L = ∫ from 0 to 2π (4|cos(θ/2)|) dθ. For 0 ≤ θ ≤ 2π, θ/2 ranges from 0 to π. In this range, cos(θ/2) is positive for 0 ≤ θ/2 ≤ π/2 (i.e., 0 ≤ θ ≤ π) and negative for π/2 < θ/2 ≤ π (i.e., π < θ ≤ 2π). So, we must split the integral: L = ∫ from 0 to π (4cos(θ/2)) dθ + ∫ from π to 2π (-4cos(θ/2)) dθ. Integrate: = [8sin(θ/2)] from 0 to π + [-8sin(θ/2)] from π to 2π. = (8sin(π/2) - 8sin(0)) + (-8sin(2π/2) - (-8sin(π/2))). = (8 * 1 - 0) + (-8 * 0 - (-8 * 1)). = 8 + 8 = 16. The final answer is B.