Problem 13 - Entrance Test

First, express x and y in terms of θ: x = rcos(θ) = 2sin(θ)cos(θ) = sin(2θ). y = rsin(θ) = 2sin²(θ). Next, find dx/dθ and dy/dθ: dx/dθ = d/dθ [sin(2θ)] = 2cos(2θ). dy/dθ = d/dθ [2sin²(θ)] = 2 * 2sin(θ)cos(θ) = 4sin(θ)cos(θ) = 2sin(2θ). Now, find dy/dx = (dy/dθ) / (dx/dθ): dy/dx = (2sin(2θ)) / (2cos(2θ)) = tan(2θ). To find d²y/dx², we use the formula d²y/dx² = d/dθ(dy/dx) / (dx/dθ): d/dθ(dy/dx) = d/dθ [tan(2θ)] = sec²(2θ) * 2 = 2sec²(2θ). So, d²y/dx² = (2sec²(2θ)) / (2cos(2θ)) = sec²(2θ) / cos(2θ) = sec³(2θ). Now, evaluate d²y/dx² at θ = π/2: At θ = π/2, 2θ = 2(π/2) = π. cos(π) = -1. sec(π) = 1/cos(π) = 1/(-1) = -1. d²y/dx² = (sec(π))³ = (-1)³ = -1. The final answer is A.