Limits, Continuity & Differentiability
Derivatives of Parametric Functions
Grade 12
Question:
<p>If <span class="math">x = \cos\theta</span> and <span class="math">y = \sin^3\theta</span>, then find <span class="math">y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2</span> at <span class="math">\theta = \frac{\pi}{2}</span>.</p>
Step-by-Step Solution
Key Concept: We need to find derivatives using parametric differentiation and chain rule, then substitute θ = π/2. The key is computing dy/dx, d²y/dx² parametrically, and carefully evaluating at the given angle where some terms become indeterminate.
<p><strong>Step 1: Find dy/dx using parametric form</strong></p><p>Given: x = cos θ, y = sin³θ</p><p>dx/dθ = -sin θ</p><p>dy/dθ = 3sin²θ cos θ</p><p>Therefore: dy/dx = (dy/dθ)/(dx/dθ) = (3sin²θ cos θ)/(-sin θ) = -3sin θ cos θ</p><p><strong>Step 2: Find d²y/dx² using parametric formula</strong></p><p>d²y/dx² = d/dx(dy/dx) = [d/dθ(dy/dx)]/(dx/dθ)</p><p>d/dθ(-3sin θ cos θ) = -3(cos²θ - sin²θ) = -3cos(2θ)</p><p>Therefore: d²y/dx² = (-3cos(2θ))/(-sin θ) = (3cos(2θ))/sin θ</p><p><strong>Step 3: Evaluate at θ = π/2</strong></p><p>At θ = π/2:</p><p>• sin(π/2) = 1</p><p>• cos(π/2) = 0</p><p>• cos(π) = -1</p><p>dy/dx|₍θ=π/₂₎ = -3(1)(0) = 0</p><p>(dy/dx)² = 0</p><p><strong>Step 4: Find y·d²y/dx²</strong></p><p>y|₍θ=π/₂₎ = sin³(π/2) = 1³ = 1</p><p>d²y/dx²|₍θ=π/₂₎ = (3cos(π))/sin(π/2) = (3(-1))/1 = -3</p><p>y·d²y/dx² = 1 × (-3) = -3</p><p><strong>Step 5: Calculate final expression</strong></p><p>y(d²y/dx²) + (dy/dx)² = -3 + 0 = -3</p><p><strong>∴ Answer: -3</strong></p>
Correct Answer: -3