Limits, Continuity & Differentiability
Higher Order Derivatives and Parametric Functions
Grade 12
Question:
<p>Given \( x = 3\tan t,\ y = 3\sec t \), find \(\left(\dfrac{d^2 y}{dx^2}\right)_{t=\pi/4}\).</p>
<p>\(\dfrac{1}{6\sqrt{2}}\)</p>
<p>\(\dfrac{1}{3\sqrt{2}}\)</p>
<p>\(\dfrac{1}{6}\)</p>
<p>\(\dfrac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: For parametric equations, use the chain rule formula d²y/dx² = [d/dt(dy/dx)]/[dx/dt], where dy/dx = (dy/dt)/(dx/dt). This avoids converting to Cartesian form which is algebraically messy.
<p><strong>Step 1: Find first derivatives</strong></p><p>x = 3tan t ⟹ dx/dt = 3sec²t</p><p>y = 3sec t ⟹ dy/dt = 3sec t tan t</p><p><strong>Step 2: Find dy/dx</strong></p><p>dy/dx = (dy/dt)/(dx/dt) = (3sec t tan t)/(3sec²t) = tan t/sec t = sin t</p><p><strong>Step 3: Find d²y/dx² using parametric formula</strong></p><p>d²y/dx² = [d/dt(dy/dx)]/(dx/dt) = (d/dt(sin t))/(3sec²t) = cos t/(3sec²t)</p><p>Since sec t = 1/cos t, we have sec²t = 1/cos²t</p><p>d²y/dx² = cos t · cos²t/3 = cos³t/3</p><p><strong>Step 4: Evaluate at t = π/4</strong></p><p>At t = π/4: cos(π/4) = 1/√2</p><p>cos³(π/4) = (1/√2)³ = 1/(2√2) = √2/4</p><p>d²y/dx²|_{t=π/4} = (√2/4)/3 = √2/12</p><p>∴ Answer: A</p>
Correct Answer: A