Limits, Continuity & Differentiability
Parametric differentiation / higher order derivatives
Grade 12
Question:
<p>If \(x = 3\tan t\) and \(y = 3\sec t\), then the value of \(\dfrac{d^2y}{dx^2}\) at \(t = \dfrac{\pi}{4}\), is:</p>
<p>\(\dfrac{1}{3\sqrt{2}}\)</p>
<p>\(\dfrac{1}{6\sqrt{2}}\)</p>
<p>\(\dfrac{3}{2\sqrt{2}}\)</p>
<p>\(\dfrac{1}{6}\)</p>
Step-by-Step Solution
Key Concept: Use parametric differentiation: find dy/dx = (dy/dt)/(dx/dt), then differentiate again using d²y/dx² = d(dy/dx)/dt · (dt/dx) = [d(dy/dx)/dt]/(dx/dt). The key is recognizing that the second derivative requires careful application of the chain rule with parametric forms.
<p><strong>Step 1:</strong> Find dx/dt and dy/dt</p><p>x = 3tan t ⟹ dx/dt = 3sec²t</p><p>y = 3sec t ⟹ dy/dt = 3sec t tan t</p><p><strong>Step 2:</strong> Find dy/dx</p><p>dy/dx = (dy/dt)/(dx/dt) = (3sec t tan t)/(3sec²t) = (tan t)/(sec t) = sin t</p><p><strong>Step 3:</strong> Find d²y/dx²</p><p>d²y/dx² = d(dy/dx)/dt · (dt/dx) = d(sin t)/dt · 1/(dx/dt)</p><p>d(sin t)/dt = cos t</p><p>d²y/dx² = cos t/(3sec²t) = (cos t · cos²t)/3 = cos³t/3</p><p><strong>Step 4:</strong> Evaluate at t = π/4</p><p>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 = <strong>√2/12</strong></p><p>∴ Answer: D</p>
Correct Answer: D