Limits, Continuity & Differentiability
Higher order derivatives / Parametric differentiation
Grade 12
Question:
<p><strong>939.</strong> If \(x = 4t^3 + 3\), \(y = 4 + 3t^4\) and \(\dfrac{\left(\dfrac{d^2x}{dy^2}\right)}{\left(\dfrac{dx}{dy}\right)^n}\) is a constant, then find the value of \(\dfrac{4}{5} + \dfrac{4}{5n} + \dfrac{4}{5n^2} + \ldots\) upto infinity.</p>
Step-by-Step Solution
Key Concept: Find n by computing d²x/dy² and dx/dy from parametric equations, then recognize that the infinite series is geometric with ratio 1/n, summing to a rational form.
<p><strong>Step 1: Find dx/dy and d²x/dy² from parametric form</strong></p><p>Given: x = 4t³ + 3, y = 4 + 3t⁴</p><p>dx/dt = 12t², dy/dt = 12t³</p><p>dx/dy = (dx/dt)/(dy/dt) = 12t²/(12t³) = 1/t</p><p><strong>Step 2: Find d²x/dy²</strong></p><p>d²x/dy² = d/dy(dx/dy) = d/dy(1/t)</p><p>Using chain rule: d/dy(1/t) = (-1/t²)·(dt/dy) = (-1/t²)·(1/(dy/dt)) = (-1/t²)·(1/(12t³))</p><p>d²x/dy² = -1/(12t⁵)</p><p><strong>Step 3: Use the constant condition</strong></p><p>(d²x/dy²)/(dx/dy)ⁿ = constant</p><p>[-1/(12t⁵)]/[1/t]ⁿ = constant</p><p>[-1/(12t⁵)]·tⁿ = constant</p><p>This requires: t⁵⁻ⁿ = constant for all t, so <strong>n = 5</strong></p><p><strong>Step 4: Find the infinite series</strong></p><p>Sum = 4/5 + 4/(5n) + 4/(5n²) + ... = (4/5)·[1 + 1/n + 1/n² + ...]</p><p>Geometric series with first term a = 4/5 and ratio r = 1/5 (since n = 5)</p><p>Sum = (4/5)·[1/(1-1/5)] = (4/5)·(5/4) = <strong>1</strong></p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1