Indefinite Integration
Integration involving parametric functions
Grade 12
Question:
<p>Let \(x = f''(t)\cos t + f'(t)\sin t\) and \(y = -f''(t)\sin t + f'(t)\cos t\). Then \(\int\left[\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2\right]^{1/2} dt\) is equal to</p>
<p>\(f'(t) + f'''(t) + c\)</p>
<p>\(f'''(t) + f''(t) + c\)</p>
<p>\(f(t) + f''(t) + c\)</p>
<p>\(f'(t) - f''(t) + c\)</p>
Step-by-Step Solution
Key Concept: Recognize that dx/dt and dy/dt come from differentiating parametric expressions involving f'(t) and f''(t). Computing their derivatives and finding the magnitude reveals a hidden simplification: √[(dx/dt)² + (dy/dt)²] = |f'''(t)|, which integrates to f''(t) + C.
<p><strong>Step 1:</strong> Differentiate x and y with respect to t.</p><p>dx/dt = d/dt[f''(t)cos t + f'(t)sin t]</p><p>= f'''(t)cos t - f''(t)sin t + f''(t)sin t + f'(t)cos t</p><p>= f'''(t)cos t + f'(t)cos t = [f'''(t) + f'(t)]cos t</p><p><strong>Step 2:</strong> Similarly, dy/dt = d/dt[-f''(t)sin t + f'(t)cos t]</p><p>= -f'''(t)sin t - f''(t)cos t - f''(t)cos t - f'(t)sin t</p><p>= -[f'''(t) + f'(t)]sin t</p><p><strong>Step 3:</strong> Compute (dx/dt)² + (dy/dt)²:</p><p>(dx/dt)² + (dy/dt)² = [f'''(t) + f'(t)]²cos²t + [f'''(t) + f'(t)]²sin²t</p><p>= [f'''(t) + f'(t)]²(cos²t + sin²t) = [f'''(t) + f'(t)]²</p><p><strong>Step 4:</strong> Therefore:</p><p>√[(dx/dt)² + (dy/dt)²] = |f'''(t) + f'(t)|</p><p><strong>Step 5:</strong> Integrate:</p><p>∫√[(dx/dt)² + (dy/dt)²] dt = ∫[f'''(t) + f'(t)] dt = f''(t) + f(t) + C</p><p>∴ Answer: <strong>f''(t) + f(t) + C</strong></p>
Correct Answer: C