Applications of Derivatives
Normals to parametric curves
Grade 12
Question:
<p>The two curves \(x = a(\cos\theta + \theta\sin\theta)\) and \(y = a(\sin\theta - \theta\cos\theta)\). The distance of the normal from the origin is</p>
<p>0</p>
<p>\(a\)</p>
<p>constant</p>
<p>\(|a|\) = constant</p>
Step-by-Step Solution
Key Concept: For a parametric curve, the normal at any point has slope -dx/dy, and the perpendicular distance from origin to a line equals |c|/√(1+m²) where the line is y = mx + c. Use the parametric derivatives and the point coordinates to find the normal equation.
<p><strong>Step 1:</strong> Find dx/dθ and dy/dθ</p><p>dx/dθ = a(-sin θ + sin θ + θ cos θ) = aθ cos θ</p><p>dy/dθ = a(cos θ - cos θ + θ sin θ) = aθ sin θ</p><p><strong>Step 2:</strong> Find dy/dx</p><p>dy/dx = (dy/dθ)/(dx/dθ) = (aθ sin θ)/(aθ cos θ) = tan θ</p><p><strong>Step 3:</strong> Slope of normal at parameter θ</p><p>Slope of normal = -1/(dy/dx) = -cot θ</p><p><strong>Step 4:</strong> Equation of normal at point (x₀, y₀)</p><p>Point: x₀ = a(cos θ + θ sin θ), y₀ = a(sin θ - θ cos θ)</p><p>Normal: y - y₀ = -cot θ(x - x₀)</p><p>y sin θ - a sin θ(sin θ - θ cos θ) = -x cos θ + a cos θ(cos θ + θ sin θ)</p><p>x cos θ + y sin θ = a cos θ(cos θ + θ sin θ) + a sin θ(sin θ - θ cos θ)</p><p>x cos θ + y sin θ = a(cos² θ + θ sin θ cos θ + sin² θ - θ sin θ cos θ)</p><p>x cos θ + y sin θ = a</p><p><strong>Step 5:</strong> Distance from origin to line x cos θ + y sin θ = a</p><p>Distance = |a|/√(cos² θ + sin² θ) = |a|/1 = a</p><p>∴ Answer: D</p>
Correct Answer: D