Applications of Derivatives
Normals to curves
Grade 12
Question:
<p>The normal to the curve \(x = a(\cos\theta + \theta\sin\theta)\), \(y = a(\sin\theta - \theta\cos\theta)\) at any point '\(\theta\)' is such that</p>
<p>it passes through the origin.</p>
<p>it makes angle \(\dfrac{\pi}{2} + \theta\) with the x-axis.</p>
<p>it passes through \(\left(a\dfrac{\pi}{2}, -a\right)\).</p>
<p>it is at a constant distance from the origin.</p>
Step-by-Step Solution
Key Concept: The normal to a parametric curve is perpendicular to the tangent. Find dy/dx using parametric derivatives, then the normal's slope is its negative reciprocal. Check if the normal passes through the origin or has a special geometric property.
<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> Find slope of normal</p><p>Slope of normal = -1/(tan θ) = -cot θ</p><p><strong>Step 4:</strong> Write equation of normal at point (x₀, y₀)</p><p>y - y₀ = -cot θ(x - x₀)</p><p>where x₀ = a(cos θ + θ sin θ) and y₀ = a(sin θ - θ cos θ)</p><p><strong>Step 5:</strong> Simplify the normal equation</p><p>y - a(sin θ - θ cos θ) = -cot θ[x - a(cos θ + θ sin θ)]</p><p>Multiplying by sin θ and simplifying:</p><p>y sin θ - a sin θ(sin θ - θ cos θ) = -cos θ[x - a(cos θ + θ sin θ)]</p><p>y sin θ + x cos θ = a sin θ sin θ + a cos θ cos θ + aθ(sin θ cos θ - sin θ cos θ)</p><p>y sin θ + x cos θ = a(sin²θ + cos²θ) = a</p><p><strong>∴ The normal passes through the origin and satisfies: x cos θ + y sin θ = a, or equivalently, the normal always has the special property that x cos θ + y sin θ = a (constant)</strong></p>
Correct Answer: D