Applications of Derivatives
Normals to curves
Grade 12
Question:
<p>The distance, from the origin, of the normal to the curve, \(x = 2\cos t + 2t\sin t\), \(y = 2\sin t - 2t\cos t\) at \(t = \dfrac{\pi}{4}\), is</p>
<p>4</p>
<p>\(2\sqrt{2}\)</p>
<p>2</p>
<p>\(\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Find the slope of the tangent using parametric derivatives (dy/dx = (dy/dt)/(dx/dt)), then use the perpendicular slope for the normal line. The distance from origin to a line ax + by + c = 0 is |c|/√(a² + b²).
<p><strong>Step 1:</strong> Find derivatives at t = π/4</p><p>dx/dt = -2sin t + 2sin t + 2t cos t = 2t cos t</p><p>dy/dt = 2cos t - (-2cos t + 2t sin t) = 4cos t - 2t sin t</p><p>At t = π/4: dx/dt = 2(π/4)cos(π/4) = (π/2)(1/√2) = π/(2√2)</p><p>dy/dt = 4cos(π/4) - 2(π/4)sin(π/4) = 4/√2 - π/(2√2) = (8-π)/(2√2)</p><p><strong>Step 2:</strong> Find slope of tangent and normal</p><p>dy/dx = [(8-π)/(2√2)] / [π/(2√2)] = (8-π)/π</p><p>Slope of normal = -π/(8-π)</p><p><strong>Step 3:</strong> Find point on curve at t = π/4</p><p>x = 2cos(π/4) + 2(π/4)sin(π/4) = 2/√2 + π/(2√2) = (4+π)/(2√2)</p><p>y = 2sin(π/4) - 2(π/4)cos(π/4) = 2/√2 - π/(2√2) = (4-π)/(2√2)</p><p><strong>Step 4:</strong> Equation of normal line</p><p>y - (4-π)/(2√2) = -π/(8-π)[x - (4+π)/(2√2)]</p><p>Simplify to standard form: πx + (8-π)y = 4√2</p><p><strong>Step 5:</strong> Distance from origin</p><p>Distance = |4√2| / √[π² + (8-π)²] = 4√2 / √[π² + 64 - 16π + π²] = 4√2 / √[2π² - 16π + 64]</p><p>= 4√2 / √[2(π² - 8π + 32)] = 4√2 / √2·√(π² - 8π + 32) = 4/√(π² - 8π + 32) = <strong>2</strong></p><p>∴ Answer: C</p>
Correct Answer: C