Applications of Derivatives
Tangent and Normal to a Curve
Grade 12
Question:
<p>The parametric form of a curve is \(x = 2\cos t + 2t\sin t\) and \(y = 2\sin t - 2t\cos t\). The distance from the origin of the normal to the curve at \(t = \pi/4\) is:</p>
<p>1</p>
<p>\(\sqrt{2}\)</p>
<p>2</p>
<p>\(2\sqrt{2}\)</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 is |c|/√(a²+b²) for line ax+by+c=0.
<p><strong>Step 1: Find derivatives</strong></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 = 2t sin t</p><p><strong>Step 2: Find dy/dx at t = π/4</strong></p><p>dy/dx = (dy/dt)/(dx/dt) = (2t sin t)/(2t cos t) = tan t</p><p>At t = π/4: dy/dx = tan(π/4) = 1</p><p><strong>Step 3: Find point on curve at t = π/4</strong></p><p>x = 2cos(π/4) + 2(π/4)sin(π/4) = 2(√2/2) + (π/2)(√2/2) = √2(1 + π/4)</p><p>y = 2sin(π/4) - 2(π/4)cos(π/4) = 2(√2/2) - (π/2)(√2/2) = √2(1 - π/4)</p><p><strong>Step 4: Equation of normal</strong></p><p>Slope of normal = -1/1 = -1</p><p>Normal equation: y - √2(1 - π/4) = -1[x - √2(1 + π/4)]</p><p>Simplifying: x + y = √2(1 - π/4) + √2(1 + π/4) = 2√2</p><p><strong>Step 5: Distance from origin</strong></p><p>Distance = |2√2|/√(1² + 1²) = 2√2/√2 = 2</p><p>∴ Answer: C</p>
Correct Answer: C