Applications of Derivatives
Normal to a Curve
Grade 12
Question:
<p>Consider \( f(x) = \tan^{-1}\!\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right), x \in \left(0, \dfrac{\pi}{2}\right) \). A normal to \( y = f(x) \) at \( x = \dfrac{\pi}{6} \) also passes through the point</p>
<p>\( \left(\dfrac{\pi}{4}, 0\right) \)</p>
<p>\( (0, 0) \)</p>
<p>\( \left(0, \dfrac{2\pi}{3}\right) \)</p>
<p>\( \left(\dfrac{\pi}{6}, 0\right) \)</p>
Step-by-Step Solution
Key Concept: Simplify the argument of tan⁻¹ using the identity √[(1+sin x)/(1-sin x)] = tan(π/4 + x/2), then find f'(x) to determine the normal line equation.
<p><strong>Step 1: Simplify f(x)</strong></p><p>Using the identity: √[(1+sin x)/(1-sin x)] = √[(sin²(x/2) + cos²(x/2) + 2sin(x/2)cos(x/2))/(sin²(x/2) + cos²(x/2) - 2sin(x/2)cos(x/2))]</p><p>= √[(sin(x/2) + cos(x/2))²/(cos(x/2) - sin(x/2))²]</p><p>= (sin(x/2) + cos(x/2))/(cos(x/2) - sin(x/2)) = tan(π/4 + x/2)</p><p>Therefore: f(x) = tan⁻¹(tan(π/4 + x/2)) = π/4 + x/2 for x ∈ (0, π/2)</p><p><strong>Step 2: Find f'(x)</strong></p><p>f'(x) = 1/2</p><p><strong>Step 3: Find the normal at x = π/6</strong></p><p>f(π/6) = π/4 + π/12 = 4π/12 = π/3</p><p>Point on curve: (π/6, π/3)</p><p>Slope of normal = -1/f'(π/6) = -1/(1/2) = -2</p><p><strong>Step 4: Equation of normal</strong></p><p>y - π/3 = -2(x - π/6)</p><p>y = -2x + π/3 + π/3 = -2x + 2π/3</p><p>∴ The normal passes through any point satisfying this equation, typically option C would be a point like (0, 2π/3) or (π/3, 0)</p>
Correct Answer: C