Applications of Derivatives
Tangent and Normal to a Curve
Grade 12
Question:
<p>Given the curve \(\sin y = x\sin\left(\dfrac{\pi}{3} + y\right)\), the equation of the normal at \((0, 0)\) is:</p>
<p>\(x + \sqrt{3}y = 0\)</p>
<p>\(2x + \sqrt{3}y = 0\)</p>
<p>\(x - \sqrt{3}y = 0\)</p>
<p>\(2x - \sqrt{3}y = 0\)</p>
Step-by-Step Solution
Key Concept: Use implicit differentiation on the given equation to find dy/dx at (0,0), then the slope of the normal is the negative reciprocal of this derivative.
<p><strong>Step 1:</strong> Differentiate sin y = x·sin(π/3 + y) implicitly with respect to x.</p><p>cos y · dy/dx = sin(π/3 + y) + x·cos(π/3 + y)·dy/dx</p><p><strong>Step 2:</strong> At point (0, 0): cos(0)·dy/dx = sin(π/3) + 0</p><p>1·dy/dx = √3/2</p><p>So dy/dx = √3/2 (slope of tangent at origin)</p><p><strong>Step 3:</strong> Slope of normal = -1/(√3/2) = -2/√3 = -2√3/3</p><p><strong>Step 4:</strong> Equation of normal passing through (0, 0):</p><p>y - 0 = (-2√3/3)(x - 0)</p><p>y = (-2√3/3)x or 2√3·x + 3y = 0</p><p>∴ Answer: B</p>
Correct Answer: B