Applications of Derivatives
Normals to curves
Grade 12
Question:
<p>The equation of a normal to the curve, \(\sin y = x\sin\left(\dfrac{\pi}{3} + y\right)\) at \(x = 0\), is</p>
<p>\(2x + \sqrt{3}y = 0\)</p>
<p>\(2y - \sqrt{3}x = 0\)</p>
<p>\(2y + \sqrt{3}x = 0\)</p>
<p>\(2x - \sqrt{3}y = 0\)</p>
Step-by-Step Solution
Key Concept: Find the point on the curve at x=0, then use implicit differentiation to get dy/dx, and finally construct the normal line using slope = -1/(dy/dx).
<p><strong>Step 1:</strong> Find the point on the curve at x = 0.</p><p>Substitute x = 0 into sin y = x sin(π/3 + y):</p><p>sin y = 0 · sin(π/3 + y) = 0</p><p>Therefore y = 0, so the point is (0, 0).</p><p><strong>Step 2:</strong> Differentiate implicitly: sin y = x sin(π/3 + y)</p><p>cos y · (dy/dx) = sin(π/3 + y) + x · cos(π/3 + y) · (dy/dx)</p><p><strong>Step 3:</strong> Evaluate at (0, 0):</p><p>cos(0) · (dy/dx) = sin(π/3) + 0</p><p>1 · (dy/dx) = √3/2</p><p>So dy/dx = √3/2</p><p><strong>Step 4:</strong> The slope of the normal is:</p><p>m_normal = -1/(dy/dx) = -2/√3 = -2√3/3</p><p><strong>Step 5:</strong> Equation of normal through (0, 0):</p><p>y - 0 = (-2√3/3)(x - 0)</p><p>y = -2x/√3 or equivalently 2x + √3y = 0</p><p>∴ Answer: B</p>
Correct Answer: B