Applications of Derivatives
Angle between curves
Grade 12
Question:
<p>The two curves \(C_1: x^3 - 3xy^2 + 2 = 0\) and \(C_2: 3x^2y - y^3 - 2 = 0\)</p>
<p>touch each other</p>
<p>cut at right angle</p>
<p>cut at an angle \(\frac{\pi}{3}\)</p>
<p>cut at an angle \(\frac{\pi}{4}\)</p>
Step-by-Step Solution
Key Concept: Recognize that C₁ and C₂ are related through implicit differentiation and represent curves whose tangents are perpendicular. Use implicit differentiation to find slopes and apply the orthogonality condition (m₁ · m₂ = -1).
<p><strong>Step 1:</strong> Find dy/dx for C₁: x³ - 3xy² + 2 = 0</p><p>Differentiating implicitly: 3x² - 3y² - 6xy(dy/dx) = 0</p><p>So dy/dx|_{C₁} = (3x² - 3y²)/(6xy) = (x² - y²)/(2xy)</p><p><strong>Step 2:</strong> Find dy/dx for C₂: 3x²y - y³ - 2 = 0</p><p>Differentiating implicitly: 6xy + 3x²(dy/dx) - 3y²(dy/dx) = 0</p><p>So dy/dx|_{C₂} = -6xy/(3x² - 3y²) = -2xy/(x² - y²)</p><p><strong>Step 3:</strong> Check orthogonality: m₁ · m₂ = [(x² - y²)/(2xy)] · [-2xy/(x² - y²)] = -1 ✓</p><p>∴ The curves are orthogonal at all intersection points. Answer: B</p>
Correct Answer: B