<p>The angle between the curves \(x^3 - 3xy^2 + 2 = 0\) and \(3x^2 y - y^3 - 2 = 0\) is:</p>
Step-by-Step Solution
Key Concept: The angle between two curves equals the angle between their tangents at the point of intersection. Find the intersection point, then compute dy/dx for each curve using implicit differentiation, and apply the angle formula tan(θ) = |(m₁ - m₂)/(1 + m₁m₂)|.
<p><strong>Step 1: Find the point of intersection</strong></p><p>Add the two equations:<br>(x³ - 3xy² + 2) + (3x²y - y³ - 2) = 0<br>x³ + 3x²y - 3xy² - y³ = 0<br>(x + y)(x² + 2xy - y²) = 0</p><p>Subtract the second from the first:<br>(x³ - 3xy² + 2) - (3x²y - y³ - 2) = 0<br>x³ - 3xy² - 3x²y + y³ + 4 = 0<br>(x - y)(x² - 2xy - y²) = 0 - 4</p><p>Testing x = 1, y = 1: First equation: 1 - 3 + 2 = 0 ✓ and Second equation: 3 - 1 - 2 = 0 ✓</p><p><strong>Step 2: Find dy/dx for curve 1: x³ - 3xy² + 2 = 0</strong></p><p>Differentiating: 3x² - 3y² - 6xy(dy/dx) = 0<br>dy/dx = (3x² - 3y²)/(6xy) = (x² - y²)/(2xy)</p><p>At (1,1): m₁ = (1 - 1)/(2) = 0</p><p><strong>Step 3: Find dy/dx for curve 2: 3x²y - y³ - 2 = 0</strong></p><p>Differentiating: 6xy + 3x²(dy/dx) - 3y²(dy/dx) = 0<br>dy/dx = -6xy/(3x² - 3y²) = -2xy/(x² - y²)</p><p>At (1,1): m₂ = -2(1)(1)/(1 - 1) = ∞ (undefined, vertical tangent)</p><p><strong>Step 4: Interpret the result</strong></p><p>m₁ = 0 (horizontal tangent) and m₂ = ∞ (vertical tangent) means m₁ · m₂ = 0 · ∞ indicates perpendicular curves.</p><p>∴ The angle between the curves is <strong>90° or π/2 radians</strong></p>
Correct Answer: A