A-Level Differentiation: worked solution

Question

The curve has a stationary point at (pa, qa) for positive constants p and q. Find the exact p and q (the nature of the point is not required).

The curve is x³ + y³ = 3axy (the folium of Descartes) and (pa, qa) is the stationary point with p, q > 0. Step 1: differentiate implicitly with respect to x. Every y term needs the chain rule, and the right-hand side needs the product rule: d/dx(x³) + d/dx(y³) = d/dx(3axy) 3x² + 3y² (dy/dx) = 3a(y + x dy/dx). Divide throughout by 3: x² + y² (dy/dx) = ay + ax (dy/dx). Step 2: make dy/dx the subject. Collect the dy/dx terms on one side: y² (dy/dx) − ax (dy/dx) = ay − x² (dy/dx)(y² − ax) = ay − x² dy/dx = (ay − x²)/(y² − ax). Step 3: impose the stationary condition. A fraction is zero exactly when its numerator is zero (its denominator being non-zero), so ay − x² = 0, that is y = x²/a. Step 4: substitute y = x²/a back into the equation of the curve. x³ + (x²/a)³ = 3ax(x²/a) x³ + x⁶/a³ = 3x³ x⁶/a³ = 2x³ x⁶ = 2a³x³. The solution x = 0 gives the origin, which does not have positive coordinates, so divide by x³: x³ = 2a³ x = 2^(1/3) a. Comparing with x = pa gives p = 2^(1/3). Step 5: find the y-coordinate. y = x²/a = (2^(1/3) a)²/a = 2^(2/3) a²/a = 2^(2/3) a. Comparing with y = qa gives q = 2^(2/3). Check that the point lies on the curve: x³ + y³ = 2a³ + 4a³ = 6a³, and 3axy = 3a(2^(1/3)a)(2^(2/3)a) = 3a³ × 2^(1/3 + 2/3) = 6a³. They agree. So p = 2^(1/3) and q = 2^(2/3).