The equation of a curve is (y = x(3 – x^{2})). Find the equation of its normal of the point where x = 2.
Explanation
Curve : (y = x(3 – x^{2}))
(y = 3x – x^{3})
Gradient function : (frac{mathrm d y}{mathrm d x} = 3 – 3x^{2})
When x = 2, y = (3(2) – (2^{3}) = -2)
Gradient of tangent : (3 – 3(-2)^{2} = -9)
Gradient of normal : (frac{1}{9})
Equation of normal through (2, -2)
(frac{y + 2}{x – 2} = frac{1}{9})
(9y + 18 = x – 2 implies x – 9y – 20 = 0)