The equation of a curve is \(y = x(3 - x^{2})\). Find the equation of its normal of the point where x = 2.
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Correct Answer: Option n
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)
\(\frac{y + 2}{x - 2} = \frac{1}{9}\)
\(9y + 18 = x - 2 \implies x - 9y - 20 = 0\)
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)
\(\frac{y + 2}{x - 2} = \frac{1}{9}\)
\(9y + 18 = x - 2 \implies x - 9y - 20 = 0\)