The equation of a curve is \(y = x(3 - x^{2})\). Find the equation of its normal of the point where x = 2.
Question 1 Report
The equation of a curve is \(y = x(3 - x^{2})\). Find the equation of its normal of the point where x = 2.
Answer Details
Expand the curve: \(y = x(3 - x^2) = 3x - x^3\).
Differentiate to get the gradient of the tangent:
\[\frac{dy}{dx} = 3 - 3x^2.\]
At \(x = 2\): gradient of tangent \(= 3 - 3(4) = -9\).
The \(y\)-coordinate there is \(y = 2(3 - 4) = -2\), giving the point \((2, -2)\).
The normal is perpendicular to the tangent, so its gradient is
\[m_{\text{normal}} = -\frac{1}{-9} = \frac{1}{9}.\]
Equation of the normal through \((2, -2)\):
\[y - (-2) = \frac{1}{9}(x - 2) \;\Rightarrow\; 9y + 18 = x - 2 \;\Rightarrow\; x - 9y - 20 = 0.\]
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