Theory question 103
DifferentiationThe equation of a curve is y = x(3 − x2). Find the equation of its normal at the point where x = 2.
Show the model answer
Model answer
y = 3x − x3, dy/dx = 3 − 3x2
At x = 2: gradient of tangent = −9, y = 2(3 − 4) = −2
Gradient of normal = 1/9
y + 2 = (1/9)(x − 2)
x − 9y − 20 = 0
Explanation
The normal is perpendicular to the tangent, so its gradient is −1 ÷ (tangent gradient).