Find the gradient to the normal of the curve (y = x^{3} – x^{2}) at the point where x = 2.
-
A.
(frac{-1}{8}) -
B.
(frac{1}{8}) -
C.
(frac{-1}{24}) -
D.
(1)
Correct Answer: Option A
Explanation
Given : (y = x^{3} – x^{2})
(frac{mathrm d y}{mathrm d x} = 3x^{2} – 2x)
(therefore text{The gradient of the tangent at point (x = 2)} = 3(2^{2}) – 2(2) )
= (12 – 4 = 8)
Recall, the tangent and the normal are perpendicular to each other and the product of the gradients of perpendicular lines = -1.
(implies text{the gradient of the normal} = frac{-1}{8})