​ A curve passes through the origin, and its gradient function is 2x – ​x^2/2 1. Find its y coordinate when x= 2.

Question

A curve passes through the origin, and its gradient function is 2x –\(\frac{x^2}{2}\)

1. Find its y coordinate when x= 2.
2. Find the equation of the tangent at x= 2.

Answer 1

1. Given;

Integrating we have; ∫dy = ∫(2x – \(\frac{x^2}{2}\))dx

⇒ y = x² – \(\frac{x^3}{6}\) + c ___(1)

Since the curve passes through (0, 0)

(1) ⇒ 0 = 0 + c ⇒ c = 0

∴ Equation of the curve is y = x² – \(\frac{x^3}{6}\)

When x = 2 ⇒ y = 2² – \(​​\frac{2^3}{6}\) = \(\frac{8}{3}\)

∴ coordinate is (2, \(\frac{8}{3}\))

2. Slope at (2, \(\frac{8}{3}\)) = 2 × 2 – \(\frac{2^2}{2}\) = 2

∴ Equation of the tangent at (2, \(\frac{8}{3}\)) is given by

y – \(\frac{8}{3}\) = 2(x – 2) ⇒ 3y – 8 = 6x – 12 ⇒ 3y = 6x – 4.