A particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the y-coordinate is changing 8 times as fast as x-coordinate.

The correct answer is $(4, 11), (-4, -\dfrac{31}{3})$

Equation of the curve is $6y = x^3 + 2$

Differentiating the above equation, we get $6\dfrac{dy}{dx}$ = $3x^2$

When the y-coordinate is changing 8 times as fast as x-coordinate, $\dfrac{dy}{dx} = 8$.

∴ $6*8$ = $3x^2$ ⇒ $x = ±4$


When $x = +4, 6y = 64 + 2$ ⇒ $y = 11$

Hence the point is $(4, 11)$

When $x = -4, 6y = -64 + 2$ ⇒ $y = -\dfrac{31}{3}$

Hence the point is $(-4, -\dfrac{31}{3})$