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 the x-coordinate.
Equation of curve is $6y=x^3+2$ $\;\; \cdots$ (1)
Differentiating equation (1) w.r.t time t gives
$6 \dfrac{dy}{dt} = 3x^2 \dfrac{dx}{dt}$
i.e. $2 \dfrac{dy}{dt} = x^2 \dfrac{dx}{dt}$ $\;\; \cdots$ (2)
Given: $\dfrac{dy}{dt} = 8 \dfrac{dx}{dt}$ $\;\; \cdots$ (3)
$\therefore$ We have from equations (2) and (3)
$2 \times 8 \dfrac{dx}{dt} = x^2 \dfrac{dx}{dt}$
i.e. $x^2 = 16$
$\implies$ $x=\pm 4$
$\therefore$ Substituting the value of x in equation (1) gives
when $x=+4$, $y=\dfrac{4^3 + 2}{6} = 11$ and
when $x=-4$, $y = \dfrac{\left(-4\right)^3 + 2}{6} = -\dfrac{31}{3}$
$\therefore$ The required points on the curve are $\left(4,11\right)$ and $\left(-4, - \dfrac{31}{3}\right)$