# A particle moves along the curve

Question:

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

Solution:

The equation of the curve is given as:

$6 y=x^{3}+2$

The rate of change of the position of the particle with respect to time (t) is given by,

$6 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}+0$

$\Rightarrow 2 \frac{d y}{d t}=x^{2} \frac{d x}{d t}$

When the y-coordinate of the particle changes 8 times as fast as the

$x$-coordinate i.e., $\left(\frac{d y}{d t}=8 \frac{d x}{d t}\right)$, we have:

$2\left(8 \frac{d x}{d t}\right)=x^{2} \frac{d x}{d t}$

$\Rightarrow 16 \frac{d x}{d t}=x^{2} \frac{d x}{d t}$

$\Rightarrow\left(x^{2}-16\right) \frac{d x}{d t}=0$

$\Rightarrow x^{2}=16$

$\Rightarrow x=\pm 4$

When $x=4, y=\frac{4^{3}+2}{6}=\frac{66}{6}=11$.

When $x=-4, y=\frac{(-4)^{3}+2}{6}=-\frac{62}{6}=-\frac{31}{3}$.

Hence, the points required on the curve are $(4,11)$ and $\left(-4, \frac{-31}{3}\right)$.