Question

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.

Answer 1

Given: $6y=x^3+2$
Let the coordinate on the curve be $(x,y)$
According to question

$\frac{dy}{dx}=8$

$\Rightarrow \frac{d}{dx}\left(\frac{x^3+2}{6}\right)=8$

$\Rightarrow \frac{1}{6}.\left(3x^2\right)=8$

$\Rightarrow x^2=16\Rightarrow x=\pm 4$

When $x=4$

$y=\left(\frac{x^3+2}{6}\right)=\left(\frac{64+2}{6}\right)=\frac{66}{6}=11$

When $x=-4$

$y=\left(\frac{x^3+2}{6}\right)=\left(\frac{-64+2}{6}\right)=\frac{-62}{6}=-\frac{31}{3}$

Therefore, required points are $(4,11)\&(-4,-\frac{31}{3})$