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

The correct answer is $(4, 11)$ and $(-4,\frac{-31}{3}).$
The equation of the curve is given as:
$6y=x^3+2$
The rate of change of the position of the particle with respect to time $(t)$ is given by,
$6\frac{dy}{dt}=3x^2 \frac{dx}{dt}+0$
$=2 \frac{dy}{dt}=x^2 \frac{dx}{dt}$
When the y-coordinate of the particle changes 8 times as fast as the
$x$-coordinate i.e.,$(\frac{dy}{dt}=8\frac{dx}{dt})$,we have:
$2(8\frac{dx}{dt})=x^2 \frac{dx}{dt}$
$\implies 16\frac{dx}{dt}=x^2 \frac{dx}{dt}$
$(x^2-16)\frac{dx}{dt}=0$
$\implies x^2=16$
$x=±4$
When $x=4,y=\frac{4^3+2}{6}=\frac{66}{6}=11$
When $x=-4,y=\frac{(-4)^3+2}{6}=\frac{-62}{2}=\frac{-31}{3}$
Hence, the points required on the curve are $(4, 11)$ and $(-4,\frac{-31}{3}).$