Question

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

Answer 1

Given that a particle moves along the curve,

$6y=x^3+2$ ………. (1)

We need to find the points on the curve at which y-coordinate is changing $8$ times as fast as the x-coordinate I.e. we need to find (x, y) for which

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

From equation (1),

$6y=x^3+2$

Diff. both sides with respect to t ,

$\frac{6dy}{dt}=3x^2\frac{dx}{dt}$

$\frac{dy}{dt}=\frac{x^2}{2}\frac{dx}{dt}$

We need to find point for which

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

Putting ,

$\frac{dy}{dx}=\frac{x^2}{2}.\frac{dx}{dt}$

$\frac{x^2}{2}\frac{dx}{dt}=8\frac{dx}{dt}$

$x^2=16$

$x=\pm 4$

$x=4,-4$

Putting the value of $x$ in equation (1),

$6y={(-4)}^3+2$

$y=11$

Point $(4,11)$

When, $x=-4$

Points is $(-4,\frac{-31}{3})$

$6y={(-4)}^3+2$

$y=\frac{-31}{3}$

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