Question

A point is moving along the curve $y^3=27x$. Find the interval of valued of $x$ in which the ordinate changes faster then abscissa is:

• A. $x\in (-1,1)$
• B. $x\in (-1,-1)-\left\{0\right\}$
• C. $x\in [-1,1]-\left\{0\right\}$
• D. $x\in (-1,0)$

Answer 1

The correct option is A $x\in (-1,1)$

$y^3=27x$

Abcissa changes at slower rate than ordinate.

$\frac{dx}{dt}<\frac{dy}{dt}.................\left(1\right)$

$y^3=27x$

$3y^2\frac{dy}{dt}=27\frac{dx}{dt}.............\left(2\right)$

Putting $\frac{dx}{dt}$ in eq $\left(1\right)$

$\frac{3y^2}{27}<\frac{dy}{dy}$

$\frac{dy}{dt}(\frac{3y^2}{27}<1)<0$

By eq$\left(2\right)$ wecan say that $\frac{dx}{dt}$ and $\frac{dy}{dt}$ will be '+ve' or '-ve'.

So, $\frac{3y^2}{27}-1<0\Rightarrow -3<y<3$ and $-1<x<1$.