Question

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

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

Answer 1

The correct option is A

$x\in \left(-1,1\right)$

Given

$$\begin{gathered} \Rightarrow y=\sqrt[3]{27x} \\ \Rightarrow y=3x^{1/3}(eq(i\left)\right) \end{gathered}$$

Step 1: Differentiating $y$with respect to $x$

Differentiate w.r.t. $x$

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

$\Rightarrow \frac{dy}{dx}=\frac{27}{3y^2}$

$\Rightarrow \frac{dy}{dx}=\frac{9}{y^2}\left(eq\right(ii\left)\right)$

Step 2: Finding the interval of the value of $x$

We know that

$x\to$ abscissa,$y\to$ ordinate

As given ordinate changes faster than abscissa.

$\Rightarrow dx<dy$

$\Rightarrow 1<\frac{dy}{dx}$

$\Rightarrow 1<\frac{9}{y^2}(fromeq(ii\left)\right)$

putting the value of $eq\left(i\right)$ in $eq\left(ii\right)$. we have,

$\Rightarrow 1<\frac{9}{{\left(3x^{1/3}\right)}^2}$

$\Rightarrow 1<\frac{9}{9x^{2/3}}$

$\Rightarrow 1<\frac{1}{x^{2/3}}$

$\Rightarrow x^{2/3}<1$

$\Rightarrow x^2<1$

$\Rightarrow -1<x<1$

so, $x\in \left(-1,1\right)$

Hence, Option A is the correct answer.