Question

A point is moving along a cubical passion $12y={{x}^{3}}$, the rate of ordinates is less than rule of abscissa when
$\begin{aligned} & a)-2 < x < 2 \\\ & b)x=\pm 2 \\\ & c)-2 \\\ & d)0 \\\ \end{aligned}$

Answer 1

We are given that a point has many long a parabola $y={{x}^{3}}$, and we have rate of ordinate is the rate of change of abscissa, to find the coordinate. We short by learning that once abscissa, what are ordinate then we unite statement rate of change of ordinate is less than rate of change of abscissa, then we differentiate our function $12y={{x}^{3}}$ to get the rate of change then we give information to find the value of solution.

Complete step by step solution:
We are given that a parabola is defined as $12y={{x}^{3}}$. Before we move forward, we learn about ordinate and abscissa. The abscissa refers to the horizontal distance of a point, it basically refers to the x-axis.
Similarly, ordinate refers to the vertical distance of a point; it basically refers to the y-axis. As given that rate of ordinate is less than that rate of abscissa. It means that the change of y-axis with time is less than the rate of change of x axis with time.
So, we get,
$\dfrac{dy}{dt} < \dfrac{dx}{dt}.......(i)$
Now we have that parabola is given as:
$12y={{x}^{3}}$
We differentiate both sides by that we get:
$\dfrac{d(12y)}{dt}=\dfrac{d({{x}^{3}})}{dt}$
$\begin{aligned} & \Rightarrow d(12y)=12dy \\\ & \Rightarrow 12\dfrac{dy}{dt}=3{{x}^{2}}\dfrac{dx}{dt} \\\ \end{aligned}$
Hence, we get,
$12\dfrac{dy}{dt}=3{{x}^{2}}\dfrac{dx}{dt}$
Simplifying we get:
$\dfrac{dy}{dt}=\dfrac{{{x}^{2}}}{4}\dfrac{dx}{dt}.............(ii)$
Using equation (ii) in equation (i) we get,
$\dfrac{{{x}^{2}}}{4}\dfrac{dx}{dt} < \dfrac{dx}{dt}\left( \dfrac{dy}{dt}=\dfrac{{{x}^{2}}}{4}\dfrac{dx}{dt} \right)$
Simplifying we get.
$\dfrac{{{x}^{2}}}{4}\dfrac{dx}{dt}-\dfrac{dx}{dt} < 0$
So,
$\Rightarrow \left( \dfrac{{{x}^{2}}}{4}-1 \right)\dfrac{dx}{dt < 0}$
So, we get
$\Rightarrow \left( \dfrac{{{x}^{2}}}{4}-1 \right) < 0$
Now using
${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$
Hence, we get,
$\left( \dfrac{x}{2}-1 \right)\left( \dfrac{x}{2}+1 \right) < 0$
Product of $2$ term is negative (less than $0$) so ensure one of them is negative and other is positive. So either
$\left( \dfrac{x}{2}-1 \right) < 0\,\,\,\,\,\,0r\,\,\,\,\,\,\left( \dfrac{x}{2}+1 \right) < 0$
So, simplifying we get.
$\dfrac{x}{2}-1 < 0\,\,\,\,\,\,and\,\,\,\,\dfrac{x}{2}+1 < 0$
So,
$\Rightarrow \dfrac{x}{2}-1 < 0\,\,\,\,x < 2$
And
$\dfrac{x}{2}+1 > 0\,\,\,\,x > -2$
So, it gives,
When $-2 < x < 2$
It gives us our answer
(ii) let $\dfrac{x}{2}-1 > 0\,\,\,\,and\,\,\,\,\dfrac{x}{2}+1 < 0$
So,
$\Rightarrow \dfrac{x}{2}-1 > 0$ gives us $x > 2$
$\Rightarrow \dfrac{x}{2}+1 < 0$ gives us $x < -2$
This is not same as using n
So, our solution is $-2 < x < 2$

So, the correct answer is “Option a”.

Note: When we compare this we need to be very careful. For example $ab > 0$ then we get a product of $2$ terms and positive it can happen when both are positive as well as when both are negative. So, we made 2 cases to check feasible regions for each option.
Over $2$ case was not true as it says we want those n which are greater than $2$ but less than $-2$
It can happen as $2$ is greater than $-2$
So, we need to look for all options which can be possible.