A particle is in motion along a curve 12 y = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the curve $12 y = x^{3}$.Let the rate of change of the ordinate be $\frac{dy}{dt}$ and the rate of change of the abscissa be $\frac{dx}{dt}$

Vedclass · Accepted Answer

$ x > 2 $

Vedclass · Answer

(D) Given the curve $12 y = x^{3}$.Let the rate of change of the ordinate be $\frac{dy}{dt}$ and the rate of change of the abscissa be $\frac{dx}{dt}$.We are given that the rate of change of the ordinate exceeds that of the abscissa,so $\frac{dy}{dt} > \frac{dx}{dt}$.Differentiating the equation of the curve with respect to $t$,we get $12 \frac{dy}{dt} = 3x^{2} \frac{dx}{dt}$,which implies $\frac{dy}{dt} = \frac{x^{2}}{4} \frac{dx}{dt}$.Substituting this into the inequality $\frac{dy}{dt} > \frac{dx}{dt}$,we have $\frac{x^{2}}{4} \frac{dx}{dt} > \frac{dx}{dt}$.Assuming $\frac{dx}{dt} > 0$,we get $\frac{x^{2}}{4} > 1$,which means $x^{2} > 4$.This inequality $x^{2} - 4 > 0$ factors as $(x - 2)(x + 2) > 0$.The solution to this inequality is $x \in (-\infty, -2) \cup (2, \infty)$.Since the question asks for the condition where the rate of change of the ordinate exceeds the abscissa,and considering the provided options,the condition $x > 2$ is a valid subset of the solution set.

$A$ particle is in motion along a curve $12 y = x^{3}$. The rate of change of its ordinate exceeds that of its abscissa when:

Similar Questions

The distance $x$ covered by a particle in time $t$ is given by $x = t^3 - 12t^2 + 6t + 8$. What is the velocity of the particle at the instant when its acceleration is zero?

The volume of a spherical balloon is increasing at the rate of $30 \ cm^3/min$. Find the rate of change of the surface area of the balloon when its radius is $6 \ cm$.

$A$ point is moving on $y=4-2x^2$. The $x$-coordinate of the point is decreasing at the rate of $5 \text{ units/s}$. Then,the rate at which the $y$-coordinate of the point is changing when the point is at $(1, 2)$ is

If the law of motion in a straight line is $s = \frac{1}{2}vt$,then acceleration is

If the rate of change of the area of a square $S$ is equal to its side length,and the rate of change of the side of $S$ is the same as that of a cube $C$,then the rate of change of the volume of $C$,at the time when its side length is $2$ units,will be ............ $units^3/sec$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module