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

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(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 $

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(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 $\frac{dy}{dt} > \frac{dx}{dt}$.Differentiating the curve with respect to $t$, we get $12 \frac{dy}{dt} = 3x^{2} \frac{dx}{dt}$, which simplifies to $\frac{dy}{dt} = \frac{x^{2}}{4} \frac{dx}{dt}$.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 implies $x^{2} > 4$.This inequality $x^{2} - 4 > 0$ factors as $(x - 2)(x + 2) > 0$.The solution is $x \in (-\infty, -2) \cup (2, \infty)$.Comparing with the given options, $x > 2$ is the correct condition.

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

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