If the rate of change of x is more than the r | vedclass

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(D) Given the curve $x^3 = 12y$. Differentiating both sides with respect to time $t$,we get: $3x^2 \frac{dx}{dt} = 12 \frac{dy}{dt}$ $\frac{dy}{dt} =

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$(0, 2)$

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(D) Given the curve $x^3 = 12y$. Differentiating both sides with respect to time $t$,we get: $3x^2 \frac{dx}{dt} = 12 \frac{dy}{dt}$ $\frac{dy}{dt} = \frac{3x^2}{12} \frac{dx}{dt} = \frac{x^2}{4} \frac{dx}{dt}$ Given that the rate of change of $x$ is more than the rate of change of $y$,i.e.,$\frac{dx}{dt} > \frac{dy}{dt}$. Substituting the value of $\frac{dy}{dt}$: $\frac{dx}{dt} > \frac{x^2}{4} \frac{dx}{dt}$ Since $x > 0$,$\frac{dx}{dt}$ is positive (assuming $x$ is increasing),so we can divide by $\frac{dx}{dt}$: $1 > \frac{x^2}{4}$ $x^2 $|x| Since the condition $x > 0$ is given,the interval for $x$ is $(0, 2)$.

If the rate of change of $x$ is more than the rate of change of $y$ on the curve $x^3 = 12y$ for $x > 0$,then $x$ lies in the interval:

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