A point is moving along the curve y^3 = 27x. | vedclass

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(C) Given the curve equation $y^3 = 27x$. Differentiating both sides with respect to time $t$,we get: $3y^2 \frac{dy}{dt} = 27 \frac{dx}{dt}$ $\frac{d

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$(-1, 1)$

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(C) Given the curve equation $y^3 = 27x$. Differentiating both sides with respect to time $t$,we get: $3y^2 \frac{dy}{dt} = 27 \frac{dx}{dt}$ $\frac{dx}{dt} = \frac{y^2}{9} \frac{dy}{dt}$. The problem states that the abscissa $(x)$ changes at a slower rate than the ordinate $(y)$,which means: $\left| \frac{dx}{dt} \right| Substituting the expression for $\frac{dx}{dt}$: $\left| \frac{y^2}{9} \frac{dy}{dt} \right| $\frac{y^2}{9} $y^2 $-3 Since $y^3 = 27x$,when $y = -3$,$x = -1$,and when $y = 3$,$x = 1$. Thus,the interval for the abscissa $x$ is $(-1, 1)$.

$A$ point is moving along the curve $y^3 = 27x$. The interval in which the abscissa changes at a slower rate than the ordinate is

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