In a Maha Kumbh, a drone camera is moving alo | vedclass

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(C) Given the equation of the path: $3y = x^3 - 3$.Differentiating both sides with respect to time $t$: $3 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$.Given t

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$(3, 8)$

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(C) Given the equation of the path: $3y = x^3 - 3$.Differentiating both sides with respect to time $t$: $3 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$.Given the condition: $\frac{dy}{dt} = 9 \frac{dx}{dt}$.Substituting the condition into the differentiated equation: $3(9 \frac{dx}{dt}) = 3x^2 \frac{dx}{dt}$.This simplifies to $27 \frac{dx}{dt} = 3x^2 \frac{dx}{dt}$, which implies $x^2 = 9$, so $x = \pm 3$.If $x = 3$, then $3y = (3)^3 - 3 = 27 - 3 = 24$, so $y = 8$. The point is $(3, 8)$.If $x = -3$, then $3y = (-3)^3 - 3 = -27 - 3 = -30$, so $y = -10$. The point is $(-3, -10)$.Comparing with the given options, $(3, 8)$ is the correct position.

In a Maha Kumbh, a drone camera is moving along the curve $3y = x^3 - 3$. The camera captures high-quality pictures when the $y$-coordinate changes $9$ times as fast as the $x$-coordinate. Which of the following is a precise position of the drone at that instant?

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