In a Mahakumbh, a drone camera is moving along $3y = x^3 - 3$. When y-coordinate changes 9 times as fast as x-coordinate, it captures good quality pictures. Then one of the precise positions of the drone at that instant is
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"Rate of change" problems almost always involve differentiating an equation with respect to time ($t$). Translating the English phrase "$A$ changes $k$ times as fast as $B$" directly into the equation $\frac{dA}{dt} = k \cdot \frac{dB}{dt}$ is the crucial first step.
The equation of the path followed by the drone camera is given by \(3y = x^3 - 3\). For simpler manipulation, we rearrange it to \(y = \frac{x^3 - 3}{3}\).
It's stated that the y-coordinate changes 9 times as fast as the x-coordinate, meaning \(\frac{dy}{dt} = 9 \frac{dx}{dt}\).
To find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) in terms of \(x\), we differentiate the equation of the path with respect to \(t\):
Given \(y = \frac{x^3 - 3}{3}\), differentiate both sides with respect to \(t\):