Class 12 Mathematics - Chapter Application of Derivatives NCERT Solutions | A particle moves along the curve 6y = x3
A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
The equation of the curve is given as:
6y = x3 + 2
The rate of change of the position of the particle with respect to time (t) is given by,
\begin{align}6\frac{dy}{dt} = 3x^2\frac{dx}{dt}+0\end{align}
\begin{align}\Rightarrow 2\frac{dy}{dt} = x^2\frac{dx}{dt}\end{align}
When the y-coordinate of the particle changes 8 times as fast as the
\begin{align}x-coordinate\; i.e.,\left(\frac{dy}{dt} = 8\frac{dx}{dt}\right), we \;have:\end{align}
\begin{align}2\left(8.\frac{dx}{dt}\right) = x^2.\frac{dx}{dt}\end{align}
\begin{align}\Rightarrow 16.\frac{dx}{dt} = x^2.\frac{dx}{dt}\end{align}
\begin{align}\Rightarrow (x^2 - 16).\frac{dx}{dt} =0 \end{align}
\begin{align}\Rightarrow x^2=16 \end{align}
\begin{align}\Rightarrow x=\pm 4 \end{align}
\begin{align}When\; x = 4, y = \frac{4^3 + 2}{6}=\frac{66}{6}=11\end{align}
\begin{align}When\; x = -4, y = \frac{(-4)^3 + 2}{6}=\frac{-62}{6}=\frac{-31}{3}\end{align}
Hence, the points required on the curve are
\begin{align} (4,11)\; and \;(-4,\frac{-31}{3}).\end{align}
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