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A particle moves along the curve $6 y=x^3+2$. Find the points on the curve at which the $y$-coordinate is changing 8 times as fast as the $\mathrm{x}$-coordinate.
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We have $6 y=x^3+2 \quad \ldots(i)$
$\Rightarrow \quad 2 \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{x}^2 \frac{\mathrm{dx}}{\mathrm{dt}} \quad \ldots(ii)$
Now, y coordinate changes 8 times as fast as $x$ coordinate i.e., $\frac{\mathrm{dy}}{\mathrm{dt}}=8 \frac{\mathrm{dx}}{\mathrm{dt}} \therefore$ in (ii), we have $x^2=16 \Rightarrow x=\pm 4$; When $x=4$, then from (i); $y=11$
When $x=-4$, then from (i); $y=\frac{1}{6}(-64+2)=-\frac{31}{3}$
Hence, the required points are $(4,11),\left(-4, \frac{-31}{3}\right)$
$\Rightarrow \quad 2 \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{x}^2 \frac{\mathrm{dx}}{\mathrm{dt}} \quad \ldots(ii)$
Now, y coordinate changes 8 times as fast as $x$ coordinate i.e., $\frac{\mathrm{dy}}{\mathrm{dt}}=8 \frac{\mathrm{dx}}{\mathrm{dt}} \therefore$ in (ii), we have $x^2=16 \Rightarrow x=\pm 4$; When $x=4$, then from (i); $y=11$
When $x=-4$, then from (i); $y=\frac{1}{6}(-64+2)=-\frac{31}{3}$
Hence, the required points are $(4,11),\left(-4, \frac{-31}{3}\right)$
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