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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 x-coordinate.

11. A particle moves along the curve $6y = x^3 + 2$ Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

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We need to find the point at which $\frac{dy}{dt} = 8\frac{dx}{dt}$
Given the equation of curve =   $6y = x^3 + 2$
Differentiate both sides w.r.t. t
$6\frac{dy}{dt} = \frac{d(x^{3})}{dx}.\frac{dx}{dt} +0$
$= 3x^{2}.\frac{dx}{dt}$
$\frac{dy}{dt} = 8\frac{dx}{dt}$   (required condition)
$6\times 8\frac{dx}{dt}= 3x^{2}.\frac{dx}{dt}$
$3x^{2}.\frac{dx}{dt} =48\frac{dx}{dt}$$\Rightarrow x^{2} = \frac{48}{3} = 16$
$x = \pm 4$
when x = 4 ,  $y = \frac{4^{3}+2}{6} = \frac{64+2}{6} = \frac{66}{6} = 11$
and
when x = -4 , $y = \frac{(-4)^{3}+2}{6} = \frac{-64+2}{6} = \frac{-62}{6} = \frac{-31}{3}$
So , the coordinates are
$(4,11) \ and \ (-4,\frac{-31}{3})$

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