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.

Answers (1)

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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