A particle moves along the curve 6y = x^3 + 2 | vedclass

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(A) The equation of the curve is given as $6y = x^3 + 2$.Differentiating both sides with respect to time $t$,we get:$6 \frac{dy}{dt} = 3x^2 \frac{dx}{

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$(4, 11)$ and $(-4, -31/3)$

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(A) The equation of the curve is given as $6y = x^3 + 2$.Differentiating both sides with respect to time $t$,we get:$6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$Dividing by $3$,we get:$2 \frac{dy}{dt} = x^2 \frac{dx}{dt}$Given that the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate,i.e.,$\frac{dy}{dt} = 8 \frac{dx}{dt}$.Substituting this into the equation:$2(8 \frac{dx}{dt}) = x^2 \frac{dx}{dt}$$16 \frac{dx}{dt} = x^2 \frac{dx}{dt}$$(x^2 - 16) \frac{dx}{dt} = 0$Assuming $\frac{dx}{dt} 
eq 0$,we have $x^2 = 16$,which gives $x = 4$ or $x = -4$.For $x = 4$,$6y = 4^3 + 2 = 64 + 2 = 66$,so $y = 11$. The point is $(4, 11)$.For $x = -4$,$6y = (-4)^3 + 2 = -64 + 2 = -62$,so $y = -62/6 = -31/3$. The point is $(-4, -31/3)$.Thus,the required points are $(4, 11)$ and $(-4, -31/3)$.

$A$ 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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