A particle moves along a curve y = frac{2x^3 | vedclass

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(D) Given the curve equation is $y = \frac{2x^3 - 1}{3}$.We are given that the rate of change of the $y$-coordinate is $18$ times the rate of change o

Vedclass · Accepted Answer

$(-3, -\frac{55}{3}), (3, \frac{53}{3})$

Vedclass · Answer

(D) Given the curve equation is $y = \frac{2x^3 - 1}{3}$.We are given that the rate of change of the $y$-coordinate is $18$ times the rate of change of the $x$-coordinate,which implies $\frac{dy}{dt} = 18 \frac{dx}{dt}$.Dividing both sides by $\frac{dx}{dt}$,we get $\frac{dy}{dx} = 18$.Now,differentiate the curve equation with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (\frac{2x^3 - 1}{3}) = \frac{1}{3} \cdot 6x^2 = 2x^2$.Equating the derivative to $18$:$2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$.For $x = 3$,$y = \frac{2(3)^3 - 1}{3} = \frac{54 - 1}{3} = \frac{53}{3}$.For $x = -3$,$y = \frac{2(-3)^3 - 1}{3} = \frac{-54 - 1}{3} = -\frac{55}{3}$.Thus,the points are $(3, \frac{53}{3})$ and $(-3, -\frac{55}{3})$.Comparing this with the given options,the correct option is $D$.

$A$ particle moves along a curve $y = \frac{2x^3 - 1}{3}$. The points on the curve at which the $y$-coordinate is changing $18$ times as fast as the $x$-coordinate are

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