  Question

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

Solution

Given that a particle moves along the curve, $$6y={{x}^{3}}+2$$       ………. (1) We need to find the points on the curve at which y-coordinate is changing $$8$$ times as fast as the x-coordinate I.e. we need to find (x, y) for which $$\dfrac{dy}{dx}=8\dfrac{dx}{dt}$$ From equation (1), $$6y={{x}^{3}}+2$$ Diff. both sides with respect to t , $$\dfrac{6dy}{dt}=3{{x}^{2}}\dfrac{dx}{dt}$$ $$\dfrac{dy}{dt}=\dfrac{{{x}^{2}}}{2}\dfrac{dx}{dt}$$ We need to find point for which $$\dfrac{dy}{dt}=8\dfrac{dx}{dt}$$ Putting , $$\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{2}.\dfrac{dx}{dt}$$ $$\dfrac{{{x}^{2}}}{2}\dfrac{dx}{dt}=8\dfrac{dx}{dt}$$ $${{x}^{2}}=16$$ $$x=\pm 4$$ $$x=4,-4$$ Putting the value of $$x$$ in equation (1), $$6y={{\left( -4 \right)}^{3}}+2$$ $$y=11$$ Point $$(4, 11)$$ When, $$x=-4$$ Points is $$\left( -4,\dfrac{-31}{3} \right)$$ $$6y={{\left( -4 \right)}^{3}}+2$$ $$y=\dfrac{-31}{3}$$ Hence, required points on the curve are $$\left( 4,11 \right)$$ and $$\left( -4,\dfrac{-31}{3} \right)$$. Mathematics

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