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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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