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Question

A particle moves along the curve by 6y=x3+2. Find the point on the curve at which the ycoordinates is changing 8 times as fast as the x coordinate.

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Solution

Given that a particle moves along the curve,

6y=x3+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

dydx=8dxdt

From equation (1),

6y=x3+2


Diff. both sides with respect to t ,

6dydt=3x2dxdt

dydt=x22dxdt

We need to find point for which

dydt=8dxdt


Putting ,

dydx=x22.dxdt

x22dxdt=8dxdt

x2=16

x=±4

x=4,4


Putting the value of x in equation (1),

6y=(4)3+2

y=11

Point (4,11)


When, x=4

Points is (4,313)

6y=(4)3+2

y=313


Hence, required points on the curve are (4,11) and (4,313).


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