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Question
A particle moves along the curve 6y = x3 +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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Solution
6y = x3 + 2 and `dy/dt = 8 dx/dt`
On differentiating with respect to t,
`dy/dt = 8 dx/dt`
`6 dy/dt = 3x^2 dx/dt + 0`
`=> 6 xx 8 dx/dt = 3x^2 dx/dt`
`=> 3x^2 = 48`
`=> x^2 = 16`
⇒ x ± 4
Taking the positive sign, 6y = 64 + 2 = 66
`=> y = 66/6 = 11`
Taking the negative sign, 6y = (-64) + 2 = -62
`=> y = (- 62)/6 = - 31/3`
`therefore` the required points are (4, 11) and `(-4, (- 31)/3)`
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