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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. - Mathematics

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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.

Sum
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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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Chapter 6: Application of Derivatives - Exercise 6.1 [Page 198]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 6 Application of Derivatives
Exercise 6.1 | Q 11 | Page 198

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