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प्रश्न
A particle moves along the curve 3y = ax3 + 1 such that at a point with x-coordinate 1, y-coordinate is changing twice as fast at x-coordinate. Find the value of a.
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उत्तर
Given:
3y = ax3 + 1
Differentiate w.r.t. t
`3dy/dt = axx3x^2 dx/dt`
`=> dy/dt = ax^2 dx/dt`
Use given condition at x = 1
`=> dy/dt = 2dx/dt`
`dy/dt = a(1^2) dx/dt adx/dt`
`adx/dt = 2dx/dt`
`dx/dt cancel= 0`
a = 2
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