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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.
उत्तर
Given, 3y = ax3 + 1 ...(i)
At x = 1, `dy/dx = 2 dx/dt` ...(ii)
Now, differentiating equation (i) w.r.t. 't', we get
`3 dy/dt = 3ax^2 dx/dt`
Put the value of `dy/dx` from (ii), we get
`3 xx 2 dx/dt = 3a xx 1^2 xx dx/dt`
`\implies 6 dx/dt = 3a dx/dt`
`\implies` 3a = 6
`\implies` a = 2
Hence the value of a = 2.
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