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The Volume of Cube is Increasing at the Rate of 9 Cm 3/S. How Fast is Its Surfacee Area Increasing When the Length of an Edge is 10 Cm?

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प्रश्न

The Volume of cube is increasing at the rate of 9 cm 3/s. How fast is its surfacee area increasing when the length of an edge is 10 cm?

योग
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उत्तर

`(dV)/dt = 9 "cm"^"3//sec x = 10 cm` (given) 

`V = x^3`

`(dV)/dt = 3x^2 dx/dt`

`9/(3x^2) = dx/dt`

Now, `s = 6x^2`

`(ds)/dt = 12x xx dx/dt`

`(ds)/dt = 12 xx 10 xx 9/(3xx10xx10)`

`= (12 xx 9)/(3xx 10) = 36/10`

`= 3.6  cm^2//sec`

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