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Question
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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Solution
`(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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