The volume of a cube is increasing at the rate of `9\ cm^3`/`sec`. How fast is the surface are increasing when the length of an edge is `10\ cm`?

Explanation

Let `V` and `S` be the volume and surface area of a cube of side `x \ cm` respectively.

Given that,

`x=10\ cm`

`(dV)/(dt)=9\ cm^3`/`sec`


Now, `V=x^3`


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


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


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


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