Question

The surface area of a cube increases at the rate of 72 cm²/sec. Find the rate of change of its volume, when the edge of the cube measures 3 cm. 

Answer 1

Let a be the side of cube, s be the surface area and v be volume of cube. 

Given, `(ds)/(dt)` = 72 cm²/sec and a = 3  cm

Surface area of cube = 6a²

⇒ `(ds)/(dt) = 12a · (da)/(dt)`

⇒ `(da)/(dt) = 72/(12a)`

= `6/a`

Now, volume of cube, v = a³

On differentiating w.r.t. t, we get

`(dv)/(dt) = 3a^2 · (da)/(dt) = 3a^2 · 6/a`

⇒ `(dv)/(dt) = 18a`

⇒ `((dv)/(dt))_(a = 3)`

= 18 × 3

= 54 cm³/sec