Question

The volume of a cube is increasing at a rate of 10 cm^3/min. How fast is the surface area increasing when the length of an edge 90 cm?

Answer 1

`12/27` cms squared per minute.

Explanation:

Let the length of a side of the cube =`x`

Then volume of cube `v` =`x^3`.........`[1]`
Surface area of cube `s` will = `6x^2`......`[2]`

Differentiating `[1]` implicitly w.r.t. `t`[ time], `[dv]/dt=3x^2dx/dt` , but we know from the question that `[dv]/dt=10`, therefore `dx/dt=[10]/[3x^2`......`[3]`

Differentiating.......`[2]` w.r.t. `t`, `[ds]/dt=12xdx/dt` and substituting for `dx/dt` from .....`[3]` we obtain,

`[ds]/dt=[12x[10]/[3x^2]]` = `[120]/[3x]`, and so we have the rate of change of the side length of the cube with respect to time, and when `x=90`, `[ds]/dt=120/[3[90]]` = `12/27`.

Hope this was helpful.