If is the side of the cube then thus taking the derivative to respect to time from both time we have , conditions of the problem says the edge is 30 thus, and the rate of change is 10, thus, . Thus, placing in these values,Originally Posted bynirva

, thus,

. But this tells you the instantenous rate of change of the side, the problem asks for the instantenous rate of change of the surface area. Since for the cube then taking the derivate to respect to time we have,

know we know that for those are the conditions of this problem and in the other problem you found that thus, substituting these values,

.

Just placing in the proper unit values to get,