# Thread:

1. ## Application of derivative

Hi guys, i have got a few question that needs help before my exams next week

The surface area of a cube with side x cm, is increasing at a rate of 16cm2/s

Whnt the total surface area of the cube reaches 216cm2
a)show that the volume of the cube is 216cm3
b) find the rate which its volume is increasing

I completed part a) but for part b, they wants dv/dt so i figure out since da/dt is given i need to find dv/da. But I am stuck ! anyone could explain to me ? THANKS IN ADVANCE !

2. ## Re: Application of derivative

Let $A$ be the surface area and $s$ be the side of the cube.

You have $\frac{dA}{dt}=16$.

So $A=16t+C$.

Taking $C=0$, $A=16t$.

$\begin{align*} A &=16t \\ \implies 6s^2 &=16t \\ \implies s^2 &=\frac{8}{3}t \\ \implies s^3 &=\left( \frac{8}{3}t \right)^{\frac{3}{2}} \\ \implies V &=\left( \frac{8}{3}t \right)^{\frac{3}{2}} \\ \implies \frac{dV}{dt} &=\frac{d\left( \frac{8}{3}t \right)^{\frac{3}{2}}}{dt} \\ \implies \frac{dV}{dt}&= \left(\frac{8\sqrt{2}}{\sqrt{3}} \right)\sqrt{t} \end{align*}$

3. ## Re: Application of derivative

$A = 6x^2$

$\frac{dA}{dt} = 12x \cdot \frac{dx}{dt}$

when $A = 216$ , $x = 6$

$16 = 12(6) \cdot \frac{dx}{dt}$

$\frac{dx}{dt} = \frac{2}{9}$

$V = x^3$

$\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}$

finish it
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