At time t seconds the length of the side of a cube is x cm, the surface area of the cube is S cm², and the volume of the cube is V cm³ - Edexcel - A-Level Maths Pure - Question 2 - 2006 - Paper 6

From the volume of the cube, we have:

$V = x^3$

Taking the derivative gives:

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

Substituting for ( \frac{dx}{dt} ):

$\frac{dV}{dt} = 3x^2 \left( \frac{2}{3x} \right) = 2x$

Also, since ( V = x^3 ), we can express this as:

( x = V^{\frac{1}{3}} ), substituting into our equation results in:

$\frac{dV}{dt} = 2V^{\frac{1}{3}}$

To solve the differential equation:

$\int \frac{1}{V^{\frac{2}{3}}} dV = \int 2 dt$

This gives:

$3V^{\frac{1}{3}} = 2t + c$

Using the condition that V = 8 when t = 0, we find:

$3(8)^{\frac{1}{3}} = c \Rightarrow 3 \times 2 = c \Rightarrow c = 6$

Thus,

$3V^{\frac{1}{3}} = 2t + 6$

Next, we need to find t when V = 8:

$3(8)^{\frac{1}{3}} = 2t + 6 \Rightarrow 6 = 2t + 6 \Rightarrow 2t = 0 \Rightarrow t = 0$

Now, for V = 16/2 = 8:

$3(8)^{\frac{1}{3}} = 2t + 6 \Rightarrow 6 = 2t + 6 \Rightarrow t = 3$