At time $t$ seconds the length of the side of a cube is $x$ cm, the surface area of the cube is $S$ cm$^2$, and the volume of the cube is $V$ cm$^3$ - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 6

The volume of a cube is given by:

$V = x^3$

2. Differentiating both sides yields:

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

3. Substituting ( \frac{dx}{dt} ) from part (a):

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

4. Using the relation ( x = V^{\frac{1}{3}} ), we have:

$\frac{dV}{dt} = 2V^{\frac{1}{3}}$. This confirms the equation.

From part (b), we have:

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

2. Integrating both sides results in:

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

3. The left-hand side integrates to ( \frac{3}{2} V^{\frac{2}{3}} + C ) and the right-hand side integrates to ( 2t + C' ), giving:

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

4. Substituting ( V = 8 ) when ( t = 0 ):

$\frac{3}{2} (8)^{\frac{2}{3}} = 0 + C \Rightarrow C = 8$

5. Therefore,

$\frac{3}{2} V^{\frac{2}{3}} = 2t + 8$

6. Now setting ( V = 8 ):

$\frac{3}{2}(8)^{\frac{2}{3}} = 2t + 8 \Rightarrow 12 = 2t + 8$

7. Thus,

$2t = 4 \Rightarrow t = 2$

8. Finally, for ( V = 16 ):

$\frac{3}{2} (16)^{\frac{2}{3}} = 2t + 8$

9. Simplifying gives:

$$$$ 12 = 2t + 8 \Rightarrow 2t = 4 \Rightarrow t = 2 $$

10. For ( V = 8 ):

It gives the same time value at ( t = 2 ).