The volume of an expanding cube is increasing with time. At any time 't' it is found that the rate of increase of the volume is directly proportion to the surface area of the cube. Show that the length of the side of the cube l is a linear function of time t, If l=1 when t = 0 and l = 3 when t = 1, find the value of 'l' when t = 8.

Solution:

Let's assume that the length of the side of the cube at time t is 'l(t)'. We're given that the rate of increase of the volume of the cube is directly proportional to the surface area of the cube, which means:

$\frac{{dV}}{{dt}}$ = k * A

where dV/dt is the rate of increase of the volume, k is a constant of proportionality, and A is the surface area of the cube.

The volume of the cube is given by V = l³, and its surface area is given by A = 6l². So we can rewrite the above equation as:

$\frac{{d{l^3}}}{{dt}}$ = k * 6l²

3l² $\frac{{dl}}{{dt}}$= 6kl²

Dividing both sides by 3l², we get:

$\frac{{dl}}{{dt}}$ = 2k

This tells us that the rate of increase of the length of the side of the cube is a constant, which means that the length of the side of the cube is a linear function of time t. 

Now, we have l = 1 when t = 0 and l = 3 when t = 1.

Using the equation l = mt + c, where m is the slope and c is the y-intercept, we can find the value of 'l' when t = 8.

From the given information, we have:

l = mt + c

à 1 = m(0) + c

à c = 1

And, when t = 1,

l = mt + c

à3 = m(1) + 1

à m = 2

Therefore, the equation for l is:

l = 2t + 1

When t = 8, we get:

l = 2(8) + 1 = 17

Hence, the value of 'l' when t = 8 is 17.