The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.

Let the surface area of the balloon be S.

S = 4πr2


According to the question,





8πr


8πrdr = ktdt


Integrating both sides, we have


8π∫rdr = k∫tdt



……(1)


Given, we have r = 1 unit when the t = 0 sec


Putting the value in equation (1)



4π (1)2 = k × 0 + c


c = 4π ……(2)


Putting the value of c in equation (1) we have,


……(3)


Given, we have r = 2 units when t = 3 sec





……(4)


Now, putting the value of k in equation (2),


We have,








1