Rates of Change

Question

A spherical balloon is released from rest and expands as it rises.After rising for t seconds, its radius is r cm and its surface area is A cm^2,where A=4πr^2.The initial radius of a balloon is 16 cm.Given that the rate of increase of the radius is constant and has the value 0.8cm s^-1, find the rate of increase of A when t=5.

 

Ans: 128π cm^2 s^-1 (please show working for me)

Answer 1

radius after 5 sec = 16cm + 5(0.8cm/s) = 20cm

 

Differentiating the formula for area wrt t, dA/dt = 8πrdr/dt

 

Substituting r = 20; dr/dt = 0.8, we get dA/dt = 8π(20)(0.8) = 128π cm²/s