A spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2cm/min. At what rates are the volume and surface area of the balloon increasing when the radius is 5cm?

Answer 1

V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
when r = 5 and dr/dt = 2
dV/dt = 4π(25)(2) = 200π cm^3/min

SA = 4πr^2
d(SA)/dt = 8πr dr/dt
= 8π(5)(2) = 80π cm^2/min

Answer 2

The radius of a circle is increase at a 3 c.m/min. Find the rate of change of the area when
A)r=2c.m B)r=3c.m