Air is being pumped into a spherical balloon so that its volume increases at a rate of 80cm^3/s. How fast is the surface area of the balloon increasing when its radius is 7cm?

Answer 1

V = (4/3)pi(r^3)
dV/dt = 4pi(r^2)dr/dt
when t=7
80 = 4pi(49)dr/dt
dr/dt = 80/(196pi)

SA = 4pi(r^2)
d(SA)/dt = 8pi(r)dr/dt
= 8pi(7)*80/(196pi)

etc

Answer 2

nmnm

Answer 3

22.85 cubic cm per second