Question

A spherical balloon is inflated at the rate of 1 cm^3 per minuter. At the instant when the radius r=1.5,
(a.) how fast is the radius increasing?
(b.) how fast is the surface area increasing?

Answer 1

V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
given : when r = 1.5 , dV/dt = 1
find dr/dt

for b),
A = 4πr^2
dA = 8πr dr/dt , you have dr/dt from a)
just plug in the given stuff

Answer 2

So dr/dt is equals to 1/9pi? or i have miscalculated again?

Answer 3

Is dA=4/3?

Answer 4

dV/dt = 4π r^2 dr/dt
1 = 4π(2.25) dr/dt
dr/dt = 1/(9π) <----- nice, you had that

dA = 8πr dr/dt
= 8π(1/(9π) = 8/9

you had 4/3, looks like a "sloppy" error.