Question

A spherical balloon is inflated with helium at the rate of 100pie ft^3/min/
a) how fast is the balloon's radius increasing at the interest the radius is 5 feet?
b) how fast is the surface area increasing at that instant.

Answer 1

Given dV/dt = 100pi

Volume of sphere, V= 4/3 pi r^3
dV/dr = 4 pi r^2

To get the rate of increase of the radius when radius is 5 ft,
note that, dr/dt = (dV/dt) * (dr/dV) multiplying it like this, the dV gets cancelled, giving you the dr/dt which is what you want.

dr/dt = 100pi / 4pi r^2 = 25/ r^2

Substitute r= 5ft,
dr/dt = 25/ 25= 1 ft/min

Rate of increase of the surface area when radius is 5ft,
Surface area, A =4pi r^2
dA/dr = 8pi r

Same as the above,
To get dA/dt = dA/dr * dr/dt
dA/dt = 8 pi r * 1
= 8 pi r

Substitute r = 5 ft.

dA/dt = 40pi ft^2 /min

Answer 2

@tiaph can you help me on one like this