A spherical balloon is inflating with helium at a rate of 64π ft/min. How fast is the balloon's radius increasing at the instant the radius is 2 ft? How fast is the surface area increasing? The balloon's radius is increasing at a rate of...

let r ,v be the radius and volume of sphere.

dv/dt=64πft³/min

r=2ft

Dr/dt=?

Step-by-step explanation:

we have

volume of sphere=4/3πr²

v=4/3πr²

differentiating both side with respect to t we get

dv/dt=4πdr²/dr×dr/dt=4π×2r×dr/dt=8/3×πr×dr/dt

substituting the given value we get

dv/dt=8/3×πr×dr/DT

64π×3/(8π×2)=dr/dt

Dr/dt=12ft/min

The balloon's radius is increasing at a rate of12 ft/min

answer:

step-by-step explanation: