Air is pumped into a spherical balloon at a rate of 4.75 ft^3 ⁄min . Find the rate of change of the radius when the diameter is 5 feet.

Answer 1

The volume of a spherical balloon is:

V = (4/3)·π·r³

To find the rate of change of the volume, take the first derivative wrt time. You'll need to apply the Chain Rule:

dV/dt = (4/3)·π·dr³/dt·(dr/dt) = (4/3)·π·3r²·(dr/dt)

dV/dt = 4·π·r²·(dr/dt)

Now dV/dt is given as 4.75 ft³/min when r = 5 ft. Plug those numbers into the above equation and solve for dr/dt, the rate of change of the radius in ft/sec

Answer 2

Air is pumped into a spherical balloon at a rate of 4.75 ft^3 ⁄min. Find the rate of change of the radius when the diameter is 5 feet.

Detailed Solution:

Given/Known: Volume of Sphere: V = (4/3)πr³, dV/dt = 4.75 ft^3 ⁄min, d = 5 feet, r = 5/2 feet = 2.5 feet, dr/dt = ?? ft ⁄min

d/dt (V = (4/3)πr³) ==> dV/dt = (4/3)π(3r²) dr/dt ==> dV/dt = 4πr² dr/dt

dV/dt = 4πr² dr/dt

4.75 = 4π(2.5)² dr/dt

4.75 = 4π(6.25) dr/dt

4.75 = 25π dr/dt

dr/dt = (4.75)/(25π) ft ⁄min <== Exact - Final Answer

dr/dt = 0.06048 ft ⁄min <== Approximation