Question A spherical balloon is inflated with helium at a rate of 140pift^3/min How fast is the balloon’s radius increasing when the radius is 7ft

Answer: dr = 0.71 ft/min the balloon’s radius is increasing at 0.71 ft/min when the radius is 7ft. Explanation: Given; Rate of inflation dV = 140pift^3/min Radius r = 7 ft Change in radius = dr Volume of a spherical balloon is; V = (4/3)πr^3 The change in volume can be derived by differentiating both sides; dV = (4πr^2)dr Making dr the subject of formula; dr = dV/(4πr^2) Substituting the given values; dr = 140π/(4π×7^2) dr = 0.714285714285 ft/min dr = 0.71 ft/min the balloon’s radius is increasing at 0.71 ft/min when the radius is 7ft. Log in to Reply

Answer:

dr = 0.71 ft/min

the balloon’s radius is increasing at 0.71 ft/min when the radius is 7ft.

Explanation:

Given;

Rate of inflation dV = 140pift^3/min

Radius r = 7 ft

Change in radius = dr

Volume of a spherical balloon is;

V = (4/3)πr^3

The change in volume can be derived by differentiating both sides;

dV = (4πr^2)dr

Making dr the subject of formula;

dr = dV/(4πr^2)

Substituting the given values;

dr = 140π/(4π×7^2)

dr = 0.714285714285 ft/min

dr = 0.71 ft/min

the balloon’s radius is increasing at 0.71 ft/min when the radius is 7ft.