Question

A spherical balloon is inflating with helium at a rate of 72 ft2/min . How fast is the​ balloon's radius increasing at the instant the radius is 3 ​ft? How fast is the surface area​ increasing?

Answer 1

The volume of the balloon is given by:

V = 4πr³/3

V = volume, r = radius

Differentiate both sides with respect to time t:

dV/dt = 4πr²(dr/dt)

Isolate dr/dt:

dr/dt = (dV/dt)/(4πr²)

Given values:

dV/dt = 72ft³/min

r = 3ft

Plug in and solve for dr/dt:

dr/dt = 72/(4π(3)²)

dr/dt = 0.64ft/min

The radius is increasing at a rate of 0.64ft/min

The surface area of the balloon is given by:

A = 4πr²

A = surface area, r = radius

Differentiate both sides with respect to time t:

dA/dt = 8πr(dr/dt)

Given values:

r = 3ft

dr/dt = 0.64ft/min

Plug in and solve for dA/dt:

dA/dt = 8π(3)(0.64)

dA/dt = 48.25ft²/min

The surface area is changing at a rate of 48.25ft²/min