It’s either movement or work or that’s what a quizzie said
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
