Air is being pumped into a balloon that is a perfect sphere at a rate of 5 cm^3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 25 cm.

Answer 1

Given the rate of change of volume with respect to time or dv/dt = 5cm³/min

Find the rate of change of the radius or dr/dt when the diameter of the sphere is 25cm or the radius is (25/2)cm.

Start with v = (4/3)πr³ and find the derivative with respect to time. Remember that (4/3)π is a constant and differentiate the r³ with respect to time (3r²dr/dt).

So, dv/dt = (4/3)π(3r²dr/dt) Now replace dv/dt with 5cm³/min and r with (25/2)cm or r² = (625/4)cm²

5cm³/min = (4/3)π(3^.(625/4)cm^{2 .} dr/dt) Now solve for dr/dt leaving out the units till the end

dr/dt = (5^.3)/(2π ^. 625) ^. cm³/(cm^{2 .} min)

dr/dt = (3/250π) cm/min