Question

A spherical balloon is inflated so that its volume is increasing at the rate of 3 ft 3/min. How fast is the radius of the balloon increasing when the radius is 1 ft?

Answer 1

Answer 2

SOLUTION :

Volume of ballon, V = 4/3 π r^3  

Differentiating w.r.t. t :

=> dV/dt = 4/3 π * 3 r^2 dr/dt 

dV/dt at r = 1 ft. Is 3 ft^3 / min 

=> 3 = 4/3 π  * 3 * (1)^2  dr/dt 

=> 3 = 4 π dr/dt 

=> dr/dt = 3/(4 π)  

=> dr/dt = 0.2387 ft / min 

So, radius is increasing at the rate of 0.2387 ft / min when radius is 1 ft . (ANSWER).