volume of a spherical balloon is increasing at rate of 3 in/s. At what rate the radius of the balloon increasing when the radius is 2 in?

Question

use defferentiation in math to solve

Answer 1

Sphere V = ⁴/₃πr³

=> Differentiate with respect to t

=> dV/dt =  (4πr²) * dr/dt 

But  dV/dt  =  3  cub in/s

=> 3  =  (4πr²) * dr/dt 

=> dr/dt  =  3/(4πr²)

when r  =  2 "

=> dr/dt  =  3/16r  =  0.0596831 in/s ................. Answer

Answer 2

3 in/s is not a rate of increase of volume

Answer 3

Sphere V = ⁴/₃πr³

dV/dr = 4πr²

dV/dt = 3

dr/dt = dV/dt / dV/dr = 3/(4πr²)

for r = 2

dr/dt = 3/(4π2²) = 3/(16π) = 0.0597 in/s