# 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?

use defferentiation in math to solve

### 3 Answers

Relevance

- Pramod KumarLv 72 months ago
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

- billrussell42Lv 72 months ago
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

Still have questions? Get your answers by asking now.