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

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  • 2 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

  • Dixon
    Lv 7
    2 months ago

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

  • 2 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

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