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Calculus about volume of sphere rate of change can someone help me?

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A ball made of ice is melting at a rate of 2ft^3 per hour. if it remains spherical, at what rate is the radius changing when the radius of the ball is 20 inches?

asked Dec 29, 2012 in CALCULUS by linda Scholar

1 Answer

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The ball made of ice is melting at at rate      2 ft^3

                                                  So             dV/dt =  -2

Volume of the sphere is      V = 4/3× × r^3

                                        dV/dt = 4/3 × ×3 × r^2 × dr/dt

                               Since    r = 20 inches

                                              = 20/12 = 5/3 ft

                   Now              -2 = 4/3 × ×3 × (5/3)^2 × dr/dt

                                         -2 = 4× 25 /9 × dr/dt

                                          (-2/4)× 9/25 = dr/dt

                                           dr/dt = - 9/50×

So, the change in radius is     -9/50×∏    feet per hour

answered Dec 29, 2012 by Johncena Apprentice

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