A spherical snowball melts so that its radius decreases at a rate of 4 in/sec. At what rate is the volume of the snowball changing when the radius is 8 in?

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Answer 1 · 2016-09-01T15:04:47

The formula for volume of a sphere is #V = 4/3r^3pi#.

Differentiating with respect to #t#, time.

#(dV)/dt = 4r^2((dr)/dt)#

The rate of change of the snowball is given by #(dV)/dt#. We know #(dr)/dt = -4#. We want to find the rate of change when #r= 8#. Hence,

#(dV)/dt = 4(8)^2(-4)#

#(dV)/dt = 4(64)(-4)#

#(dV)/dt = -1024#

Thus, the volume of the snowball is decreasing at a rate of #-1024" in"^3/"sec"#.

Hopefully this helps!