Using implicit differentiation, it is found that the radius is increasing at a rate of 0.0081 cm per minute.
The volume of a sphere of radius r is given by:
[tex]V = \frac{4\pi r^3}{3}[/tex]
Applying implicit differentiation, the rate of change is given by:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
In this problem, we have that:
[tex]\frac{dV}{dt} = 500, r = 70[/tex]
Hence the rate of change of the radius is given as follows:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
[tex]19600\pi\frac{dr}{dt} = 500[/tex]
[tex]\frac{dr}{dt} = \frac{500}{19600\pi}[/tex]
[tex]\frac{dr}{dt} = 0.0081[/tex]
The radius is increasing at a rate of 0.0081 cm per minute.
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