Explanation:
We have volume of sphere given by
[tex]V=\frac{4}{3}\pi r^3[/tex]
Differentiating with respect to time
[tex]\frac{dV}{dt}=\frac{4}{3}\pi \times\frac{dr^3}{dt}\\\\\frac{dV}{dt}=4\pi r^2\times\frac{dr}{dt}[/tex]
Given that
[tex]r=10cm\\\\\frac{dV}{dt}=-8cm^3/hr[/tex]
Substituting
[tex]\frac{dV}{dt}=4\pi r^2\times\frac{dr}{dt}\\\\-8=4\pi \times 10^2\times\frac{dr}{dt}\\\\\frac{dr}{dt}=-6.37\times 10^{-3}cm/hr[/tex]
The radius of the snowball is decreasing at a rate 6.37 x 10⁻³ cm/hr
