The volume of the snowball is given by: V=π6D3V = \dfrac{\pi}{6}D^3\\V=6πD3 .
The rate of volume change: dVdt=dVdD⋅dDdt=π2D2dDdt\dfrac{dV}{dt} = \dfrac{dV}{dD}\cdot\dfrac{dD}{dt}=\dfrac{\pi}{2}D^2\dfrac{dD}{dt}\\dtdV=dDdV⋅dtdD=2πD2dtdD .
From the problem's statement: dDdt=0.4cmmin, D=13 cm.\dfrac{dD}{dt} = 0.4\dfrac{cm}{min}, \space D = 13 \space \text{cm}.dtdD=0.4mincm, D=13 cm.
dVdt=\dfrac{dV}{dt} =dtdV= π2⋅132⋅0.4≈106cm3min\dfrac{\pi}{2}\cdot13^2\cdot 0.4 \approx 106 \dfrac{cm^3}{min}2π⋅132⋅0.4≈106mincm3
Answer: dVdt=106cm3min\dfrac{dV}{dt} = 106 \dfrac{cm^3}{min}dtdV=106mincm3
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