Question #103911
A spherical snowball is melting in such a way that its diameter is decreasing at rate of 0.4 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 13 cm. (Note the answer is a positive number).
1
Expert's answer
2020-03-02T09:47:55-0500

The volume of the snowball is given by: V=π6D3V = \dfrac{\pi}{6}D^3\\ .

The rate of volume change: dVdt=dVdDdDdt=π2D2dDdt\dfrac{dV}{dt} = \dfrac{dV}{dD}\cdot\dfrac{dD}{dt}=\dfrac{\pi}{2}D^2\dfrac{dD}{dt}\\ .

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}.

dVdt=\dfrac{dV}{dt} =  π21320.4106cm3min\dfrac{\pi}{2}\cdot13^2\cdot 0.4 \approx 106 \dfrac{cm^3}{min}

Answer: dVdt=106cm3min\dfrac{dV}{dt} = 106 \dfrac{cm^3}{min}


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