Answer to Question #103911 in Trigonometry for Martin

Question

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).

Answer 1

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

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

From the problem's statement: $\dfrac{dD}{dt} = 0.4\dfrac{cm}{min}, \space D = 13 \space \text{cm}.$

$\dfrac{dV}{dt} =$  $\dfrac{\pi}{2}\cdot13^2\cdot 0.4 \approx 106 \dfrac{cm^3}{min}$

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