Question

(1 pt) 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 12 cm. (Note the answer is a positive number).

Answer 1

Rate at which volume of the snowball decreasing = 90.48 cm³/min

Explanation:

`\texttt{Volume of sphere, V =}(4)/(3)\pi r^3=(4)/(3)\pi \left ((D)/(2) \right )^3=(1)/(6)\pi D^3`

Differentiating with respect to time,

`(dV)/(dt)=(d)/(dt)\left ((1)/(6)\pi D^3 \right )=(1)/(6)\pi * 3D^2(dD)/(dt)\\\\(dV)/(dt)=(1)/(2)\pi D^2(dD)/(dt)`

Substituting

D = 12 cm and rate of change of diameter = 0.4 cm/min.

`(dV)/(dt)=(1)/(2)\pi D^2(dD)/(dt)\\\\(dV)/(dt)=(1)/(2)\pi * 12^2* 0.4=90.48cm^3/min`

Rate at which volume of the snowball decreasing = 90.48 cm³/min