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

Answer:

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

Step-by-step explanation:

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

Differentiating with respect to time,

                $\frac{dV}{dt}=\frac{d}{dt}\left (\frac{1}{6}\pi D^3 \right )=\frac{1}{6}\pi \times 3D^2\frac{dD}{dt}\\\\\frac{dV}{dt}=\frac{1}{2}\pi D^2\frac{dD}{dt}$

Substituting

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

             $\frac{dV}{dt}=\frac{1}{2}\pi D^2\frac{dD}{dt}\\\\\frac{dV}{dt}=\frac{1}{2}\pi \times 12^2\times 0.4=90.48cm^3/min$

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