To determine the rate at which the volume of a spherical snowball is decreasing when its radius is 16 cm, given that the radius is decreasing at a rate of [tex]\(0.3 \text{ cm/min}\)[/tex], we can use the formula for the volume of a sphere and apply the chain rule from calculus.
1. Volume of a Sphere: The volume [tex]\(V\)[/tex] of a sphere with radius [tex]\(r\)[/tex] is given by:
[tex]\[
V = \frac{4}{3} \pi r^3
\][/tex]
2. Rate of Change of Volume: To find the rate at which the volume is changing with respect to time, we need to take the derivative of the volume with respect to time ([tex]\(\frac{dV}{dt}\)[/tex]). Using the chain rule, we get:
[tex]\[
\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}
\][/tex]
3. Derivative of Volume with Respect to Radius: First, we find [tex]\( \frac{dV}{dr} \)[/tex] by differentiating [tex]\( V \)[/tex] with respect to [tex]\( r \)[/tex]:
[tex]\[
V = \frac{4}{3} \pi r^3
\][/tex]
[tex]\[
\frac{dV}{dr} = 4 \pi r^2
\][/tex]
4. Given Rate of Change of Radius: We are given that:
[tex]\[
\frac{dr}{dt} = -0.3 \, \frac{\text{cm}}{\text{min}}
\][/tex]
(The negative sign indicates that the radius is decreasing.)
5. Substitute the Known Values: Now, we substitute [tex]\( r = 16 \, \text{cm} \)[/tex] into [tex]\( \frac{dV}{dr} \)[/tex]:
[tex]\[
\frac{dV}{dr} = 4 \pi (16)^2
\][/tex]
(We do not explicitly calculate [tex]\( 4 \pi (16)^2 \)[/tex] here, assuming it has already been calculated.)
6. Calculate the Rate of Change of Volume: Finally, substitute [tex]\(\frac{dV}{dr}\)[/tex] and [tex]\(\frac{dr}{dt}\)[/tex] into the rate of change of volume formula:
[tex]\[
\frac{dV}{dt} = 4 \pi (16)^2 \cdot (-0.3)
\][/tex]
The numeric result obtained for [tex]\( 4 \pi (16)^2 \)[/tex] is approximately [tex]\( 3216.99 \)[/tex]:
[tex]\[
\frac{dV}{dt} \approx 3216.99 \cdot -0.3
\][/tex]
[tex]\[
\frac{dV}{dt} \approx -965.10 \, \frac{\text{cm}^3}{\text{min}}
\][/tex]
Since the problem asks for the rate of decrease as a positive number:
[tex]\[
\left| \frac{dV}{dt} \right| \approx 965.10 \, \frac{\text{cm}^3}{\text{min}}
\][/tex]
Therefore, the volume of the snowball is decreasing at a rate of approximately [tex]\( 965.10 \, \frac{\text{cm}^3}{\text{min}} \)[/tex] when its radius is 16 cm.
