The rate at which the volume is increasing is [tex]10,048cm^3/s[/tex]
The formula for calculating the volume of a sphere is expressed as:
[tex]V=\frac{4}{3} \pi r^3[/tex]
r is the radius of the sphere.
The rate at which the volume is increasing is expressed as:
[tex]\frac{dV}{dt} =\frac{dV}{dr} \cdot \frac{dr}{dt} \\ \frac{dV}{dt} =4\pi r^2 \cdot \frac{dr}{dt}[/tex]
Given the following parameters:
[tex]r =\frac{d}{2} =20mm\\\rac{dr}{dt} = 2mm/s[/tex]
Substitute into the formula:
[tex]\frac{dV}{dt} =4\pi r^2 \cdot \frac{dr}{dt}\\ \frac{dV}{dt} =4(3.14)(20)^2 \cdot 2\\ \frac{dV}{dt} =10,048cm^3/s[/tex]
Hence the rate at which the volume is increasing is [tex]10,048cm^3/s[/tex]
Learn more on rate of change here: https://brainly.com/question/8728504
