Answer:
The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.
Step-by-step explanation:
The volume of the sphere ([tex]V[/tex]), measured in cubic feet, is represented by the following formula:
[tex]V = \frac{4\pi}{3}\cdot r^{3}[/tex]
Where [tex]r[/tex] is the radius of the sphere, measured in feet.
The change in volume is obtained by means of definition of total difference:
[tex]\Delta V = \frac{\partial V}{\partial r}\Delta r[/tex]
The derivative of the volume as a function of radius is:
[tex]\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}[/tex]
Then, the change in volume is expanded:
[tex]\Delta V = 4\pi \cdot r^{2}\cdot \Delta r[/tex]
If [tex]r = 40\,ft[/tex] and [tex]\Delta r = 40\,ft-40.05\,ft = 0.05\,ft[/tex], the change in the volume of the sphere is approximately:
[tex]\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)[/tex]
[tex]\Delta V \approx 1005.310\,ft^{3}[/tex]
The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.
