Answer:
a)
[tex]\displaystyle\frac{dV}{dr} = 50.24~ft^3/ft[/tex]
b)
[tex]dV = 10.048\text{ cubic feet}[/tex]
Step-by-step explanation:
We are given the following information in the question:
Volume of sphere =
[tex]V = \displaystyle\frac{4}{3}\pi r^3[/tex]
a) We have to find the rate at which the volume change with respect to the radius when r = 2 ft.
Differentiating, we get,
[tex]\displaystyle\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3}\pi r^3) = 4\pi r^2[/tex]
Putting r = 2
[tex]\displaystyle\frac{dV}{dr} = 4\pi (2)^2 = 50.24~ft^3/ft[/tex]
b) Approximate change in volume when radius changes from 2 to 2.2 feet
[tex]\displaystyle\frac{dV}{dr} = 4\pi r^2\\dV = 4\pi r^2~dr[/tex]
[tex]dr = 2.2-2 = 0.2~ft[/tex]
Putting the values, we get:
[tex]dV = 4\pi (2)^2(0.2) = 10.048\text{ cubic feet}[/tex]
