Question

The volume V = (4/3)π r^3 of a spherical balloon changes with the radius.a. At what rate (ft^3/ft) does the volume change with respect to the radius when r = 2 ft?b. By approximately how much does the volume increase when the radius changes from 2 to 2.2 ft?

Answer 1

Answer:

a)

$\displaystyle\frac{dV}{dr} = 50.24~ft^3/ft$

b)

$dV = 10.048\text{ cubic feet}$

Step-by-step explanation:

We are given the following information in the question:

Volume of sphere =

$V = \displaystyle\frac{4}{3}\pi r^3$

a) We have to find the rate at which the volume change with respect to the radius when r = 2 ft.

Differentiating, we get,

$\displaystyle\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3}\pi r^3) = 4\pi r^2$

Putting r = 2

$\displaystyle\frac{dV}{dr} = 4\pi (2)^2 = 50.24~ft^3/ft$

b) Approximate change in volume when radius changes from 2 to 2.2 feet

$\displaystyle\frac{dV}{dr} = 4\pi r^2\\dV = 4\pi r^2~dr$

$dr = 2.2-2 = 0.2~ft$

Putting the values, we get:

$dV = 4\pi (2)^2(0.2) = 10.048\text{ cubic feet}$