Question

homework6: problem 1
(3 points)
a spherical balloon is to be deflated so that its radius decreases at a constant rate of 13 cm/min. at what rate must air be removed when the radius is 7 cm?
air must be removed at cm³/min.
preview my answers submit answers
you have attempted this problem 0 times.
you have unlimited attempts remaining.
email instructor
page generated october 19, 2025, 8:16:40 pm cdt
webwork © 1996 - 2024 | theme: math4 - ttu | ww_version 2.19 | pg_version 2.19
the webwork project

Explanation:

Step1: Recall the volume formula for a sphere

The volume \( V \) of a sphere is given by \( V=\frac{4}{3}\pi r^{3}\), where \( r \) is the radius.

Step2: Differentiate the volume formula with respect to time \( t \)

Using the chain - rule \(\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}\).
First, find \(\frac{dV}{dr}\): \(\frac{dV}{dr}=\frac{d}{dr}(\frac{4}{3}\pi r^{3}) = 4\pi r^{2}\).
We are given that \(\frac{dr}{dt}=- 13\) cm/min (negative because the radius is decreasing).

Step3: Substitute the values of \( r \) and \(\frac{dr}{dt}\) into the derivative formula

When \( r = 7\) cm, \(\frac{dV}{dt}=4\pi r^{2}\cdot\frac{dr}{dt}\).
Substitute \( r = 7\) and \(\frac{dr}{dt}=-13\):
\(\frac{dV}{dt}=4\pi\times(7)^{2}\times(-13)\)
\(=4\pi\times49\times(-13)\)
\(=-2548\pi\)

Answer:

\(-2548\pi\approx - 8007.26\) \(cm^{3}/min\). The rate at which air is removed is \(2548\pi\approx8007.26\) \(cm^{3}/min\) (we take the magnitude since rate of removal is a positive quantity representing the speed of volume decrease).