homework6: problem 1\n(3 points)\na spherical balloon is to be deflated so that its radius decreases at a…

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).