Question

1. a spherical hot air balloon is being inflated. air is blown into the balloon at the rate of 2 ft³/sec.

a.) how fast is the radius of the balloon changing when the radius is 3 feet?
b.) how fast is the surface area of the balloon changing at the same time?

Explanation:

Step1: Recall the volume formula for a sphere

The volume formula for a sphere is \(V=\frac{4}{3}\pi r^{3}\). Differentiate both sides with respect to time \(t\):
\(\frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}\)
We know that \(\frac{dV}{dt}=2\) (the rate at which air is blown into the balloon) and \(r = 3\) (the radius of the balloon at the given instant).
Substitute the values into the equation:
\(2=4\pi\times(3)^{2}\times\frac{dr}{dt}\)

Step2: Solve for \(\frac{dr}{dt}\)

First, simplify the right - hand side of the equation: \(4\pi\times9\times\frac{dr}{dt}=36\pi\frac{dr}{dt}\)
Then, solve for \(\frac{dr}{dt}\): \(\frac{dr}{dt}=\frac{2}{36\pi}=\frac{1}{18\pi}\) feet per second.

Step3: Recall the surface - area formula for a sphere

The surface - area formula for a sphere is \(A = 4\pi r^{2}\). Differentiate both sides with respect to time \(t\):
\(\frac{dA}{dt}=8\pi r\frac{dr}{dt}\)
We already know that \(r = 3\) and \(\frac{dr}{dt}=\frac{1}{18\pi}\)
Substitute the values into the equation: \(\frac{dA}{dt}=8\pi\times3\times\frac{1}{18\pi}\)

Step4: Solve for \(\frac{dA}{dt}\)

Simplify the right - hand side of the equation:
\(\frac{dA}{dt}=\frac{24\pi}{18\pi}=\frac{4}{3}\) square feet per second

Answer:

a. The radius is changing at a rate of \(\frac{1}{18\pi}\approx0.0177\) feet per second.
b. The surface area is changing at a rate of \(\frac{4}{3}\approx1.33\) square feet per second.