A spherical balloon is being inflated by a compressor that is pumping air at a rate of $4.5 \mathrm{ft}^{3} / \mathrm{min}$. At what rate is the radius of the balloon increasing when the radius is $1.5 \mathrm{ft}$ ? Round to three decimal places.
\[
\frac{d r}{d t}=
\]

Step 1 ：We are given a problem of related rates in calculus. The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^3\). The rate at which the volume is changing, \(\frac{dV}{dt}\), is given as 4.5 ft³/min. We are asked to find the rate at which the radius is changing, \(\frac{dr}{dt}\), when the radius, r, is 1.5 ft.

Step 2 ：We can use the chain rule to differentiate the volume with respect to time. This will give us an equation involving \(\frac{dr}{dt}\).

Step 3 ：Taking the derivative of the volume with respect to r, we get \(\frac{dV}{dr} = 4.0\pi r^2\).

Step 4 ：We can then solve this equation for \(\frac{dr}{dt}\) when r = 1.5 ft. Doing so, we find that \(\frac{dr}{dt} = 0.159\).

Step 5 ：Final Answer: The rate at which the radius of the balloon is increasing when the radius is 1.5 ft is \(\boxed{0.159 \, \text{ft/min}}\).