Question

a spherical hot - air balloon has a diameter of 55 feet. when the balloon is inflated, the radius increases at a rate of 1.5 feet per minute. approximately how long does it take to inflate the balloon to \\( \frac { 2 } { 3 } \\) of its maximum volume? use \\( \pi = 3.14 \\) and \\( v = \frac { 4 } { 3 } \pi r ^ { 3 } \\).
16 minutes
18 minutes
23 minutes
26 minutes

Explanation:

Step1: Calculate the maximum radius

The diameter of the balloon is \(d = 55\) feet, so the maximum radius \(r_{max}=\frac{d}{2}=\frac{55}{2}=27.5\) feet.

Step2: Find the volume formula

The volume of a sphere is \(V=\frac{4}{3}\pi r^{3}\). Let the radius at time \(t\) be \(r = r_{0}+kt\), where \(r_{0} = 0\) (assuming it starts from deflated state) and \(k = 1.5\) (rate of change of radius). So \(r=1.5t\). The maximum volume \(V_{max}=\frac{4}{3}\pi r_{max}^{3}\). We want to find \(t\) when \(V=\frac{2}{3}V_{max}\).
Substitute \(V=\frac{4}{3}\pi(1.5t)^{3}\) and \(V_{max}=\frac{4}{3}\pi(27.5)^{3}\) into \(V = \frac{2}{3}V_{max}\):

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Another way:
Since \(V=\frac{4}{3}\pi r^{3}\), when \(V=\frac{2}{3}V_{max}\), we have \(\frac{4}{3}\pi r^{3}=\frac{2}{3}\times\frac{4}{3}\pi r_{max}^{3}\), then \(r^{3}=\frac{2}{3}r_{max}^{3}\), \(r = r_{max}\times(\frac{2}{3})^{\frac{1}{3}}\)
\(r_{max} = 27.5\), \(r=1.5t\)
\(1.5t=27.5\times(\frac{2}{3})^{\frac{1}{3}}\)
\((\frac{2}{3})^{\frac{1}{3}}\approx0.8736\)
\(1.5t\approx27.5\times0.8736\)
\(1.5t\approx24.024\)
\(t=\frac{24.024}{1.5}\approx16\)

Answer:

16 minutes