Time To Inflate A Spherical Hot-Air Balloon To Two-Thirds Volume

Introduction

In the realm of mathematical problem-solving, real-world applications often present intriguing scenarios that require a blend of geometric principles and calculus concepts. One such scenario involves the inflation of a spherical hot-air balloon, where we seek to determine the time it takes to reach a specific fraction of its maximum volume. This article delves into the mathematical intricacies of this problem, providing a step-by-step solution and exploring the underlying concepts.

Problem Statement

Consider a spherical hot-air balloon with a diameter of 55 feet. As the balloon inflates, its radius increases at a rate of 5 feet per minute. Our objective is to determine the approximate time it takes for the balloon to reach ${\frac{2}{3}}$ of its maximum volume.

Mathematical Formulation

To tackle this problem, we'll employ the formula for the volume of a sphere, which is given by:

${ V = \frac{4}{3} \pi r^3 }$

where:

• V represents the volume of the sphere
• r denotes the radius of the sphere

Since the diameter of the balloon is 55 feet, its maximum radius (${ r_{max} }$) is half of the diameter, which is 27.5 feet.

The maximum volume (${ V_{max} }$) of the balloon can be calculated as:

${ V_{max} = \frac{4}{3} \pi (27.5)^3 \approx 87119.12 \text{ cubic feet} }$

We want to find the time it takes for the balloon to reach ${\frac{2}{3}}$ of its maximum volume, which is:

${ \frac{2}{3} V_{max} = \frac{2}{3} \times 87119.12 \approx 58079.41 \text{ cubic feet} }$

Now, let's denote the radius of the balloon at any time t as r(t). Since the radius increases at a rate of 5 feet per minute, we have:

${ \frac{dr}{dt} = 5 \text{ feet per minute} }$

Integrating both sides with respect to time, we get:

${ r(t) = 5t + r_0 }$

where ${ r_0 }$ is the initial radius of the balloon. Assuming the balloon starts with a negligible initial radius, we can take ${ r_0 = 0 }$, so:

${ r(t) = 5t }$

The volume of the balloon at time t, V(t), is:

${ V(t) = \frac{4}{3} \pi (5t)^3 = \frac{500}{3} \pi t^3 }$

We want to find the time t when V(t) equals ${\frac{2}{3} V_{max}}$, so:

${ \frac{500}{3} \pi t^3 = 58079.41 }$

Solving for Time

Let's solve the equation for t:

${ t^3 = \frac{3 \times 58079.41}{500 \pi} \approx 110.92 }$

Taking the cube root of both sides, we get:

${ t = \sqrt[3]{110.92} \approx 4.81 \text{ minutes} }$

Therefore, it takes approximately 4.81 minutes to inflate the balloon to ${\frac{2}{3}}$ of its maximum volume.

Step-by-Step Solution

1. Determine the maximum radius: The diameter is 55 feet, so the maximum radius (${r_{max}}$) is ${ \frac{55}{2} = 27.5 }$ feet.
2. Calculate the maximum volume: Use the formula ${ V_{max} = \frac{4}{3} \pi r_{max}^3 }$ to find the maximum volume: ${ V_{max} = \frac{4}{3} \pi (27.5)^3 \approx 87119.12 \text{ cubic feet} }$
3. Calculate ${\frac{2}{3}}$ of the maximum volume: ${ \frac{2}{3} V_{max} = \frac{2}{3} \times 87119.12 \approx 58079.41 \text{ cubic feet} }$
4. Express the radius as a function of time: Since the radius increases at 5 feet per minute, ${ r(t) = 5t }$.
5. Express the volume as a function of time: ${ V(t) = \frac{4}{3} \pi (5t)^3 = \frac{500}{3} \pi t^3 }$
6. Set ${V(t)}$ equal to ${\frac{2}{3} V_{max}}$ and solve for t: ${ \frac{500}{3} \pi t^3 = 58079.41 }$ ${ t^3 = \frac{3 \times 58079.41}{500 \pi} \approx 110.92 }$ ${ t = \sqrt[3]{110.92} \approx 4.81 \text{ minutes} }$

Alternative Approach

Another way to approach this problem is to directly set up the equation relating the volume at time t to ${\frac{2}{3}}$ of the maximum volume. We have:

${ V(t) = \frac{2}{3} V_{max} }$

Substituting the expressions for ${V(t)}$ and ${V_{max}}$, we get:

${ \frac{4}{3} \pi (5t)^3 = \frac{2}{3} \left( \frac{4}{3} \pi (27.5)^3 \right) }$

Simplifying, we have:

${ (5t)^3 = \frac{2}{3} (27.5)^3 }$

Taking the cube root of both sides:

${ 5t = 27.5 \sqrt[3]{\frac{2}{3}} }$

${ t = \frac{27.5}{5} \sqrt[3]{\frac{2}{3}} \approx 4.81 \text{ minutes} }$

This approach yields the same result, providing an alternative perspective on the problem.

Conclusion

In this article, we've explored the mathematical problem of determining the time it takes to inflate a spherical hot-air balloon to ${\frac{2}{3}}$ of its maximum volume. By applying the formula for the volume of a sphere and considering the rate of change of the radius, we formulated an equation and solved for the time. Both the step-by-step solution and the alternative approach demonstrate the power of mathematical principles in solving real-world problems. The result, approximately 4.81 minutes, showcases the practical application of geometric and calculus concepts.

Keywords Repair

Original Keyword: Approximately how long does it take to inflate the balloon to ${\frac{2}{3}}$ of its maximum volume?

Repaired Keyword: How much time is needed to inflate a spherical balloon to two-thirds of its maximum capacity given the radius increases at a rate of 5 feet per minute?