Problem 48 - Calculus

Question

The functions \(r=f(t)\) and \(V=g(r)\) give the radius and the volume of a commercial hot air balloon being inflated for testing. The variable \(t\) is in minutes, \(r\) is in feet, and \(V\) is in cubic feet. The inflation begins at \(t=0 .\) In each case, give a mathematical expression that represents the given statement. The volume of the balloon \(t\) minutes after inflation began.

Step-by-Step Solution

Answer

The volume at time \( t \) is \( V = g(f(t)) \).

Step 1: Understand the Functions

We are given two functions: \( r = f(t) \) which represents the radius of the balloon in feet at time \( t \) in minutes, and \( V = g(r) \) which represents the volume of the balloon in cubic feet in terms of its radius \( r \). Our task is to determine the volume in terms of \( t \).

Step 2: Compose the Functions

To find the volume in terms of \( t \), we need to combine the two functions that we have. Specifically, we substitute the first function into the second: \( V = g(f(t)) \). This composition will give us the volume as a function of time.

Step 3: Formulate the Expression

The expression \( V = g(f(t)) \) represents the volume of the balloon \( t \) minutes after the inflation began. This expression combines both the change in radius over time and the dependency of volume on radius.

Key Concepts

Volume Calculation

Understanding the concept of volume calculation is essential, especially when dealing with changing physical properties like those of a balloon being inflated. The volume of an object tells us how much space it occupies. In this exercise, the volume is being calculated based on the radius of a hot air balloon that changes over time as it is inflated.

The main idea here is to understand that the volume of a sphere (similar in shape to a balloon) is calculated using the formula: \[V = \frac{4}{3} \pi r^3\]
where \( r \) is the radius. When inflation occurs, the radius is not constant, hence affecting the volume calculation. As the radius increases over time, so does the volume. This changing volume must be represented in terms of a mathematical expression that accounts for the dependency on the time-variable radius.

Mathematical Expression

To express the volume of the balloon in a way that captures its dependency on time, we need to work with mathematical expressions. This involves functions, which are an essential part of capturing relationships between different variables.

In this case, the function compositions are the key. We have:

• Function 1: \( r = f(t) \), showing how radius \( r \) changes with time \( t \).
• Function 2: \( V = g(r) \), representing how volume \( V \) depends on radius \( r \).

To find the composition, we substitute one function into another, forming \( V = g(f(t)) \). This final expression allows us to determine the volume directly based on time, capturing both time and spatial changes in one concise formula.

Time Dependency

Time dependency in mathematical functions speaks to how a certain property changes as time progresses. It's crucial in situations where parameters are not static but evolve, such as during the inflation of a balloon. In our exercise, the radius changes with time, making the volume time-dependent.

To understand how this works, consider the nature of the function \( f(t) \). This function provides the radius at any given time \( t \), capturing how the physical dimension of the balloon grows.

When creating the composite function \( V = g(f(t)) \), we see this time dependency explicitly. The function \( g \) applies the relationship of volume to radius, while \( f(t) \) provides how radius varies over time, thus effectively linking everything back to time. Understanding this aids in comprehending both the immediate and cumulative effects of changes over time on the volume.