Problem 55 - Applied 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 if its radius were twice as big.

Step-by-Step Solution

Answer

The volume with twice the radius is given by \( V = g(2f(t)) \).

Step 1: Understand the Problem

We need to express the volume of the balloon, knowing it is given by a function \( V = g(r) \), but the radius is twice the current size \( r = f(t) \). So, the new radius will be \( 2f(t) \).

Step 2: Substitute the New Radius into the Volume Equation

The volume with the new radius becomes \( V = g(2f(t)) \). This is because we are directly substituting the expression for the doubled radius into the function \( g \) that calculates volume based on the radius.

Step 3: Present the Mathematical Expression

The mathematical expression for the volume of the balloon with a radius twice the original is \( V = g(2f(t)) \). This expression shows the dependence on the new, larger radius doubled from its original radius at time \( t \).

Key Concepts

Functions

In applied calculus, functions play a fundamental role. A function is essentially a relationship or rule that uniquely determines one output value for each valid input value. In this context, we have two primary functions:

• The first function, \( r = f(t) \), relates time \( t \) to the radius \( r \) of the balloon as it is inflated.
• The second function, \( V = g(r) \), connects the radius \( r \) of the balloon to its volume \( V \).

These functions not only help us understand the changes over time but also provide a way to model and predict the balloon's behavior at various stages of inflation. By analyzing \( f \) and \( g \), one can determine how the balloon's physical characteristics, like the radius and volume, evolve as it fills with air.

Radius

The term 'radius' refers to the distance from the center of a circle to any point on its perimeter. In our balloon problem, the radius \( r \) changes over time according to the function \( r = f(t) \). This function describes how the radius of the balloon increases as more air is added.Understanding the effect of doubling the radius is crucial. If the radius is doubled, as described by the expression \( 2f(t) \), it doesn't just change by twice; this doubling impacts other properties, such as the volume, in specific, nonlinear ways. Thus, analyzing the radius via functions helps us explore these compounded effects on the balloon's volume during inflation.

Volume

Volume measures the amount of space an object occupies, and in our exercise, it relates to how much air fits inside the hot air balloon. The relationship between the balloon's radius and its volume is defined by the function \( V = g(r) \).When the radius is doubled to \( 2f(t) \), the volume function becomes \( V = g(2f(t)) \). This substitution allows us to calculate how the total space within the balloon increases as the radius changes. Generally, for spherical objects, the volume scales up significantly more than double even if the radius is merely doubled, illustrating non-linear growth characteristic of volumes of spheres.

Mathematical Expression

A mathematical expression succinctly captures complex scenarios using a combination of numbers, variables, and operations. For this exercise, the challenge is to express the new volume of the balloon given a changed condition: a doubled radius.Initially, we have the volume function \( V = g(r) \). By substituting \( r \) with \( 2f(t) \) to reflect the doubled radius, we get \( V = g(2f(t)) \). This expression captures how the balloon's volume responds when the radius is increased. Mathematical expressions like these enable precise calculations and predictions about physical systems just by understanding the input relationships.