Problem 62 - Algebra and Trigonometry

Question

Inflating a Balloon A spherical balloon is being inflated. The radius of the balloon is increasing at the rate of 1 \(\mathrm{cm} / \mathrm{s}\) . (a) Find a function \(f\) that models the radius as a function of time. (b) Find a function \(g\) that models the volume as a function of the radius. (c) Find \(g \circ f .\) What does this function represent?

Step-by-Step Solution

Answer

\( (g \circ f)(t) = \frac{4}{3} \pi t^3 \) represents the volume of the balloon over time.

Step 1: Find the Radius as a Function of Time

Given that the radius of the balloon increases at a rate of 1 cm/s, we can model the radius as a linear function of time. If we let \( r(t) \) represent the radius at time \( t \), then since the radius increases by 1 cm each second, we have:\[ r(t) = t \]Here, \( r(t) \) is the radius in centimeters when \( t \) is in seconds.

Step 2: Find Volume as a Function of Radius

The volume \( V \) of a sphere is given by the formula:\[ V = \frac{4}{3} \pi r^3 \]To model the volume as a function of the radius, define:\[ g(r) = \frac{4}{3} \pi r^3 \]Thus, \( g(r) \) expresses the volume of the sphere (balloon) in terms of its radius \( r \).

Step 3: Find the Composition of Functions

To find \((g \circ f)(t)\), substitute the function \( r(t) = t \) into \( g(r) = \frac{4}{3} \pi r^3 \). This gives the volume as a function of time:\[ (g \circ f)(t) = g(r(t)) = \frac{4}{3} \pi (t)^3 \]Thus:\[ (g \circ f)(t) = \frac{4}{3} \pi t^3 \]\( g \circ f \) represents the volume of the balloon as a function of time, showing how the volume changes with time as the balloon is inflated.

Key Concepts

Volume of a Sphere

The volume of a sphere is a fundamental concept in geometry and mathematics used to describe the amount of space a spherical object occupies. It's particularly useful when dealing with spherical objects like balloons, planets, or even bubbles. The formula for the volume of a sphere with radius \( r \) is given by:
\[ V = \frac{4}{3} \pi r^3 \]This formula tells us how the volume increases as the radius gets larger. Notice that the radius \( r \) is raised to the power of three (cubed). This indicates that even a small increase in the radius will result in a much larger increase in volume.

• The factor \( \frac{4}{3} \pi \) is a constant derived from the geometry of a sphere.
• Since the radius is cubed, the growth in volume is not linear—it's exponential.

This property is crucial in real-world applications, such as calculating the capacity of storage tanks, balloons, or anything that is spherical. Understanding this equation helps predict how much air is needed to inflate a balloon as its radius increases.

Function Composition

Function composition is a way of combining two functions where the output of one function becomes the input of another. This is an integral part of calculus and helps in building complex relationships from simple ones.
In our exercise, we have two functions: one for the radius of a sphere changing over time \( f(t) = t \), and another for the volume as a function of the radius \( g(r) = \frac{4}{3} \pi r^3 \).

• First, \( f(t) = t \) describes how the radius grows with time.
• Then, \( g(r) \) uses this radius to compute the sphere's volume.

The composition \( g \circ f \) means we take \( f(t) \), plug it into \( g \), and find the resulting volume over time:
\[ (g \circ f)(t) = g(f(t)) = \frac{4}{3} \pi t^3 \]This function gives us the sphere's volume as it changes with time, providing insight into the dynamic nature of growth patterns.

Linear Function

A linear function is one in which the relation between two variables is a straight line when plotted on a graph. Its general form is \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept.
In the given exercise, the radius of the balloon increases linearly over time. This is modeled with a simple linear function:
\[ r(t) = t \]Here:

• \( m = 1 \), indicates the slope, which in this case means the radius increases by 1 cm per second.
• \( b = 0 \), implies the radius is zero when time starts.

Linear functions are straightforward to understand because changes are consistent, reflecting a constant rate of growth or decline. This concept simplifies complex situations and creates manageable models, allowing predictions of future behavior based on current rates. In our balloon problem, it helps predict the future size of the balloon at any given time, given the steady rate of inflation.