Problem 58 - 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

(a) \(f(t) = t\); (b) \(g(r) = \frac{4}{3} \pi r^3\); (c) \(g \circ f = \frac{4}{3} \pi t^3\) represents volume over time.

Step 1: Determine the Radius Function

To find the function \(f\) modeling the radius \(r\) of the balloon as a function of time \(t\), we note that the radius is increasing at a constant rate of 1 cm/s. Therefore, the radius as a function of time \(f(t)\) can be expressed as:\[ f(t) = r(t) = t \]This equation tells us that for each second that passes, the radius increases by 1 cm.

Step 2: Determine the Volume Function

The volume \(V\) of a sphere with radius \(r\) is given by the formula:\[ V = \frac{4}{3} \pi r^3 \]Thus, the function \(g\) that models the volume as a function of the radius is:\[ g(r) = \frac{4}{3} \pi r^3 \]

Step 3: Find the Composite Function

To find the composite function \(g \circ f\), we substitute \(f(t)\) into \(g(r)\). We know from Step 1 that \(f(t) = t\), so:\[ g(f(t)) = g(t) = \frac{4}{3} \pi (t)^3 \]This results in a function that gives the volume as a function of time, \(V(t) = \frac{4}{3} \pi t^3\).

Step 4: Interpretation of the Composite Function

The function \(g \circ f\), or \(V(t) = \frac{4}{3} \pi t^3\), represents the volume of the balloon as it changes over time. It tells us how the volume increases cubicly as the time increases, due to the relationship between the radius and the volume of a sphere.

Key Concepts

Composite Functions

Composite functions are a way of combining two functions, resulting in a new function. Imagine you have two functions: a function that describes the radius as a function of time, and a function that describes volume as a function of radius. By creating a composite function, you can link these two together to understand how the volume changes directly over time.

• The primary function, often called the inner function, is applied first. This is represented as \(f(t) = t\) in our problem, showing the radius as a function of time.
• The secondary function, the outer function, takes the result from the first function and then applies its own rule. This is \(g(r) = \frac{4}{3} \pi r^3\). In our case, by plugging in \(f(t)\) into \(g(r)\), you form a single function: \(g(f(t)) = \frac{4}{3} \pi t^3\).

This composite function captures the full picture by showing how the balloon's volume increases as time goes on, without needing to consider the radius separately in each step.

Rate of Change

The rate of change conceptually tells us how fast something changes over the effect of another variable. For our problem, understanding the rate of change is quite simple because the radius grows at a consistent rate. This rate of increase is 1 cm per second.

In practical terms, this means that for every second that passes, the balloon's radius grows by 1 cm. Rates of change can be constant, as in our example, or they can vary. Knowing the rate of change helps us predict future states of an object. It simplifies calculations as well, letting us easily write functions describing motion or growth over time, such as \(f(t) = t\) for the radius.

Recognizing and using rates of change is crucial when modeling real-world situations, where many factors might grow or decrease over time.

Volume of a Sphere

The volume of a sphere is an important concept in geometry. For any sphere with radius \(r\), the volume \(V\) can be calculated using the formula \(V = \frac{4}{3} \pi r^3\). This formula essentially sums up how a sphere's volume scales with the cube of its radius.

The cubic relationship is vital because it means that changes in the radius lead to much larger changes in volume. For instance, if the radius of a balloon doubles, its volume increases eightfold (since \(2^3 = 8\)).

In our exercise, this relationship is captured by creating the function \(g(r) = \frac{4}{3} \pi r^3\). This function gives us insight into how dramatically the balloon's volume grows as it inflates, which is reflected in the composite function's cubic term.

Radius as a Function of Time

In the problem of inflating our balloon, the radius being a function of time is central to understanding how both the radius and volume change as a function of time. The given rate of 1 cm per second simplifies our function for the radius greatly, leading to \(f(t) = t\).

This linear relationship is straightforward because the radius simply equals the time in seconds, given the rate is 1 cm/s. If we plugged another number into our rate, say 2 cm/s, then our function would adjust to \(f(t) = 2t\), doubling the amount the radius increases per second.

Thinking about the radius as a time-dependent function forms the foundation for modeling more complex scenarios where the radius potentially changes at different rates. It allows for easily updating predictions and understanding how quickly a balloon might reach a certain size given different inflation rates.