Problem 37 - Calculus Early Transcendentals

Question

A spherical balloon is being inflated. (a) Find a general formula for the instantaneous rate of change of the volume \(V\) with respect to the radius \(r\) given that \(V=\frac{4}{3} \pi r^{3}\) (b) Find the rate of change of \(V\) with respect to \(r\) at the instant when the radius is \(r=5\)

Step-by-Step Solution

Answer

The general formula is \(\frac{dV}{dr} = 4 \pi r^2\) and when \(r = 5\), the rate is \(100 \pi\).

Step 1: Understanding the Volume Formula

The problem gives the formula for the volume of a sphere: \(V = \frac{4}{3} \pi r^3\). We recognize that \(V\) represents the volume, and \(r\) is the radius of the sphere.

Step 2: Differentiate to Find the Instantaneous Rate of Change

To find the instantaneous rate of change of the volume with respect to the radius, we take the derivative of the volume formula with respect to \(r\). This gives us:\[\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3} \pi r^3\right)\]Using the power rule for differentiation (\(\frac{d}{dr} (r^n) = n r^{n-1}\)), we find:\[\frac{dV}{dr} = 4 \pi r^2\]

Step 3: Calculate the Rate of Change at a Specific Radius

Now that we have the formula \(\frac{dV}{dr} = 4 \pi r^2\), we can find the specific rate of change when \(r = 5\). Substitute \(r = 5\) into the derivative:\[\frac{dV}{dr} = 4 \pi (5)^2 = 100 \pi\]

Step 4: Interpret Results

The result from Step 3, \(100\pi\), is the instantaneous rate of change of the volume when the radius is 5 units. This means that at the instant when the radius is 5, the volume of the balloon is increasing at a rate of \(100\pi\) cubic units per unit increase in radius.

Key Concepts

Instantaneous Rate of Change

In calculus, the instantaneous rate of change is crucial for understanding how quantities shift at any specific moment. It's much like asking how fast a car is going at precisely one second along its journey. For a sphere, as in our exercise, we talk about how quickly the volume of the sphere changes as the radius changes. This tells us how fast the volume inflates when we slightly increase the radius.

To find this, we use differentiation, focusing on calculating the derivative of our sphere's volume formula with respect to its radius. This derivative gives us the rate at which the volume changes instantaneously as the radius shifts by a tiny bit. It's a fundamental concept that helps us understand dynamic systems in real-world scenarios, like the inflation of a balloon. It provides a lens to zoom in on the moment when the radius grows by an infinitesimal amount while precisely measuring the volume's growth rate.

Spherical Volume Formula

The formula for the volume of a sphere might appear a little intimidating at first glance, but breaking it down makes it clearer. The spherical volume formula, expressed as:

• \( V = \frac{4}{3} \pi r^3 \)

encapsulates how the volume of a sphere depends on its radius, capturing the essence of three-dimensional shapes.

- **\(\pi\)**: This mathematical constant hints at how circles are intertwined with spheres, as spheres contain numerous circular cross-sections.
- **\(r^3\)**: The radius cubed indicates a three-dimensional quantity we're considering. This shows that the sphere extends in three directions.
- **\(\frac{4}{3}\)**: This fraction accounts for the spherical shape's unique way of storing space, different from a simple cube.

Understanding this fundamental volume formula isn't just academic. It sets the base for more calculations, like determining rates of change or analyzing space-efficiency of spherical designs.

Power Rule for Differentiation

The power rule for differentiation simplifies the process of finding the derivative of powers of variables. This rule is essential when dealing with polynomial functions and plays a significant role when working with the volume formula, such as in our example with a sphere. The power rule states:

• For a function \( f(x) = x^n \), its derivative is \( f'(x) = n x^{n-1} \)

This means we multiply by the current power and decrease the power by one.

Applying this rule to our spherical volume formula, \( V = \frac{4}{3} \pi r^3 \), allows us to smoothly find the rate of change formula. By using the power rule, we differentiate:

• \(\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3} \pi r^3\right) = 3 \cdot \frac{4}{3} \pi r^{3-1} = 4 \pi r^2\)

The power rule simplifies complex differentiation into manageable steps. It's a powerful tool enabling us to handle various calculus problems efficiently and accurately.