Question

A sporting goods manufacturer designs a golf ball having a volume of 2.48 cubic inches. (a) What is the radius of the golf ball? (b) The volume of the golf ball varies between 2.45 cubic inches and 2.51 cubic inches. How does the radius vary? (c) Use the \(\varepsilon-\delta\) definition of limit to describe this situation. Identify \(\varepsilon\) and \(\delta .\)

Short answer

The radius of the golf ball is approximately \(r = \sqrt[3]{\frac{3*2.48}{4\pi}}\). The radius varies between \(r_{min} = \sqrt[3]{\frac{3*2.45}{4\pi}}\) and \(r_{max} = \sqrt[3]{\frac{3*2.51}{4\pi}}\). The concepts of \(\epsilon\) and \(\delta\) represent the variation in volume and resulting variation in radius, respectively.

Step-by-step solution

Finding the radius

We have the volume of the ball as 2.48 cubic inches. We could equate this to the volume formula of a sphere \(\frac{4}{3}\pi r^3\) and solve for the radius \(r\). \[2.48 = \frac{4}{3}\pi r^3\] Solving for \(r\), we find that \(r = \sqrt[3]{\frac{3*2.48}{4\pi}}\]

Calculating radius variation

To find out how does the radius vary, plug in the minimum and maximum values of volume (2.45 and 2.51 cubic inches respectively) into the same formula and solve for the radius. \[r_{min} = \sqrt[3]{\frac{3*2.45}{4\pi}}\] \[r_{max} = \sqrt[3]{\frac{3*2.51}{4\pi}}\] The difference (\(\delta r\)) between \(r_{max}\) and \(r_{min}\) gives us how much the radius varies.

Applying the \(\varepsilon-\delta\) definition

For part c, using the \(\varepsilon-\delta\) definition of a limit, we can interpret \(\varepsilon\) as the variation in the volume (±0.03 cubic inches) and \(\delta r\) as the resulting variation in the radius. We are basically stating that for every small variation \(\varepsilon\) in volume results a change \(\delta r\) in radius.

Key concepts

Volume of a Sphere Calculation

Understanding how to calculate the volume of a sphere is essential in various fields, including manufacturing and design. The formula for the volume of a sphere is given by the equation \[ V = \frac{4}{3}\pi r^3 \. \] To find the volume of a sphere, you'll need to know the radius and then cube it (multiply it by itself three times), multiply that result by π (approximately 3.14159), and finally multiply by 4/3. For example, a golf ball with a volume of 2.48 cubic inches suggests the manufacturer designed it with precision, as each part of this formula affects the end result.

In the exercise provided, we would solve for the radius \(r\) by rearranging the formula to isolate \(r\). First, multiply both sides by \(\frac{3}{4}\pi\) to get \[ r^3 = \frac{3*2.48}{4\pi} \. \] Then take the cube root of both sides to solve for \(r\). The cube root can be found using a calculator or other computational tools. This calculation is essential not only in answering the exercise but also in real-world applications such as determining the amount of material needed to make the ball, or to fit the sphere into a specific space.

Epsilon-Delta Definition of Limit

The epsilon-delta (\(\varepsilon-\delta\)) definition of a limit is a formal mathematical concept used to give a precise meaning to the idea of a function approaching a limit as its input gets close to some value. It involves two values that can be thought of as 'wiggle room': \(\varepsilon\) represents how much the function's output can vary (the 'output wiggle'), and \(\delta\) represents how much the function's input can vary (the 'input wiggle').

In simpler terms, if we can make the function's output (volume of the sphere) as close as we like to a specific limit (the desired volume) by making the input (the radius) sufficiently close to some value, then we say the limit exists. Using the \(\varepsilon-\delta\) definition in the context of a sphere's volume and radius, we can describe how precisely we must control the radius to ensure the volume stays within a given range. For example, a manufacturer may want to ensure that a golf ball’s volume stays within ± 0.03 cubic inches of the desired volume, and by using this definition of limit, they can determine the necessary precision for the ball's radius.

Radius Variation in Spheres

When a sphere's volume varies, its radius must also vary, as the two are directly related by the volume formula. In applications such as manufacturing a golf ball, it’s crucial to understand how changes in the volume affect the radius. From the exercise, we explore how the radius changes when the volume fluctuates between 2.45 and 2.51 cubic inches. We calculate the minimum and maximum radius possible within this volume range by plugging these values into the sphere's volume formula and solving for \(r\).

To illustrate, if the radius is too small, even by a tiny fraction, the ball will be undersized. Conversely, a radius that is slightly too large will result in an oversized ball. Each variation in radius, denoted as \(\delta r\), impacts the final product. Therefore, understanding the relationship between volume and radius is vital for quality control in production processes. The practice of calculating this variation aids students in grasping the sensitivity of spherical dimensions to changes in volume.