Question

Sports A sporting goods manufacturer designs a golf ball with a volume of 2.48 cubic inches. (a) What is the radius of the golf ball? (b) If the ball's volume can vary between 2.45 cubic inches and 2.51 cubic inches, how can the radius vary? (c) Use the \(\varepsilon-\delta\) definition of a limit to describe this situation. Identify \(\varepsilon\) and \(\delta\).

Short answer

The radius of the golf ball when the volume is 2.48 cubic inches is calculated using the formula for the volume of a sphere. The range of the radius in correspondence with the volumes 2.45 and 2.51 cubic inches is calculated likewise. Using the \(\varepsilon-\delta\) definition, \(\varepsilon\) stands for any change in volume, while \(\delta\) is the corresponding change in radius.

Step-by-step solution

Calculate the radius from the given volume

Use the formula \(V=\frac{4}{3}\pi r^3\) to find out the radius. Rearranging for r we get \(r=\left(\frac{3V}{4\pi}\right)^{\frac{1}{3}}\). Substituting \(V=2.48\) cubic inches, calculate the radius r.

Calculate the range of radius

Here, it's necessary to calculate the radius for both the lower and higher volume limits. Substituting \(V=2.45\) cubic inches and \(V=2.51\) cubic inches in \(r=\left(\frac{3V}{4\pi}\right)^{\frac{1}{3}}\), calculate the radii for these two scenarios. The result will give a range of radii for the golf ball.

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

The \(\varepsilon-\delta\) definition is used to describe limits in calculus. Here it can be used to describe the situation by identifying \(\varepsilon\) as the change in volume i.e., \(\varepsilon = |V-2.48|\) and \(\delta\) as the corresponding change in radius i.e., \(\delta = |r-r_{0}|\), where \(r_{0}\) is the radius calculated in step 1 for \(V=2.48\) cubic inches. The goal is to ensure that for any \(\varepsilon > 0\) (i.e., any change in volume), there exists some \(\delta > 0\) (corresponding change in radius) such that if \(0<|V-2.48|<\varepsilon\) then \(|r-r_{0}|<\delta\).