Problem 55 - Thomas Calculus

Question

Grinding engine cylinders Before contracting to grind engine cylinders to a cross-sectional area of 9 in^{2} , \text { you need to know how } much deviation from the ideal cylinder diameter of \(c=3.385\) in. you can allow and still have the area come within 0.01 in \(^{2}\) of the required 9 in \(^{2} .\) To find out, you let \(A=\pi(x / 2)^{2}\) and look for the interval in which you must hold \(x\) to make \(|A-9| \leq 0.01\) What interval do you find?

Step-by-Step Solution

Answer

The cylinder diameter \( x \) must be in the interval \([3.377, 3.393]\).

Step 1: Express Area as a Function

The area of the cylinder is given by the formula \( A = \pi \left( \frac{x}{2} \right)^2 \). Simplify this formula to \( A = \frac{\pi x^2}{4} \). This expresses the area in terms of the diameter \( x \).

Step 2: Set Up Inequality

We want the area to be within 0.01 in\(^2\) of 9 in\(^2\). Thus, we need to solve \(|A - 9| \leq 0.01\). Substitute for \( A \) from step 1 to get \( \left| \frac{\pi x^2}{4} - 9 \right| \leq 0.01 \).

Step 3: Solve the Inequality

Start by solving the inequality \( \left| \frac{\pi x^2}{4} - 9 \right| \leq 0.01 \). This gives two inequalities to solve separately: \( \frac{\pi x^2}{4} - 9 \leq 0.01 \) and \( 9 - \frac{\pi x^2}{4} \leq 0.01 \).

Step 4: Solve Both Inequalities

For the first inequality, \( \frac{\pi x^2}{4} \leq 9.01 \). Multiplying through by 4 gives \( \pi x^2 \leq 36.04 \). Divide by \( \pi \) to get \( x^2 \leq \frac{36.04}{\pi} \), and take square roots to find \( x \leq \sqrt{\frac{36.04}{\pi}} \).For the second inequality, \( \frac{\pi x^2}{4} \geq 8.99 \). Similar steps yield \( \pi x^2 \geq 35.96 \), which simplifies to \( x^2 \geq \frac{35.96}{\pi} \), and taking square roots results in \( x \geq \sqrt{\frac{35.96}{\pi}} \).

Step 5: Find the Interval for Diameter x

The solution to the combined inequalities from Step 4 gives \( \sqrt{\frac{35.96}{\pi}} \leq x \leq \sqrt{\frac{36.04}{\pi}} \). Calculate these expressions to find the interval for \( x \).

Step 6: Calculate Numerical Values

Calculate \( \sqrt{\frac{35.96}{\pi}} \approx 3.377 \) and \( \sqrt{\frac{36.04}{\pi}} \approx 3.393 \). Hence, the interval for \( x \) is approximately \([3.377, 3.393]\).

Key Concepts

Inequality Solving

Inequality solving is a fundamental concept in mathematics used to determine the set of values that fulfill certain conditions, often expressed in the form of inequalities. In this exercise, the task is to ensure that the area of a cylinder remains within 0.01 square inches of a target value. This requires setting up and solving an inequality.

• Begin by writing the equation for the area of a circle: \( A = \pi \left( \frac{x}{2} \right)^2 \) and simplify it to \( A = \frac{\pi x^2}{4} \).


• To keep the area close to 9 in\(^2\), the expression \( |A - 9| \leq 0.01 \) is used. This inequality allows us to find the acceptable values for the diameter \( x \).

This inequality splits into two separate inequalities. For the problem:

• \( \frac{\pi x^2}{4} - 9 \leq 0.01 \)
• \( 9 - \frac{\pi x^2}{4} \leq 0.01 \)

By solving each individually, using arithmetic operations, we find the minimum and maximum values of \( x \) that will ensure the area remains close to the desired measure. Inequality solving can sound daunting, but it's simply a matter of carefully balancing numbers on both sides until they fit within the limits you've set.

Geometry

Geometry, the branch of mathematics dealing with shapes, sizes, and the properties of space, plays a key role in understanding this exercise. The cylinder's cross-sectional area is a crucial factor requiring us to comprehend motion from dimensions to area calculation.

• In a cycle like grinding engine cylinders, understanding the relationship between diameter and area is important.


• We're not just analyzing numbers; we're working to provide a physical object with precision.

The formula for the area of a circle stems from basic geometric properties which apply here because every cross-section of the cylinder can be likened to a circle:

• Formula for area \( A = \frac{\pi x^2}{4} \) highlights how diameter \( x \) affects this area in relation to \( \pi \).

Accuracy in geometry ensures that pieces fit perfectly in real-world applications like engines, emphasizing precision engineering. This connection between theoretical geometry and practical application is at the heart of this exercise.

Pi Calculation

Pi (\( \pi \)) is one of the most famous constants in mathematics, responsible for linking a circle's circumference to its diameter.

• In this problem, \( \pi \) is vital for converting the diameter of the cylinder to its area.


• Its appearance in the area formula \( A = \frac{\pi x^2}{4} \) showcases its role in calculations involving circles and circular objects.

Even though \( \pi \) is often approximated as 3.14159, it's an irrational number—meaning it has a non-repeating, non-terminating decimal sequence. This infinite nature makes precise calculation both challenging and fascinating.

• In applications like this exercise, engineers often need just enough precision (or significant figures) from \( \pi \) to maintain functional accuracy without overcomplicating the calculations.

Understanding \( \pi \)'s contribution reveals how mathematical constants create a bridge from abstract ideas to tangible outcomes—in this case, retaining an engine cylinder's optimal function.