Problem 38 - Basic Technical Mathematics with Calculus

Question

Solve the given applied problem. When mineral deposits form a uniform coating \(1 \mathrm{mm}\) thick on the inside of a pipe of radius \(r\) (in \(\mathrm{mm}\) ), the cross-sectional area \(A\) through which water can flow is \(A=\pi\left(r^{2}-2 r+1\right) .\) Sketch \(\bar{A}=f(r)\).

Step-by-Step Solution

Answer

The graph is a symmetric parabola centered at \(r = 1\) opening upwards.

Step 1: Understand the Problem

The problem involves finding the cross-sectional area \(A\) through which water can flow when mineral deposits form a uniform coating of \(1 \text{ mm}\) inside a pipe. The inside of the pipe originally has a radius \(r\) mm. The formula given for the area is \(A = \pi(r^2 - 2r + 1)\). We need to sketch \(\bar{A} = f(r)\).

Step 2: Analyze the Formula

The formula given is \(A = \pi(r^2 - 2r + 1)\), which can be further simplified. Recognize that \( r^2 - 2r + 1 \) is the perfect square \((r-1)^2\). Thus, the formula becomes \(A = \pi (r-1)^2\).

Step 3: Identify the Function Behavior

The function \(f(r) = \pi (r-1)^2\) is a parabola that is centered at \(r = 1\). It opens upwards since the coefficient of \((r-1)^2\) is positive (\(\pi\)). The minimum point of the parabola is at \(r = 1\).

Step 4: Plot Key Points and Sketch the Graph

Identify key values. At \(r = 1\), the area \(A = \pi \times 0 = 0\). For \(r < 1\), the area becomes a very small positive value. As \(r\) grows much larger than 1, \(A\) increases significantly due to the \((r-1)^2\) term. Plot points such as \(r = 0\), \(r = 1\), and \(r = 2\) to help sketch the parabola centered at \(r = 1\). Known points: \((1, 0)\), \((0, \pi)\) and \((2, \pi)\).

Step 5: Draw the Graph

Sketch the function using the identified points and behavior. The graph should be a symmetric parabola opening upwards, centered at \(r = 1\), with its vertex touching the horizontal axis at this point and rising on both sides.

Key Concepts

Cross-sectional area

In this problem, we are given a formula to calculate the cross-sectional area, which is crucial for determining how much water can flow through the pipe. The cross-sectional area is simply the area of the pipe's interior surface through which the fluid passes. This is affected by the pipe's radius and any material build-up. In our scenario,

• the pipe initially has a radius
• the mineral deposits reduce the effective radius by a slim uniform coating of 1 mm (or applicable unit).

The given formula for the cross-sectional area is: \[ A = \pi(r^2 - 2r + 1) \]. On simplifying, this becomes \[ A = \pi(r-1)^2 \]. This represents an area that will vary as the radius of the pipe changes, particularly due to build-ups.
A smaller cross-sectional area restricts water flow, performing like a bottleneck in plumbing. Understanding this helps in applications where flow rate adjustments are crucial, like irrigation or pipe maintenance.

Parabola

The function \(f(r) = \pi(r-1)^2\) represents a specific mathematical shape— a parabola. Parabolas are ubiquitous in mathematics and science because of their unique curve. Specifically, this parabola:

• is centered at \(r = 1\)
• opens upwards, meaning it increases as \(r\) moves away from 1, due to the positive coefficient \(\pi\)
• has a vertex (the lowest point in this case) at \(r = 1\), where the area is \(0\)

Since the formula is derived from a perfect square, the graph is symmetric around \(r = 1\). As you trace the function away from the vertex (both to the left and right), the function value rapidly increases because \[ (r-1)^2 \] becomes larger. This symmetrical growth pattern is a key feature of parabolas. Learning to recognize and sketch them can be helpful in numerous disciplines, ranging from physics to engineering.

Function behavior

The behavior of the function \(f(r) = \pi(r-1)^2\) is predictable due to its parabolic form. Knowing how this function operates over different values of \(r\) is crucial for understanding the cross-sectional area:

• At \(r = 1\), the area function reaches its minimum value, 0, where the mineral deposit fills the pipe completely at this stage.
• For \(r < 1\), the function reflects smaller pipe radii where the opening is gradually reducing while the area remains positive.
• As \(r\) grows (\(r > 1\)), \(A\) increases significantly as the pipe's effective radius is less obstructed by the mineral ring.

The function's graph, a "U"-shaped parabola, is symmetric around \(r = 1\) and illustrates these changes vividly. Recognizing how such functions behave gives insight into the potential flow through the pipe under different conditions, which is applicable in fields like fluid mechanics and resource management.