Problem 39 - Thomas Calculus

Question

In Exercises \(35-40,\) use a CAS to perform the following steps for the given graph of the function over the closed interval. \begin{equation}\begin{array}{l}{\text { a. Plot the curve together with the polygonal path approxima- }} \\ {\text { tions for } n=2,4,8 \text { partition points over the interval. (See }} \\ {\text { Figure } 6.22 . )} \\\ {\text { b. Find the corresponding approximation to the length of the }} \\\ {\text { curve by summing the lengths of the line segments. }} \\ {\text { c. Evaluate the length of the curve using an integral. }} \\ {\text { Compare your approximations for } n=2,4,8 \text { with the }} \\ {\text { actual length given by the integral. How does the actual }} \\ {\text { length compare with the approximations as } n \text { increases? }} \\ {\text { Explain your answer. }}\end{array}\end{equation} $$f(x)=\frac{x-1}{4 x^{2}+1},-\frac{1}{2} \leq x \leq 1$$

Step-by-Step Solution

Answer

As \(n\) increases, the polygonal approximation converges to the curve's actual length.

Step 1: Plot the Curve and Approximations

To plot the given function \(f(x) = \frac{x-1}{4x^{2} + 1}\) over the interval \(-\frac{1}{2} \leq x \leq 1\), use a CAS (Computer Algebra System). First, plot the function. Then, overlay the polygonal path approximations. For \(n=2\), calculate the function values at the endpoints and midpoint, then connect these points linearly. Repeat this process for \(n=4\) and \(n=8\), dividing the interval into 4 and 8 equidistant points respectively, and draw the connecting lines.

Step 2: Calculate the Polygonal Path Length

To approximate the curve length for \(n=2, 4, 8\), sum the lengths of the line segments. For any segment connecting \(x_i\) and \(x_{i+1}\), use the distance formula: \(d = \sqrt{(x_{i+1} - x_i)^2 + (f(x_{i+1}) - f(x_{i}))^2}\). Here, \(x_i\) are the partition points. Calculate and sum the lengths for each value of \(n\).

Step 3: Evaluate the Length of the Curve using an Integral

The actual curve length over \([-1/2, 1]\) is given by the integral \( L = \int_{-1/2}^{1} \sqrt{1 + \left(\frac{d}{dx}{\frac{x-1}{4x^{2} + 1}}\right)^2} \, dx\). Compute the derivative of \(f(x)\), substitute it into the integral, and evaluate using a CAS for the precise curve length.

Step 4: Compare Approximations with Actual Length

Compare the lengths from Step 2 with the exact length obtained in Step 3. As \(n\) increases, the approximation should become more accurate, showing convergence toward the actual length. Discuss how smaller partitions give a finer approximation, meaning the discrepancies between the polygonal path and the curve reduce as \(n\) grows.

Key Concepts

Polygonal Path

A polygonal path is a series of straight line segments that join a sequence of points on a curve. It's a common method used to approximate the length of a curve, especially when calculating an exact length can be complex. In mathematics, these polygonal paths are often used to estimate the length of a curve by connecting discrete points and then summing the lengths of the segments between these points.

For example, if we have points \(A, B, C\) on a curve, we can connect these with line segments \(AB\) and \(BC\). The length of each segment can be calculated using the distance formula: \(\) \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]where \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of two consecutive points.

In the given exercise, polygonal paths were utilized to approximate the curve length of the function \(f(x) = \frac{x-1}{4x^{2} + 1}\) over the interval \(-\frac{1}{2} \leq x \leq 1\). By increasing the number of line segments, we can improve the accuracy of the approximation, converging towards the actual curve length as the number of segments increases.

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals and their properties. It is essential for calculating areas under curves, volumes, and other quantities where aggregation over intervals is required. In the context of finding curve lengths, integral calculus is used to determine the exact length of a curve defined by a function over a certain interval.

To compute the length of a curve \(y = f(x)\) from \(x = a\) to \(x = b\), we use the formula: \[L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]This formula accounts for both the horizontal distance and the change in the function values (slope). The integral aggregates this continuous change to provide the precise curve length.

In our exercise, we applied this formula to find the length of the function \(f(x) = \frac{x-1}{4x^{2} + 1}\). After calculating the derivative of the function, we substituted it into the integral equation to calculate the exact length over the given interval.

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is a fundamental concept in calculus, capturing the rate of change or the slope of a function at any given point. In the context of curve length, the derivative is used to find the slope of the function, which is incorporated into the integral to calculate the exact curve length.

For the function \(f(x) = \frac{x-1}{4x^{2} + 1}\), the derivative is calculated using standard differentiation rules. Specifically, the quotient rule comes into play for this function: \[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\]where \(u = x-1\) and \(v = 4x^{2}+1\). Here, we find the derivative and then substitute it into the integral calculus formula from the previous section.

The derivative allows us to understand the 'steepness' or 'flatness' of the curve at any given point, which is crucial for determining exact lengths in integral calculus.

Computer Algebra System

A Computer Algebra System (CAS) is a software tool that facilitates symbolic mathematics. It can perform algebraic operations, differentiation, integration, and even plot functions. With the complexity of calculus problems often involving intricate functions, using a CAS makes it easier to handle calculations and visualize results without manual errors.

In practical terms, a CAS helps students and professionals solve complex calculus problems by automating tedious algebraic manipulation. For the given exercise, a CAS was used to:

• Plot the curve \(f(x) = \frac{x-1}{4x^{2} + 1}\) and its polygonal approximations.
• Calculate derivatives required for integral calculus.
• Evaluate integrals to find the exact length of the curve.

This technology not only speeds up computation but also provides visual aids that help understand the behavior of functions, supporting a deeper insight into integral and differential calculus.