Problem 48 - Calculus of a Single Variable

Question

Use a graphing utility to (a) graph the function \(f\) on the given interval, (b) find and graph the secant line through points on the graph of \(f\) at the endpoints of the given interval, and (c) find and graph any tangent lines to the graph of \(f\) that are parallel to the secant line. \(f(x)=x-2 \sin x, \quad[-\pi, \pi]\)

Step-by-Step Solution

Answer

The graph of \(f(x)=x-2 \sin x\) is sketched on the interval \([-π, π]\). A secant line with the equation \(y=x-π\) is produced. The tangent lines that are parallel to the secant line have the equations \(y=x-\frac{π}{2}\) and \(y=x+\frac{π}{2}\), and these lines are also graphed.

Step 1: Graphing the Given Function

Use a graphing utility to graph the function \(f(x)=x-2\sin x\) on the interval \([-π, π]\).

Step 2: Find and Graph the Secant Line

To find the secant line through points on the graph of \(f\) at the endpoints \([-π, π]\), first calculate the coordinates of these points. Plugging \(-π\) and \(π\) in the given function, the obtained points are \((-π, -π+2\), \((π, π-2)\). The slope of the secant line passing through these points is \(\frac{(π-2)-(-π+2)}{π-(-π)}=1\). Use point slope form to find the equation of the secant line: \(y-(-π+2)=1(x+π)\) which simplifies to \(y=x-π\). Now graph this line using the graphing tool.

Step 3: Find and Graph the Tangent Lines

Tangent lines to the graph of \(f\) that are parallel to the secant line will have the same slope as the secant line, that is \(1\). The derivative \(f'(x)\) is needed to find these lines. Calculate the derivative of \(f(x)=x-2\sin x\) to get \(f'(x)=1-2\cos x\). Setting this equal to \(1\) (the slope of the secant line), we find that \(2\cos x=0\), resulting in \(x=\frac{π}{2}+nπ\), where \(n\) is an integer. Since \(n\) can be \(0\) or \(-1\) given the domain, the tangent points are \((-\frac{π}{2}, -\frac{π}{2}+2)\) and \((\frac{π}{2}, \frac{π}{2}-2)\). The equations of the tangent lines at these points will be \(y=x-\frac{π}{2}\) and \(y=x+\frac{π}{2}\) respectively. Now, graph these lines using the graphing utility.

Key Concepts

Secant Line

A secant line is a straight line connecting two points on a curve. In the context of graphing functions, it represents the average rate of change between these two points. This concept is important in calculus because it provides a way to approximate the slope of a curve between two points.
In the given exercise, the function \(f(x) = x - 2 \sin x\) is examined over the interval \([-\pi, \pi]\). The endpoints \(-\pi\) and \(\pi\) represent two points on this curve. By calculating the function values at these endpoints, we obtain the points \((-\pi, -\pi+2)\) and \((\pi, \pi-2)\).
The secant line through these points has a slope of \(1\), determined by the formula for the slope of a line: \(\frac{(\pi-2) - (-\pi+2)}{\pi - (-\pi)}\). The equation of the line is then derived using the point-slope form: \(y - (-\pi + 2) = 1(x + \pi)\). This simplifies to \(y = x - \pi\). Graphing this line provides a visual representation of how the secant line approximates the behavior of the function over the specified interval.

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it at that point. Unlike a secant line, which connects two distinct points, a tangent line only has a single point in common with the curve. The tangent line represents the instantaneous rate of change or the derivative at that specific point.
In the exercise, tangent lines that are parallel to the secant line are identified. Since these tangent lines must have the same slope as the secant line to be parallel, we use the previously calculated slope of \(1\).
To find where these tangent lines occur, the derivative of the function \(f(x) = x - 2 \sin x\) is calculated to be \(f'(x) = 1 - 2 \cos x\). Setting the derivative \(f'(x)\) equal to the slope of the secant line \(1\), we solve for \(x\), obtaining \(2 \cos x = 0\). This implies \(x = \frac{\pi}{2} + n\pi\), where \(n\) is an integer.
The only valid \(x\) values within the interval are \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). The corresponding tangent lines have equations \(y = x - \frac{\pi}{2}\) and \(y = x + \frac{\pi}{2}\). These lines, when graphed, illustrate how the curve of the function \(f\) behaves at these specific points.

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It's often interpreted as the slope of the tangent line to the curve of the function at a specific point.
In practice, the derivative gives the rate at which the function's value changes with respect to a change in the input. Calculating the derivative involves finding the limit of the slope of the secant line as the two points on the function come infinitesimally close to each other.
For the function \(f(x) = x - 2 \sin x\), the derivative is calculated as \(f'(x) = 1 - 2 \cos x\). This derivative tells us how the function is changing at any given point \(x\). For instance, where \(f'(x)\) equals zero, the function has horizontal tangents, meaning it's neither increasing nor decreasing at those points.
The derivative aids in finding points where tangent lines are parallel to a given line (as in this problem, parallel to the secant line) by equating it to the slope of that line. In this way, the derivative is a powerful tool for understanding and graphing functions.

Slope of a Line

The slope of a line is a measure of its steepness or incline and is given by the ratio of the vertical change to the horizontal change between two points on the line. This concept is crucial in understanding both secant and tangent lines.
For any line, the slope \(m\) can be calculated using the formula:

• \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

In terms of graphing functions, the slope of a secant line represents the average rate of change of the function between two endpoints. Meanwhile, the slope of a tangent line provides the instantaneous rate of change at a single point.
In the exercise context, the secant line through the endpoints of the given function has a slope of \(1\) because moving horizontally from \(-\pi\) to \(\pi\) results in a vertical change from \(-\pi + 2\) to \(\pi - 2\). This consistency in slope is what allows us to identify the tangent lines that are parallel to the secant line.
Understanding the slope helps in analyzing and predicting the behavior of mathematical functions and is a key element of calculus and algebra.