Problem 199 - Calculus Volume 1

Question

Let \(f(x)=\cot x\) . Determine the points on the graph of \(f\) for \(0 < x < 2 \pi\) where the tangent line(s) is (are) parallel to the line \(y=-2 x\).

Step-by-Step Solution

Answer

The points are at \( x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4} \).

Step 1: Differentiate the Function

To find where the tangent is parallel to a given line, we need the derivative of the function to equal the slope of the line. The derivative of \( f(x) = \cot x \) is \( f'(x) = -\csc^2 x \).

Step 2: Identify the Slope of the Given Line

The equation of the line given is \( y = -2x \). Since the equation is in slope-intercept form \( y = mx + b \), the slope \( m \) is \(-2\). Our task is to find \( x \) such that the slope of the tangent line to \( f \) equals \(-2\).

Step 3: Set the Derivative Equal to the Slope

Set \(-\csc^2 x = -2\). Simplifying gives \( \csc^2 x = 2 \).

Step 4: Solve for x

Since \( \csc x = \frac{1}{\sin x} \), \( \csc^2 x = 2 \) implies \( \sin^2 x = \frac{1}{2} \), hence \( \sin x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \). Now, solve \( \sin x = \frac{\sqrt{2}}{2} \) and \( \sin x = -\frac{\sqrt{2}}{2} \) for \( 0 < x < 2\pi \).

Step 5: Find Specific Points

For \( \sin x = \frac{\sqrt{2}}{2} \), \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \). For \( \sin x = -\frac{\sqrt{2}}{2} \), \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \). These are within the interval \( 0 < x < 2\pi \).

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus and represents the rate at which a function changes. For any function, the derivative with respect to a variable, like x, illustrates how sensitive the function is to small changes in x. In practical terms, it's the slope of the tangent line at any point on the graph of the function.
For the function \( f(x) = \cot x \), the derivative is found using trigonometric differentiation rules. Differentiating, we get \( f'(x) = -\csc^2 x \).

This derivative tells us that at any point x, the slope of the tangent line to \( f(x) \) is \(-\csc^2 x \). By finding the points where this slope equals the given condition (like a specific slope), we can solve for specific x values where certain qualities in the graph occur.

Trigonometric Functions

Trigonometric functions like \( \cot x \), \( \sin x \), \( \cos x \), and \( \csc x \) play crucial roles in understanding periodic phenomena. These functions relate the angles and sides of triangles and have deep roots in analysis, especially when dealing with derivatives.

In this exercise, the focus is on the cotangent function, \( \cot x \), which is the reciprocal of the tangent function. It's represented as \( \cot x = \frac{\cos x}{\sin x} \). When differentiating \( \cot x \), we use the identity that its derivative is \( -\csc^2 x \), where \( \csc x \) is the cosecant function, equivalent to \( \frac{1}{\sin x} \). Understanding these relationships is essential for solving problems that involve the differentiation of trigonometric functions.

Slope of a Line

The slope of a line indicates its steepness and direction. For example, in the equation \( y = mx + b \), the term m represents the slope.

If the slope is positive, the line rises as it moves from left to right; if it is negative, it falls. In the given line \( y = -2x \), the slope is \(-2\). This means that for every unit increase in x, y decreases by 2 units.

In calculus, comparing the derivative of a function with the slope of a line helps identify points where tangent lines to the function are parallel to that line. This involves setting the derivative equal to the given slope and solving for the variable, often x, to find where this condition occurs on the graph.

Tangent Lines

A tangent line touches a curve at a single point and reflects the instantaneous direction of the curve at that point. It's essentially the best linear approximation of the function near that point.

Imagine zooming in on a small part of a curve until it appears as a straight line—that line is, in essence, the tangent line.
To find a tangent line that's parallel to another line, such as \( y = -2x \), the slope of the tangent (given by the derivative) must equal the slope of that line.

This forms the basis of many calculus problems where one is tasked to find specific points on a curve with particular tangential properties. In this exercise, solving \( -\csc^2 x = -2 \) provides the x-values where the tangent to \( f(x) = \cot x \) is parallel to \( y = -2x \).