Problem 41 - University Calculus: Early Transcendentals

Question

Do the graphs of the functions have any horizontal tangents in the interval \(0 \leq x \leq 2 \pi ?\) If so, where? If not, why not? Visualize your findings by graphing the functions with a grapher. $$y=x-\cot x$$

Step-by-Step Solution

Answer

The function has no horizontal tangents in the interval.

Step 1: Understand the Horizontal Tangent Concept

A function has a horizontal tangent at a point where its derivative equals zero. Therefore, to find where the function \( y = x - \cot x \) has horizontal tangents, we need to find where \( \frac{dy}{dx} = 0 \).

Step 2: Find the Derivative

Differentiate the function \( y = x - \cot x \) with respect to \( x \). The derivative of \( x \) is 1, and the derivative of \( \cot x \) is \(-\csc^2 x\). Thus, the derivative is: \[ \frac{dy}{dx} = 1 + \csc^2 x. \]

Step 3: Set the Derivative Equal to Zero

To find horizontal tangents, set the derivative equal to zero: \[ 1 + \csc^2 x = 0. \] Solve for \( x \).

Step 4: Solve the Equation

The equation \( 1 + \csc^2 x = 0 \) implies \( \csc^2 x = -1 \). However, \( \csc^2 x \) is always non-negative for real \( x \), so there are no solutions where this equation holds true.

Step 5: Conclusion

Since the equation \( \csc^2 x = -1 \) has no real solutions, the function \( y = x - \cot x \) has no horizontal tangents in the interval \( 0 \leq x \leq 2\pi \).

Key Concepts

Understanding Derivatives in Calculus

Derivatives are fundamental to calculus, acting as a tool to measure how a function changes at any given point. Think of them as the slope of a curve at a specific point. When you're asked to find where a function has horizontal tangents, you're looking for places where the derivative equals zero. This means that the slope of the tangent line at that point is zero, indicating a flat or horizontal line.
In mathematical terms, if a function is given as \( y = f(x) \), its derivative is written as \( \frac{dy}{dx} \). For the function \( y = x - \cot x \), we calculated that \( \frac{dy}{dx} = 1 + \csc^2 x \).

• The derivative \( \frac{dy}{dx} = 1 + \csc^2 x \) shows that the rate of change of the function is influenced by both a constant slope of 1 and the trigonometric term \( \csc^2 x \).
• A key takeaway is that understanding how derivatives behave can help simplify complex problems, such as determining whether horizontal tangents exist on a particular interval.

Exploring Trigonometric Functions

Trigonometric functions like sine, cosine, and cosecant (\( \csc x \)) are crucial in calculus, defining the relationships between angles and sides in triangles and extending that logic to describe periodic phenomena. The cosecant function, \( \csc x \), is the reciprocal of the sine function and plays a crucial role in this exercise.

• When differentiating functions involving \( \cot x \), it helps to remember, \( \cot x = \frac{1}{\tan x} \) and \( \frac{d (\cot x)}{dx} = -\csc^2 x \).
• Thus, in our problem, differentiating \( \cot x \) correctly led us to use \( \csc^2 x \) in the derivative.

Understanding the properties of trigonometric functions helps predict their behavior. For instance, \( \csc^2 x \) is always non-negative because it's based on the square of the sine function. This property directly led us to conclude that \( \csc^2 x = -1 \) has no real solutions in the interval \( 0 \leq x \leq 2 \pi \).
Recognizing these trigonometric properties can often make or break a calculus problem-solving session.

Strategic Calculus Problem Solving

When solving calculus problems, especially those involving derivatives and trigonometric functions, follow a strategic approach:
1. **Understand the Problem:** Identify key concepts—like horizontal tangents in this problem—then translate that into mathematical terms, usually involving derivatives.

2. **Compute the Derivative:** Use rules for derivatives appropriately. When trigonometric functions are involved, recall their specific derivatives. For instance, the derivative of \( \cot x \) is \(-\csc^2 x\).

3. **Evaluate Your Findings:** After finding the derivative, solve the mathematical equations it leads to. In this example, setting \( 1 + \csc^2 x = 0 \) allowed us to conclude no horizontal tangents exist because the cosecant squared term is non-negative.

4. **Verify Results Visually:** Use graphing tools to confirm your analytical findings visually. This helps solidify the theoretical work with practical inspection of the graph’s shape.

Approaching problems step-by-step not only ensures accuracy but also deepens comprehension of the intricate dance between different mathematical concepts involved.