Chapter 3: Problem 41
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. $$y=x-\cot x$$
Short Answer
Expert verified
The function has no horizontal tangents in the interval.
Step by step solution
01
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 \( \frac{dy}{dx} = 0 \).
02
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: \[ \frac{dy}{dx} = 1 + \csc^2 x. \]
03
Set the Derivative Equal to Zero
To find horizontal tangents, set the derivative equal to zero: \[ 1 + \csc^2 x = 0. \] Solve for \( x \).
04
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.
05
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 \( 0 \leq x \leq 2\pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( \frac{dy}{dx} = 1 + \csc^2 x \).
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.
Recognizing these trigonometric properties can often make or break a calculus problem-solving session.
- 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.
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.
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.