Problem 69 - Thomas Calculus

Question

\begin{equation} \begin{array}{l}{\text { a. The function } y=\cot x-\sqrt{2} \csc x \text { has an absolute maxi- }} \\ {\text { mum value on the interval } 0 < x < \pi . \text { Find it. }} \\ {\text { b. Graph the function and compare what you see with your }} \\ {\text { answer in part (a). }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The absolute maximum value of the function on \( 0 < x < \pi \) is -1 at \( x = \frac{\pi}{4} \).

Step 1: Find the derivative

To find the maximum value of the function, we'll first find its derivative. The function is given by \( y = \cot{x} - \sqrt{2}\csc{x} \). The derivative \( y' \) is \( y' = -\csc^2{x} + \sqrt{2}\csc{x}\cot{x} \).

Step 2: Set the derivative to zero

To find critical points where a maximum might occur, set the derivative equal to zero: \[ -\csc^2{x} + \sqrt{2}\csc{x}\cot{x} = 0 \] Factor out \( \csc{x} \): \[ \csc{x}(-\csc{x} + \sqrt{2}\cot{x}) = 0 \] Solving this gives \( \csc{x} = 0 \) or \( -\csc{x} + \sqrt{2}\cot{x} = 0 \). Since \( \csc{x} = 0 \) is undefined on the interval \( 0 < x < \pi \), solve for \( \cot{x} \): \[ \cot{x} = \frac{\csc{x}}{\sqrt{2}} \] Which simplifies to \( \sqrt{2}\sin{x} = \cos{x} \).

Step 3: Solve for x

Solving \( \sqrt{2}\sin{x} = \cos{x} \) leads us to \( \tan{x} = \frac{1}{\sqrt{2}} \). The solution to \( \tan{x} = \frac{1}{\sqrt{2}} \) in the interval \( 0 < x < \pi \) is \( x = \frac{\pi}{4} \).

Step 4: Evaluate function at critical point

Evaluate \( y \) at \( x = \frac{\pi}{4} \): \[ y = \cot\left(\frac{\pi}{4}\right) - \sqrt{2}\csc\left(\frac{\pi}{4}\right) \] \( \cot\left(\frac{\pi}{4}\right) = 1 \) and \( \csc\left(\frac{\pi}{4}\right) = \sqrt{2} \), so \[ y = 1 - \sqrt{2} \times \sqrt{2} = 1 - 2 = -1 \].

Step 5: Analyze boundary points

Check the boundaries \( x \to 0^+ \) and \( x \to \pi^- \). As \( x \to 0^+ \), \( \cot{x} \to \infty \) and \( \csc{x} \to \infty \), resulting in a negative number. As \( x \to \pi^- \), \( \cot{x} \to -\infty \) and \( \csc{x} \to \infty \), resulting in a non-positive number. The maximum value on this interval must be at the critical point.

Step 6: Graph the function for verification

Graph the function \( y = \cot{x} - \sqrt{2}\csc{x} \) over the interval \( 0 < x < \pi \). The graph should show that the maximum value of -1 occurs at \( x = \frac{\pi}{4} \), confirming our analytical solution.

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in understanding oscillatory behavior in functions, and are pivotal in calculus optimization problems. For our exercise, the function given is \( y = \cot{x} - \sqrt{2} \csc{x} \). Here, \( \cot{x} \) (cotangent) is the ratio of the adjacent side to the opposite side in a right triangle, or \( \frac{\cos{x}}{\sin{x}} \). The \( \csc{x} \) (cosecant) is the reciprocal of sine, i.e., \( \csc{x} = \frac{1}{\sin{x}} \).
Knowing these relationships is crucial for understanding how the function behaves over different intervals. Especially since both \( \cot{x} \) and \( \csc{x} \) have undefined values at certain points, like when \( x = 0 \) or \( x = \pi \). Therefore, these functions directly influence the continuity and differentiability vital for finding critical points and extrema on a specified interval.

Derivative Calculation

Derivative calculation is essential in determining the rate of change of a function. By differentiating, we can find critical points that tell us where functions have potential maxima or minima. For \( y = \cot{x} - \sqrt{2} \csc{x} \), the derivative is calculated as:

• The derivative of \( \cot{x} \) is \( -\csc^2{x} \).
• The derivative of \( \sqrt{2} \csc{x} \) involves the chain rule and is \( \sqrt{2} \csc{x}\cot{x} \).

Putting it all together, we have the derivative: \( y' = -\csc^2{x} + \sqrt{2}\csc{x}\cot{x} \). This derivative is crucial for pinpointing where the function's rate of change is zero, indicating potential maxima or minima on the interval \( 0 < x < \pi \). By setting the derivative equal to zero, we can further analyze these points.

Critical Points Analysis

Critical points analysis involves understanding where the function's derivative equals zero or is undefined. These points are where possible maxima or minima occur. After calculating the derivative for \( y = \cot{x} - \sqrt{2} \csc{x} \), we set it to zero:
\[ -\csc^2{x} + \sqrt{2}\csc{x}\cot{x} = 0 \]
By factoring, we discover that solving \( -\csc{x} + \sqrt{2}\cot{x} = 0 \) leads us to the equation \( \tan{x} = \frac{1}{\sqrt{2}} \). Thus, in the interval \( 0 < x < \pi \), the solution \( x = \frac{\pi}{4} \) emerges as a critical point. Here, evaluating \( y \) shows the function's maximum at -1, confirming that \( x = \frac{\pi}{4} \) is indeed critical for this optimization problem.

Interval Evaluation

Interval evaluation is a powerful tool in calculus for verifying extremum conditions. Analyzing how a function behaves towards the endpoints of an interval ensures that we account for all maxima and minima, especially when these intervals don't include boundary points.
In this context, we checked the function behavior as \( x \to 0^+ \) and \( x \to \pi^- \):- As \( x \to 0^+ \), both \( \cot{x} \) and \( \csc{x} \) tend towards infinity, resulting in negative values.- As \( x \to \pi^- \), \( \cot{x} \) becomes \(-\infty\) and \( \csc{x} \) stays positive, indicating non-positive outcomes.
This bound-inspection reveals no maxima at interval edges, validating our earlier finding that the maximum -1 occurs at the critical point \( x = \frac{\pi}{4} \). Confirming analytical answers through graphical representation consolidates these insights, offering a visual verification.