Problem 58 - Thomas Calculus

Question

a. The function \(y=\tan x+3 \cot x\) has an absolute minimum value on the interval \(0

Step-by-Step Solution

Answer

The absolute minimum value is \(2\sqrt{3}\) at \(x = \frac{\pi}{3}\).

Step 1: Find the First Derivative

Start by differentiating the function \( y = \tan x + 3 \cot x \) with respect to \( x \). The derivative of \( \tan x \) is \( sec^2 x \) and the derivative of \( \cot x \) is \(-csc^2 x\). Thus, the derivative \( y' = \sec^2 x - 3 \csc^2 x \).

Step 2: Find Critical Points

Set the first derivative equal to zero to find the critical points. Solve \( \sec^2 x - 3 \csc^2 x = 0 \). This equation can be rewritten as \( \sec^2 x = 3 \csc^2 x \) or \( \frac{1}{\cos^2 x} = 3 \frac{1}{\sin^2 x} \). Simplifying gives \( \sin^2 x = 3 \cos^2 x \).

Step 3: Simplify to Find Possible Values

From \( \sin^2 x = 3 \cos^2 x \), rewrite as \( 3 \cos^2 x + \sin^2 x = 1 \). This can be solved to find \( \sin x = \sqrt{\frac{3}{4}} \) and \( \cos x = \frac{1}{2} \). So, \( \sin x = \frac{\sqrt{3}}{2} \) and \( \cos x = \frac{1}{2} \).

Step 4: Calculate the Critical Point

This occurs at \( x = \frac{\pi}{3} \) where \( \sin x = \frac{\sqrt{3}}{2} \) and \( \cos x = \frac{1}{2} \). Verify this by substituting into \( \tan x = \frac{\sqrt{3}}{1/2} = \sqrt{3} \); \( \cot x = \frac{1/\sqrt{3}}{1}= \frac{1}{\sqrt{3}} \).

Step 5: Evaluate the Function at Critical Points

Substitute \( x = \frac{\pi}{3} \) into the original function: \( y = \sqrt{3} + 3 \times \frac{1}{\sqrt{3}} = \sqrt{3} + \sqrt{3} = 2\sqrt{3} \).

Step 6: Confirm Absolute Minimum

Check the endpoints of the interval \((0, \frac{\pi}{2})\) don't offer smaller values by evaluating \( y \) at \( x \to 0 \) and \( x \to \frac{\pi}{2} \). Both lead towards infinity. Thus, \( y = 2\sqrt{3} \) is the absolute minimum.

Step 7: Graph the Function for Visualization

Plot the graph of \( y = \tan x + 3 \cot x \) on the interval \( 0 < x < \frac{\pi}{2} \) and verify visually. The graph should show the lowest point at \( x = \frac{\pi}{3} \) confirming the analytic solution.

Key Concepts

Critical Points

To locate the critical points of a function, we begin by considering its first derivative. Critical points occur where this derivative equals zero or does not exist. These points are where the function either has a potential maximum, minimum, or a point of inflection.
In the problem at hand, the derivative of the function is set to zero as:

• \( \sec^2 x - 3 \csc^2 x = 0 \)

This equation leads us to find specific values of \( x \) within the given interval. Solving it by manipulating trigonometric expressions gives us insights into which points need further evaluation to confirm whether they are indeed minima.

First Derivative

The first derivative is a crucial tool in calculus used to determine the slope of a function at any point. It helps in identifying the function's behavior, such as where it increases, decreases, or if it levels out.
In this exercise, we differentiate the function \( y = \tan x + 3 \cot x \). Using known derivative results of trigonometric functions, we find:

• The derivative of \( \tan x \) being \( \sec^2 x \)
• The derivative of \( \cot x \) leading to \(-\csc^2 x \)

Hence, the overall first derivative becomes:

• \( y' = \sec^2 x - 3 \csc^2 x \)

The first derivative plays a significant role in finding critical points, which are necessary for determining the absolute minimum value of the function on the specified interval.

Absolute Minimum

An absolute minimum is the lowest point on a function over a specific interval, representing the smallest value that the function will achieve on that interval.
This value is determined by evaluating the function at critical points and boundary values of the interval. In our case, for the function \( y = \tan x + 3 \cot x \), we evaluate it at the critical point \( x = \frac{\pi}{3} \). Here:

• Substitute \( x = \frac{\pi}{3} \) into the function.
• We find it returns \( 2\sqrt{3} \).

Checking the boundary values, for instance, as \( x \to 0 \) or \( x \to \frac{\pi}{2} \), results in \( y \) trending towards infinity. Hence, neither boundary provides a value lower than \( 2\sqrt{3} \), confirming that this is indeed the absolute minimum within \(0 < x < \frac{\pi}{2}\).

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. They are vital in simplifying expressions and solving complex equations.
In the context of this problem, identities such as:

• \( \sec x = \frac{1}{\cos x} \)
• \( \csc x = \frac{1}{\sin x} \)

assist in establishing relationships between various derivatives and critical points.
Furthermore, by applying the Pythagorean identity, \( \sin^2 x + \cos^2 x = 1 \), we ease the solution steps. For instance, transforming \( \sin^2 x = 3 \cos^2 x \) provides a path to find the necessary values for \( \sin x \) and \( \cos x \). Understanding and leveraging these identities can streamline the process of solving trigonometric equations and optimizing functions.