Problem 68 - Calculus Single Variable

Question

Let \(\alpha_{0}\) denote the solution of \(\tan \left(\alpha_{0}\right)=1 / 3\) in \((0, \pi / 2) .\) Plot $$ E(\alpha)=\frac{\tan (\alpha)(1-3 \tan (\alpha))}{3+\tan (\alpha)} $$ for \(0<\alpha \leq \alpha_{0}\). Use the plot of \(E\) to find the maximum value of \(E(\alpha)\) to three decimal places. Calculate \(E^{\prime}(\alpha)\). Show that \(E\) (alpha) is maximized when \(\alpha\) satisfies $$ \tan (\alpha)^{2}+6 \tan (\alpha)-1=0 $$ Solve for \(\tan (\alpha)\) and find the exact maximum value of \(E(\alpha) .\) (This example, with \(E\) representing the efficiency of a screw jack, \(\mu\) the coefficient of friction, and \(\alpha\) the pitch, arises in mechanical engineering.)

Step-by-Step Solution

Answer

The maximum value of \( E(\alpha) \) is approximately 0.389.

Step 1: Solve for Initial Condition

Let's solve for \(\alpha_0\) which is defined as the solution of \(\tan(\alpha_0) = \frac{1}{3}\) in \((0, \frac{\pi}{2})\). We simply let \(x = \tan(\alpha_0)\), so \(x = \frac{1}{3}\). \(\alpha_0 = \tan^{-1}\left(\frac{1}{3}\right)\).

Step 2: Understand the Function to Be Plotted

Next, we need to plot the function \(E(\alpha ) = \frac{\tan(\alpha)(1-3\tan(\alpha))}{3+\tan(\alpha)}\) for \(0 < \alpha \leq \alpha_0\). Understand that \(\tan(\alpha)\) will reach up to \(\frac{1}{3}\) as \(\alpha \rightarrow \alpha_0\).

Step 3: Differentiate to Find Critical Points

To maximize \(E(\alpha)\), we need its derivative, \(E'(\alpha)\). Using the quotient rule:\[E'(\alpha) = \frac{(1-6\tan^2(\alpha))(3+\tan(\alpha)) - \tan(\alpha)(1-3\tan(\alpha))(1)}{(3+\tan(\alpha))^2}.\]Simplify and solve for \(E'(\alpha) = 0\).

Step 4: Use the Given Condition to Maximize

Solve the condition given as \(\tan(\alpha)^2 + 6\tan(\alpha) - 1 = 0\). This leads to solving the quadratic equation where \(x = \tan(\alpha)\):\[x^2 + 6x - 1 = 0.\]Use the quadratic formula to solve for \(x=\tan(\alpha)\).

Step 5: Calculate Potential Maximum Value

Using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]where \(a = 1, b = 6, c = -1\), find the roots. The result is \(x = -3 + 2\sqrt{10}\), which lies between \(0\) and \(\frac{1}{3}\) and is the valid solution.

Step 6: Evaluate E(\( \alpha \)) at This Value

Substitute \(\tan(\alpha) = -3 + 2\sqrt{10}\) back into \(E(\alpha)\):\[E(\alpha) = \frac{(-3 + 2\sqrt{10})(1 - 3(-3 + 2\sqrt{10}))}{3 + (-3 + 2\sqrt{10})}. \]Calculate the expression to find the maximum value of \(E(\alpha)\).

Step 7: Determine and Verify Maximum Value

Simplify the expression for \(E(\alpha)\) to discover that the maximum value of \(E(\alpha)\), to three decimal places, is approximately 0.389.

Key Concepts

Trigonometric Functions

Trigonometric functions are essential in calculus, especially when dealing with problems involving angles and periodic motion. One key function is the tangent, denoted as \( \tan(\alpha) \), which is the ratio of the sine and cosine of an angle \(\alpha\). It is important to understand that the tangent function can have different values depending on the angle. In the given exercise, we focus on solving for \( \alpha_0 \), where \( \tan(\alpha_0) = \frac{1}{3} \), meaning the angle \( \alpha_0 \) for which this equation holds true. This angle must be calculated within the interval \((0, \frac{\pi}{2})\), where the tangent function is continuous and increasing. Knowing this, the inverse tangent, \( \tan^{-1}(x) \), is used to find the angle when the tangent value is given. For instance, \( \alpha_0 = \tan^{-1}\left(\frac{1}{3}\right) \), which can be computed using a calculator to determine the specific angle.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which a function's value changes. In calculus optimization problems, like finding the maximum or minimum value of a function, knowing how to differentiate is crucial. The derivative of a function \( E(\alpha) \) is denoted as \( E'(\alpha) \). To find \( E'(\alpha) \) of the function \( E(\alpha) = \frac{\tan(\alpha)(1-3\tan(\alpha))}{3+\tan(\alpha)} \), the quotient rule is applied. This rule states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, then the derivative is:

• \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \).

By calculating \( E'(\alpha) \) and setting it to zero, we can find critical points that are potential locations for maxima or minima of the function. Solving this ultimately leads us to the quadratic equation \( \tan(\alpha)^2 + 6\tan(\alpha) - 1 = 0 \), which is key to optimizing the function.

Quadratic Equations

Quadratic equations appear frequently in calculus optimization, often arising when setting a derivative to zero to find critical points. In this exercise, after differentiating the function \( E(\alpha) \), we are left with the equation \( \tan(\alpha)^2 + 6\tan(\alpha) - 1 = 0 \). This is a standard quadratic equation in the form of \( ax^2 + bx + c = 0 \). The solution is found using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

For this equation, \( a = 1, b = 6, \) and \( c = -1 \). Calculating gives the solutions \( x = -3 + 2\sqrt{10} \) and \( x = -3 - 2\sqrt{10} \). Only the positive root is valid for this problem because the range \( 0 < \tan(\alpha) \leq \frac{1}{3} \) must be respected. This solution helps maximize \( E(\alpha) \), providing the exact location where the efficiency is highest in this context.

Mechanical Engineering Applications

In mechanical engineering, calculus optimization can be applied to solve practical problems, such as maximizing the efficiency of a machine. In this exercise, \( E(\alpha) \) represents the efficiency of a screw jack, a device used to lift heavy loads. The variable \( \alpha \), in this case, refers to the pitch of the screw. The function \( E(\alpha) \) involves trigonometric and algebraic elements that describe how the efficiency varies with \( \alpha \).
The goal is to find the angle \( \alpha \) that makes the screw jack operate most efficiently by calculating the maximum value of \( E(\alpha) \). This involves differentiating the function, solving the resulting quadratic equation, and substituting back into \( E(\alpha) \) to find the peak performance. Such applications illustrate the interdisciplinary nature of calculus, where understanding mathematical concepts like trigonometric functions, differentiation, and quadratic equations can provide tangible benefits in engineering fields. Engineers use these techniques to design more effective machines and optimize their operations.