Problem 128 - Calculus Volume 1

Question

For the following exercises, find the local and/or absolute maxima for the functions over the specified domain. $$y=4 \sin \theta-3 \cos \theta \text { over }[0,2 \pi]$$

Step-by-Step Solution

Answer

The local and absolute maximum is at \( \theta \approx 2.21 \) with \( y \approx 4.472 \).

Step 1: Find the Derivative

First, find the derivative of the function \( y = 4 \sin \theta - 3 \cos \theta \) with respect to \( \theta \). Using differentiation rules, we have: \( \frac{dy}{d\theta} = 4 \cos \theta + 3 \sin \theta \).

Step 2: Set the Derivative to Zero

To find the critical points, set the derivative equal to zero: \( 4 \cos \theta + 3 \sin \theta = 0 \). Solve for \( \theta \) by using the trigonometric identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Rearranging, we get \( \tan \theta = -\frac{4}{3} \).

Step 3: Solve for Critical Points

Solving \( \tan \theta = -\frac{4}{3} \), we find \( \theta \approx -0.93 \) and \( \theta \approx 2.21 \). We must adjust these solutions to lie within \( [0, 2\pi] \). Thus, \( \theta \approx 2.21 \) and \( \theta \approx 5.35 \).

Step 4: Calculate the Function Values at Critical Points

Substitute the critical points back into the original function: \( y = 4 \sin(2.21) - 3 \cos(2.21) \approx 4.472 \) and \( y = 4 \sin(5.35) - 3 \cos(5.35) \approx -4.472 \).

Step 5: Evaluate Endpoints of the Interval

Evaluate the function at the endpoints of the interval: For \( \theta = 0 \), \( y = 4 \cdot 0 - 3 \cdot 1 = -3 \). For \( \theta = 2\pi \), \( y = 4 \cdot 0 - 3 \cdot 1 = -3 \).

Step 6: Determine Maximum Values

Compare all function values: At \( \theta = 2.21 \), \( y \approx 4.472 \) is the maximum. At \( \theta = 5.35 \), \( y \approx -4.472 \) is a minimum. At endpoints \( \theta = 0 \) and \( \theta = 2\pi \), \( y = -3 \). Thus, the local and absolute maximum occurs at \( \theta \approx 2.21 \).

Key Concepts

Derivative

The derivative is a powerful tool in calculus that helps us determine how a function changes as its input changes. In this exercise, we are given a trigonometric function, which involves both sine and cosine functions:

• Original Function: \( y = 4 \sin \theta - 3 \cos \theta \)
• To find the derivative, apply standard rules for differentiation of trigonometric functions:
• The derivative of \( \sin \theta \) with respect to \( \theta \) is \( \cos \theta \).
• The derivative of \( \cos \theta \) is \(-\sin \theta \).
• Using these rules, the derivative becomes \( \frac{dy}{d\theta} = 4 \cos \theta + 3 \sin \theta \).

This derivative helps us identify where the function reaches its maximum and minimum values, which are important in many practical applications such as physics, engineering, and economics.

Critical Points

Critical points are values of \( \theta \) where the derivative of the function equals zero, or where the derivative does not exist. These points are potential locations for local maxima or minima. In our problem, we set the derivative to zero:

• Equation: \( 4 \cos \theta + 3 \sin \theta = 0 \)
• To solve for \( \theta \), we rearrange the equation using the trigonometric identity for tangent: \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
• This simplifies our equation to \( \tan \theta = -\frac{4}{3} \).

By solving this, we find specific values of \( \theta \) within the interval \( [0, 2\pi] \): approximately \( 2.21 \) radians and \( 5.35 \) radians. These critical points are where we will check for potential maximum or minimum values.

Maxima and Minima

Maxima and minima represent the highest or lowest points in a function over a given interval. These are of two types, local and absolute:

• Local Maxima and Minima: Points that are higher or lower than nearby points.
• Absolute Maxima and Minima: The highest or lowest points over the entire interval.

For this function, we substituted the critical points \( \theta = 2.21 \) and \( \theta = 5.35 \) back into the original equation to find function values:

• At \( \theta = 2.21 \): \( y \approx 4.472 \) (maximum value).
• At \( \theta = 5.35 \): \( y \approx -4.472 \) (minimum value).
• Additionally, at the endpoints \( \theta = 0 \) and \( \theta = 2\pi \), \( y = -3 \).

Thus, the function has an absolute maximum value of approximately 4.472 at \( \theta = 2.21 \).

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variable. They are invaluable in simplifying functions and solving equations:

• In this exercise, we used \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) to solve the equation for critical points.
• This identity allows us to eliminate one of the trigonometric functions, making it easier to solve for \( \theta \).
• Identities like these can transform complex problems into more manageable ones.

Being comfortable with these trigonometric identities is key to solving calculus problems that involve trigonometric functions. They help not only in calculus but also in understanding patterns in waves and oscillations that occur in physics and engineering.
Familiarity with these identities is crucial for becoming adept at solving a wide range of real-world problems.