Problem 97 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the system of equations for \(u\) and \(v\). While solving for these variables, consider the transcendental functions as constants. (Systems of this type are found in a course in differential equations.) \(\left\\{\begin{array}{l}u \sin x+v \cos x=0 \\ u \cos x-v \sin x=\sec x\end{array}\right.\)

Step-by-Step Solution

Answer

The solution for the system of equations is \(u = \frac{\sec x}{\sin x + \cos x}\) and \(v = -\frac{\sec x \sin x}{\cos x (\sin x + \cos x)}\).

Step 1: Sum up the equations

Add the two equations to eliminate \(v\). This gives \(u \sin x + u \cos x = \sec x\).

Step 2: Solve for \(u\)

Rearrange the equation and isolate \(u\) on one side: \(u = \frac{\sec x}{\sin x + \cos x}\). This is the solution for \(u\).

Step 3: Solve for \(v\)

Substitute the obtained value of \(u\) into one of the original equations. Choosing the first one: \(v = -\frac{u \sin x}{\cos x} = -\frac{\sec x \sin x}{\cos x (\sin x + \cos x)}\). This is the solution for \(v\).

Key Concepts

Transcendental Functions

Transcendental functions are a class of functions that are not algebraic. They cannot be expressed as finite polynomials, roots, or any combination thereof. Instead, they include things like exponential, logarithmic, and trigonometric functions. In our given system, although we are dealing with sine, cosine, and secant (which are all trigonometric functions), we treat these functions as constants while solving the system of equations.
This approach simplifies our process. By considering transcendental functions as constants, we're reducing the system to a simpler algebraic form. The idea is to bypass the complexity of these functions when focusing on finding unknown variables like \(u\) and \(v\). The main goal is to isolate these variables, treating the transcendental parts as fixed coefficients we work with.

Differential Equations

Differential equations involve derivatives and are used to describe phenomena with rates of change, such as motion, growth, or decay. Systems that contain transcendental functions often appear in differential equations courses, as seen in the original exercise.
When solving such systems, understanding the relational dynamics between the functions and their derivatives is crucial. In this context, solutions might not be straightforward, requiring unique techniques such as integrating factors or substitution methods to find relationships between variables. While the original task simplifies this by treating transcendental functions as constants, understanding the nature of differential equations broadens the comprehension of how complex systems behave and evolve over time.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any value of the variables. In the context of our problem, identities like \( \sin x \) and \( \cos x \) play critical roles.

• Using identities can help simplify systems of equations containing trigonometric terms.
• They allow us to express one trigonometric function in terms of others, aiding in elimination or substitution processes.

For instance, knowing that \( \sec x = \frac{1}{\cos x} \) helps us substitute and simplify equations effectively when solving for variables like \(u\) and \(v\). Understanding these identities is key to solving more complex trigonometric systems that might otherwise seem daunting.

Elimination Method

The elimination method is a straightforward technique used to solve systems of equations. It focuses on eliminating one variable to solve for another, leading to a simpler equation. In our exercise, adding both original equations helps eliminate \(v\) which results in a simpler expression solely in terms of \(u\).
This method is particularly useful when dealing with linear systems where straightforward addition or subtraction can simplify terms. The process typically involves:

• Aligning equations to identify terms that can be directly cancelled.
• Adding or subtracting equations to eliminate a variable.
• Solving the resultant equation for the remaining variable.

Once one variable is found, it’s substituted back into any original equation to find the value of the other variable, completing the solution.