Problem 70 - Precalculus with Limits

Question

Advanced Applications In Exercises 69 and 70 , solve the system of equations for \(u\) and \(v .\) While solving for these variables, consider the transcendental functions as constants. (Systems of this type appear in a course in differential equations.) $$ \left\\{\begin{aligned} u \cos 2 x+& v \sin 2 x=& 0 \\ u(-2 \sin 2 x)+v(2 \cos 2 x) &=\csc 2 x \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \( u = -\frac{1}{2}\tan 2x \) and \( v = \frac{1}{2\sin^2 2x} \)

Step 1: Identify Variables and Equations

Based on the system of equations, we have two unknowns \( u \) and \( v \). The two equations are: \[ u \cos 2 x+ v \sin 2 x = 0 \] and \[ u(-2 \sin 2 x)+v(2 \cos 2 x) = \csc 2 x \]

Step 2: Simplify the Second Equation

Using the trigonometric identity \( \csc \theta = \frac{1}{\sin \theta} \), we can rewrite the second equation: \[ u(-2 \sin 2 x)+v(2 \cos 2 x) =\frac{1}{\sin 2x} \]

Step 3: Solve for u

Rearrange the first equation to isolate \( u \): \[ u = -\frac{v \sin 2x}{\cos 2x} \]

Step 4: Substitute u into the Second Equation

Substitute the expression we found for \( u \) into the second equation: \[ \left(-\frac{v \sin 2x}{\cos 2x}\right)(-2 \sin 2 x)+v(2 \cos 2 x) =\frac{1}{\sin 2x} \] This simplifies to: \[ v = \frac{1}{2\sin^2 2x} \]

Step 5: Substitute v into the First Equation

Finally, substitute \( v \) from step 4 into the first equation to solve for \( u \): \[ u = -\frac{\left(\frac{1}{2\sin^2 2x}\right) \sin 2x}{\cos 2x} \] This simplifies to: \[ u = -\frac{1}{2}\tan 2x \]

Key Concepts

Transcendental Functions

Transcendental functions are a category of functions that go beyond the realm of algebraic functions, as they cannot be expressed simply in terms of polynomials. These functions include trigonometric functions, exponential functions, and logarithms. In the context of the given system of equations, we treat these transcendental functions—such as sine, cosine, and cosecant—as constants while solving for the unknowns. This simplifies the process and allows us to focus on algebraic manipulation. Understanding transcendental functions is crucial when dealing with systems of equations that appear in differential equations. These functions frequently model natural phenomena such as waves, growth, and decay.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and hold true for any angle. They are incredibly useful in simplifying expressions and solving equations involving trigonometric functions. In our problem, we use some key trigonometric identities:

• \( \csc \theta = \frac{1}{\sin \theta} \)
• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

These identities help us simplify and solve the system of equations. For example, by substituting the identity for cosecant, we transform the second equation to one that's more manageable. This allows us to find solutions for the variables \(u\) and \(v\) in terms of \(\sin 2x\) and \(\cos 2x\). Understanding and using trigonometric identities is similar to having a set of useful tools when working with trigonometry and calculus.

Differential Equations

Differential equations involve functions and their derivatives and are used to model various real-world systems. Even though the exercise itself does not directly solve a differential equation, it introduces a type of system commonly encountered in differential equations courses. In many cases, systems of equations like the one provided are solved to set the stage for more complex problems that involve derivatives and integrals.
Understanding systems of equations is fundamental for solving differential equations, as they often form the backbone of these problems. The transcendental functions and trigonometric identities we deal with are just components of the broader applications necessary to understand the behavior of physical systems, engineering models, or even biological processes governed by differential equations.