Problem 75 - Essential Calculus

Question

A string stretched between the two points (0,0) and (2,0) is plucked by displacing the string \(h\) units at its midpoint. The motion of the string is modeled by a Fourier Sine Series whose coefficients are given by \(b_{n}=h \int_{0}^{1} x \sin \frac{n \pi x}{2} d x+h \int_{1}^{2}(-x+2) \sin \frac{n \pi x}{2} d x\) Find \(b_{n}\)

Step-by-Step Solution

Answer

The coefficient \(b_{n}\) is calculated by breaking down the given integral expression into simpler integral parts, using the method of integration by parts in order to obtain integrals in terms of cosine functions, calculating these integrals and finally multiplying the outcome with \(h\).

Step 1: Break down expressions in integral

Begin by breaking down the expression contained in the integral into two separate integrals and write it as follows: \(b_{n}=h \left( \int_{0}^{1} x \sin \frac{n \pi x}{2} d x + \int_{1}^{2}(-x+2) \sin \frac{n \pi x}{2} d x \right)\)

Step 2: Integrate the two integral parts

To process the integrals, you need to use the method of integration by parts, which is defined as \(\int u \, dv = u v - \int v \, du\). For the first integral, choose \(u=x\) and \(dv= \sin(n \pi x/2) dx\). The resulting integral is as follows: \[\int_{0}^{1} x \sin(n \pi x / 2) dx = [- \frac{2}{n \pi} x \cos(n \pi x / 2)]_{0}^{1} + \frac{2}{n \pi} \int_{0}^{1} \cos(n \pi x / 2) dx\]The same operation is applied for the second integral \(\int_{1}^{2}(- x + 2) \sin(n \pi x / 2) dx \) with the selection of \(u = -x + 2\) and \(dv = \sin(n \pi x / 2) dx\). Both integrals now only include cosine functions.

Step 3: Calculate the integrals with cosine functions

Now that you have integrals with cosine functions, proceed with calculating them. Generally, the integral of the cosine function is the sine function. After processing the integrals with cosine functions, add them together according to the direct relation first stated.

Step 4: Calculate the coefficient \(b_{n}\)

After calculation of the integrals and plugging in respective values, proceed to determine the coefficient \(b_{n}\) by multiplying the result by \(h\).

Key Concepts

Integration by Parts

Understanding the concept of integration by parts is paramount when dealing with calculus problems that involve the product of two functions. It's essentially a technique used to integrate the product of two functions, and it's based on the product rule for differentiation. The rule is usually given by the formula:
\[\begin{equation}\int u \thinspace dv = uv - \int v \thinspace du \. \end{equation}\]
When applying this to our Fourier Sine Series example, we identify parts of the function that can simplify the integration when differentiated or integrated individually. We chose \(u=x\) and \(dv = \thinspace \sin(\frac{n \pi x}{2}) dx\) for the first integral, then we differentiate \(u\) to find \(du\), and integrate \(dv\) to find \(v\). This method transforms a possibly complex integral into simpler parts, hopefully leading to an easier solution. Integration by parts often requires strategic choices of \(u\) and \(dv\) to simplify the integral and may require the use of other integration techniques, such as trigonometric integrals, to fully solve the problem.

Trigonometric Integrals

Trigonometric integrals are integrals that involve trigonometric functions such as sine, cosine, tangent, etc. In the Fourier Sine Series problem, after applying integration by parts, we are left with integrals that contain cosine functions.
\[\begin{equation}\int \cos(\alpha x) \thinspace dx = \frac{1}{\alpha} \sin(\alpha x) + C \. \end{equation}\]
To integrate these, we use the fundamental antiderivatives of trigonometric functions. In our example, we simplify the integrands to simply \(\cos(\frac{n \pi x}{2})\) by performing integration by parts on the initial integrals, and then straightforwardly integrate to find the resulting sine terms. Being comfortable with these types of integrals is essential since they frequently arise in calculus problems, particularly in series expansions and differential equations. Mastery of trigonometric identities can greatly assist in evaluating these integrals more efficiently.

Calculus Problems

Calculus problems, like the Fourier Sine Series, often require a sequence of integration and differentiation techniques. The ability to recognize which method to apply is a key skill.

In our problem, integration by parts was the chosen strategy, followed by evaluating trigonometric integrals. Each step relies on a solid understanding of foundational calculus concepts such as limits, derivatives, integrals, and the properties of trigonometric functions. Sometimes, calculus problems also involve piecing together several smaller results to find a final solution, as we did by first calculating the integrals individually then adding them to find the Fourier coefficient \(b_n\). Patience and attention to detail are crucial, as seemingly small errors in earlier steps can lead to significant mistakes in the final answer.

Mathematical Coefficients

Mathematical coefficients in problems like the Fourier Sine Series represent specific constants that multiply functions within a series. In our example, the coefficient \(b_n\) is crucial for determining the amplitude of each sine component in the series.

Coefficients are derived through integration, showcasing their fundamental connection to calculus. These constants are not just arbitrary numbers; they hold the essence of the function's behavior and characteristics, dictating how it behaves over its domain. In the physical representation of our example, they represent the way a string vibrates over time when plucked. Understanding the role coefficients play in mathematical series and equations is key to unlocking more complex areas of study, such as Fourier analysis, where these coefficients are used to transform functions into a sum of sine and cosine terms for easier analysis and computation.