Problem 34 - Essential Calculus

Question

Solve the differential equation. $$ \frac{d s}{d \alpha}=\sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2} $$

Step-by-Step Solution

Answer

The solution to the differential equation is \( s = -\frac{1}{2} \cos \alpha + C \), where \( C \) is the constant of integration.

Step 1: Rewriting the equation

Rewrite the differential equation as:\[ \frac{d s}{d \alpha}=\frac{1}{2} \sin \alpha \]by using the double-angle identity \( \sin(\alpha) = 2 \sin(\alpha/2)\cos(\alpha/2) \).

Step 2: Integrating both sides

Integrate both sides with respect to \( \alpha \), thus obtaining:\[ \int ds = \frac{1}{2}\int \sin \alpha\, d \alpha \]The integral of \(ds\) with respect to \( \alpha \) is simply \(s\), and the integral of \( \sin \alpha \) with respect to \( \alpha \) is \(- \cos \alpha\).

Step 3: Apply integration results

Apply the result of the integrals to get\[ s = -\frac{1}{2} \cos \alpha + C \]where \(C\) is the constant of integration. This is the general solution of the differential equation.

Key Concepts

Integration

Integration is a fundamental concept in calculus, which is essentially the process of finding the area under a curve. When we have an equation that describes a rate of change, like a differential equation, integration allows us to reverse that process and find the original function from its derivative.

For example, if you're given the derivative of a function \( \frac{ds}{d\text{alpha}} \) and asked to find \( s \), you'd perform integration. In the exercise, the integration of \( \frac{ds}{d\text{alpha}} = \frac{1}{2}\text{sin} \text{alpha} \) leads us to find the function \( s \) by calculating the area under the curve \( \frac{1}{2}\text{sin} \text{alpha} \). Simple integration of sine and cosine functions is usually one of the first techniques learned in calculus.

Double-Angle Identity

The double-angle identities are trigonometric identities that express trigonometric functions of double angles in terms of single angles. These identities are particularly useful when simplifying expressions and solving equations involving trigonometric functions.

One of these is the identity \( \text{sin}(2\text{alpha}) = 2\text{sin}(\text{alpha})\text{cos}(\text{alpha}) \), which was used in the step 1 of our exercise. By recognizing the given expression \( \text{sin}^2\frac{\text{alpha}}{2}\text{cos}^2\frac{\text{alpha}}{2} \) as part related to the double-angle identity, we could rewrite it as \( \frac{1}{2} \text{sin} \text{alpha} \) and simplify the integration process.

Constant of Integration

The constant of integration is a critical concept when dealing with indefinite integrals. When we integrate a function, we are finding all possible antiderivatives, which include any constant term that would disappear when the function is differentiated.

In our problem, after performing the integration \( s = -\frac{1}{2} \text{cos} \text{alpha} \), we add the constant of integration \( C \) to represent any vertical shift in the function that wouldn’t be apparent from the derivative alone. This constant \( C \) is vital for the complete family of solutions to the differential equation.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are relationships between the angles and sides of a triangle, and they extend to cyclical patterns, which can represent a multitude of phenomena, ranging from sound waves to the movement of the planets.

In the context of our exercise, the trigonometric function \( \text{sin} \) occurs as part of the differential equation. Understanding the basic properties and graphs of these functions assists in predicting their behavior when integrated or differentiated. Recognizing how these functions behave is key to solving many calculus problems.