Problem 35 - Thomas Calculus in SI Units

Question

Find all possible functions with the given derivative. a. \(y^{\prime}=\sin 2 t\) b. \(y^{\prime}=\cos \frac{t}{2} \quad\) c. \(y^{\prime}=\sin 2 t+\cos \frac{t}{2}\)

Step-by-Step Solution

Answer

a) \( y = -\frac{1}{2} \cos 2t + C \); b) \( y = 2 \sin \frac{t}{2} + C \); c) \( y = -\frac{1}{2} \cos 2t + 2 \sin \frac{t}{2} + C \).

Step 1: Identify the Integral

To find the original function from its derivative, we need to integrate the given derivative. For the first function, identify that we need to find: \( y = \int \sin 2t \, dt \).

Step 2: Use Substitution for Integration (a)

To integrate \( \sin 2t \), use the substitution method. Let \( u = 2t \), then \( du = 2 \, dt \) or \( dt = \frac{du}{2} \). The integral becomes: \( \int \sin u \cdot \frac{du}{2} = \frac{1}{2} \int \sin u \, du \).

Step 3: Solve the Integral (a)

Integrate \( \sin u \): \( \int \sin u \, du = -\cos u + C \). Therefore, the function becomes: \( y = -\frac{1}{2} \cos 2t + C \).

Step 4: Identify the Integral (b)

Identify the integral for the second part as follows: \( y = \int \cos \frac{t}{2} \, dt \).

Step 5: Use Substitution for Integration (b)

For the function \( \cos \frac{t}{2} \), use the substitution \( u = \frac{t}{2} \), then \( du = \frac{1}{2} \, dt \) or \( dt = 2 \, du \). The integral becomes: \( \int \cos u \cdot 2 \, du = 2 \int \cos u \, du \).

Step 6: Solve the Integral (b)

Integrate \( \cos u \): \( \int \cos u \, du = \sin u + C \). Therefore, the function becomes: \( y = 2 \sin \frac{t}{2} + C \).

Step 7: Combine the Derivatives for (c)

For function (c), use the linearity of integration to combine the integrals as follows: \( y = \int \left( \sin 2t + \cos \frac{t}{2} \right) \, dt = \int \sin 2t \, dt + \int \cos \frac{t}{2} \, dt \).

Step 8: Solve the Separate Integrals (c)

Using previous results:For \( \int \sin 2t \, dt \) we found \( -\frac{1}{2} \cos 2t \), and for \( \int \cos \frac{t}{2} \, dt \) we have \( 2 \sin \frac{t}{2} \). Combine these results:\( y = -\frac{1}{2} \cos 2t + 2 \sin \frac{t}{2} + C \).

Key Concepts

Integration Techniques

Integration is a fundamental technique in calculus and is used to find the antiderivative or integral of a function. This process involves reversing the operation of differentiation to retrieve the original function from its derivative. In this exercise, we are dealing with trigonometric functions, specifically integrating derivatives that include sine and cosine terms.

One of the techniques used here is substitution. Substitution is helpful when integrating functions involving composites, such as sine and cosine with arguments like \( 2t \) or \( \frac{t}{2} \). This involves setting \( u \) equal to the argument of the trigonometric function. For example, when we substitute \( u = 2t \), it simplifies the integration process as follows:

• Convert the integral into simpler terms by substituting \( u = 2t \)
• Adjust the differential: \( du = 2 \, dt \) leading to \( dt = \frac{du}{2} \)
• The integral becomes \( \int \sin u \cdot \frac{du}{2} \)

Ultimately, substitution simplifies the integration of complex functions and is a key technique for solving integrals of derivatives from functions involving trigonometric expressions.

Function Derivatives

Derivatives are crucial in calculus, representing the rate at which a function changes. They serve as the foundation for integration, as antiderivatives \( (or \ integrals) \) effectively reverse the differentiation process. The given exercise involves finding functions whose derivatives are given.

To find a function's derivative, it's essential to understand the underlying rules of differentiation. In the examples provided, derivatives like \( y^{\prime} = \sin 2t \) and \( y^{\prime} = \cos \frac{t}{2} \) hint toward the kind of trigonometric antiderivatives we need to compute. These are determined using integral calculus, specifically reversing the derivative rules for sine and cosine:

• For \( \sin u \), its antiderivative is \(-\cos u + C\)
• For \( \cos u \), its antiderivative is \( \sin u + C\)

The constant \( C \) is crucial as it accounts for the indefinite nature of integration, indicating that there are infinite functions with the same derivative.

Trigonometric Functions

Trigonometric functions, such as sine \( (\sin) \) and cosine \( (\cos) \), are periodic functions widely used in mathematics, particularly in calculus. They are essential when dealing with problems involving rates and oscillatory motion because of their cyclic properties.

Understanding how to integrate these functions, as shown in this exercise, is fundamental in solving calculus problems. For instance, integrating \( \sin 2t \) requires recognizing that its rate of change is proportional to \(-\cos 2t\), and following integration rules. For \( \cos \frac{t}{2} \), the integration results in \( 2 \sin \frac{t}{2}\), after applying the substitution technique.

These integrations demonstrate that the amplitudes or periods of sine and cosine functions heavily influence the integration result. Mastering these integrations allows us to comprehensively understand how these functions behave over specified intervals and is crucial for applications in physics and engineering.