Problem 31 - Thomas Calculus

Question

Find all possible functions with the given derivative. a. \(y^{\prime}=\sin 2 t \quad\) 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: Understanding the Problem

We are required to find the functions whose derivatives match those given in the exercise. This involves integrating each given function.

Step 2: Solve for Part (a)

The derivative is given as \(y' = \sin 2t\). To find the original function \(y\), we need to integrate \(\sin 2t\) with respect to \(t\).

Step 3: Integration for Part (a)

The integral of \(\sin 2t\) is \(-\frac{1}{2}\cos 2t\). Thus, the function is \(y = -\frac{1}{2}\cos 2t + C\), where \(C\) is the constant of integration.

Step 4: Solve for Part (b)

The derivative is given as \(y' = \cos \frac{t}{2}\). We will find \(y\) by integrating this expression.

Step 5: Integration for Part (b)

The integral of \(\cos \frac{t}{2}\) is \(2\sin \frac{t}{2}\). So, the function is \(y = 2\sin \frac{t}{2} + C\), where \(C\) is the constant of integration.

Step 6: Solve for Part (c)

The derivative is given as \(y' = \sin 2t + \cos \frac{t}{2}\). We need to find the integral of this combined function.

Step 7: Integration for Part (c)

Integrate \(\sin 2t + \cos \frac{t}{2}\) to get \(-\frac{1}{2}\cos 2t + 2\sin \frac{t}{2}\). Hence, the function is \(y = -\frac{1}{2}\cos 2t + 2\sin \frac{t}{2} + C\), where \(C\) is the constant of integration.

Key Concepts

Integration

Integration is an essential concept in calculus that allows us to find the original function from its derivative. Think of it as the reverse process of differentiation.

• When you differentiate a function, you find its rate of change.
• Integration lets you find what the function was before differentiation.

In our exercise, when given a derivative, we're asked to perform integration to determine the original function. This process involves integrating the given expression with respect to the variable, commonly turning a derivative like \( y' = \, \text{expression} \, \) into \( y = \, \text{integrated expression} + C \, \) where \( C \, \) is the constant of integration.
This constant represents any constant value that could have been part of the original function, which nullifies its derivative.

Antiderivatives

The antiderivative, or the indefinite integral of a function, is the function you get when you integrate. It's the opposite of finding the derivative.

• Every function can have multiple antiderivatives, differing by a constant.
• The general formula for an antiderivative adds a \( + C \, \) where \( C \, \) is any real number.

For instance, the antiderivative of \( \sin 2t \, \) is \( -\frac{1}{2}\cos 2t + C \, \) because when you differentiate \( -\frac{1}{2}\cos 2t + C \, \) with respect to \( t \, \), you get \( \sin 2t \, \) back.
The process of finding an antiderivative allows us to reconstruct functions when provided with their rates of change, often encountered in real-world applications.

Trigonometric Functions

Trigonometric functions like \( \sin t \, \) and \( \cos t \, \) are periodic functions that repeat in regular intervals, and they play a significant role in calculus, especially when dealing with integration.

• These functions have known integrals that help us solve many problems in calculus.
• Integrating \( \sin kx \, \) often involves obtaining a result of the form \( -\frac{1}{k}\cos kx + C \, \).

Similarly, integrating \( \cos kx \, \) gives \( \frac{1}{k}\sin kx + C \, \).
In the exercise, the expressions \( \sin 2t \, \) and \( \cos \frac{t}{2} \, \) required integration rules for trigonometric functions. These rules simplify the process of determining the original function, especially when trigonometric terms are involved in the derivative.