Problem 32 - Thomas Calculus

Question

Find all possible functions with the given derivative. \begin{equation}\quad \text { a. }y^{\prime}=2 x \quad \text { b. } y^{\prime}=2 x-1 \quad \text { c. } y^{\prime}=3 x^{2}+2 x-1\end{equation}

Step-by-Step Solution

Answer

a: \(y = x^2 + C\); b: \(y = x^2 - x + C\); c: \(y = x^3 + x^2 - x + C\).

Step 1: Understand the Problem

We are given derivatives of functions and asked to find the original functions. This is a problem of finding antiderivatives or indefinite integrals.

Step 2: Finding Antiderivative for Part a

For part a, find a function whose derivative is given by \( y' = 2x \). The integral of \( 2x \) with respect to \( x \) is \( \int 2x \, dx = x^2 + C \), where \( C \) is the constant of integration.

Step 3: Finding Antiderivative for Part b

For part b, find a function whose derivative is \( y' = 2x - 1 \). The integral is \( \int (2x - 1) \, dx = x^2 - x + C \), where \( C \) is the constant of integration.

Step 4: Finding Antiderivative for Part c

For part c, find a function whose derivative is \( y' = 3x^2 + 2x - 1 \). Integrating, we have \( \int (3x^2 + 2x - 1) \, dx = x^3 + x^2 - x + C \), where \( C \) is the constant of integration.

Step 5: Summarize the Solutions

For each part, there are infinite solutions due to the constant of integration \( C \) which can be any real number: - Part a: \( y = x^2 + C \) - Part b: \( y = x^2 - x + C \) - Part c: \( y = x^3 + x^2 - x + C \)

Key Concepts

Indefinite Integrals

Indefinite integrals are an important concept in calculus, representing the antiderivative of a function. When finding indefinite integrals, you're essentially reversing the process of differentiation. While differentiation tells you how quickly a function is changing, integration helps you find the original function from its derivative.
For example, when you have a derivative like \( y' = 2x \), you find the indefinite integral by integrating the expression with respect to \( x \). The result, \( \int 2x \, dx = x^2 + C \), gives you the family of functions that could all have \( 2x \) as their derivative. The "family" is due to the "\(+ C\)", which represents the constant of integration.
Remember, integrating is like putting together the composite pieces of a puzzle. Once combined, you reveal the larger picture or function, reflecting how the pieces (derivatives) connect.

Constant of Integration

The constant of integration, typically denoted by \( C \), is a crucial part of indefinite integrals. It signifies that when you reverse the differentiation process, there can be an infinite number of functions that share the same derivative.
Consider the simple example of the derivative \( y' = 2x \). By integrating, you obtain \( y = x^2 + C \).

• Each different value of \( C \) represents a different horizontal shift of the function \( x^2 \).
• This constant arises because when you differentiate \( x^2 + C \), the "\(+ C\)" vanishes, leaving \( 2x \).

Understanding this concept is key in solving problems involving differential equations, where identifying the constant can help determine specific solutions under certain conditions. It underscores the flexibility and breadth of potential solutions in indefinite integration, illustrating the idea that many paths can lead to the same rate of change.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. They are fundamental in expressing how quantities change and often model real-world systems, like population growth, motion, or chemical reactions.
When you're asked to find the original function from its derivative, like in the exercise above, you're dealing with a basic differential equation. For each derivative given in the exercise:

• \( y' = 2x \)
• \( y' = 2x - 1 \)
• \( y' = 3x^2 + 2x - 1 \)

You're reversing the derivative process to solve the differential equation by integration. These simple forms are often called first-order differential equations because they consist of the first derivative of a function.
By learning how to solve these equations through integration, you unlock the ability to describe systems and explore mathematical relationships in a structured, precise way. Each solution uncovered not only answers the posed problem but broadens the understanding of the behaviors and properties of functions, ultimately linking mathematics with its practical applications.