Problem 34 - Thomas Calculus in SI Units

Question

Find all possible functions with the given derivative. a. \(y^{\prime}=\frac{1}{2 \sqrt{x}}\) b. \(y^{\prime}=\frac{1}{\sqrt{x}} \quad\) c. \(y^{\prime}=4 x-\frac{1}{\sqrt{x}}\)

Step-by-Step Solution

Answer

a. \(y=\sqrt{x}+C\); b. \(y=2\sqrt{x}+C\); c. \(y=2x^{2}-2\sqrt{x}+C\).

Step 1: Understand the Problem

We are given derivatives of functions and need to find the original functions. This is an integration problem where we need to find the antiderivatives.

Step 2: Integrate Part (a)

We need to find the function whose derivative is \( y^{\prime}=\frac{1}{2\sqrt{x}} \). By integrating \( \frac{1}{2\sqrt{x}} \), we rewrite it as \( \frac{1}{2}x^{-1/2} \). The antiderivative of \( x^{-1/2} \) is \( 2x^{1/2} \), so integrate to get \( y=\frac{1}{2}\cdot2x^{1/2}+C=\sqrt{x}+C \).

Step 3: Integrate Part (b)

For \( y^{\prime}=\frac{1}{\sqrt{x}} \), rewrite as \( x^{-1/2} \). Integrating \( x^{-1/2} \) gives \( 2x^{1/2} \). Thus, the function is \( y=2x^{1/2}+C \).

Step 4: Integrate Part (c)

For \( y^{\prime}=4x-\frac{1}{\sqrt{x}} \), split into two separate parts: \( 4x \) and \( -x^{-1/2} \). Integrate \( 4x \) to get \( 2x^{2} \) and integrate \( -x^{-1/2} \) to get \( -2x^{1/2} \). Therefore, the function is \( y=2x^{2}-2x^{1/2}+C \).

Key Concepts

Antiderivative

The antiderivative is an essential concept in calculus. It involves finding a function whose derivative is a given function. This is also known as finding the indefinite integral. In our original exercise, you needed to find the antiderivatives for three different derivatives.

Let's understand this with an example. Suppose you have the derivative function \( y' = \frac{1}{2\sqrt{x}} \). The task is to find the original function \( y \). You do this by integrating the given derivative. In mathematical terms, we write this as:

• Rewriting the expression: \( \frac{1}{2\sqrt{x}} \) becomes \( \frac{1}{2}x^{-1/2} \).
• Finding the antiderivative: Integrate to get \( \sqrt{x} \).
• Don't forget +C: All antiderivatives include an arbitrary constant \( C \), since derivatives of constant values are zero.

This results in the original function being \( y = \sqrt{x} + C \).

Antiderivatives help us in solving many real-world problems, especially in scenarios involving rates and accumulations.

Derivative

The derivative is one of the core building blocks of calculus. It measures how a function changes as its input changes, essentially capturing the idea of rate of change or slope of the function.

In the context of our exercise, the given functions were already in their derivative form. For example, \( y' = \frac{1}{\sqrt{x}} \) shows how the function's output changes as \( x \) changes. This derivative leads us back to figuring out the original function through integration.

Some key points about derivatives include:

• Instantaneous Rate of Change: Derivatives tell us the immediate rate of change of a function.
• Relation with Tangents: The derivative at a point gives the slope of the tangent line to the curve at that point.
• Sensitivity Analysis: In various contexts, derivatives can show how sensitive a system or process is to change.

Understanding derivatives is crucial as they pave the way to finding antiderivatives, thus solving integration problems.

Calculus

Calculus is a branch of mathematics that studies continuous change. It has two major branches: differential calculus and integral calculus.

Differential calculus is primarily concerned with derivatives, focusing on the concept of change. Integral calculus deals with integrals or antiderivatives and is used to find areas, volumes, and other quantities under curves.

In our exercise:

• We performed integral calculus by integrating the derivatives to determine the original functions.
• This allowed us to move from the derivative back to the initial function, using the antiderivative and adding an arbitrary constant \( C \) to reflect all possible solutions.

Calculus is fundamental for fields that require an understanding of dynamic change, such as physics, engineering, economics, and many more. It's a tool that helps in solving complex problems by breaking them down into simpler elements and understanding both the small-scale changes and the larger accumulations.