Problem 32 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

a. \(y = \tan \theta + C\); b. \(y = \frac{2}{3}\theta^{3/2} + C\); c. \(y = \frac{2}{3}\theta^{3/2} - \tan \theta + C\).

Step 1: Identify the Problem

We need to find functions whose derivatives match the given functions. These are integration problems, requiring us to find the indefinite integrals of the given derivatives.

Step 1: Integrate Part (a)

For part (a), the derivative is given as \(y' = \sec^2\theta\). We recognize that the derivative of the tangent function is \(\sec^2\theta\). Thus:\[y = \int \sec^2\theta \, d\theta = \tan \theta + C\]where \(C\) is the constant of integration.

Step 2: Integrate Part (b)

For part (b), the derivative is \(y' = \sqrt{\theta}\). This can be rewritten as a power function \(\theta^{\frac{1}{2}}\). We can integrate this as follows:\[y = \int \theta^{\frac{1}{2}} \, d\theta = \frac{2}{3}\theta^{\frac{3}{2}} + C\]where \(C\) is the constant of integration.

Step 3: Integrate Part (c)

For part (c), we need to integrate \(y' = \sqrt{\theta} - \sec^2\theta\). We can split this into two separate integrals:\[y = \int (\sqrt{\theta} - \sec^2\theta) \, d\theta = \int \theta^{\frac{1}{2}} \, d\theta - \int \sec^2\theta \, d\theta\]Using the results from previous steps, we can write:\[y = \frac{2}{3}\theta^{\frac{3}{2}} - \tan \theta + C\]where \(C\) is the constant of integration.

Key Concepts

Indefinite Integrals

When we work with differential equations, one common task is to find an indefinite integral, essentially reversing the process of differentiation. This task requires finding a function whose derivative matches a given function. For instance, if you're given a derivative like \( y' = \sec^2 \theta \), you're looking to find a function so that when you differentiate it, you get \( \sec^2 \theta \).

An indefinite integral is like asking, "what function originally produced this derivative?" Such a function includes a constant, represented as \( C \), because «differentiation» erases constant terms. This constant is known as the "constant of integration."

Finding indefinite integrals involves applying known rules or formulas to reverse differentiation. This is a foundational technique in calculus, utilized primarily to solve differential equations.

Integration Techniques

Different functions necessitate various integration techniques, each tailored to specific types of problems. Let’s explore some basic strategies you’d employ for the given problems:

• Power Rule for Integration: This rule is helpful when dealing with polynomial expressions. If you have \( y' = \sqrt{\theta} \), you can rewrite it as \( \theta^{\frac{1}{2}} \), and apply the power rule. You increase the exponent by one and divide by the new exponent. So, \( \int \theta^{\frac{1}{2}} \, d\theta = \frac{2}{3}\theta^{\frac{3}{2}} + C \).


• Trigonometric Integrals: When facing trigonometric derivatives like \( \sec^2 \theta \), recognize common integral forms. Here, the integration yields \( \tan \theta \), because the derivative of \( \tan \theta \) is \( \sec^2 \theta \).


• Combination of Techniques: Sometimes functions are combinations of different expressions, like \( y' = \sqrt{\theta} - \sec^2 \theta \). In this case, break down the function into simpler parts, integrate each separately, and combine the results.

Effective integration relies on recognizing the structure of the function you need to integrate and matching it with an appropriate technique.

Calculus Problems

In calculus, solving problems can involve finding functions with specific derivatives using techniques of integration. Such problems are prevalent in differential equations, where the goal is to reverse-engineer functions from their rates of change.

For example, problems like finding a function whose derivative is \( \sec^2 \theta \) require recognition of differentiation rules, then applying integration techniques to solve them. Integrating indefinitely means calculating an answer with a general form including an integration constant. This is a vital step in solving general calculus problems.

Confronted with these scenarios, the key steps typically involve identifying the type of function you're dealing with, applying the most suitable integration method, and lastly, adjusting for constants of integration. These problems form the basis of numerous applications in calculus, from physics to engineering, demonstrating how differentiation and integration are powerful tools in understanding changes and patterns in various systems.