Finding F(π/4) Given F''(x) = -sin(x) And Initial Conditions

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Introduction

In this article, we will explore a classic calculus problem involving finding the value of a function at a specific point, given its second derivative and some initial conditions. Specifically, we are given that the second derivative of a function f(x) is f''(x) = -sin(x), and we also know that the first derivative at x = 0 is f'(0) = -1 and the function's value at x = 0 is f(0) = -3. Our goal is to determine the value of the function at x = π/4, that is, f(π/4). This problem demonstrates the fundamental concepts of integration and how initial conditions are used to uniquely determine a function.

Problem Statement

We are tasked with the following:

Given:

  • f''(x) = -sin(x)
  • f'(0) = -1
  • f(0) = -3

Find:

  • f(π/4)

Solution

Step 1: Integrate f''(x) to find f'(x)

To find the first derivative, f'(x), we need to integrate the second derivative, f''(x), with respect to x. Given that f''(x) = -sin(x), we have:

∫ f''(x) dx = ∫ -sin(x) dx

The integral of -sin(x) is cos(x), plus a constant of integration, which we'll call C₁. So:

f'(x) = cos(x) + C₁

Now, we use the given initial condition f'(0) = -1 to find the value of C₁:

f'(0) = cos(0) + C₁ = -1

Since cos(0) = 1, we have:

1 + C₁ = -1

Solving for C₁, we get:

C₁ = -2

Thus, the first derivative of the function is:

f'(x) = cos(x) - 2

Step 2: Integrate f'(x) to find f(x)

Next, we integrate f'(x) to find f(x). We have:

∫ f'(x) dx = ∫ (cos(x) - 2) dx

The integral of cos(x) is sin(x), and the integral of -2 is -2x. We add another constant of integration, C₂. So:

f(x) = sin(x) - 2x + C₂

Now, we use the given initial condition f(0) = -3 to find the value of C₂:

f(0) = sin(0) - 2(0) + C₂ = -3

Since sin(0) = 0, we have:

0 - 0 + C₂ = -3

Thus,

C₂ = -3

Therefore, the function f(x) is:

f(x) = sin(x) - 2x - 3

Step 3: Evaluate f(π/4)

Finally, we evaluate f(x) at x = π/4:

f(π/4) = sin(π/4) - 2(π/4) - 3

We know that sin(π/4) = √2 / 2, so:

f(π/4) = (√2 / 2) - (π/2) - 3

This is the exact value of f(π/4). We can approximate it numerically:

f(π/4) ≈ (1.414 / 2) - (3.1416 / 2) - 3

f(π/4) ≈ 0.707 - 1.5708 - 3

f(π/4) ≈ -3.8638

Thus, the value of f(π/4) is approximately -3.8638.

Conclusion

We have successfully found the value of f(π/4) by integrating the second derivative f''(x) twice and using the given initial conditions to determine the constants of integration. The final value of f(π/4) is (√2 / 2) - (π/2) - 3, which is approximately -3.8638. This problem underscores the importance of initial conditions in determining the unique solution of a differential equation. The steps involved integration, substitution using known values, and finally, the arithmetic calculation to get to the final answer. This methodology is fundamental in solving a wide array of problems in calculus and differential equations.

Detailed Explanation of Each Step

To provide a comprehensive understanding, let's break down each step in more detail.

Step 1: Integrating f''(x) to Find f'(x) – A Closer Look

When we are given the second derivative f''(x) = -sin(x), our primary goal is to find f'(x). The process of finding the original function from its derivative is known as integration, or finding the antiderivative. The antiderivative of -sin(x) is cos(x) because the derivative of cos(x) is -sin(x). However, integration always introduces a constant of integration, often denoted as C. This is because the derivative of any constant is zero, meaning many functions could have the same derivative. Thus, f'(x) = cos(x) + C₁. To find the precise function, we need additional information, which is provided through initial conditions. We were given that f'(0) = -1. Substituting x = 0 into f'(x) gives us cos(0) + C₁ = -1. Since cos(0) = 1, the equation simplifies to 1 + C₁ = -1. Solving for C₁, we subtract 1 from both sides, yielding C₁ = -2. Therefore, the first derivative is f'(x) = cos(x) - 2. This step highlights the crucial role of initial conditions in narrowing down the infinite possibilities of antiderivatives to a single, specific function.

Step 2: Integrating f'(x) to Find f(x) – A Detailed Analysis

Having found f'(x) = cos(x) - 2, the next step is to find the original function f(x). We again perform integration, this time on f'(x). The integral of cos(x) is sin(x), and the integral of the constant -2 with respect to x is -2x. As before, we introduce another constant of integration, C₂, because the derivative of any constant is zero. Therefore, f(x) = sin(x) - 2x + C₂. To determine the specific value of C₂, we use the initial condition f(0) = -3. Substituting x = 0 into f(x) gives us sin(0) - 2(0) + C₂ = -3. Since sin(0) = 0, the equation simplifies to 0 - 0 + C₂ = -3, which immediately yields C₂ = -3. Thus, the original function is f(x) = sin(x) - 2x - 3. This step further demonstrates the significance of initial conditions in uniquely defining the function, without which we would have a family of functions differing only by a constant.

Step 3: Evaluating f(π/4) – A Step-by-Step Calculation

Now that we have found the original function f(x) = sin(x) - 2x - 3, the final step is to evaluate it at x = π/4. This involves substituting π/4 into the function: f(π/4) = sin(π/4) - 2(π/4) - 3. We know that sin(π/4) is √2 / 2, which is a standard trigonometric value derived from the unit circle or special right triangles. The term 2(π/4) simplifies to π/2. So, the expression becomes f(π/4) = (√2 / 2) - (π/2) - 3. This is the exact value of f(π/4). To get a numerical approximation, we use the approximate values √2 ≈ 1.414 and π ≈ 3.1416. Thus, f(π/4) ≈ (1.414 / 2) - (3.1416 / 2) - 3 ≈ 0.707 - 1.5708 - 3 ≈ -3.8638. This approximation provides a clearer sense of the function's value at x = π/4. The process of evaluating the function involves not only algebraic substitution but also understanding trigonometric values and performing arithmetic calculations to arrive at the final answer. This step completes the problem, showing how to move from derivatives and initial conditions to a specific functional value.

Importance of the Problem

This problem is important for several reasons:

  1. Fundamental Calculus Concepts: It reinforces the understanding of integration as the reverse process of differentiation. Students grasp how to find antiderivatives and use initial conditions to solve for constants of integration.
  2. Application of Trigonometry: The problem involves trigonometric functions and their integrals, ensuring students are comfortable with these functions in a calculus context.
  3. Problem-Solving Skills: It requires a step-by-step approach, teaching students to break down a complex problem into smaller, manageable parts.
  4. Real-World Applications: These types of problems have applications in physics, engineering, and other fields where differential equations are used to model real-world phenomena. For example, in physics, the acceleration of an object (the second derivative of position) can be used to find the velocity and position functions, given initial conditions.
  5. Mathematical Rigor: It emphasizes the importance of mathematical precision and the careful application of rules and theorems.

By working through this problem, students solidify their understanding of core calculus principles and develop valuable problem-solving skills.

Common Mistakes and How to Avoid Them

When tackling problems like this, several common mistakes can occur. Being aware of these pitfalls can significantly improve accuracy and understanding.

Forgetting the Constant of Integration

Mistake: One of the most frequent errors is forgetting to add the constant of integration (C) after each integration step. Remember, the antiderivative of a function is not unique; there is a family of functions that differ only by a constant term.

How to Avoid: Always include + C after every integration. This constant represents an infinite number of possible vertical shifts of the function. Only by using initial conditions can we determine the specific value of C.

Incorrectly Integrating Trigonometric Functions

Mistake: Another common error is misremembering the integrals of trigonometric functions. For instance, the integral of -sin(x) is cos(x), not -cos(x). A sign error here can propagate through the rest of the problem.

How to Avoid: Review the basic integrals of trigonometric functions. Create a quick reference sheet or use flashcards to memorize these. Double-check the sign when integrating trigonometric functions.

Misusing Initial Conditions

Mistake: Initial conditions are essential for finding the constants of integration, but they must be used correctly. A common mistake is plugging the initial conditions into the wrong derivative or the original function at the wrong stage.

How to Avoid: Clearly identify when and where to use each initial condition. Use f'(0) = -1 after integrating f''(x) to find C₁ in f'(x), and use f(0) = -3 after integrating f'(x) to find C₂ in f(x). Label your steps clearly to avoid confusion.

Arithmetic Errors

Mistake: Simple arithmetic errors can derail the entire solution, especially when dealing with fractions, negative signs, and trigonometric values.

How to Avoid: Write out each step clearly and double-check your calculations. Pay close attention to signs and fractions. When evaluating trigonometric functions, recall the values from the unit circle or special triangles (30-60-90 and 45-45-90 triangles).

Incorrectly Evaluating Trigonometric Functions

Mistake: Misremembering trigonometric values for common angles like π/4, π/2, etc., is a frequent mistake. For example, confusing sin(π/4) with cos(π/4) or assigning the wrong value can lead to incorrect results.

How to Avoid: Memorize the values of common trigonometric functions for key angles. Use the unit circle or special triangles as a reference. Practicing with trigonometric functions regularly can reinforce these values.

Algebraic Mistakes

Mistake: Algebraic errors, such as incorrect simplification or distribution, can lead to an incorrect final answer.

How to Avoid: Write out each algebraic step clearly and double-check your work. Use parentheses to avoid errors with distributing negative signs. If a step seems complex, break it down into smaller, more manageable steps.

By being mindful of these common mistakes and taking steps to avoid them, students can increase their accuracy and confidence when solving calculus problems involving integration and initial conditions.

Summary

In conclusion, finding f(π/4) given f''(x) = -sin(x), f'(0) = -1, and f(0) = -3 involves several key steps:

  1. Integrate f''(x) to find f'(x): ∫-sin(x) dx = cos(x) + C₁. Using f'(0) = -1, we find C₁ = -2, so f'(x) = cos(x) - 2.
  2. Integrate f'(x) to find f(x): ∫(cos(x) - 2) dx = sin(x) - 2x + C₂. Using f(0) = -3, we find C₂ = -3, so f(x) = sin(x) - 2x - 3.
  3. Evaluate f(π/4): f(π/4) = sin(π/4) - 2(π/4) - 3 = (√2 / 2) - (π/2) - 3. This value is approximately -3.8638.

This problem illustrates the fundamental concepts of integration and the importance of initial conditions in determining a unique solution. By avoiding common mistakes and carefully executing each step, students can confidently solve similar problems in calculus and related fields. Understanding these concepts is crucial for various applications in physics, engineering, and other scientific disciplines, where differential equations and their solutions play a vital role.