Problem 43 - Calculus: Early Transcendental Functions

Question

Find all functions satisfying the given conditions. $$f^{\prime \prime}(x)=3 \sin x+4 x^{2}$$

Step-by-Step Solution

Answer

The functions that meet the specified condition are of the form \(f(x) = -3 \sin x + x^4 + C_1x + C_5\), where \(C_1\) and \(C_5\) are constants.

Step 1: Integration of the function

We start by integrating \(3 \sin x + 4x^2\) with respect to \(x\). This gives us the first derivative of the function. Let's denote the first derivative as \(f'(x)\). The integral of \(3 \sin x\) is \(-3 \cos x + C_1\) where \(C_1\) is the constant of integration, and the integral of \(4x^2\) is \(\frac{4}{3} x^3 + C_2\), where \(C_2\) is the constant of integration. So, the first derivative \(f'(x) = -3 \cos x + \frac{4}{3} x^3 + C_1\).

Step 2: Further Integration

Now, we need to integrate the resulting function (the first derivative) to obtain the original function. So, we integrate \(-3 \cos x + \frac{4}{3} x^3 + C_1\). The integral of \(-3 \cos x\) is \(-3 \sin x + C_3\) where \(C_3\) is the constant of integration, and the integral of \(\frac{4}{3} x^3\) is \(x^4 + C_4\) where \(C_4\) is the constant of integration. So, the function \(f(x) = -3 \sin x + x^4 + C_1x + C_5\), where \(C_5 = C_3 + C_4\).

Step 3: Geometry Analysis

Note that \(C_1\) represents the initial velocity of the function because it is the constant term in the first derivative and \(C_5\) represents the initial position of the function because it is the constant term in the original function \(f(x)\).

Key Concepts

Integration

Integration is a fundamental concept in calculus. It's the process of finding an integral, which can be thought of as the opposite of differentiation. This means it helps us determine the function from its derivative. For our exercise, we're given the second derivative and asked to find the original function.

• We perform integration to step back through each level of derivative.
• This way, we start from the second derivative and end up with the function itself.

Each integration step gives us an integral constant often represented as \( C \), which represents the myriad functions that can have the same derivative. In simple terms, integration helps us uncover the hidden function from its rate of change, the derivative.

Differential equations

Differential equations involve equations that relate a function to its derivatives. In essence, they tell us how a function is changing. Solving a differential equation means finding a function that satisfies this relationship. In this problem, the second derivative \( f''(x) \) is provided.

• The task is to "reverse-engineer" this into a function \( f(x) \).
• This often involves multiple integration steps.

An important part of this is the initial conditions, represented by constants of integration. They allow us to account for shifts or rotations of the function curve, giving an entire family of solutions.

Trigonometric functions

Trigonometric functions, like sine and cosine, represent periodic behaviors frequently seen in calculus problems. They describe oscillations, rotations, and can be integrated to find other functions. In our exercise, the \( \sin x \) function is part of the expression we need to integrate.

• Integration of \( 3 \sin x \) results in a \( -3 \cos x \), a standard result.
• The integral reflects how trigonometric functions often translate into other forms of themselves when integrated or differentiated.

Understanding how these functions behave and transform is crucial, as they often appear in the solution to differential equations, especially in physics and engineering.

Polynomial integration

Polynomial integration is typically straightforward, involving well-known rules. For this exercise, the polynomial part is \( 4x^2 \). When integrating polynomials, each term must be integrated separately and results in an increase of the power by one.

• The integral of \( 4x^2 \) becomes \( \frac{4}{3}x^3 \).
• Each power of \( x \) is raised by one degree, and the coefficient is adjusted accordingly.

Integrating polynomials forms the basic foundation of solving more complex calculus problems, supporting both the understanding and solving of differential equations.