Problem 28 - Essential Calculus Early Transcendentals

Question

Find \(f\) $$f^{\prime}(x)=4 / \sqrt{1-x^{2}}, \quad f\left(\frac{1}{2}\right)=1$$

Step-by-Step Solution

Answer

The function is \( f(x) = 4 \arcsin(x) + 1 - \frac{2\pi}{3} \).

Step 1: Recognize the Integral Form

To find the function \( f(x) \), we need to integrate the given derivative \( f'(x) = \frac{4}{\sqrt{1 - x^2}} \). This resembles the derivative of \( \arcsin(x) \), so we identify that the antiderivative is related to \( \arcsin(x) \).

Step 2: Calculate the Indefinite Integral

Integrate \( \frac{4}{\sqrt{1 - x^2}} \) with respect to \( x \):\[ f(x) = \int \frac{4}{\sqrt{1 - x^2}} \, dx = 4 \arcsin(x) + C\]Here, \( C \) is the constant of integration.

Step 3: Apply the Initial Condition

Use the given condition \( f\left(\frac{1}{2}\right) = 1 \) to find the constant \( C \):\[ 1 = 4 \arcsin\left(\frac{1}{2}\right) + C\]The value of \( \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \). Substituting this, we get:\[ 1 = 4 \times \frac{\pi}{6} + C \Rightarrow 1 = \frac{2\pi}{3} + C\]

Step 4: Solve for the Constant \( C \)

Rearrange the equation to solve for \( C \):\[ C = 1 - \frac{2\pi}{3}\]Now, substitute \( C \) back into the expression for \( f(x) \).

Step 5: Write the Final Expression for \( f(x) \)

The function \( f(x) \) is given by:\[ f(x) = 4 \arcsin(x) + 1 - \frac{2\pi}{3}\]This function satisfies both the derivative condition and the initial value \( f\left(\frac{1}{2}\right) = 1 \).

Key Concepts

Arcsin Function

The arcsin function, denoted as \( \arcsin(x) \), is the inverse of the sine function restricted to \([-\frac{\pi}{2}, \frac{\pi}{2}]\). It returns the angle whose sine is the given number, \(x\). In trigonometry, the arcsin function plays a crucial role because it allows us to find angles based on sine values.

For example, if \( \sin(\theta) = x \), then \( \theta = \arcsin(x) \). This is used in solving problems where the outcome is an angle, especially in calculus and geometry.

In the context of integration, the derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \). Thus, when you encounter this form in an integral, recognizing it can quickly lead you to identify the integral of the function. This derivation is a nice link between trigonometry and calculus. Often, spotting such relationships simplifies the process of finding antiderivatives.

Initial Value Problem

An initial value problem (IVP) in calculus involves finding a function that satisfies a given differential equation and an initial condition. Specifically, you know both a derivative and one specific value of the function, and you use this to solve the function altogether.

In this exercise, we had \( f^{\prime}(x)=\frac{4}{\sqrt{1-x^{2}}} \) as our derivative and needed to find \( f(x) \), subject to the condition \( f\left(\frac{1}{2}\right)=1 \).

To solve an IVP:

• First, find the antiderivative of the given derivative to determine the general form of the function.
• Then, use the initial condition to solve for any integration constants involved.

This two-step method allows you to find the particular solution that not only satisfies the differential equation but also fits the initial value condition given in the problem.

Integration Technique

Integration is a fundamental concept in calculus, used to find antiderivatives or the area under curves. One common integration technique involves recognizing the forms that match well-known derivatives, as seen with the arcsin function.

In this problem, integrating \( \frac{4}{\sqrt{1 - x^2}} \) is simplified by recognizing it as a derivative form of \( \arcsin(x) \). This recognition allows us to directly apply:

• \( \int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x) + C \)
• Then, multiplying by 4 gives us the integral \( 4\arcsin(x) + C \)

Once the antiderivative is found, apply the initial condition to solve for the constant \( C \).

By learning to spot these forms and use properties of inverse trigonometric functions, integration can become straightforward. This efficiency is vital in solving calculus problems rapidly and accurately.