Problem 54 - Calculus Volume 1

Question

Find the linear approximation \(L(x)\) to \(y=f(x)\) near \(x=a\) for the function. \(f(x)=x \sin x, a=2 \pi\)

Step-by-Step Solution

Answer

Linear approximation is \(L(x) = 2\pi x - 4\pi^2\).

Step 1: Find the Derivative of f(x)

To find the linear approximation, we need the function's derivative. The given function is \(f(x) = x \sin x\). Use the product rule for differentiation: \(\frac{d}{dx}[uv] = u'v + uv'\) with \(u = x\) and \(v = \sin x\). This gives: \(f'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x\).

Step 2: Evaluate f(a) and f'(a)

We need the function and its derivative evaluated at \(x = a = 2\pi\). Firstly, substitute in the original function: \(f(2\pi) = 2\pi \sin(2\pi) = 2\pi \cdot 0 = 0\). For the derivative, substitute in \(f'(x) = \sin x + x \cos x\): \(f'(2\pi) = \sin(2\pi) + 2\pi \cos(2\pi) = 0 + 2\pi \cdot 1 = 2\pi\).

Step 3: Write the Linear Approximation Formula

The linear approximation \(L(x)\) of \(f(x)\) at \(x = a\) is given by the formula \(L(x) = f(a) + f'(a)(x - a)\). Substitute \(f(a) = 0\), \(f'(a) = 2\pi\) and \(a = 2\pi\) into this formula: \(L(x) = 0 + 2\pi(x - 2\pi)\).

Step 4: Simplify the Linear Approximation

Simplify the expression for \(L(x)\). This results in \(L(x) = 2\pi x - 4\pi^2\). Thus, the linear approximation of \(f(x)\) near \(x = 2\pi\) is \(L(x) = 2\pi x - 4\pi^2\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to determine the rate at which a function changes at any given point. It provides a way to understand how functions behave and is essential in finding linear approximations. In simple terms, differentiation helps us calculate the slope of a tangent line to a curve at a particular point. This slope is represented by the derivative of the function.

For the function provided, that is, \(f(x) = x \sin x\), differentiation aids us in determining how the function behaves around the point \(x = a\). To find this, we need to perform differentiation by applying specific rules like the product rule.

Product Rule

The product rule is a vital tool used in differentiation, especially when dealing with functions that are products of two other functions. It stipulates that if you have a function that is a product of two functions, say \(u(x)\) and \(v(x)\), then the derivative is given by:

• \(\frac{d}{dx}[uv] = u'v + uv'\)

For the exercise at hand, we have the function \(f(x) = x \sin x\), where we identify \(u = x\) and \(v = \sin x\).

To find \(f'(x)\), apply the product rule, resulting in:

• \(f'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x\)

The product rule reveals how the combination of two functions impacts the overall derivative.

Derivative Evaluation

Derivative evaluation involves substituting a specific value into the derivative to find how the function's rate of change behaves at that particular point. This is a key step in linear approximation, as it tells us the slope of the tangent line at that point of interest.

In our exercise, after computing \(f'(x)\), we need to evaluate it at \(x = 2\pi\). This step involves:

• Calculating \(f'(2\pi) = \sin(2\pi) + 2\pi \cos(2\pi)\)
• Knowing that \(\sin(2\pi) = 0\) and \(\cos(2\pi) = 1\), we have \(f'(2\pi) = 2\pi\)

This derivative evaluation tells us that the rate of change of our function at \(x = 2\pi\) is \(2\pi\).

Calculus Applications

Calculus applications are abundant in both theoretical and practical contexts. One such application is using derivatives and linear approximations to estimate function values nearby a given point. This has beneficial applications in fields such as physics, economics, and engineering, where exact values may be unknown or difficult to compute.

Through linear approximation, the function \(f(x) = x \sin x\) is represented linearly near \(x = 2\pi\) using:

• The approximation formula \(L(x) = f(a) + f'(a)(x - a)\)
• Given \(f(2\pi) = 0\) and \(f'(2\pi) = 2\pi\), we have \(L(x) = 0 + 2\pi(x - 2\pi)\)
• Simplifying yields \(L(x) = 2\pi x - 4\pi^2\)

This approximation provides a linear function that is easier to handle, especially in computations, while still closely representing the original function near the specified point.