Problem 37 - Calculus Single Variable

Question

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=\cos (x), c=\pi / 3 $$

Step-by-Step Solution

Answer

The linearization of \( f(x) = \cos(x) \) at \( x = \frac{\pi}{3} \) is \( L(x) = \frac{1}{2} - \frac{\sqrt{3}}{2}(x - \frac{\pi}{3}) \).

Step 1: Identify the Function and the Point of Interest

The function provided is \( f(x) = \cos(x) \) and the point at which we need to calculate the linearization is \( c = \frac{\pi}{3} \).

Step 2: Compute the Value of the Function at \( c \)

Calculate \( f(c) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).

Step 3: Determine the Derivative of the Function

The derivative of \( f(x) = \cos(x) \) is \( f'(x) = -\sin(x) \).

Step 4: Evaluate the Derivative at \( c \)

Compute the value of the derivative at \( c = \frac{\pi}{3} \): \( f'(c) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \).

Step 5: Formulate the Linearization Equation

Using the linearization formula \( L(x) = f(c) + f'(c)(x - c) \), substitute the values: \( L(x) = \frac{1}{2} - \frac{\sqrt{3}}{2}(x - \frac{\pi}{3}) \).

Step 6: Simplify the Linearization Expression

Distribute and simplify the expression if necessary to get: \( L(x) = \frac{1}{2} - \frac{\sqrt{3}}{2}x + \frac{\sqrt{3}\pi}{6} \).

Key Concepts

The Basics of Calculus

Calculus is a fundamental branch of mathematics used to study change. It encompasses two main concepts: differentiation and integration.

In our specific problem, we focus on differentiation. Calculus allows us to find instantaneous rates of changes and linear approximations. Linearization, as seen in the exercise, is one of these approximations.

Linearization provides a way to approximate a function near a point using a straight line. This is helpful because working with simple linear equations is easier than dealing with complex trigonometric functions directly.

Understanding Derivatives

The derivative of a function is a measure of how the function's output changes as its input changes. For a given function, the derivative indicates the rate of change or slope at any point.

In the context of our exercise, we consider the function \( f(x) = \cos(x) \). Its derivative is \( f'(x) = -\sin(x) \). This shows that as \( x \) changes slightly, \( f(x) \) will change at a rate adjusted by \(-\sin(x)\).

• Derivatives are essential for calculating slopes of curves.
• They help in finding maxima, minima, and points of inflection.
• Derivatives are fundamental to understanding motion and change.

In Step 4 of the exercise, we find the derivative's value at \( c = \frac{\pi}{3} \). This gives us \(-\frac{\sqrt{3}}{2}\), indicating how steep the cosine curve is at that point.

Exploring Trigonometric Functions

Trigonometric functions are periodic and describe relationships between angles and side lengths in right triangles. The most common trigonometric functions include sine, cosine, and tangent.

In the given exercise, \( f(x) = \cos(x) \) represents the cosine function. Cosine relates to the adjacent side over the hypotenuse in a right triangle. The function \( \cos(x) \) has a range from -1 to 1, and it oscillates as \( x \) changes.

• Cosine is an even function, meaning \( \cos(-x) = \cos(x) \).
• Its period is \( 2\pi \), meaning the function repeats every \( 2\pi \) units.
• At \( x = \frac{\pi}{3} \), \( \cos(x) = \frac{1}{2} \), a common trigonometric value related to a 60-degree angle.

These properties greatly assist in simplifying calculations and understanding the behavior of waves and cycles in various fields, like physics and engineering.