Problem 11 - Thomas Calculus

Question

In Exercises \(11-14\) , find the linearization of \(f\) at \(x=a\) . Then graph the linearization and \(f\) together. $$ f(x)=\sin x \quad \text { at } \quad(\text { a }) x=0, \quad \text { (b) } x=\pi $$

Step-by-Step Solution

Answer

Linearization at \( x=0 \) is \( x \); at \( x=\pi \) is \( \pi-x \).

Step 1: Understand the Problem

The problem asks us to find the linearization of the function \( f(x) = \sin x \) at two points, \( x = 0 \) and \( x = \pi \). Linearization involves finding the tangent line to the curve at a given point.

Step 2: Determine the Derivative

To find the linearization, we first need the derivative of \( f(x) = \sin x \). The derivative is \( f'(x) = \cos x \). This derivative will help calculate the slope of the tangent line at any given point.

Step 3: Calculate the Tangent Line at \( x=0 \)

For the point \( x=0 \), compute the function value \( f(0) = \sin 0 = 0 \) and the derivative \( f'(0) = \cos 0 = 1 \). The equation of the tangent line (linearization) at \( x=0 \) is found using the point-slope form: \( L(x) = f(a) + f'(a)(x-a) \). Thus, \( L(x) = 0 + 1(x-0) = x \).

Step 4: Calculate the Tangent Line at \( x=\pi \)

For the point \( x=\pi \), compute the function value \( f(\pi) = \sin \pi = 0 \) and the derivative \( f'(\pi) = \cos \pi = -1 \). The equation of the tangent line at \( x=\pi \) becomes \( L(x) = 0 - 1(x-\pi) = \pi - x \).

Step 5: Graph \( f(x) \) and Linearizations

Plot the original function \( f(x) = \sin x \) and the linearizations \( L(x) = x \) at \( x = 0 \) and \( L(x) = \pi - x \) at \( x = \pi \) on the same graph. The linearizations should appear as tangent lines to the curve at the specified points.

Key Concepts

Linearization

Linearization is a method used in calculus to approximate a complex function with a simple linear one. It is particularly helpful when you want to make quick predictions without complex calculations. The technique involves creating a tangent line at a specific point of a function. This tangent line approximates how the function behaves near that point. To linearize a function at a point, you use the formula for the tangent line:

• \[ L(x) = f(a) + f'(a)(x-a) \]

The above equation helps you find the linearization of a function \( f \) at a particular point \( x = a \). The term \( f(a) \) is the function's value at that point, and \( f'(a) \) is the slope of the tangent line at that point.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are used to describe the relationships of angles in right-angled triangles and periodic phenomena. In the given exercise, we deal with the function \( f(x) = \sin x \). Sine function attributes:

• The sine function ranges from -1 to 1.
• The sine wave repeats every \( 2\pi \).
• Key points include: \( \sin 0 = 0 \), \( \sin \frac{\pi}{2} = 1 \), \( \sin \pi = 0 \).

Understanding these key values can help simplify computations and graph plotting, especially when combined with techniques like linearization.

Tangent Line

A tangent line is a straight line that just touches a curve at one point. It represents the instantaneous direction or slope of the curve at that specific point. For effective linearization, this tangent line acts as the best straight-line approximation to a curve near a given point.In the exercise, we used tangent lines to approximate \( \sin x \) at \( x = 0 \) and \( x = \pi \):

• At \( x = 0 \), the tangent line was \( L(x) = x \).
• At \( x = \pi \), it became \( L(x) = \pi - x \).

These linearizations simplify the sine function to either a positive or negative straight line, effectively showing how the function might behave near those points.

Derivative

The derivative of a function tells us about the rate at which one quantity changes with respect to another. It is crucial for finding the slope of a tangent line, which is used in the linearization process.To linearize \( f(x) = \sin x \), we need its derivative \( f'(x) = \cos x \). This derivative function gives us the slope of the sine function at any point \( x \):

• At \( x = 0 \), the slope is \( f'(0) = \cos 0 = 1 \).
• At \( x = \pi \), the slope is \( f'(\pi) = \cos \pi = -1 \).

The computed slopes are then used to construct the linear equations for the tangent lines, reflecting how \( \sin x \) changes locally near \( x=0 \) and \( x=\pi \).