Problem 14 - Thomas Calculus

Question

In Exercises \(11-14\) , find the linearization of \(f\) at \(x=a\) . Then graph the linearization and \(f\) together. $$ f(x)=\tan x \quad \text { at } \quad(\text { a }) x=0, \quad \text { (b) } x=\pi / 4 $$

Step-by-Step Solution

Answer

The linearization at \( x = 0 \) is \( L(x) = x \) and at \( x = \pi/4 \) is \( L(x) = 2x - \pi/2 + 1 \).

Step 1: Understanding Linearization

Linearization approximates a function near a given point using the tangent line to the function at that point. The linearization of a function \( f(x) \) at \( x = a \) is given by the formula \( L(x) = f(a) + f'(a)(x-a) \). We need this linearization at two points: \( x = 0 \) and \( x = \pi/4 \).

Step 2: Calculating Linearization at x=0

First, find \( f(x) = \tan x \) evaluated at \( x = 0 \), which gives \( f(0) = \tan(0) = 0 \). Next, find the derivative \( f'(x) = \sec^2 x \), and evaluate it at \( x=0 \), giving \( f'(0) = \sec^2(0) = 1 \). Substitute into the linearization formula: \( L(x) = 0 + 1(x-0) = x \). Thus, the linearization at \( x = 0 \) is \( L(x) = x \).

Step 3: Calculating Linearization at x=\pi/4

Find \( f(x) = \tan x \) at \( x = \pi/4 \), which gives \( f(\pi/4) = \tan(\pi/4) = 1 \). Using the derivative \( f'(x) = \sec^2 x \), evaluate it at \( x = \pi/4 \), yielding \( f'(\pi/4) = \sec^2(\pi/4) = 2 \). Substitute into the linearization formula: \( L(x) = 1 + 2(x-\pi/4) \) or \( L(x) = 2x - \pi/2 + 1 \). Thus, the linearization at \( x = \pi/4 \) is \( L(x) = 2x - \pi/2 + 1 \).

Step 4: Graphing the Functions

On a graph, plot the function \( f(x) = \tan x \) and the linearizations \( L(x) = x \) at \( x = 0 \), along with \( L(x) = 2x - \pi/2 + 1 \) at \( x = \pi/4 \). You will see that the tangent lines (linearizations) closely approximate the \( \tan x \) graph near the points of tangency, illustrating the local linearization.

Key Concepts

Tangent Line Approximation

Tangent line approximation is a powerful tool in calculus for estimating the values of a function near a specific point. The process involves using the tangent line at a given point to approximate the function itself. This approach is known as linearization.
For a function \( f(x) \), the linear approximation at \( x = a \) can be represented as \( L(x) = f(a) + f'(a)(x-a) \), where \( f'(a) \) is the derivative of \( f \) at point \( a \). This line approximates \( f(x) \) when \( x \) is close to \( a \).

• Use linearization for functions that are difficult to compute exactly but are smooth enough for a linear approximation to be accurate over a small interval.
• It simplifies complex functions to straight lines at points, aiding in quick calculations and graph sketching.

The beauty of tangent line approximation lies in its simplicity and the fact that it gives insight into the behavior of a function at a specific location.

Derivative of Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent have derivatives that are used extensively in calculus. For the tangent function, \( f(x) = \tan x \), the derivative is \( f'(x) = \sec^2 x \). This derivative is crucial for evaluating the slope of the tangent line used in linearization.
When we find the derivative \( f'(x) = \sec^2 x \), we explore how the tangent function changes as \( x \) changes. Understanding these derivatives helps us determine how steep the tangent line is at different points on the graph of the function.

• The derivative \( \sec^2 x \) describes how fast the tangent function is changing. It's essential for calculating linearizations accurately.
• When \( x = 0 \) or \( x = \pi/4 \), substituting the value into \( \sec^2 x \) will give you the exact slope of the tangent line at these points.

Therefore, practicing derivatives of trigonometric functions not only supports solving linearization problems but also deepens one's understanding of calculus principles.

Graphing Functions

Graphing functions is a vital skill in mathematics, allowing us to visualize a function's behavior and its linear approximation. By plotting both the function and its tangent line, we can see how well the linear approximation matches the original function near a given point.

• For example, when graphing \( f(x) = \tan x \), it’s important to note how the function and its tangent line \( L(x) \) intersect at \( x = 0 \) and \( x = \pi/4 \).
• The tangent line at \( x = 0 \) is simply \( L(x) = x \) and at \( x = \pi/4 \), the line is \( L(x) = 2x - \pi/2 + 1 \).

By plotting these on the same graph, students visually understand how linearization provides an approximation of the tangent function near these points.
Graphing helps students see the "local" nature of approximations. It emphasizes how curves and lines intersect, revealing deeper insights into calculus. Understanding this relationship between functions and their graphs is essential for interpreting mathematical results effectively.