Problem 188 - Calculus Volume 1

Question

For the following exercises, find the equation of the tangent line to each of the given functions at the indicated values of \(x .\) Then use a calculator to graph both the function and the tangent line to ensure the equation for the tangent line is correct. $$[\mathbf{T}] f(x)=\sec x, x=\frac{\pi}{4}$$

Step-by-Step Solution

Answer

The tangent line at \( x = \frac{\pi}{4} \) is \( y = \sqrt{2}x - \frac{\pi \sqrt{2}}{4} + \sqrt{2} \).

Step 1: Understand the Problem

To find the equation of the tangent line to the function \( f(x) = \sec x \) at \( x = \frac{\pi}{4} \), we need to determine the slope of the tangent line at this point and the function's value at this point.

Step 2: Find the Derivative

The derivative of \( f(x) = \sec x \) is \( f'(x) = \sec x \tan x \). We'll use this derivative to find the slope of the tangent line at \( x = \frac{\pi}{4} \).

Step 3: Evaluate the Derivative at \( x = \frac{\pi}{4} \)

Substitute \( x = \frac{\pi}{4} \) into the derivative: \( f'(\frac{\pi}{4}) = \sec(\frac{\pi}{4}) \tan(\frac{\pi}{4}) \). We know \( \sec(\frac{\pi}{4}) = \sqrt{2} \) and \( \tan(\frac{\pi}{4}) = 1 \), so \( f'(\frac{\pi}{4}) = \sqrt{2} \cdot 1 = \sqrt{2} \). This is the slope \( m \) of the tangent line.

Step 4: Find \( f(\frac{\pi}{4}) \)

Calculate the function value at the point: \( f(\frac{\pi}{4}) = \sec(\frac{\pi}{4}) = \sqrt{2} \). This gives us the point \( (\frac{\pi}{4}, \sqrt{2}) \) on the tangent line.

Step 5: Write the Equation of the Tangent Line

Use the point-slope form of a line equation \( y - y_1 = m(x - x_1) \). Substituting the point \((x_1, y_1) = (\frac{\pi}{4}, \sqrt{2})\) and \( m = \sqrt{2} \), we get: \[ y - \sqrt{2} = \sqrt{2}(x - \frac{\pi}{4}) \].Simplifying, \[ y = \sqrt{2}x - \frac{\pi \sqrt{2}}{4} + \sqrt{2}. \]

Step 6: Verify with a Graph

Use a calculator or graphing software to plot both \( f(x) = \sec x \) and the tangent line equation \( y = \sqrt{2}x - \frac{\pi \sqrt{2}}{4} + \sqrt{2} \). Ensure they touch at \( x = \frac{\pi}{4} \) without crossing, confirming the tangent line is correct.

Key Concepts

Secant Function

The secant function, denoted by \( \sec x \), is one of the six trigonometric functions and can be defined as the reciprocal of the cosine function. Mathematically, it is expressed as \( \sec x = \frac{1}{\cos x} \). This implies that wherever \( \cos x \) equals zero, the secant function is undefined, leading to vertical asymptotes on the graph at those points.

Key features of the secant function include:

• Periodicity: Like most trigonometric functions, secant is periodic with a period of \( 2\pi \).
• Range: The function value of secant is always greater than or equal to 1 or less than or equal to -1, thus it does not touch or stay within \([-1, 1]\).
• Vertical Asymptotes: Occur at every point where \( \cos x = 0 \), commonly found at odd multiples of \( \frac{\pi}{2} \).

Understanding secant helps in solving trigonometric equations and in analyzing waveforms, oscillations, and other phenomena modeled by trigonometric relationships.

Derivatives

Derivatives are a core concept in calculus used to determine the rate at which a function is changing at any given point. They are fundamentally the slope of the tangent line at any point on a curve.

For the secant function specifically, the derivative \( f'(x) = \sec x \tan x \) can be derived by applying the rules of differentiation to \( \sec x = \frac{1}{\cos x} \). The derivative provides crucial information about how the secant graph behaves, such as where it is increasing or decreasing.

Applications of derivatives include:

• Finding critical points where the slope is zero or undefined.
• Analyzing the concavity of functions which gives insight into how the graph curves.
• Optimizing functions to find maximum or minimum values in real-world problems.

Point-Slope Form

Point-slope form is an essential formula for writing the equation of a line when you know a point on the line and the slope. It is expressed as \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the given point and \(m\) is the slope.

This form is particularly helpful for finding tangent lines to curves. Once you have the slope from the derivative and a point from the function value, you can apply this formula to determine the precise equation of the tangent line at that particular point.

Benefits of point-slope form include:

• Simplicity: It connects directly the concept of slope to the line's equation.
• Flexibility: Easily convert to other forms like slope-intercept or general form.
• Practicality: Useful in various fields like physics for instantaneous velocity and other dynamic measurements.

Using this form, we crafted the tangent line equation \( y = \sqrt{2}x - \frac{\pi \sqrt{2}}{4} + \sqrt{2} \). This provides a clear understanding and validation of how the line behaves at the specified point on the secant function.