Problem 188 - Calculus Volume 1

Question

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. $$ f(x)=\sec x, x=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

Tangent line equation: \(y = \sqrt{2}x - \frac{\pi}{2} + \sqrt{2}\).

Step 1: Find the derivative of the function

To find the equation of the tangent line, we first need to find the derivative of the function. The derivative of \(f(x) = \sec x\) is given by \(f'(x) = \sec x \cdot \tan x\). This will be used to find the slope of the tangent line at \(x = \frac{\pi}{4}\).

Step 2: Calculate the slope at the given point

Substitute \(x = \frac{\pi}{4}\) into the derivative to find the slope of the tangent line. Thus, \(f'(\frac{\pi}{4}) = \sec(\frac{\pi}{4}) \cdot \tan(\frac{\pi}{4}) = \sqrt{2} \cdot 1 = \sqrt{2}\). So, the slope \(m\) of the tangent line is \(\sqrt{2}\).

Step 3: Find the y-coordinate of the function at the given point

Next, we find the y-coordinate of the point by substituting \(x = \frac{\pi}{4}\) into the original function: \(f(\frac{\pi}{4}) = \sec(\frac{\pi}{4}) = \sqrt{2}\). Thus, the point of tangency is \( (\frac{\pi}{4}, \sqrt{2}) \).

Step 4: Write the equation of the tangent line

Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is \((\frac{\pi}{4}, \sqrt{2})\) and \(m = \sqrt{2}\), substitute these values into the formula: \[ y - \sqrt{2} = \sqrt{2}(x - \frac{\pi}{4}) \] Simplifying gives:\[ y = \sqrt{2}x - \frac{\pi}{2} + \sqrt{2} \] as the equation for the tangent line.

Step 5: Graph the function and the tangent line

Use a graphing calculator or software to plot \(f(x) = \sec x\) and the line \(y = \sqrt{2}x - \frac{\pi}{2} + \sqrt{2}\). Ensure the tangent line only touches the curve at \(x = \frac{\pi}{4}\) confirming the line's correctness.

Key Concepts

Derivative of a Function

To find a tangent line to a curve at a specific point, the derivative of the function comes into play. The derivative gives us a powerful tool by telling us the slope of the tangent line at any given point on the function. In simpler terms, the derivative denotes how fast a function is changing at any given point.

For the function \(f(x) = \sec x\), the derivative is \(f'(x) = \sec x \cdot \tan x\). This result follows known trigonometric derivatives and helps us find the exact slope of the curve at any point \(x\).

Here, the specific value \(x = \frac{\pi}{4}\) was given, so substituting \(x = \frac{\pi}{4}\) into the derivative \(f'(x)\), you find the slope of the tangent line is \(\sqrt{2}\). This slope will be vital for forming the equation of the tangent line. Derivatives not only show us the tangent's slope but also highlight points of increase or decrease on the graph.

Trigonometric Functions

Trigonometric functions like secant, tangent, sine, and cosine are periodic and are among the primary building blocks in mathematics for analyzing cycles and waves. In this particular problem, we are interested in the secant function, \(\sec x = \frac{1}{\cos x}\).

At \(x = \frac{\pi}{4}\), a popular angle in trigonometry, \(\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\), hence \(\sec(\frac{\pi}{4}) = \sqrt{2}\). Similarly, the \(\tan(\frac{\pi}{4}) = 1\). These calculated values make \(f'(\frac{\pi}{4}) = \sqrt{2}\) when using the derivative \(f'(x) = \sec x \cdot \tan x\).

Trigonometric evaluations at specific angles such as \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), and 0 are frequently used in calculus problems, making them critical for anyone studying the subject to understand fully. They simplify expressions during calculations and are often taught with the unit circle.

Point-Slope Form

To write the equation of a tangent line, the point-slope form is a handy tool. The point-slope form of a line is expressed as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a known point on the line.

In our scenario, the slope is \(\sqrt{2}\) derived from the derivative at \(x = \frac{\pi}{4}\), and the point is \((\frac{\pi}{4}, \sqrt{2})\), taken from evaluating the secant function at this angle.

Plug these into the point-slope formula to find the tangent line:

• \(x_1 = \frac{\pi}{4}\)
• \(y_1 = \sqrt{2}\)
• \(m = \sqrt{2}\)

This results in the equation \(y - \sqrt{2} = \sqrt{2}(x - \frac{\pi}{4})\). Simplifying leads to \(y = \sqrt{2}x - \frac{\pi}{2} + \sqrt{2}\). This tangent line gives us a straight line that just gently "kisses" the secant curve exactly at \(x = \frac{\pi}{4}\).