Problem 187 - 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. $$[\mathbf{T}] f(x)=1+\cos x, x=\frac{3 \pi}{2}$$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = x - \frac{3\pi}{2} + 1 \).

Step 1: Find the Derivative of the Function

To find the equation of the tangent line, you first need to find the derivative of the function, which gives the slope of the tangent line. The function is \( f(x) = 1 + \cos x \). The derivative of \( \cos x \) is \( -\sin x \), so the derivative of \( f(x) \) is \( f'(x) = -\sin x \).

Step 2: Calculate the Slope of the Tangent Line

Now, evaluate the derivative at the given point \( x = \frac{3\pi}{2} \) to find the slope of the tangent line. Substituting into the derivative, we get \( f'(\frac{3\pi}{2}) = -\sin(\frac{3\pi}{2}) \). Since \( \sin(\frac{3\pi}{2}) = -1 \), we have \( f'(\frac{3\pi}{2}) = 1 \).

Step 3: Calculate the Function Value at the Given Point

Next, find the value of the function at \( x = \frac{3\pi}{2} \) to determine the point of tangency. Substituting into the original function, \( f(\frac{3\pi}{2}) = 1 + \cos(\frac{3\pi}{2}) \). Since \( \cos(\frac{3\pi}{2}) = 0 \), we have \( f(\frac{3\pi}{2}) = 1 \).

Step 4: Write the Equation of the Tangent Line

The equation of a line can be written as \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. Using the point-slope form, \( y - y_1 = m(x - x_1) \), and substituting \( m = 1 \), \( x_1 = \frac{3\pi}{2} \), and \( y_1 = 1 \), we get \( y - 1 = 1\left(x - \frac{3\pi}{2}\right) \), simplifying to \( y = x - \frac{3\pi}{2} + 1 \).

Step 5: Verify by Graphing

Graph both the function \( f(x) = 1 + \cos x \) and the tangent line \( y = x - \frac{3\pi}{2} + 1 \) on a calculator. Confirm that the tangent line touches the function at the point \( (\frac{3\pi}{2}, 1) \). This ensures that the calculated tangent line equation is correct.

Key Concepts

Derivative

In calculus, the derivative of a function gives us the slope of the tangent line at any point on the graph of the function. It provides a way to determine the rate at which a function is changing. For the function given in the exercise, which is \( f(x) = 1 + \cos x \), finding the derivative helps us understand how the function moves at any particular value of \(x\). The derivative of \( \cos x \) is \(-\sin x\), so the derivative of \( f(x) \) becomes \( f'(x) = -\sin x \). This derivative tells us that the slope of the tangent line changes depending on the \(x\)-value chosen. By calculating the derivative at a specific point, such as \( x = \frac{3\pi}{2} \), we find the precise slope of the tangent line at that point. This slope is crucial for determining the exact equation of the tangent line.

Sine and Cosine Functions

The sine and cosine functions are fundamental in trigonometry and appear frequently in calculus. They are periodic functions and possess specific properties that make them unique. The cosine function, \( \cos x \), which is used in the original exercise, is known for its wave-like pattern with a period of \(2\pi\). It reaches its maximum value of 1 and its minimum value of -1. In the exercise, the function \( f(x) = 1 + \cos x \) shifts the cosine graph upwards by one unit.
For \( x = \frac{3\pi}{2} \), the value of \( \cos x \) is 0, indicating that it is at a zero-crossing point of the wave. Understanding these values helps in evaluating functions at specific points; in this problem, \( f(\frac{3\pi}{2}) = 1 + 0 = 1 \). This information not only locates us at a specific point on the graph but also aids in computing the slope and forming the tangent line equation.

Point-Slope Form

The point-slope form is a convenient way to write the equation of a line when you know the slope and a point through which the line passes. The formula is \( y - y_1 = m(x - x_1) \), where \(m\) is the slope and \((x_1, y_1)\) is the point. This form is especially helpful when dealing with tangent lines, as you will often have precisely this information. For our exercise:
- Slope \(m = 1\),
- Point \( (x_1, y_1) = \left( \frac{3\pi}{2}, 1 \right) \).
Plug these values into the point-slope form to find the equation of the tangent line: \( y - 1 = 1(x - \frac{3\pi}{2}) \). Simplifying this yields the line equation \( y = x - \frac{3\pi}{2} + 1 \). Knowing this allows us to draw the tangent line on the graph accurately.

Graphing Calculators

Graphing calculators are powerful tools for visualizing mathematical equations and checking analytical solutions. When given a function and its tangent line, as in this exercise, graphing both using a calculator can confirm the accuracy of your tangent line's equation. You input:\
- The function: \(f(x) = 1 + \cos x\)
- The tangent line: \( y = x - \frac{3\pi}{2} + 1 \)
On the graph, you should see that the line perfectly 'touches' the curve at the point \( (\frac{3\pi}{2}, 1) \). This touch point is the point of tangency, and it visually verifies that your calculations are correct. By seeing the tangent line on the graph, students can also gain a deeper understanding of how change is represented visually with the function's slope at a given point.