Problem 30 - Thomas Calculus

Question

In Exercises \(27-30\) , graph the curves over the given intervals, together with their tangents at the given values of \(x\) . Label each curve and tangent with its equation. $$ \begin{array}{l}{y=1+\cos x, \quad-3 \pi / 2 \leq x \leq 2 \pi} \\ {x=-\pi / 3,3 \pi / 2}\end{array} $$

Step-by-Step Solution

Answer

The equations for the tangent lines are \( y = \frac{\sqrt{3}}{2}(x + \frac{\pi}{3}) + \frac{3}{2} \) and \( y = x - \frac{3\pi}{2} + 1 \).

Step 1: Find the Derivative of the Function

The function given is \( y = 1 + \cos x \). To find the tangents, we need the derivative of \( y \), which represents the slope of the tangent lines. The derivative of \( y = 1 + \cos x \) is \( y' = -\sin x \).

Step 2: Compute the Slope of the Tangent Lines

Evaluate the derivative at the given points \( x = -\frac{\pi}{3} \) and \( x = \frac{3\pi}{2} \) to find the slope of the tangent lines.- For \( x = -\frac{\pi}{3} \), the slope \( m = - \sin(-\frac{\pi}{3}) = - (-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} \).- For \( x = \frac{3\pi}{2} \), the slope \( m = - \sin(\frac{3\pi}{2}) = 1 \).

Step 3: Determine the Tangent Line Equations

To find the equation of the tangent lines, use the point-slope form: \( y - y_1 = m(x - x_1) \).- At \( x = -\frac{\pi}{3} \): The point is \( \left(-\frac{\pi}{3}, 1 + \cos(-\frac{\pi}{3}) = 1 + \frac{1}{2} = \frac{3}{2}\right) \). The tangent line is \( y - \frac{3}{2} = \frac{\sqrt{3}}{2}(x + \frac{\pi}{3}) \).- At \( x = \frac{3\pi}{2} \): The point is \( \left(\frac{3\pi}{2}, 1 + \cos(\frac{3\pi}{2}) = 1 \right) \). The tangent line is \( y - 1 = 1(x - \frac{3\pi}{2}) \), which simplifies to \( y = x - \frac{3\pi}{2} + 1 \).

Step 4: Sketch the Graph

Plot the curve \( y = 1 + \cos x \) over the interval \( -\frac{3\pi}{2} \leq x \leq 2\pi \). Draw the tangent lines:- The tangent line at \( x = -\frac{\pi}{3} \) with equation \( y = \frac{\sqrt{3}}{2}(x + \frac{\pi}{3}) + \frac{3}{2} \).- The tangent line at \( x = \frac{3\pi}{2} \) with equation \( y = x - \frac{3\pi}{2} + 1 \).Label each curve and tangent with its respective equation.

Key Concepts

Derivative

Understanding derivatives is crucial in calculus, as they give us the rate at which a function is changing at any given point. For a function, the derivative is essentially the function's slope or steepness. If you imagine the graph of a function as a mountain range and you're climbing at a specific point, the derivative tells you how steep the slope is right at your position.
In our problem, the function given is \( y = 1 + \cos x \). To find how fast this function changes or "grows" at any point \( x \), we calculate its derivative. The derivative of \( \cos x \) is \( -\sin x \). Hence, the derivative of our function \( y = 1 + \cos x \) becomes \( y' = -\sin x \). This tells us how the function will increase or decrease at any particular point on the graph based on the value of \( x \).
Derivatives are fundamental when working to find tangent lines to a function's graph as they provide the necessary slopes.

Tangent Line

Tangent lines are straight lines that just touch a curve at a single point without crossing it. They are essential in calculus as they help illustrate how a function behaves locally around a given point.
When you graph a function and pick any point \( x \) on the curve, the tangent line at this point gives the best linear approximation to the graph around that point. The slope of this tangent line is determined by the derivative at that point. So if the function is constantly rising, the tangent line will have a positive slope; if it's falling, the slope will be negative.
For our example, the tangent lines at two points, \( x = -\frac{\pi}{3} \) and \( x = \frac{3\pi}{2} \), were found by using the derivative \( y' = -\sin x \). Calculating \( -\sin(-\frac{\pi}{3}) \), we get a slope of \( \frac{\sqrt{3}}{2} \). At \( x = \frac{3\pi}{2} \), the slope is \( 1 \). These slopes help in constructing the equations of the tangent lines using the point-slope form from algebra.

Graphing Functions

Graphing functions is a core skill in calculus that allows visualizing how functions behave over specific intervals. It shows where functions rise and fall and provides insights into their periodicity and symmetry.
In the given exercise, the function \( y = 1 + \cos x \) needs to be graphed over the interval \( -\frac{3\pi}{2} \leq x \leq 2\pi \). This helps us visualize where the two tangent lines intersect the curve and illustrates the periodic nature of cosine functions, which exhibit repeating patterns.
By graphing both the main function and tangent lines, we're not only finding immediate solutions to the exercise but also learning to interpret the shape and features of the curve. This skill is crucial for higher-level mathematics and physics where function graphs model real-world phenomena.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are periodic and extremely common in several branches of mathematics and its applications. They are vital in physics, engineering, and even economics for modeling cyclical patterns.
The cosine function, \( \cos x \), maps an angle \( x \) to the x-coordinate of a point on the unit circle. It is periodic, meaning it repeats its values over regular intervals. This property is represented in trigonometric graphs, with cosine cycles every \( 2\pi \).

• At \( x = 0 \), \( \cos x = 1 \).
• At \( x = \pi \), \( \cos x = -1 \).
• The cycle undergoes a full repetition as \( x \) increases from \( 0 \) to \( 2\pi \).

In the problem, the function \( y = 1 + \cos x \) demonstrates how a standard trigonometric function can be shifted vertically. This adjustment simply moves the baseline of the cosine function up by one unit. Observing these changes and patterns is key to mastering trigonometric applications.