Problem 27 - 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. $$ \begin{array}{l}{y=\sin x, \quad-3 \pi / 2 \leq x \leq 2 \pi} \\ {x=-\pi, 0,3 \pi / 2}\end{array} $$

Step-by-Step Solution

Answer

Graph \( y = \sin x \) and its tangents at \( x = -\pi, 0, 3\pi/2 \) with equations: \( y = -x - \pi \), \( y = x \), and \( y = -1 \).

Step 1: Find Derivative of the Curve

The curve given is \( y = \sin x \). The derivative of a curve represents the slope of the tangent at any point on the curve. For \( y = \sin x \), the derivative is \( y' = \cos x \).

Step 2: Calculate Slope of Tangents

We need to find the slope of the tangents at \( x = -\pi, 0, 3\pi/2 \). Using the derivative from Step 1:- At \( x = -\pi \): \( y' = \cos(-\pi) = -1 \).- At \( x = 0 \): \( y' = \cos(0) = 1 \).- At \( x = 3\pi/2 \): \( y' = \cos(3\pi/2) = 0 \).

Step 3: Determine Point on Curve

For each \( x \) value, we calculate \( y \):- At \( x = -\pi \): \( y = \sin(-\pi) = 0 \).- At \( x = 0 \): \( y = \sin(0) = 0 \).- At \( x = 3\pi/2 \): \( y = \sin(3\pi/2) = -1 \).

Step 4: Write Equations of Tangents

Using the slope-point form \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point and \( m \) is the slope:- At \( x = -\pi \): The point is \((-\pi, 0)\) and the slope is \(-1\), giving the equation: \( y - 0 = -1(x + \pi) \Rightarrow y = -x - \pi \).- At \( x = 0 \): The point is \((0, 0)\) and the slope is \(1\), giving the equation: \( y - 0 = 1(x - 0) \Rightarrow y = x \).- At \( x = 3\pi/2 \): The point is \((3\pi/2, -1)\) and the slope is \(0\), giving the equation: \( y + 1 = 0(x - 3\pi/2) \Rightarrow y = -1 \).

Step 5: Graph the Curve and Tangents

Draw the graph of \( y = \sin x \) over the interval \(-3\pi/2 \leq x \leq 2\pi \). On the same graph, draw the tangent lines:- \( y = -x - \pi \) at \( x = -\pi \)- \( y = x \) at \( x = 0 \)- \( y = -1 \) at \( x = 3\pi/2 \). Each tangent should touch the curve at its respective x-coordinate and follow the line described by its equation.

Key Concepts

Derivative of Sine Function

When studying the trigonometric function \( y = \sin x \), one of the fundamental concepts is understanding its derivative. The derivative of a function is crucial because it tells us the slope of the tangent line to the curve at any given point along the curve. For the sine function, the derivative is \( y' = \cos x \). This result comes from a fundamental rule of differentiation in calculus related to trigonometric functions.
To understand this more clearly, consider the behavior of the sine wave. As the function \( y = \sin x \) transitions through its periodic cycle, the slope of the tangent line at any point on the curve is simply the cosine of that same point. This means:

• Where the sine function has peaks and troughs (at \( x = \frac{\pi}{2}, \frac{3\pi}{2} \)), the derivative \( \cos x \) is zero, indicating a horizontal tangent line.
• At points where the slope of the sine wave is steepest (at \( x = 0, \pi, 2\pi \)), the derivative reaches its maximum or minimum value (\( +1 \) or \( -1 \)).

Recognizing this derivation provides insight not only into the nature of the sine wave but also into the application of derivatives in graphing and understanding function behavior.

Tangent Line Equations

The equation of a tangent line is derived from the slope-intercept form, \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point of tangency on the curve, and \( m \) is the slope, obtained using the derivative.
Let's break down how to find tangent line equations for the sine function \( y = \sin x \) at specific points:

• At \( x = -\pi \), the sine value is zero, so the point is \((-\pi, 0)\). With a slope \(-1\), the tangent line equation is \( y = -x - \pi \).
• At \( x = 0 \), where the curve crosses the origin \((0, 0)\), and the slope is \(1\), the tangent line simplifies to \( y = x \).
• For \( x = \frac{3\pi}{2} \), \( \sin x = -1 \) at this point, and the slope is \(0\), indicating a horizontal line, giving us the equation \( y = -1 \).

Understanding tangent line equations reinforces the relationship between the function's geometry and its rate of change. Each tangent provides a linear approximation of the curve near its point of tangency.

Slope of Tangent

The concept of the slope of a tangent is pivotal when analyzing the behavior of curves, such as \( y = \sin x \). The slope informs us how rapidly the function is increasing or decreasing at a particular point. This slope is calculated using the derivative of the function.
In our example of \( y = \sin x \), using the derivative \( y' = \cos x \), we compute slopes at specific points:

• At \( x = -\pi \), the slope is \( -1 \), indicating a descending tangent direction.
• At \( x = 0 \), the slope comes out as \( 1 \), showing an upward incline at the origin.
• At \( x = \frac{3\pi}{2} \), the slope is \( 0 \), reflecting a flat, horizontal tangent.

Understanding these slopes is akin to understanding the behavior of the curve locally at each point. It helps to visualize how the function bends and twists across its domain, anticipating future changes and how these changes align with observed data. This makes slopes invaluable in both mathematical explorations and real-world applications, where predicting trends and changes can be critical.