Problem 23 - Calculus

Question

Draw the graph of \(y=\sin \frac{1}{2} x .\) State its slope at \(x=0, \pi / 2\), \(\pi,\) and \(2 \pi .\) Does \(\frac{1}{2} \sin x\) have the same slopes?

Step-by-Step Solution

Answer

The slopes are \( \frac{1}{2}, \frac{\sqrt{2}}{4}, 0, -\frac{1}{2} \); \( \frac{1}{2} \sin x \) has different slopes.

Step 1: Understand the Function

The function given is a sinusoidal wave, specifically \( y = \sin \frac{1}{2}x \). This represents a sine wave that is horizontally stretched relative to the standard \( y = \sin x \) because the argument of the sine function, \( x \), is divided by 2.

Step 2: Identify the Period

The standard period of a sine function \( y = \sin x \) is \( 2\pi \). Since we are dealing with \( y = \sin \frac{1}{2}x \), the period becomes \( 4\pi \). This is due to the fact that the period is calculated as \( \frac{2\pi}{\frac{1}{2}} = 4\pi \).

Step 3: Determine the Derivative

To find the slope of the curve, we take the derivative of the function. The derivative of \( y = \sin \frac{1}{2}x \) is given by \( y' = \frac{1}{2}\cos \frac{1}{2}x \). This tells us how the slope of the tangent line changes as \( x \) changes.

Step 4: Calculate Slopes at Specific Points

Now calculate the slopes at \( x=0, \frac{\pi}{2}, \pi, \text{and} \;2\pi \) using \( y' = \frac{1}{2}\cos \frac{1}{2}x \).- At \( x=0 \): \( y' = \frac{1}{2}\cos 0 = \frac{1}{2} \)- At \( x=\frac{\pi}{2} \): \( y' = \frac{1}{2}\cos \frac{\pi}{4} = \frac{1}{2}\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4} \)- At \( x=\pi \): \( y' = \frac{1}{2}\cos \frac{\pi}{2} = 0 \)- At \( x=2\pi \): \( y' = \frac{1}{2}\cos \pi = -\frac{1}{2} \)

Step 5: Compare with \( \frac{1}{2} \sin x \)

The function \( \frac{1}{2} \sin x \) would have the derivative \( \frac{1}{2} \cos x \). Comparing their slopes at 0, \( \frac{\pi}{2} \), \( \pi \), and \( 2\pi \), these are:- At \( x = 0 \): Slope is \( \frac{1}{2} \)- At \( x = \frac{\pi}{2} \): Slope is \( 0 \)- At \( x = \pi \): Slope is \( -\frac{1}{2} \)- At \( x = 2\pi \): Slope is \( \frac{1}{2} \)These differ from \( y = \sin \frac{1}{2}x \) as demonstrated in step 4.

Key Concepts

Understanding the Sine Function

The sine function, denoted as \( y = \sin x \), is a fundamental trigonometric function that creates a wave-like pattern when graphed. This function oscillates between -1 and 1, repeating its cycle every \(2\pi\). It is commonly used to model periodic phenomena due to its smooth and continuous nature.

• The basic sine wave starts rising from zero, reaches a maximum at \(\pi/2\), returns to zero at \(\pi\), dips to a minimum at \(3\pi/2\), and completes its cycle back at zero at \(2\pi\).
• Modifications to the standard form \(y = \sin x\) often involve changes to amplitude, period, or phase shift, affecting the wave's shape, duration, or position.

In this context, understanding the form \( y = \sin \frac{1}{2}x \) is crucial as it signifies a horizontal stretching of the wave. This modification relates to the frequency and the distance spanned by one complete cycle of the sine wave.

Calculating Derivatives of Functions

Derivatives are vital in understanding how functions change. For a trigonometric function like sine, the derivative indicates the slope of the tangent line at any point, showing how steep or flat the graph is. The general derivative of \( y = \sin x\) is \( y' = \cos x\). For the modified function \( y = \sin \frac{1}{2}x \), the chain rule helps us find the new derivative.

• Chain Rule: Differentiate the outer function, then multiply by the derivative of the inner function.
• In this case, \( y' = \frac{1}{2} \cos \frac{1}{2}x \), stemming from differentiating \( \sin \frac{1}{2}x \).

This formula helps us calculate the slopes at various points, indicating the direction and velocity of the graph's movement. Exploring slopes at specified points helps determine how the graph of \( y = \sin \frac{1}{2}x \) behaves at those moments.

Determining the Function Period

Functions like \( y = \sin x \) have a set period, which is the length over which they repeat. For the standard sine function, this period is \(2\pi\). Modifications such as \( y = \sin \frac{1}{2}x \) affect the function's period and are crucial in analyzing the graph's stretch along the x-axis.

• The formula used to determine the new period is \( \frac{2\pi}{b} \), where \( b \) is the coefficient of \( x \) within the sine function.
• For \( \sin \frac{1}{2}x\), the period becomes \( 4\pi \), obtained from \( \frac{2\pi}{1/2} \).

Thus, the graph of \( y = \sin \frac{1}{2}x \) completes one cycle over a span of \(4\pi\), making it twice as long as the standard sine wave. Recognizing this difference in period is fundamental when graphing or analyzing trigonometric functions.

Graphing Trigonometric Functions

Graphing trigonometric functions like sine involves understanding their characteristics and transformations. To accurately depict \( y = \sin \frac{1}{2}x \), one must consider its period, amplitude, and phase shift.

• With a period of \(4\pi\), each cycle of the graph stretches horizontally compared to \(2\pi\) of the standard \( y = \sin x \).
• The amplitude remains 1, meaning the peaks and troughs reach maximums of 1 and -1.

To manually graph this function:

• Mark every \(4\pi\) along the x-axis to indicate complete cycles.
• Identify key points within one cycle: start at zero, reach a peak at \(\pi\), return to zero at \(2\pi\), decline to a trough at \(3\pi\), and return to zero at \(4\pi\).
• Plot these points, maintaining consistent amplitude, to visualize the wave.

Graphing not only enhances comprehension but provides visual confirmation of calculations and theoretical understanding, enabling more intuitive learning of trigonometric behaviors.