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Magazine Chapter 5: Problem 19
Determine the amplitude and phase shift for each function, and sketch at least one cycle of the graph. Label five points as done in the examples. $$y=-3 \sin x$$
Short Answer
Expert verified
Amplitude: 3, Phase shift: 0. Key points: (0,0), \((\pi/2,-3)\), \((\pi,0)\), \((3\pi/2,3)\), \((2\pi,0)\).
Step by step solution
01
Identify the general form of the sine function
The general form of the sine function is given by \[ y = a \, \text{sin}(bx - c) + d \]. In this case, the function is \( y = -3 \, \text{sin}(x) \).
02
Determine the amplitude
The amplitude of the sine function is represented by the absolute value of \( a \). Here, \( a = -3 \), so the amplitude is \( \left| -3 \right| = 3 \).
03
Determine the phase shift
The phase shift of the sine function is given by \( \frac{c}{b} \). In this function, \( b = 1 \) and \( c = 0 \), so the phase shift is \( \frac{0}{1} = 0 \). There is no phase shift.
04
Plot key points and one cycle
Since the sine function has a period of \( 2\pi \) and no phase shift, the key points to plot for one cycle are at: \[ x = 0, \pi/2, \pi, 3\pi/2, 2\pi \]. Evaluate the function at these points: \[ y(0) = -3\sin(0) = 0 \], \[ y(\pi/2) = -3\sin(\pi/2) = -3 \], \[ y(\pi) = -3\sin(\pi) = 0 \], \[ y(3\pi/2) = -3\sin(3\pi/2) = 3 \], \[ y(2\pi) = -3\sin(2\pi) = 0 \].
05
Sketch the graph
Using the key points \((0,0), (\pi/2,-3), (\pi,0), (3\pi/2,3), (2\pi,0)\), sketch one cycle of the sine function starting at \( x = 0 \) and ending at \( x = 2\pi \). Set the range from \(-3\) to \(3\) on the y-axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Amplitude
The amplitude of a trigonometric function, such as the sine function, indicates the peak value the function reaches above and below its midline. For the general form of the sine function, \(y = a \, \text{sin}(bx - c) + d\), the amplitude is denoted by the absolute value of parameter \(a\). Here, amplitude shows how far the wave oscillates from the center line (midline). In our example, \(y = -3\, \, \text{sin}(x)\), \(a = -3\).
Since amplitude is always a positive value, we take the absolute value: \left| -3 \right| = 3\. Therefore, the wave will reach up to 3 units above and 3 units below the midline of the graph.
Key points to remember:
Since amplitude is always a positive value, we take the absolute value: \left| -3 \right| = 3\. Therefore, the wave will reach up to 3 units above and 3 units below the midline of the graph.
Key points to remember:
- Amplitude tells you the height of the wave
- It鈥檚 derived from the absolute value of parameter \(a\)
- In our problem, it鈥檚 3 units above and below the midline
Phase Shift
Phase shift refers to the horizontal movement of the wave along the x-axis. This tells us how much the function has shifted from its usual position. For the general sine function, \( y = a \, \text{sin}(bx - c) + d \), the phase shift is calculated using the formula: \frac{c}{b}\.
In our given function, \(y = -3 \, \text{sin}(x)\), \(b = 1\) and \(c = 0\). Using the formula gives us: \frac{0}{1} = 0\. This indicates there is no phase shift, meaning the wave starts at the same point it usually does (x = 0).
Key points to remember:
In our given function, \(y = -3 \, \text{sin}(x)\), \(b = 1\) and \(c = 0\). Using the formula gives us: \frac{0}{1} = 0\. This indicates there is no phase shift, meaning the wave starts at the same point it usually does (x = 0).
Key points to remember:
- Phase shift indicates the horizontal shift
- It is calculated as \frac{c}{b}\
- In our problem, there is no shift (phase shift is 0)
Sine Function
The sine function is a fundamental trigonometric function that describes a smooth periodic oscillation. The general form in trigonometry is given by \( y = a \, \text{sin}(bx - c) + d \). Let's break down the components of the function:
For our given function, \( y = -3 \, \text{sin}(x) \):
This results in a sine wave centered around y = 0, with peaks at -3 and troughs at 3, and completing one cycle every \(2\backslashpi\).
- \(a\): This parameter affects the amplitude
- \(b\): This parameter affects the frequency (and thus, the period)
- \(c\): This parameter affects the phase shift
- \(d\): This parameter affects the vertical shift
For our given function, \( y = -3 \, \text{sin}(x) \):
- The negative sign in front of \(a\) indicates the wave is inverted
- Since \(a = -3\), it means the wave鈥檚 amplitude is 3
- There is no \(b\) term affecting the period, so it鈥檚 the standard \(2\backslashpi\)
- There is no \(c\) term so phase shift is zero
- There is no \(d\) term so there is no vertical shift
This results in a sine wave centered around y = 0, with peaks at -3 and troughs at 3, and completing one cycle every \(2\backslashpi\).
Period
The period of a trigonometric function refers to the length of one complete cycle of the wave along the x-axis. For the sine function in its general form, \( y = a \, \text{sin}(bx - c) + d \), the period can be determined by the formula
In our example, \( y = -3 \, \text{sin}(x)\), \(b = 1\). Therefore, the period is: \frac{2\backslashpi}{1} = 2\backslashpi\. This means it takes \ 2\backslashpi\ units along the x-axis for the sine wave to complete one full cycle.
Key points to remember:
This period length is crucial when sketching the graph, as you can use this to locate key points in one full cycle.
Period = \frac{2\backslashpi}{b}\
In our example, \( y = -3 \, \text{sin}(x)\), \(b = 1\). Therefore, the period is: \frac{2\backslashpi}{1} = 2\backslashpi\. This means it takes \ 2\backslashpi\ units along the x-axis for the sine wave to complete one full cycle.
Key points to remember:
- The period is the length of one complete wave cycle
- It鈥檚 calculated using the formula \frac{2\backslashpi}{b}\
- In this problem, the period is \ 2\backslashpi\
This period length is crucial when sketching the graph, as you can use this to locate key points in one full cycle.
