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Chapter 1: Problem 80

Give an example of: A cosine function which oscillates between values of 1200 and 2000

Short Answer

Expert verified
The function is \( f(x) = 400 \cdot \cos(x) + 1600 \).

Step by step solution

01

Determine the Midline

The midline of a trigonometric function is the average of its maximum and minimum values. Here, the maximum value is 2000 and the minimum value is 1200.Calculate the midline using the formula:\[ Midline = \frac{{2000 + 1200}}{2} = 1600 \]So, the midline of the function is 1600.
02

Calculate the Amplitude

The amplitude of a trigonometric function is the distance from the midline to either the maximum or minimum value. It can be calculated as:\[ Amplitude = \frac{{2000 - 1200}}{2} = 400 \]Thus, the amplitude of the function is 400.
03

Write the General Cosine Function

The general form of a cosine function is:\[ f(x) = A \cdot \cos(Bx + C) + D \]Where:- \( A \) is the amplitude,- \( D \) is the midline (vertical shift),- \( B \) affects the period,- \( C \) affects the phase shift.Given the amplitude is 400 and the midline is 1600, the function partially takes the form:\[ f(x) = 400 \cdot \cos(Bx) + 1600 \]
04

Determine the Period and Define B

The period of a cosine function is given by \( \frac{2\pi}{B} \). For simplicity, assume a standard period of \( 2\pi \) so that the cosine function completes one cycle over this interval.This implies \( B = 1 \). Thus, the function becomes:\[ f(x) = 400 \cdot \cos(x) + 1600 \]
05

Finalize the Function

Since we've assumed standard period and no phase shift for simplicity, our cosine function that oscillates between 1200 and 2000 is:\[ f(x) = 400 \cdot \cos(x) + 1600 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Midline of Trigonometric Functions
The midline of a trigonometric function serves as the horizontal center about which the function oscillates. For a cosine function, this midline is crucial as it determines the vertical shift. To find the midline, you need to calculate the average of the function's maximum and minimum values.

In this problem, the cosine function oscillates between a maximum value of 2000 and a minimum value of 1200. Thus, the calculation for the midline is straightforward:
  • Add the maximum and minimum values: 2000 + 1200 = 3200
  • Divide the sum by 2 to find the average: 3200 / 2 = 1600
Therefore, the midline of the function is 1600. This means all oscillations will occur about this central value, shifting the whole graph vertically on the y-axis.
Amplitude Calculation
The amplitude is an important aspect of a cosine function, determining the height of the wave from its midline. You can think of the amplitude as how far the function extends above or below the midline.

To calculate the amplitude of a trigonometric function, you subtract the minimum value from the maximum value, then divide by 2:
  • Find the difference between the maximum and minimum: 2000 - 1200 = 800
  • Divide the difference by 2: 800 / 2 = 400
Thus, the amplitude of the function is 400. This means that from the midline of 1600, the function peaks to a maximum of 2000 and dips to a minimum of 1200.
Graphing Trigonometric Functions
Graphing a trigonometric function like the cosine function involves understanding its basic structure and transformations. A general cosine function is depicted in the form:
  • \[ f(x) = A \cdot \cos(Bx + C) + D \]
Where:
  • \( A \) is the amplitude, determining the height of oscillations.
  • \( B \) affects the period of the wave, modifying how frequently it oscillates within an interval.
  • \( C \) shifts the wave horizontally, known as phase shift.
  • \( D \) is the midline, moving the whole graph vertically.
In our example, the function is:
  • \[ f(x) = 400 \cdot \cos(x) + 1600 \]
Here:
  • Amplitude \( A = 400 \)
  • Midline \( D = 1600 \)
  • Standard period implying \( B = 1 \), making the period \( 2\pi \)
Thus, the waveform will oscillate between 1200 and 2000, repeating every \( 2\pi \) units along the x-axis. Understanding these parameters helps in sketching an accurate graph of the function.

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Most popular questions from this chapter

Cyanide is used in solution to isolate gold in a mine. \(^{49}\) This may result in contaminated groundwater near the mine, requiring the poison be removed, as in the following table, where \(t\) is in years since 2012. (a) Find an exponential model for \(c(t),\) the concentration, in parts per million, of cyanide in the groundwater. (b) Use the model in part (a) to find the number of years it takes for the cyanide concentration to fall to 10 ppm. (c) The filtering process removing the cyanide is sped up so that the new model is \(D(t)=c(2 t) .\) Find \(D(t).\) (d) If the cyanide removal was started three years earlier, but run at the speed of part (a), find a new model, \(E(t).\) $$\begin{array}{c|c|c|c} \hline t \text { (years) } & 0 & 1 & 2 \\ \hline c(t)(\mathrm{ppm}) & 25.0 & 21.8 & 19.01 \\ \hline \end{array}$$

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