Applying Trigonometric Functions to Real-World Situations

This is only one example portion of the graph that could be chosen as one cycle. The horizontal length of one cycle is 4 units. Because 1 unit corresponds to 0.15 seconds, the period can be determined by multiplying these numbers. 4 * 0.15 = 0.6seconds The period is 0.6 seconds. This means that one heartbeat lasts 0.6 seconds. Let p be the pulse rate of the person. Therefore, p is the number of times that the heart beats in one minute. If 0.6 is multiplied by p, the result is 60 seconds.

The pulse rate of the accident victim is 100 beats per minute.

\begin{gathered} A = \dfrac{y_\text{max} - y_\text{min}}{2} \end{gathered} Davontay points out that each repeating portion of the graph is identical, therefore it is sufficient to analyze only one cycle. Determine the difference between the maximum and minimum electrical activity in the graph.

Because one vertical unit corresponds to 1 millivolt, the difference y_\text{max} - y_\text{min} is 3 millivolts. Divide this value by 2 to find the amplitude of the function. A = 3/2 mV ⇔ A= 1.5 mV

It is given that the point (0,0) lies on the sinusoid and is halfway between the lowest and the highest point. Therefore, the sine function without any translation should be chosen for the equation of the sinusoid. Recall the format of a sine function. y = a sin bx Here, | a| and 2π| b| are the amplitude and the period of the graph, respectively. Since it is given that the amplitude is 5 millimeters, | a| is equal to 5. Therefore, there are two possible values for a — positive and negative. | a| = 5 ⇓ a = 5 or a = -5 For simplicity, the positive value a=5 will be chosen. Next, the period of the function will be determined. Recall that the period of a periodic function is the reciprocal of its frequency. Period = 1/Frequency By knowing that the frequency is 16, the period can be found.

The period of the sinusoid is 6. To calculate the value of b in the function rule, use the fact that the period is 2π| b|.

Therefore, b can be either positive or negative. For simplicity, keep the positive value of b. Finally, the equation of the sinusoid can be written. Keep in mind that the input of the function is t. y = a sin bx ⇓ d(t) = 5 sin π/3t

1. Find the amplitude, period, and translation of the function
2. Draw the midline
3. Plot some key points on the graph
4. Draw the graph

It is given that the amplitude is 5 millimeters. In Part A, it was also obtained that the period is 6 seconds. Equation:& d(t) = 5sin π3t Amplitude:& 5millimeters Period:& 6seconds Looking at the equation, it is seen that its graph is not translated in any direction. Next, since the sine function is not translated, the midline of the graph is the horizontal line y = 0. Graph this line on a coordinate plane.

Now, key points ranging over at least one cycle will be plotted. These points are the maximums, minimums, and intersections with the midline. The maximums of y=sin x occur once every cycle at the following x-coordinates. x = -3π/2, π/2, 5π/2, ... The period of the considered sinusoid is 6, which is 3π times the period 2π of the parent sine function. Because the graph of d(t) is not translated horizontally, the maximums are not shifted to the right or to the left. This means that the t-coordinates of the maximums are as follows. t = 3/π (-3π/2), 3/π( π/2), 3/π( 5π/2), ... ⇕ t = -9/2, 3/2, 15/2, ... Since the midline is y=0 and the amplitude is 5, the maximum value of the function is 0 + 5 = 5. Plot the maximum points on the graph with the midline.

Similarly, the minimums of d(t) are located halfway between the maximums at the following t-coordinates. t = - 15/2, - 3/2, 9/2, ... The minimum value of the function is 0 - 5 = -5. Plot the minimums in the same coordinate plane.

Next, between every neighboring maximum and minimum are the intersections with the midline. t = - 6, - 3, 0, 3, 6, ... Because these points lie on the midline, their y-coordinate is 0.

Finally, by connecting the points with a smooth curve and continuing it periodically in both directions, the sinusoid can be drawn.

The absolute displacement of the ground after 28 seconds is the absolute value of - 5sqrt(3)2. |d(28)| = | - 5sqrt(3)/2 | ⇓ |d(28)|=5sqrt(3)/2 Therefore, the absolute displacement is 5sqrt(3)2 millimeters.

y = a cos b(x- h) + k Because it is given that the function is not translated, both h and k are equal to 0. In this function rule | a| is the amplitude and 2π| b| is the period. The amplitude of a sound wave is 60 decibels. This means that | a| is equal to 60. | a| = 60 ⇓ a = 60 or a = -60 There are two possible values of a. For example, the positive value a=60 will be chosen. Next, the period of the function will be determined. Recall that the period of a periodic function is the reciprocal of its frequency. Period = 1/Frequency Because the given frequency is 440 hertz, the period of the function is 1440 seconds. To calculate the value of b in the function rule, use the fact that the period is 2π| b|.

This answer indicates that b can be either positive or negative. Because a cosine function that is not translated is an even function, both values give the same function. Not to complicate the formula, keep the positive value of b. Finally, the function rule for the cosine function can be written. y = a cos bx ⇓ y = 60 cos ( 880πx)

1. Find the amplitude, period, and translation of the function.
2. Draw the midline.
3. Plot some key points on the graph.
4. Draw the graph.

It is given that the amplitude is 60 decibels. In Part A, it was also obtained that the period is 1440 seconds. Next, since the cosine function is not translated, the midline of this function is y = 0. Graph this line on a coordinate plane.

Now, key points ranging over one cycle will be plotted. These points are the maximums, minimums, and intersections with the midline. The maximums of y=cos x occur once every cycle at the following x-coordinates. x = 0, 2π, ... The period of the considered sinusoid is 1440, which is 1880π of the period 2π of the parent function. Because the function is not translated horizontally, the maximums are not shifted to the right or to the left. This means that the x-coordinates of the maximums are as follows. x = 1/880π* 0, 1/880π* 2π, ... ⇓ x = 0, 1/440, ... Since the midline is y=0 and the amplitude is 60, the maximum value of the function is 0 + 60 = 60. Plot the maximum points on the graph with the midline.

The minimum of the function is located halfway between the maximums at x = 1880. The minimum value is 0 - 60 = -60. Plot the minimum in the same coordinate plane.

Next, between every neighboring maximum and minimum the sinusoid intersects with the midline. x = 1/1760, 3/1760, ... Because these points lie on the midline, their y-coordinate is 0.

Finally, by connecting the points with a smooth curve, the sinusoid can be drawn. Although the graph may not represent the whole sound wave, keep in mind that only one cycle of each graph is given in the options.

This graph corresponds to option B.

Period

As mentioned in Part A, the period of a periodic function is the reciprocal of its frequency. Period = 1/Frequency Therefore, the period and frequency represent an inverse variation where the constant of variation is 1. This means that when the frequency decreases two times, the period doubles. Then, the pitch of the sound is noticeably lower.

Amplitude

The amplitude of a periodic function is half the difference between the maximum and minimum values, while the period is the length of one cycle.

The amplitude is related to neither period nor frequency. Therefore, when the frequency decreases 2 times, the amplitude does not change. This means that the pitch of a sound does not influence its volume, which makes perfect sense. The guitar player is thankful for this understanding.

Because compressions are started at t=0, the maximum height is also at t=0. This means that a cosine function will be used. Cosine h(t) = acos b(t- h) + k In this function, | a| is the amplitude, 2π| b| is the period, h is the horizontal shift, and k is the vertical shift. Now, write the equation for the midline. Recall that the midline lies between the maximum and minimum values. It is given that the maximum height is centimeters and the depth is 6 centimeters. Maximum Height:& cm Minimum Height:& - 6 = -6cm The equation for the midline is given by the average of these values. This equation gives the vertical translation of the function. Midline y = 0 + (-6)/2 ⇔ y= -3 Since the midline is y=-3, the vertical shift is k = -3. Next, find the amplitude of the function by calculating the difference between the midline value and the maximum value.

Since compressions are started at t=0 and then the height is maximum, choose a positive value of a. The rate of 120 compressions per minute means that there are 2 compressions every second, which gives a frequency of 2 hertz. The period of the function is the reciprocal of its frequency. Therefore, the period is 12 seconds. Write the equation for the period and solve it for b.

Therefore, b can be either positive or negative. Because not translated cosine function is an even function, both values give the same function. Not to complicate the formula, keep the positive value of b. Because the horizontal shift has already been fixed, there is all the information needed to write an equation of the function. h(t) = 3cos 4π(t- 0) + ( -3) ⇕ h(t) = 3 cos 4π t - 3

y = cos x ↓ h(t) = 3 cos 4π t - 3 Start by identifying the transformations given by the factors next to the input t and the cosine function.

In the given function rule, both 3 and 4π are greater than 1. This means that they represent a vertical stretch by a factor of 3, and a horizontal shrink by a factor of 4π, respectively. Also, Davontay points out that 3 is subtracted from the whole function rule. h(t) = 3 cos 4π t - 3 This operation represents a vertical translation 3 units down of the graph. With this information, the graph of the parent cosine function can be transformed to h(t) = 3 cos 4π t - 3.

This graph corresponds to option D.