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Health The function \(P=100-20 \cos \frac{5 \pi t}{3}\) approximates the blood pressure \(P\) (in millimeters of mercury) at time \(t\) (in seconds) for a person at rest. (a) Find the period of the function. (b) Find the number of heartbeats per minute.

Short Answer

Expert verified
The period of the function is \(\frac{6}{5}\) seconds and there are 50 heartbeats per minute.

Step by step solution

01

Identify the Model

The model is an oscillatory function given by \(P=100-20 \cos \frac{5 \pi t}{3}\). This is a cosine function shifted vertically by 100, and compressed horizontally by a factor of \(\frac{5 \pi}{3}\). The amplitude is 20.
02

Calculate the Period

The period of a cosine function is given by the formula \(T=\frac{2\pi}{B}\), where B is the coefficient of t in the cosine function. In this case, B is \(\frac{5 \pi}{3}\), therefore, the period \(T=\frac{2\pi}{\frac{5 \pi}{3}}=\frac{6}{5}\) seconds.
03

Convert Period to Heartbeats per Minute

One complete cycle of the function represents one heartbeat. So, we need to convert this period from seconds to minutes and take reciprocal, since we know there are 60 seconds in a minute and want to find heartbeats per minute. Therefore, the number of heartbeats per minute is \(\frac{60}{\frac{6}{5}}=50 \) beats per minute.

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Key Concepts

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

Period of a Cosine Function
Understanding the period of a cosine function is essential when dealing with oscillatory behavior in trigonometric problems. The period refers to the length of one complete cycle of the waveform. In a general form, a cosine function can be represented as \( f(t) = A \times \text{cos}(Bt + C) + D \), where:
  • \( A \) is the amplitude,
  • \( B \) affects the period,
  • \( C \) is the horizontal shift, and
  • \( D \) is the vertical shift.

In the given exercise, \(P=100-20 \text{cos} \frac{5 \text{Ï€} t}{3}\) models blood pressure over time, indicating a vertical shift of 100 units and an amplitude of 20. To find the period \( T \) of this specific cosine function, we use the formula \(T=\frac{2\text{Ï€}}{B}\), with \(B\) being the coefficient of \(t\). Here, \(B = \frac{5 \text{Ï€}}{3}\), giving us the period \( T = \frac{2\text{Ï€}}{\frac{5\text{Ï€}}{3}} = \frac{6}{5} \) seconds. This means the blood pressure completes one oscillation every \( \frac{6}{5} \) seconds. Simplifying complex formulas through such examples assists students in grasping the concept of periodicity in trigonometric functions.
Oscillatory Mathematical Models
Oscillatory mathematical models are invaluable for describing phenomena that repeat over regular intervals. They provide a framework for representing periodic behavior in a variety of real-world contexts, from the predictable swing of a pendulum to the rhythms of the human heart.

An oscillatory model often involves trigonometric functions like sine and cosine, which are inherently periodic. These functions, due to their wave-like properties, are adept at capturing the cyclic nature of these occurrences. Key features of these models include amplitude, which measures the size of the oscillations; phase shift, which describes any horizontal displacement of the waveform; and, as discussed previously, period, which is the duration of a full cycle.

These models can be customized to fit specific data by adjusting these properties. In our exercise, the model \( P=100-20 \text{cos} \frac{5 \text{Ï€} t}{3} \) represents a person's blood pressure at rest, which naturally fluctuates in an oscillatory manner due to the beating of the heart.
Heart Rate Calculation
Calculating heart rate from oscillatory functions is a practical application of trigonometry in the field of health. The number of heartbeats per minute can be determined by analyzing the period of a function that models the heartbeat.

In this exercise, one complete cycle of our cosine function equates to one heartbeat. To convert the period into heartbeats per minute, we simply find the reciprocal of the period (in seconds) and then scale it up to a minute. As explained in the solution, the period of the given function is \( \frac{6}{5} \) seconds. There are 60 seconds in a minute, so we calculate the number of heartbeats per minute as \( \frac{60}{\frac{6}{5}} = 50 \) beats per minute.

This approach gives us an estimate of the resting heart rate, which is a vital sign checked routinely in medical assessments. Such calculations underscore how mathematical concepts like trigonometry have significant implications in health and medicine.

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