Problem 82
Question
Everybody's blood pressure varies over the course of the day. In a certain individual the resting diastolic blood pressure at time \(t\) is given by \(B(t)=80+7 \sin (\pi t / 12),\) where \(t\) is measured in hours since midnight and \(B(t)\) in \(\mathrm{mmHg}\) (millimeters of mercury). Find this person's diastolic blood pressure at (a) 6: 00 A.M. (b) 10: 30 A.M. (c) Noon (d) 8: 00 P.M.
Step-by-Step Solution
Verified Answer
(a) 87 mmHg; (b) 86.467 mmHg; (c) 80 mmHg; (d) 73.938 mmHg
1Step 1: Understand the Function
The given function is \( B(t) = 80 + 7 \sin \left( \frac{\pi t}{12} \right) \). It represents the diastolic blood pressure as a function of time \( t \), where \( t \) is measured in hours since midnight.
2Step 2: Find the Blood Pressure at 6:00 A.M.
For 6:00 A.M., \( t = 6 \). Substitute \( t = 6 \) into the function: \[ B(6) = 80 + 7 \sin \left( \frac{\pi \times 6}{12} \right) \]Calculate: \[ B(6) = 80 + 7 \sin \left( \frac{\pi}{2} \right) = 80 + 7 \times 1 = 87 \text{ mmHg} \]
3Step 3: Find the Blood Pressure at 10:30 A.M.
For 10:30 A.M., \( t = 10.5 \). Substitute \( t = 10.5 \) into the function: \[ B(10.5) = 80 + 7 \sin \left( \frac{\pi \times 10.5}{12} \right) \]Calculate: \[ B(10.5) = 80 + 7 \sin \left( \frac{10.5\pi}{12} \right) = 80 + 7 \sin \left( \frac{7\pi}{8} \right) \]Using a calculator, find \( \sin \left( \frac{7\pi}{8} \right) \approx 0.9239, \)\[ B(10.5) = 80 + 7 \times 0.9239 \approx 86.467 \text{ mmHg} \]
4Step 4: Find the Blood Pressure at Noon
For Noon, \( t = 12 \). Substitute \( t = 12 \) into the function: \[ B(12) = 80 + 7 \sin \left( \frac{\pi \times 12}{12} \right) \]Calculate: \[ B(12) = 80 + 7 \sin (\pi) = 80 + 7 \times 0 = 80 \text{ mmHg} \]
5Step 5: Find the Blood Pressure at 8:00 P.M.
For 8:00 P.M., \( t = 20 \). Substitute \( t = 20 \) into the function: \[ B(20) = 80 + 7 \sin \left( \frac{\pi \times 20}{12} \right) \]Calculate: \[ B(20) = 80 + 7 \sin \left( \frac{5\pi}{3} \right) \]Using a calculator, find \( \sin \left( \frac{5\pi}{3} \right) = -\frac{\sqrt{3}}{2} \approx -0.866 \),\[ B(20) = 80 + 7 \times (-0.866) \approx 80 - 6.062 = 73.938 \text{ mmHg} \]
Key Concepts
Sine FunctionDiastolic Blood PressureMathematical Modeling
Sine Function
The sine function is a fundamental element of trigonometry. It describes how a point on a unit circle moves around as an angle changes. In basic terms, the sine of an angle represents the ratio of the length of the side of a right triangle opposite the angle to its hypotenuse.
In our exercise, we have the function \( \sin(\pi t / 12) \), a periodic function that oscillates between -1 and 1, since the sine wave cycles every \( 2\pi \). This describes natural periodic phenomena, like the rhythm of a day or biological processes, beautifully.
Using sine in modeling time-based functions, like blood pressure, lets us represent rhythmic changes smoothly. This is particularly useful in capturing daily fluctuations in diastolic blood pressure as it ebbs and flows throughout the day.
In our exercise, we have the function \( \sin(\pi t / 12) \), a periodic function that oscillates between -1 and 1, since the sine wave cycles every \( 2\pi \). This describes natural periodic phenomena, like the rhythm of a day or biological processes, beautifully.
Using sine in modeling time-based functions, like blood pressure, lets us represent rhythmic changes smoothly. This is particularly useful in capturing daily fluctuations in diastolic blood pressure as it ebbs and flows throughout the day.
Diastolic Blood Pressure
Diastolic blood pressure is the pressure in the arteries when the heart rests between beats. It is an essential indicator of cardiovascular health. High diastolic pressure can indicate health problems such as hypertension.
In our context, we see how diastolic blood pressure changes through a day for a particular individual. The equation \( B(t) = 80 + 7 \sin(\pi t / 12) \) includes a baseline pressure of 80 mmHg. The sine term describes the variation from this baseline due to daily biological rhythms.
In our context, we see how diastolic blood pressure changes through a day for a particular individual. The equation \( B(t) = 80 + 7 \sin(\pi t / 12) \) includes a baseline pressure of 80 mmHg. The sine term describes the variation from this baseline due to daily biological rhythms.
- At specific times, like 6:00 A.M., the pressure might peak, representing the body's natural increase in preparation for waking up.
- At noon, the sine term becomes zero, representing the blood pressure baseline.
- In the evening, the pressure might decrease, reflecting the body's relaxation phase.
Mathematical Modeling
Mathematical modeling is the process of representing real-world scenarios using mathematical equations and expressions. This approach is vital in various fields like biology, finance, and engineering.
The problem explores how mathematical modeling describes the diastolic blood pressure's daily fluctuations. Using the equation \( B(t) = 80 + 7 \sin(\pi t / 12) \), we encapsulate the natural, cyclical pattern observed in many biological processes.
The problem explores how mathematical modeling describes the diastolic blood pressure's daily fluctuations. Using the equation \( B(t) = 80 + 7 \sin(\pi t / 12) \), we encapsulate the natural, cyclical pattern observed in many biological processes.
- This model uses a sine function to predict blood pressure throughout a day. It shows how real-world phenomena can be understood and predicted with precision.
- It highlights how patterns and periodicities in nature can translate into mathematical language, offering insights and solving practical problems.
Other exercises in this chapter
Problem 80
Variable Stars Variable stars are ones whose brightness varies periodically. One of the most visible is R Leonis; its brightness is modeled by the function $$ b
View solution Problem 81
Compositions Involving Trigonometric Functions This exercise explores the effect of the inner function \(g\) on a composite function \(y=f(g(x))\) (a) Graph the
View solution Problem 83
After the switch is closed in the circuit shown, the current \(t\) seconds later is \(I(t)=0.8 e^{-3 t}\) sin \(10 t\) Find the current at the times (a) \(t=0.1
View solution Problem 80
Determine whether the function is even, odd, or neither. $$f(x)=\cos (\sin x)$$
View solution