Question

Blood Pressure Blood pressure is a way of measuring the amount of force exerted on the walls of blood vessels. It is measured using two numbers: systolic (as the heart beats) blood pressure and diastolic (as the heart rests) blood pressure. Blood pressures vary substantially from person to person, but a typical blood pressure is 120/80, which means the systolic blood pressure is 120 mmHg and the diastolic blood pressure is 80 mmHg. Assuming that a person’s heart beats 70 times per minute, the blood pressureP of an individual after t seconds can be modeled by the function$P\left(t\right)=100+20\sin \left(\frac{7\mathrm{\pi }}{3}t\right)$

(a) In the interval [0, 1], determine the times at which the

blood pressure is 100 mmHg.

(b) In the interval [0, 1], determine the times at which the

blood pressure is 120 mmHg.

(c) In the interval[0, 1], determine the times at which the

blood pressure is between 100 and 105 mmHg.

Short answer

Part a. In the interval $\left[0,1\right],$ the times at which the blood pressure is 100mmHg are$0,0.43,0.86\mathrm{seconds}.$

Part b. In the interval $\left[0,1\right],$ time at which the blood pressure is 120mmHg is $0.21\mathrm{seconds}.$

Part c. In the interval$\left[0,1\right],$time at which the blood pressure is between100and105mmHg is$\left(0,0.03\right)\cup \left(0.39,0.43\right)\cup \left(0.86,0.89\right).$

Step-by-step solution

Part (a) Step 1. Given Information 

The given function for blood pressure of an individual after tseconds is $P\left(t\right)=100+20\sin \left(\frac{7\mathrm{\pi }}{3}t\right).$

We have to determine the time at which the blood pressure is 100mmHg in the interval $\left[0,1\right]$.

Part (a) Step 2. Finding the time

To find the time we will use the given function.

$$\begin{gathered} P\left(t\right)=100+20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ 100=100+20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ 0=20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ 0=\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ \mathrm{Sine}\mathrm{angle}\mathrm{is}0\mathrm{at}0,\mathrm{\pi },2\mathrm{\pi } \\ \sin \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=0 \\ \sin \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=\sin 0 \\ \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=0 \\ \mathrm{t}=0 \\ \mathrm{Now},\mathrm{sin\pi }=0 \\ \sin \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=0 \\ \sin \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=\mathrm{sin\pi } \\ \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=\mathrm{\pi } \\ \mathrm{t}=0.43 \\ \mathrm{Now},\sin 2\mathrm{\pi }=0 \\ \sin \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=0 \\ \sin \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=\sin 2\mathrm{\pi } \\ \left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right)=2\mathrm{\pi } \\ \mathrm{t}=0.86 \end{gathered}$$

So, in the interval $\left[0,1\right]$, the times at $0,0.43,0.86\mathrm{seconds}$ time blood pressure is 100mmHg.

Part (b) Step 1. Finding the time

To find the time we will use the given function.

$$\begin{gathered} P\left(t\right)=100+20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ 120=100+20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ 20=20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ 1=\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ t=0.21 \end{gathered}$$

Thus, in the interval$\left[0,1\right]$, the time at$0.21\mathrm{seconds}$the blood pressure is120mmHg.

Part (c) Step 1. Finding the time

To find the time we will use the given function.

$$\begin{gathered} 100<P\left(t\right)<105 \\ P\left(t\right)=105 \\ P\left(t\right)=100+20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ 105=100+20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ 5=20\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \\ \frac{1}{4}=\sin \left(\frac{7\mathrm{\pi }}{3}t\right) \end{gathered}$$

The roots of $\frac{1}{4}=\sin \left(\frac{7\mathrm{\pi }}{3}t\right)$in $[0,2\mathrm{\pi })$are

$$\begin{gathered} \frac{1}{4}=\left(\frac{7\mathrm{\pi }}{3}t\right) \\ t=0.03s\mathrm{econds}..........\left(\mathrm{a}\right) \\ \mathrm{\pi }-\frac{1}{4}=\left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right) \\ \mathrm{t}=0.39\mathrm{seconds}..........\left(\mathrm{b}\right) \\ 2\mathrm{\pi }+\frac{1}{4}=\left(\frac{7\mathrm{\pi }}{3}\mathrm{t}\right) \\ \mathrm{t}=0.89\mathrm{seconds}..........\left(\mathrm{c}\right) \end{gathered}$$

From (a), (b), (c), and the time we get from part (a) the time for the blood pressure between100and150is$\left(0,0.03\right)\cup \left(0.39,0.43\right)\cup \left(0.86,0.89\right).$