Chapter 6.3, Problem 53E

(a)

To determine

To Find: The blood pressure at $t=0,t=0.25,t=0.5,t=0.75,t=1$

Answer to Problem 53E

  $100,120,100,80,100$

Explanation of Solution

Given: $P=100+20\sin 2\pi rt$

Given: $P=100+20\sin 2\pi rt$

Substitute $t=0,t=0.25,t=0.5,t=0.75,t=1$

We get

  $\begin{array}{l}P=100+20\sin 2\pi t\\ P=100+20\sin 2\pi (0)=100\\ P=100+20\sin 2\pi (0.25)=120\\ P=100+20\sin 2\pi (0.5)=100\\ P=100+20\sin 2\pi (0.75)=80\\ P=100+20\sin 2\pi (1)=100\end{array}$

Hence the blood pressures are $100,120,100,80,100$ .

(b)

To determine

To Find: The blood pressure at maximum during the first seconds.

Answer to Problem 53E

  $t=0.25$

Explanation of Solution

Given: $P=100+20\sin 2\pi rt$

Given: $P=100+20\sin 2\pi rt$

Substitute $t=0,t=0.25,t=0.5,t=0.75,t=1$

We get

  $\begin{array}{l}P=100+20\sin 2\pi t\\ P=100+20\sin 2\pi (0)=100\\ P=100+20\sin 2\pi (0.25)=120\\ P=100+20\sin 2\pi (0.5)=100\\ P=100+20\sin 2\pi (0.75)=80\\ P=100+20\sin 2\pi (1)=100\end{array}$

Hence the blood pressure is maximum when $t=0.25$ .

(b)

To determine

To Find: The blood pressure at minimum during the first seconds.

Answer to Problem 53E

  $t=0.75s$

Explanation of Solution

Given: $P=100+20\sin 2\pi rt$

Given: $P=100+20\sin 2\pi rt$

Substitute $t=0,t=0.25,t=0.5,t=0.75,t=1$

We get

  $\begin{array}{l}P=100+20\sin 2\pi t\\ P=100+20\sin 2\pi (0)=100\\ P=100+20\sin 2\pi (0.25)=120\\ P=100+20\sin 2\pi (0.5)=100\\ P=100+20\sin 2\pi (0.75)=80\\ P=100+20\sin 2\pi (1)=100\end{array}$

Hence the blood pressure is maximum when $t=0.75s$ .