Question

Suppose that Keisha's blood pressure can be modeled by the following function:

p(t)=83-18sin(71*pi*t)

Keisha's blood pressure increases each time her heart beats, and it decreases as her heart rests in between beats. In this equation, p(t)is the blood pressure in mmHg (millimeters of mercury), and t is the time in minutes.

Find the following. If necessary round to the nearest hundredth.

Amplitude of p

Number of heartbeats per minute

Time for one full cycle of p

Answer 1

For the amplitude, figure out the minimum and maximum values, and take the difference. Remember, you don't have to find out the value(s) of \(t\) where they take place, just the maximum and minimum.

To find the frequency of a sinusoid \(\sin(\omega t)\), use\[f = \frac{\omega}{2\pi}\]

To find the time for one full cycle of p(t), use the relationship
\[P = \frac{1}{f}\]Remember that \(t\) is in minutes, so \(f\) is in \(\text{min}^{-1}\) and \(P\) is in minutes.

Answer 2

amplitude 18
time for one full cycle: 0.281
heartbeats: 2130

is this right?

Answer 3

Nope. What is the amplitude of the sine function? Just plain old \(\sin x\)

Answer 4

Never mind, you've got the amplitude correct, we don't need to review that.

Let's do some sanity checking on the rest of the answers. Get out your watch and take your pulse for 10 seconds. Multiply by 6. You probably got a number somewhere between 50 and 100.

Now, if 1 heartbeat takes only 0.281 seconds for a full cycle, that means in 1 minute (60 seconds), Keisha's going to have \(\dfrac{60}{0.281} = 214\) heartbeats! Does that seem reasonable? Perhaps at full-intensity exercise, if she's young, but it should be making you skeptical of your answer's correctness.

Answer 5

Similarly, 2130 heartbeats per minute?

Answer 6

it did that's why I asked....lol

Answer 7

So, amplitude is just the absolute value of the coefficient of the sin function. 18 is fine.

Now, the argument to the sin function here is \(\omega t = 71\pi t\).

Answer 8

So, amplitude is just the absolute value of the coefficient of the sin function. 18 is fine.

Now, the argument to the sin function here is \(\omega t = 71\pi t\).

Answer 9

That means \(\omega = 71\pi\)

Every time the value of the argument of \(\sin x\) goes through \(2\pi\), we get one complete trip around the unit circle (one complete cycle from 0 to 1 to 0 to -1 back to 0). If we multiply the argument by a number whose magnitude is greater than 1, we get more wiggles in a fixed amount of time. If the argument is multiplied by a number whose magnitude is less than 1, we get fewer wiggles. Clear so far?

Answer 10

\[71\pi*0.281=62.68?\]

Answer 11

A picture may be helpful. Here I've plotted \(y = \sin(t)\) in blue, \(y = \sin(2t)\) in purple, and \(y= \sin(0.5t)\) in olive.

Answer 12

This \(\omega\) factor is how we convert from "real world" measurements to the trig world. Because \(\theta\) needs to go from \(0\) to \(2\pi\) for \(\sin \theta\) to make one complete trip around the unit circle, a frequency of 1 corresponds to multiplying \(t\) by \(2\pi\). If we are plotting \(\sin(2\pi t)\) with \(t\) in seconds, we will get exactly one full sine cycle for each 1 second along the \(t\) axis.

Answer 13

Now, given that \(\omega t = 2\pi t\) gives us 1 full cycle in 1 unit of \(t\), what will \(\omega t = 4\pi t\) give us in 1 unit of \(t\)?

Answer 14

2 full cycles

Answer 15

Good. What if we had \(\omega t = 71\pi t\)? How many full cycles in 1 unit of \(t\) ?

Answer 16

\[\frac{71\pi}{2\pi} = \frac{71}{2} = 35.5\]So we'll get 35.5 heartbeats in 1 unit of \(t\), or 35.5 heartbeats per minute. Pretty low for someone who isn't a serious endurance athlete, but not impossibly so.

Answer 17

way better than the answer I had.

Answer 18

or more realistic..how do I find time for one full cycle of p?

Answer 19

Okay, to get the time, or period, we just take the reciprocal of the frequency.

\[P = \frac{1\text{ minute}}{35.5\text{ cycle}} = 0.028169 \text{ minutes/cycle} * 60 \text{ seconds/minute} \]\[\qquad=\text{ __________ seconds/cycle}\]

Answer 20

Hope that makes it clear for you, time for me to hit the sack!

Answer 21

18 amplitude
35.5 heartbeats per minute
1.69 minutes a cycle?


is this right @arabpride, @ganeshie8, @kc_kennylau @BangkokGarrett @luigi0210

Answer 22

i havnt a clue.....o.o

Answer 23

once check units in ur period

Answer 24

amplitude and frequency are right !
u need to fix units for period

Answer 25

I even told you the units!

If you get 35.5 heartbeats in 1 minute, with each heartbeat being a full cycle, how could you think a full cycle takes more than a minute and a half?!?

Answer 26

rounded to the nearest hundredth its 0.03. the cycle is the cycle of one heartbeat?

Answer 27

Punch the numbers I gave you into your calculator. And think about your answer!

Answer 28

@whpalmer4 what I thought you were telling me to evaluate is this

P=1 minute/35.5 cycle=0.028169 minutes/1∗60 seconds/minute which is how I go the first answer.

the only other thing I can imagine substituting is 0.028169/35.3*60/1= 0.047609577465 which is 0.05 which still does seem right

Answer 29

*doesnt

Answer 30

\[P = \frac{1\text{ minute}}{35.5\text{ cycle}} = 0.028169 \text{ minutes/cycle} * 60 \text{ seconds/minute}\]\[\qquad=0.028169*60 \frac{\cancel{ \text{minutes}}} { \text{cycle}}* \frac{\text{seconds}}{\cancel{\text{minutes}}} = 1.69 \frac{\text{seconds}}{\text{cycle}}\]