ANOTHER sinusoid graph problem

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In summary, the conversation was about a project involving sinusoid graph problems and the need for help. The conversation then shifted to discussing an equation to represent Frank's high blood pressure and normal blood pressure, and graphing both functions on the same coordinate plane. The conversation explored the variables in the equation, including amplitude, period, phase shift, and vertical shift. The group discussed the importance of understanding these variables and how they relate to blood pressure readings. Ultimately, the conversation concluded with appreciation for the helpful advice provided.
  • #1
We are doing a project with sinusoid graph problems. My group has done absolutely nothing, and it is due tomorrow. I am in desperate need of help. You have to know a bit about blood reading blood pressure, and relating it to heart rate. My teacher told me that I could use 60 beats per minute as my heart rate.

1. Frank has high blood pressure, and it is usually measured to be around 150/90mm. Normal blood pressure is usually considered to be around 120/80mm. Assume that both Frank and a "normal" person have a "normal" heart rate.

a. Find an equation, F(t), to represent Frank's blood pressure.
b. Find an equation, N(t), to represent normal blood pressure.
c. Graph both functions on the same coordinate plane.
d. When are the two blood pressures the same?
e. What are the first two times Frank's blood pressure hits 120mm?



This is the model equation : y = A sin or cos (B(x-c)) + D



He says he gave us "A", but I don't even know what number in the problem would be "A". I know that "B" is the period. Remember that 2pi/B is your period on the graph. C is what I think is the min or max. D would be your vertical shift.
Some of the variables may not be used. I am honestly not sure. I didn't do well when we studied sinusoid graphs, so this project is beyond impossible for me.
Thank you (:
 
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  • #2
Thanks again (:
Any help would be appreciated. I am completely lost.
 
  • #3
Welcome to PF! you will find a lot of useful and informative posts here as well as some
helpful advice. Next time, make sure you allow enough time to interact with our forum (the more
lead time the better).

I am seeing your amplitude (A) implied in your blood pressure readings. http://www.blood-pressure-hypertension.com/high-blood-pressure/definition-2.shtml
I see your frequency (1/period) as a rate of change over time. (were you given some data with units like that?)
Try plotting that so far and see if you can deduce C in your expression (the change in phase or time relative to t=0), and the vertical shift D. The vertical shift may be found using your maximum & minimum pressures.
 
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  • #4
Alright I will try that. I know time (period) is 1 second. My teacher told me to draw a heartbeat over 1 seconds' length. It would be a full heart beat because the average is 60 beats per minute, making a full heartbeat one second.
 
  • #5
Okay, so we when we used to solve sinusoid equations, we had to find :

Amplitude (A) - number in front that tells you how high and how low your curves will go. it is always a positive number.

Period (P) - is 2pi/B. Tells you where your graph will end ; endpoint

Phase Shift (PS) - action happening inside parentheses. Is C/B. Tells you where your graph will begin ; startpoint. The number you use on your graph is opposite the number in the equation because you are looking for what will make the number in the equation equal to zero.

Vertical Shift (D) - number outside parentheses that tells you where up or down your graph will be shifted.I figured that if I worked my way backwards, I could make the equation using these things.
I am having the most trouble with Amplitude. But really with all of them :/
 
  • #6
Okay I got the different amplitudes. Now I do not understand how to find a vertical shift, if there even is one .. ?
 
  • #7
Don't be distracted by the fancy word problem, it's still just a drawing a sine wave problem

The highest point his blood pressure reaches is 150. The lowest point is 90.(for the sake of drawing it all pretty don't forget the units. It's mmHg, a nifty pressure unit you'll learn about in chemistry, but don't worry about it, it's just letters to you right now)

So if you have a sine(or cosine) wave, not even caring about frequency or anything right now, that's oscillating between 150 and 90, you could just draw it and figure out the vertical shift and amplitude

There are specific equations for it you may or may not have memorized, but just look at it. The vertical shift is how high the "center" is(the y value it's oscillating around)and the amplitude is how high or low you go from that y value. You can read it right off the picture

60 beats per minute is of course 1 beat per second, like you said, so you know the period.

So now you have this graph of a wave, you know how big it is and how high it is, but not how fast. You know that every second, it gets back to where it started. Well there you go

Phase shift is the one that may be more confusing

I don't see anything that tells you where you should start(so at time = 0 you could start in the middle of a heart beat for all you care, it doesn't say)So I would of course start at the value that eliminates the phase shift. What's that mean? IF you use a sine wave, and it's not shifted anywhere left or right, what value does it start at?(at time =0?)How about a cosine wave? Note that you can actually draw the exact same picture using a sine or cosine wave, the difference will be the phase shift. That is in fact the only difference between sine and cosine, they're shifted pi/2 apart
 
  • #8
You were a TON help. Thank you so much (:
 
  • #9
I probably sound stupid, but I just can't tell where a vertical shift would come from.
I made a sine graph, and I know it's completely terrible :/
With the normal heart rate, I began at zero .. brought my curve up to 120 at the third block .. crossed through zero at pi .. brought my curve down to 80 .. and back up to zero at 2pi.
With Franks pressure, I began in the same place and moved up to 150 at the third block and crossed through zero at pi and brought my curve down to 90 and back to zero at 2pi.
I know that is completely wrong :/
 
  • #10
So if the wave were bouncing between, say, 50 and -50, what's the point it's oscillating around/in the middle/equilibrium point/whatever you want to call it? 0. So what's the vertical shift? 0

You don't have a single negative number in this problem, it's going to be pretty tough for 0 to be in the middle of that wave when it should never go below 0. Or even get close
 
  • #11
OH !
i see now !
oh wow yes yes yes !
I was going down to NEGATIVE 80 and NEGATIVE 90 !
wow i feel so dumb hahaha.
thank you thank you thank you
 
  • #12
My Graph

okay, here is my graph.
i am a little stuck on the equation.
i just don't know how i would write the vertical shift.

xmms81.jpg



this is my guess at the equations :

F(t) = 150/90sinx +

N(t) = 120/80sinx +

i know i need to add something at the send for the vertical shift. and i feel like its correct on my graph (although it may not be).
but i don't know what to write. would i say "+ 100" or ?
i don't know.
 
  • #13
Ok, so let's look at Frank's graph, not the normal one, first

The highest point should be 150, good, lowest point should be 90...I think you mixed up your own lines? Looks like you have Frank's lowest point at...70?

So the amplitude is going to be the furthest it gets away from that equilibrium middle point. How far away that middle point is from 0 is also going to be your vertical shift. So the point exactly between 150 and 90 is...100? No that's too close to 90. The point exactly between them will actually be the average, it'll be more than 100.

You used sine without a phase shift, since you apparently thought 110 was the middle point, you started the waves at 110, and that would've been right except it's not 110 like I just said. RIGHT IDEA though, you understood that fine enough

Once you find that middle point, the amplitude is going to be the furthest it gets away from it. Let's PRETEND we have a wave centered at 100, and it went up to 150 and down to 50. The amplitude is 50, so it'd be y=50*sin(...)+100. It goes without saying that even drawn incorrectly, the amplitude of these waves isn't going to be 1.666... >_> (150/90)

Ok so that's vertical shift, amplitude, and phase shift, what about frequency/wavelength?

Your x-axis should be time, you drew a sine wave with a very typical period of 2pi, which is saying that the heart beats one time in...6.28 seconds(though I don't think you even considered units there and just forgot)? Remember, the period was 1 second!
 
  • #14
On the other hand you drew the normal blood pressure graph exactly right(except for the period of 2pi again)though in writing the equation you made the same mistake with the amplitude

One thing that the problem didn't state that I assumed you knew, when you measure blood pressure, you get those two numbers, and that's not like a fraction or anything, one's the blood pressure when your heart is pumping(the highest it will be)and one when it's sucking(the lowest it will be). Or maybe that's backwards. Or biologically really wrong, but one's the high, one's the low is what you're given there
 
  • #15
I see blochwave has been helping you out here.. thanks bloch. I agree that since no specific points were given through a period, for simplicity, the phase term can be zero, as long as you begin a sine function at a minimum or cosine function at a maximum.

So summarizing, for the normal pressure mmHg (from my reference) systolic/diastolic=120/80.. (lets say it were 125/85, then maximum = 125 & minimum = 85, the average = (max-min)/2 + min = 40/2 + 85 = 105mmHg ..The central axis of the sinusoid is 105mmHg above the zero horizonal axis. in this side example, that is positive offset D.

For period ([itex]\tau[/itex]), the frequency of beats is 60beats/min = 1beat/sec and frequency f = 1/[itex]\tau[/itex]. Therefore the period [itex]\tau[/itex] is 1/(1 beat/sec) = 1 sec/beat ..The B term in simplyshannon's expression is often called [itex] \omega [/itex]= 2[itex] \pi[/itex]f = (2[itex]\pi[/itex])(1/[itex]\tau[/itex]) For Frank's high blood pressure, you would follow similar logic but using his systolic/diastolic values.

I agree there is some confusion in SS's minimum for Frank's pressure and also in labeling the horizontal axis (which should be units of time). The vertical axis should be labelled better as well (i don't see any description or units)

You've now done a,b&c. Part d when are the two BPs the same? If you have drawn both F(t) and N(t) with the same phase, there are no intersections between the two graphs, so there are no points where the BPs are the same. If you interpret that question to mean at what times does the normal pressure equal Franks pressure, they will occur at separate times. For part e, when are the first two times Franks pressure = 120 mmHg, you can read those values right off your graph.
 
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  • #16
Yah, I implied that 1 beat/second was the period, when in fact that was the frequency, the period would've been 1/f which is 1 second/beat

Fortunately 1/1 is still 1, but that was luck. I also for d) I don't see how(if drawn correctly)they intersect unless they're out of phase, and I don't see why you would assume they are unless there's more to the problem than typed.
 

What is a sinusoid graph?

A sinusoid graph, also known as a sine wave, is a mathematical function that describes a smooth repetitive oscillation. It is characterized by a curve that oscillates above and below a center line and follows a specific pattern.

What is the equation for a sinusoid graph?

The equation for a sinusoid graph is y = A sin(Bx + C) + D, where A is the amplitude, B is the frequency, C is the phase shift, and D is the vertical shift.

How do I graph a sinusoid function?

To graph a sinusoid function, first identify the amplitude, frequency, phase shift, and vertical shift from the equation. Then, plot the points on a graph and connect them with a smooth curve. You can also use a graphing calculator or software to graph the function.

What are some real-life applications of sinusoid graphs?

Sinusoid graphs have many real-life applications, including modeling sound and light waves, analyzing data in fields such as economics and engineering, and predicting natural phenomena such as tides and weather patterns.

How do I solve problems involving sinusoid graphs?

To solve problems involving sinusoid graphs, you can use the equation y = A sin(Bx + C) + D to find the amplitude, frequency, phase shift, and vertical shift. You can also use trigonometric identities and properties to simplify the equation and solve for specific values or variables.

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