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Help with grade 12 Trignometry project

  1. Dec 31, 2004 #1
    Help with grade 12 Trignometry project!!!a.s.a.p plzzz

    Hiii,,
    I am a grade 12 student ,and i have a math project related to sinosoidal functions to complete...My only problem at the moment is that ,with the notes my teacher gave to us it says that the sinosoidal regression on our calculators is in the form of :y= a sin(bx+c)+d , rather than the form :
    y= a sin[b(x-c)]+d ,which we are required to use when solving all the questions related to this project ,and so my question is is there any difference between those two equatios and if there is how would i be able to go around solving them especially when it comes to describing traslations and streches,and i want to know what is the difference in these two forms of equations so that i can know if there is a difference in the a,b,c,d values that i obtain from the calculator or the translations.Your soonest reply would be greately appreciated since i only have a limited time,as well as having the information about using a texas instrument which is TI-83 to graph such functions since i have no idea of how to do that at all!
    thanking you in advance for your help.
     
    Last edited: Dec 31, 2004
  2. jcsd
  3. Dec 31, 2004 #2
    To go from the first form to the other, factor [itex]b[/itex]:
    [tex]y_{1}=a\sin{(bx+c)}+d=a\sin{\left[b\left(x+\frac{c}{b}\right)\right]}+d[/tex]

    To avoid confusion, we can use different variables in the second equation:
    [tex]y_{2}=A\sin{[B(x-C)]}+D[/tex]

    When you convert between the two forms, a=A, b=B, c/b=-C, d=D. The second form ([itex]y_{2}[/itex]) is more useful when considering transformations (such as stretching, etc.) as the variable C can now also be used:

    For transformations of [itex]y=f(x)[/itex] to [itex]y=af[b(x-c)]+d[/itex], the original function is stretched by a factor of [itex]a[/itex] with respect to the x-axis and by a factor of [itex]\frac{1}{b}[/itex] with respect to the y-axis, and translated [itex]c[/itex] units in the horizontal (x) direction and [itex]d[/itex] units in the vertical (y) direction.

    What exactly are you having trouble with as far as graphing on the calculator is concerned? To graph any of these function, enter them (in any form) in the "y=" menu accessed through the button with that label in the top lefthand corner.
     
  4. Dec 31, 2004 #3
    Thank you very much for your help Dirus. The part with the calculator is that whenever i put the function as u said y= and i press graph it dosnt draw a graph at all so i was wondering could it be the window setting and if it is how would i set it to graph such functions
     
  5. Dec 31, 2004 #4
    More help !!

    Thank you very much for your help sirus. The part with the calculator is that whenever i put the function as u said y= and i press graph it dosnt draw a graph at all so i was wondering could it be the window setting and if it is how would i set it to graph such functions
     
  6. Dec 31, 2004 #5
    If you enter something in the form we discussed above, the calculator will graph it. The problem might be, as you suggested, related to the window settings. Consider the function you are trying to graph: where does it intersect either axis, what are the approximate domain and range? Now try to set your window so that your screen displays a large enough area to allow you to see the graph (use the 'WINDOW' button right next to the 'y=' button).

    If this is not the problem, you will have to tell me exactly what function you are trying to graph so we can figure out what's going on. If all else fails, a good way to return all your settings back to default is to clear your calculator's memory, if you don't need anything stored in it.
     
  7. Dec 31, 2004 #6
    Extra Help

    Well ya ,the function i was trying to graph is the sinosoidal regression one but the problem is i have problems with determing the domain and range of the function if u could kindly provide some help on that as well as what is meant by standard deviation and how would it be used wether it be on calculator or other ways.And could u plz specify how i would know the domain and range as well as the window range for graphing lets say
    E(n)=4.0685sin0.107(x+78.512)+12.151 And regarding this equation i cgot a value of negative 78.512 and so when u put it into the equation y=a sin[b(x-c)]+d it becomes positive right ??thanks in advance for your help
     
  8. Dec 31, 2004 #7
    Info on standard deviation here. Google can also help you get a general grasp on the topic.

    Regarding your equation, if you mean you have a value for x of -78.512, then yes, if you substitute, you find that the function is positive at that point.

    To find domain/range, consider what the variables in your general formula (a, b, c, d) represent. Are you still having trouble seeing the graph?
     
  9. Jan 1, 2005 #8
    thanx alot sirus , but actually yes i am still having trouble seeing the graph its probably about the window settings that i am battling with ????!!! :cry: so i wonder if u mind helping me out on this graphing part ??? :redface:
     
  10. Jan 1, 2005 #9
    Here is a window setting that allows you to see a couple wavelengths' worth of this function:

    WINDOW
    Xmin=-4000
    Xmax=4000
    Xscl=1
    Ymin=0
    Ymax=20
    Yscl=1
    Xres=1

    Are you seeing it with these window settings?
     
  11. Jan 1, 2005 #10
    Idk if this helps but sometimes my calculator is in degree mode from physics and I forget to change to radians when I do trig and calculus. That can screw up the graph if your used to it one way.
     
  12. Jan 1, 2005 #11
    Help !!!

    hii,
    thanx so much yapper for your help ,but ya i did check if it was on radian or degree! sirus i did put these values in but the one value for Xmin=-4000 dosent go in like i punch it in and then there would be an error for that value only ?wat do u think the problem is ??
    Another question is how would u find a line where a strech has been performed about if it an equation like the regression form ??? :shy:

    And then my last question says this : For Mexico city , consider june 20 (day 172 ),with 13.32 hours of daylight ,as the day on which there is the maximum number of hours of daylight and december 21 (day 356),with 10.97 hours of daylight ,as the day on which there is the minimum number of hours of daylight .using the data from these days (not the regression data ),algebraically determine a cosine equation in the form of h(n)=a cos [b(n-c)]+d for the number of hours of daylight.Explain how you determined each of the parametres a,b,c,d to the nearest a thousandths .

    Your help to these last questions is greatly appreciated as soon as possible please thanking you soooo much .... :smile:
     
  13. Jan 1, 2005 #12
    One last Extra help other than the above ones

    i was just wondering if u have an example of a periodic function with its data table as well as what i would put into the lists in my calculator for that graph e.g. time versus distance etc. in each list so that it graphs it as periodic ???????
     
  14. Jan 1, 2005 #13
    That's weird...so you type the number, then press enter or go to your graph screen ('y='), and what exactly happens?

    Can you figure this out using the clues in my first post in this thread?

    Think about what those parameters represent graphically, and how the function corresponds to the data (i.e. how the data can be used to graph the function and find its equation).
     
  15. Jan 2, 2005 #14
    I think you can play around with this on your own. Have you graphed simple trigonometric functions in class? Take y=sinx and make a data table (using the unit circle as reference if needed), then graph it. This may help you answer some of your earlier questions as well.

    To graph such a function using data points, you can enter the data into your lists in two columns (one for x and one for y), then perform sinusoidal regression on the point set.
     
    Last edited: Jan 2, 2005
  16. Jan 2, 2005 #15
     
  17. Jan 2, 2005 #16
    Think about what those parameters represent graphically, and how the function corresponds to the data (i.e. how the data can be used to graph the function and find its equation).[/QUOTE]

    Sorry but i still dont get the question and how with the data given i can graph it :cry: could u provide more help with this question a.s.a.p because i dont get it at all ????
     
  18. Jan 2, 2005 #17
    with the question i asked about an example of a periodic function i meanta real life phenomena like tidal data for example ?????so for an example like that what would u put into lists like would it be the times ,low tide and high tide ??? and which of those would u use to graph ???? :confused:
     
  19. Jan 2, 2005 #18
    Let me answer both questions at once by working an example with you.

    Let's take a real-life scenario: you have a bike, with pedals of length 20 cm and wheel radius 30 cm (assume the pedals are attached at the same height as the wheels). The problem tells you the angular velocity of the wheels is 40/3 rad/sec and the pedals 10/3 rad/sec, and the bike is moving at 4 m/sec. Give a cosine graph of the height of a pedal as a function of time.

    Let's use the general equation for the function

    [tex]h(t)=a\cos{[k(t-p)]}+q[/tex]

    We know that the minimum height of the pedal is 10 cm, because the wheel has a 30-cm radius and the length of the pedal is 20 cm. This also tells us that the maximum height will be 50 cm. Start by plotting these points on a graph and thinking about how the graph could look (we'll set the minimum height to occur at t=0). This allows us to already fill in values for a and q. a is the amplitude, which is half the vertical range of the wave function here. In this case that's just the length of the pedal, 20 cm. Also, q is the vertical displacement of the graph from its original form, [itex]h(t)=\cos{t}[/itex]. As you can probably see, q is located graphically at the midway point between the max and min values, so 30 cm. So now we have:

    [tex]h(t)=20\cos{[k(t-p)]}+30[/tex]

    Using the angular velocity of the pedals and the fact that h(0)=10 cm, we can find that the peak height (50 cm) occurs at t=3/20 sec. Therefore, we can assign a few values to the t-axis and see that the period of the wave function is 3/10 sec. We know that

    [tex]\mbox{period}=\frac{2\pi}{k}[/tex]

    so we can find that

    [tex]k=\frac{3\pi}{5}[/tex]

    So now we have

    [tex]h(t)=20\cos{\left[\frac{3\pi}{5}(t-p)\right]}+30[/tex]

    Now all we need is p. In this form, p is the horizontal displacement of the graph from the original [itex]h(t)=\cos{t}[/itex]. Since that graph (the original) normally peaks on the h-axis, we can simply take the displacement of the peak from that axis, which we can see is 3/20 sec. on the t-axis.

    So we have the final equation for the height of the pedal as a function of time:

    [tex]h(t)=20\cos{\left[\frac{3\pi}{5}\left(t-\frac{3}{20}\right)\right]}+30[/tex]

    Nemo, this is how you must approach your assignment. To return to your lists question, for this question you would input height values in one list and time in another, then plot them with height as the vertical axis. Thus you are plotting h as a function of t, or h(t).
     
    Last edited: Jan 2, 2005
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