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Im stuck! Question a bout graphing sin and plotting points

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Im doing a question where I have to graphc y = -4 sin x

    Ok well I got the period and amplitude, so I have the interval of [0, 2pi]

    Now, in order to plot points, I have to divide the interval into 4 sub intervals.

    I dont understand how I do that! This is where I get stuck, and dont know how to plot the points now.

    Any help would be great!

    2. Relevant equations



    3. The attempt at a solution

    In my book it says " We divide the interval [0, 2pi] into 4 sub intervals, each of length 2pi/4 = pi/2

    I dont get that.
     
  2. jcsd
  3. Jun 26, 2011 #2

    Integral

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    Do just what it says, divide the interval [0,2[itex]\pi[/itex] ] into 4 intervals. They even tell you what points to use. Now just evaluate sin at each of those points, this will tell how to draw the graph.
     
  4. Jun 26, 2011 #3
    I am not confident on how to divide the intervals.

    I know for this one I can just refer to the unit circle, and get,

    pi/2, pi, 3pi/2, 2pi

    And those are my x cords right? and my y cords would just be -4, 0, 4 0

    But given the fact that im not 100% confident in dividing the intervals, how would I divide the following interval into 4 sub intervals?

    [pi/2, 13pi/2]
     
  5. Jun 26, 2011 #4

    HallsofIvy

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    When they say divide into four intervals, they mean four equal intervals. Mark 0 and [itex]2\pi[/itex] on our x-axis, then mark the midpoint- which divide the interval from 0 to [itex]2\pi[/itex] into two intervals, then mark the midpoint of each of those to get four intervals.

    Look at a graph of sin(x) itself. It starts at (0, 0), goes up to [itex](\pi/2, 1)[/itex], down to [itex](\pi, 0)[/itex], continues down to [itex](3\pi/2, -1)[/itex], then goes back up to [itex](2\pi, 0)[/itex]- and then repeats.

    So when you have decided where you want 0 to [itex]2\pi[/itex] line, mark first the midpoint. That will be, of course, at [itex](2\pi)/2= \pi[/itex]. That divides the line into two intervals, from 0 to [itex]\pi[/itex] and from [itex]\pi[/itex] to [itex]2\pi[/itex]. Now mark the midpoint of each those intervals, at [itex]\pi/2[/itex] and [itex]3\pi/2[/itex]. At those last two points, mark 4 units below and 4 units above the axis.

    Your graph will start at (0, 0), go down to [itex](\pi/2, -4)[/itex] at that first "quarter point" you marked,up to the axis at the middle point, [itex](\pi, 0)[/itex], up to [itex](3\pi/2, 4)[/itex] and finally back down to [itex](2\pi, 0)[/itex]

    For [itex]\pi/2[/itex] to [itex]13\pi/2[/itex], the midpoint will be the average of the two numbers: [itex](\pi/2+ 13\pi/2)/2= 14\pi/4= 7\pi/2[/itex] so mark that point. Now average [itex]\pi/2[/itex] and [itex]7\pi/2[/itex], [itex](\pi/2+ 7\pi/2)/2= 8\pi/4= 2\pi[/itex], and average [itex]7\pi/2[/itex] and [itex]13\pi/2[/itex], [itex](7\pi/2+ 13\pi/2)/2= 20\pi/4= 5\pi[/itex].

    BUT when you are drawing the graphs, what you really want to do is "eyeball" those points. It should be pretty easy to mark the midpoint of an interval without calculating anything.
     
    Last edited: Jun 26, 2011
  6. Jun 26, 2011 #5
    HallsofIVY - Great, thanks! But I have a question.

    in regards to: For π/2 to 13π/2, the midpoint will be the average of the two numbers: (π/2+13π/2)/2=14π/4=7π/2 so mark that point. Now average π/2 and 7π/2, (π/2+7π/2)/2=8π/4=2π, and average 7π/2 and 13π/2, (7π/2+13π/2)/2=20π/4=5π.

    As you stated, the midpoint wuold be 7pi/2 - How and where do I know to put that point??
     
  7. Jun 26, 2011 #6

    HallsofIvy

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    How do you know where to put "0" and "[itex]13\pi/2[/itex]"? In order to graph anything, you have to have a coordinate system which means you must have and x-axis, marked with x-values.

    Are you simply saying that you don't know what number [itex]7\pi/2[/itex]?

    [itex]\pi= 3.1715926...[/itex] so [itex]7\pi= 21.991148575128552669238503682957[/itex] and [itex]7\pi/2=10.995574287564276334619251841478[/itex]

    (Using the Windows calculator.)
     
  8. Jun 26, 2011 #7

    SteamKing

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    You can still use a calculator can't you?
     
  9. Jun 26, 2011 #8

    I would just put 7pi/2 half way between the 2 intervals. And to get the other 2 intervals I just average the starting point and midpoint, and the midpoint and the end interval?
     
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