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Math Help - Urgent Help...Graphs of Sine and Cosine

  1. #1
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    Urgent Help...Graphs of Sine and Cosine

    Please help! I need to find the equation for each of these....I do not understand how to do so...please help...
    The Amplitude on 67 looks like it is 3...
    How do I solve for the period...If I know that, then I can put it in the equation...

    On 68, the amplitude is 2
    And 70 also appears to be 2

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    Please help! I need to find the equation for each of these....I do not understand how to do so...please help...
    The Amplitude on 67 looks like it is 3...
    How do I solve for the period...If I know that, then I can put it in the equation...

    On 68, the amplitude is 2
    And 70 also appears to be 2

    look for when the graph begins to repeat itself. you will realize that for 67 the graph repeats itself twice between 0 and 2 \pi. use that to figure out the period. in 68, the graph has only one oscillation between 0 and 2 \pi, and for 70, we have one and a half periods between 0 and 2 \pi
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    To figure out the period, I would do 2(pi)/b
    But I do not understand where to get this from...
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    Quote Originally Posted by aikenfan View Post
    To figure out the period, I would do 2(pi)/b
    But I do not understand where to get this from...
    you can't get it from the picture. just look at the graph and see how long it takes to repeat
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    For 67, the five points are: (0,0) (pi/4, -4) (pi/2, 0 ) (3pi/4, 4) (pi, 0)
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    Quote Originally Posted by Jhevon View Post
    look for when the graph begins to repeat itself. you will realize that for 67 the graph repeats itself twice between 0 and 2 \pi. use that to figure out the period. in 68, the graph has only one oscillation between 0 and 2 \pi, and for 70, we have one and a half periods between 0 and 2 \pi
    Well, since you have the amplitude, to figure out the period, as someone already said, you figure out where it repeats. In 67, it looks like it stops at pi. I got this because it repeates twice between 0 and 2pi, so divide that in half, and you get pi for the period. However, when you put it into you equation you must use:

    2pi/K = Period (K is the number in your equation)
    2pi/K = pi
    Therefore, k = 2, because the two's would cancel out leaving you with just pi. So your equation would be y = 3sin(pi)x . Hope that helps and makes the rest of them easier.
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    Quote Originally Posted by mayflower29 View Post
    Well, since you have the amplitude, to figure out the period, as someone already said, you figure out where it repeats. In 67, it looks like it stops at pi. I got this because it repeates twice between 0 and 2pi, so divide that in half, and you get pi for the period. However, when you put it into you equation you must use:

    2pi/K = Period (K is the number in your equation)
    2pi/K = pi
    Therefore, pi = 2, because the two's would cancel out leaving you with just pi. So your equation would be y = 3sin(pi)x . Hope that helps and makes the rest of them easier.
    you mean k = 2

    i don't think the poster was having problems with writing the formula. it was actually identifying the period here that was the issue
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    Quote Originally Posted by Jhevon View Post
    you mean k = 2

    i don't think the poster was having problems with writing the formula. it was actually identifying the period here that was the issue
    Yup, you're right, I meant K.
    I didn't mean to quote you there, I meant to quote the original question.
    And yes, it may have been the period, but I always forget or get a little confused once trying to figure out K, so I thought I would throw that in there.
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    Wait, where did the 2 come from...
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    Quote Originally Posted by aikenfan View Post
    Wait, where did the 2 come from...
    The two came from:

    In your graph, you can tell that the period is pi.
    but you can't just put pi in your equation, because the "K" value in your equation is not the period, to get the period it is 2pi DIVIDED by K.

    2pi/k = pi (you know pi from your graph)
    What value of K gives you a period of pi?
    2 does, because the twos cancel out and your left with pi as your period.
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    This is what I have got for the others...Do they seem correct?
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    Quote Originally Posted by aikenfan View Post
    This is what I have got for the others...Do they seem correct?
    they are incorrect
    Last edited by Jhevon; November 18th 2007 at 03:54 PM.
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    Do you know what it is that I am doing wrong? I put the period in the equation right?
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    Quote Originally Posted by aikenfan View Post
    Do you know what it is that I am doing wrong? I put the period in the equation right?
    your amplitude is wrong as well as the period

    recall, the general form is y = a \sin k (x - b) + c

    where |a| is the amplitude

    b is the phase shift

    c is the vertical shift

    \frac {2 \pi}{|k|} is the period, and

    \left[ b, b + \frac {2 \pi}{|k|}\right] is the appropriate graphing interval

    thus, if your period is 4 \pi, it means that,

    4 \pi = \frac {2 \pi}{k} \implies k = \frac 12

    also, your amplitude is 3 here. there is no vertical or phase shifts, thus, you equation for 68 is:

    y = 3 \sin \frac 12 x


    now try 70 again
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    I'm sorry, I'm a bit confused...On 68, I thought that the Amplitude came from the graph...the line goes up to 2 (but the grid goes up to 3)
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