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Math Help - Basic Trigonometric Function (amplitude, period)

  1. #1
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    Basic Trigonometric Function (amplitude, period)

    I've got to find the period, amplitude and graph a trig function similar to the one below and I'm not sure if I'm on the right track. I haven't posted the exact question so if something doesn't look right with the function that's why.

    f(d) = 8 - 3\cos(2\pi d)

    From that I think it's:

    Amplitude: 3

    Period: 1 ie \frac{2\pi}{2\pi}

    Is raised up 8 on the y/sin axis so the max and minimum values are 11 and 5 with nothing to shift it right so graphing will pass through the point (0, 11)

    Does that all seem right?

    And what changes by there being a minus sign between the 8 and 3 rather than an addition?
    Would the y axis start at the minimum point (5) rather than the maximum (11)?

    Thanks
    Last edited by Welkin; July 26th 2013 at 07:18 PM.
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  2. #2
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    Re: Basic Trigonometric Function (amplitude, period)

    Quote Originally Posted by Welkin View Post
    I've got to find the period, amplitude and graph a trig function similar to the one below and I'm not sure if I'm on the right track. I haven't posted the exact question so if something doesn't look right with the function that's why.

    f(d) = 8 - 3\cos(2\pi d)

    From that I think it's:

    Amplitude: 3

    Period: 1 ie \frac{2\pi}{2\pi}

    Is raised up 8 on the y/sin axis so the max and minimum values are 11 and 5 with nothing to shift it right so graphing will pass through the point (0, 11)

    Does that all seem right?

    And what changes by there being a minus sign between the 8 and 3 rather than an addition?
    Would the y axis start at the minimum point (5) rather than the maximum (11)?

    Thanks
    Does this look like a period of 1?
    Basic Trigonometric Function (amplitude, period)-cos.jpg

    -Dan
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    Re: Basic Trigonometric Function (amplitude, period)

    If \frac {\pi}{2} is ~ 1.57... then yes?
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    Re: Basic Trigonometric Function (amplitude, period)

    Quote Originally Posted by Welkin View Post
    If \frac {\pi}{2} is ~ 1.57... then yes?
    I guess I'm looking at another instance of where I should have been in bed earlier.

    Your answer of 1 is correct. If I hadn't reflexively put the x axis in terms of \pi then I would have seen that your answer was correct.

    -Dan
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    Re: Basic Trigonometric Function (amplitude, period)

    You had me worried I was lost again.... well more lost


    And it does start (or pass through) at (0, 5) doesn't it?
    and I should read it as -3cos?


    * Actually I'm still lost with this one, I thought that -cos means the y(sin) intercept is the low point rather than the high point but rereading my question here it also states that t = 0 corresponds with the temp being maxed (maxima?) which would be (0, 11) for the above example and so I'm still lost as to how the negative cos affects the graph?





    Thanks for the help I'll add another one here doing the reverse:


    If w cycles through twice in an hour, has a maximum of 100 and a minimum of 50. w = 0 corresponds to the maximum.

    Then with the x (cos) axis representing minutes, I think the function would be:


     f(w) = 50 +25\cos(\frac{\pi}{15}w)
    Last edited by Welkin; July 26th 2013 at 09:49 PM.
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    Re: Basic Trigonometric Function (amplitude, period)

    Quote Originally Posted by Welkin View Post
    You had me worried I was lost again.... well more lost


    And it does start (or pass through) at (0, 5) doesn't it?
    and I should read it as -3cos?


    * Actually I'm still lost with this one, I thought that -cos means the y(sin) intercept is the low point rather than the high point but rereading my question here it also states that t = 0 corresponds with the temp being maxed (maxima?) which would be (0, 11) for the above example and so I'm still lost as to how the negative cos affects the graph?
    The f-intercept (read as y-intercept) is indeed (0, 5). The -3 is no problem...it simply means that we use -cos() instead of cos(). So the function moves from f = 5 to f = 11. Thus the amplitude is still 3.

    -Dan
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    Re: Basic Trigonometric Function (amplitude, period)

    Quote Originally Posted by Welkin View Post
    Thanks for the help I'll add another one here doing the reverse:


    If w cycles through twice in an hour, has a maximum of 100 and a minimum of 50. w = 0 corresponds to the maximum.

    Then with the x (cos) axis representing minutes, I think the function would be:


     f(w) = 50 +25\cos(\frac{\pi}{15}w)
    Since cos() varies between -1 and 1 your f(w) varies between 50 - 25 and 50 + 25. Does this match your conditions?

    -Dan
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    Re: Basic Trigonometric Function (amplitude, period)

    25 and 75, no was trying to write one with an amplitude of 25 and maxima of 100 and minima of 50. thanks see it now need to add 75

    Think im sorted with the negative cos. had it mixed up its the second question that had the t= 0 is maximum heat qualifier (not the first), so positive cos.
    Last edited by Welkin; July 26th 2013 at 10:45 PM.
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