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Math Help - cosine function

  1. #1
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    cosine function

    Find an equation for the cosine function that has one maximum at the point (4,6) and its next minimum at the point (5.5,1).
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mr_Green View Post
    Find an equation for the cosine function that has one maximum at the point (4,6) and its next minimum at the point (5.5,1).
    y(x) = A*cos(k*x + (phi))

    y'(x) = -A*k*sin(k*x + (phi))

    y''(x) = -A*k^2*cos(k*x + (phi)) = -k^2*y(x)

    So we know that (4, 6) is a maximum point. That is, we know that
    y'(4) = -A*k*sin(4k + (phi)) = 0
    y''(4) = -k^2*y(4) = -A*k^2*cos(4k + (phi)) < 0 (So y(4) > 0)

    This means:
    sin(4k + (phi)) = 0

    4k + (phi) = n*(pi) <-- n is some integer.

    We know that (11/2, 1) is a minimum. That is, we know that
    y'(11/2) = -A*k*sin(11k/2 + (phi)) = 0
    y''(11/2) = -k^2*y(11/2) = -A*k^2*cos(11k/2 + (phi)) > 0 (So y(11/2) < 0)

    Thus
    sin(11k/2 + (phi)) = 0

    11k/2 + (phi) = m*(pi) <-- m is some integer.

    So we have two equations in k and (phi):
    4k + (phi) = n*(pi)
    11k/2 + (phi) = m*(pi)

    The solution is:
    k = 2(m - n)(pi)/3
    (phi) = (11n - 8m)(pi)/3

    So
    y(x) = A*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi))
    Where
    y(4) > 0 and y(11/2) < 0

    There are many many possibilities for this. Simply by playing around with m and n I find that m = 2 and n = 1, with A = -1 gives a good function. That is, one solution is
    y(x) = -cos((2(pi)/3)*x - (5(pi)/3))

    -Dan
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    if the minimum is (5.5,1) then it is not possible for the graph to go below the x axis. the equation that you derived does not work.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mr_Green View Post
    if the minimum is (5.5,1) then it is not possible for the graph to go below the x axis. the equation that you derived does not work.
    try topsquark's method using y(x) = A*cos(k*x + (phi)) + C as the general form of the cosine function. finding that vertical shift C should solve your problems

    or just alter topsquark's formula. say his minimum was at (5.5, -1) then just add 2 to his formula. that would cause a vertical shift of 2 units up, making the min point (5.5,1), of course that depends on whether or not his max point would fall in the right place
    Last edited by Jhevon; April 22nd 2007 at 07:49 AM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mr_Green View Post
    if the minimum is (5.5,1) then it is not possible for the graph to go below the x axis. the equation that you derived does not work.
    (Ahem.) I knew that. Really.

    I don't have time to redo this tonight. I'll try to get to it tomorrow morning.

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Okay, let's do this again. Most of my previous solution still holds, it just doesn't finish the job.
    y(x) = A*cos(k*x + (phi))

    y'(x) = -A*k*sin(k*x + (phi))

    y''(x) = -A*k^2*cos(k*x + (phi)) = -k^2*y(x)

    So we know that (4, 6) is a maximum point. That is, we know that
    y'(4) = -A*k*sin(4k + (phi)) = 0
    y''(4) = -k^2*y(4) = -A*k^2*cos(4k + (phi)) < 0 (So y(4) > 0)

    This means:
    sin(4k + (phi)) = 0

    4k + (phi) = n*(pi) <-- n is some integer.

    We know that (11/2, 1) is a minimum. That is, we know that
    y'(11/2) = -A*k*sin(11k/2 + (phi)) = 0
    y''(11/2) = -k^2*y(11/2) = -A*k^2*cos(11k/2 + (phi)) > 0 (So y(11/2) < 0)

    Thus
    sin(11k/2 + (phi)) = 0

    11k/2 + (phi) = m*(pi) <-- m is some integer.

    So we have two equations in k and (phi):
    4k + (phi) = n*(pi)
    11k/2 + (phi) = m*(pi)

    The solution is:
    k = 2(m - n)(pi)/3
    (phi) = (11n - 8m)(pi)/3

    So
    y(x) = A*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi))
    Where
    y(4) > 0 and y(11/2) < 0
    The m = 2, n = 1 part still works because it gives the location of the x values where we have the max and mins, though in reality x = 4 is a minimum point and x = 11/2 is a maximum. (That's why I had a negative A in my first solution.)

    We need to modify my solution a bit so that we have
    y(x) = A*cos(k*x + (phi)) + C
    as Jhevon suggested.

    Specifically, I'm going to work with:
    y(x) = A*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi))
    + C

    Now, the cosine function varies between -1 and 1. We need it to vary between 1 and 6. To do this we need to adjust C such that it is right in the middle of the range [1, 6], that is we need C to be the average of 1 and 6:
    C = (1/2)*(1 + 6) = 7/2

    Now we need to "stretch" the cosine function so that it will go below the line y = 7/2 down to 1 and above the line y = 7/2 up to 6. Thus we need |A| to be:
    |A| = (7/2) - 1 = 6 - (7/2) = 5/2

    As I said before I need A to be negative given my choice of m and n, so I get:
    y(x) = -(5/2)*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi)) + (7/2)

    -Dan
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