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Math Help - Help with a sinusoidal function

  1. #1
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    Help with a sinusoidal function

    Michaelís weight fluctuates according to a sinusoidal function of time. Let x = 0 correspond to the beginning of the year (January, 1st). Because of his new year resolutions, Michael loses weight and reaches a minimum of 205 pounds at x = 60 days. Michael then stops exercising and dieting and his weight increases to a maximum of 250 pounds at x = 140 days.

    If summer starts on the day x = 171 and ends on the day x = 265, how many days of summer will Michaelís weight be below 220 pounds (these are the days he fits into his bathing suit)?

    Here is what I got:
    Using the sinusoidal function Asin((2pi/B)(x-C))+D
    I got 22.5sin((2pi/160)(x-100))+227.5


    So where do I go after I go the equation?
    ALSO, another curious question (if anyone got time or wants to answer it), how do you get rid of a sin in an equation.

    Thanks!
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  2. #2
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    Re: Help with a sinusoidal function

    Quote Originally Posted by Chaim View Post
    Here is what I got:
    Using the sinusoidal function Asin((2pi/B)(x-C))+D
    I got 22.5sin((2pi/160)(x-100))+227.5
    I agree, though I prefer replacing x - 100 by x + 60. This stresses that the sine wave begins at x = -60 and reaches its first minimum at x = 60. However, replacing x - 100 by x + 60 does not change the function itself. Let's denote this function by w(x).

    Quote Originally Posted by Chaim View Post
    So where do I go after I go the equation?
    First draw a graph. We see that summer takes place during the second dip. Moreover, w(171) and w(265) are greater than 220 (and even greater than the average weight 227.5), so the interval when Michaelís weight is below 220 during the second dip is entirely inside summer. This means that we just need to find the length of this interval (and do not subsequently need to take intersection with the summer interval).

    Since (220 - 227.5) / 22.5 = 1/3, we need to find the length of the interval inside the second dip when w(x) deviates more than 1/3 of its amplitude from the average 227.5. This is the same as the length of the interval inside the first wave of the function w'(x) = sin(2pi/160*x) where w'(x) > 1/3. Solve 2pi / 160 * x = sin-1(1/3). Then the length in question is semi-period minus 2 * x.
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  3. #3
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    Re: Help with a sinusoidal function

    What do you mean by "get rid of sin"? (I was tempted to answer "turn your pencil around and use the eraser!")
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  4. #4
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    Re: Help with a sinusoidal function

    Quote Originally Posted by HallsofIvy View Post
    (I was tempted to answer "turn your pencil around and use the eraser!")
    And I wanted to say, "Repent!"
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