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Math Help - solving for t?

  1. #1
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    solving for t (daylight hours problem)?

    i'm trying to solve the following problem for t: 720 = 11.895 + 2.545 sin [(2pi/366)(t-80)]. my formula is using the format y = d + a cos [b(t-c)]


    I found a similar problem/solution online but don't quite understand how they solved the steps in bold... could someone explain how they got those answers in more detail?:

    In Philadelphia the number of hours of daylight on day t (where t is the number of days after Jan. 1) is modeled by the function
    L(t)= 12+2.83sin(2pi/365(t-365))

    A) Which days of the year have about 10 hours of daylight?
    B) Which days of the year have more than 10 hours of daylight?


    • math - drwls, Saturday, August 13, 2011 at 10:13pm(A) Solve
      10 = 12 + 2.83 sin[(2*pi/365*(t-365)]

      -2 = 2.83 sin[(2*pi*/365)(t-365)]
      -0.7067 = sin[(2*pi*/365)(t-365)]
      sin[(2*pi*/365)(t-365)] = -0.7848
      Solve for t.

      [(2*pi*/365)(t-365)] = -0.90244
      Use the first value

      t-365 = -52
      t = 313 days

      Oct 1 is day 303. So Oct 11 is one answer. The other day will be 52 days after the winter solstice, or about Feb 11.
    Last edited by nyago; November 25th 2012 at 05:48 PM.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: solving for t?

    if  720 = 11.895 + 2.54 sin(\frac{2\pi}{366(t-80)}) is your equation. Sin(x) reaches a max value of 1. So there is no way your equality could be right.
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    Re: solving for t?

    Quote Originally Posted by jakncoke View Post
    if  720 = 11.895 + 2.54 sin(\frac{2\pi}{366(t-80)}) is your equation. Sin(x) reaches a max value of 1. So there is no way your equality could be right.
    @jakncoke
    Correct LaTeX code makes a more professional post.

    [tex]720 = 11.895 + 2.54 \sin\left(\frac{2\pi}{366(t-80)}\right) [/tex] gives  720 = 11.895 + 2.54 \sin\left(\frac{2\pi}{366(t-80)}\right)
    Click on the “go advanced” tab. On the toolbar you will see \boxed{\Sigma} clicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them.
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    Re: solving for t?

    ..........
    Last edited by nyago; November 25th 2012 at 08:22 PM.
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    Re: solving for t?

    that is not how i wrote my equation.... it's 720 = 11.895 + 2.545 sin [(2pi/366)(t-80)]. my equation is modelling daylight hours, I need to discuss solving the inequality y(t) ≥ 720 for t

    this is my original formula:
    y= 11.895 +2.545 sin [(2pi/366)(t-80.5)
    one more time ... no way the right side of the equation is > the left side.

    maybe you should post the original problem as it is presented to you ... you've more than likely made a mistake in your set-up.
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    Re: solving for t?

    This is the original problem:
    Daylight is seasonal (there are more daylight hours in the summer and fewer in the winter).
    According to the Unites States Naval Observatory, the duration of daylight in Los Altos Hills each day
    of 2012 is given in the accompanying table (6-Daylight Los Altos Hills 2012.pdf), also available at
    Astronomical Applications Department.
    Your job is to analyze y, the minutes of daylight in Los Altos Hills, as a function of t, the number of
    days since the beginning of 2012. In particular, you should
    a) Produce an appropriately-labeled and appropriately-scaled graph of the daylight function
    covering a time period of at least three calendar years including 2012;
    b) Find a formula for the daylight function using constants determined from the data;
    e) In the context of daylight duration, discuss solving the inequality y(t) ≥ 720 for t

    This is the formula I got for problem b: y= 11.895 + 2.545 sin [(2pi/366)(t-80)] using the format y= d + a sin [b (t-c)]

    a= amplitude
    b = wavelength
    d = centerline
    c = phase shift
    Last edited by nyago; November 25th 2012 at 05:27 PM.
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    Re: solving for t?

    let me ask this ...

    y= 11.895 + 2.545 sin [(2pi/366)(t-80)]

    what are the units for these two numbers?
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    Re: solving for t?

    Amplitude:
    A= 1/2 (14.44-9.35)
    = 2.545

    Centerline:
    D = 1/2 (14.44+9.35)
    = 2.545
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  9. #9
    Forum Admin topsquark's Avatar
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    Re: solving for t?

    Another thought. You derived an equation to fit the data? I couldn't find data for this. How did you get your equation?

    -Dan
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    Re: solving for t?

    The data was from the US Naval Observatory... I chose the maximum/minimum amount of daylight from a one year set of data. Have you done these types of problems before??
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    Re: solving for t?

    You didn't answer my question about the units.

    Fyi, I believe your units are in hours ... the 720 is in minutes.
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    Re: solving for t?

    the data set was in hours and minutes. that 's how i wrote the units for 11.895 + 2.545 ... and I thought the 720 was in hours


    so, what should i do? i'm confused now
    Last edited by nyago; November 25th 2012 at 06:04 PM.
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    Re: solving for t?

    should i have rounded up my numbers? i didn't do that , otherwise the high and low points were off a bit when i graphed with an online grapher (desmos)
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    Re: solving for t?

    read the problem again ...

    Your job is to analyze y, the minutes of daylight in Los Altos Hills, as a function of t, the number of
    days since the beginning of 2012. In particular, you should
    a) Produce an appropriately-labeled and appropriately-scaled graph of the daylight function
    covering a time period of at least three calendar years including 2012;
    b) Find a formula for the daylight function using constants determined from the data;
    e) In the context of daylight duration, discuss solving the inequality y(t) ≥ 720 for t
    your amplitude and vertical shift was in hours, so you need to modify your equation since y is in minutes ...

    y= 60(11.895) + 60(2.545) sin [(2pi/366)(t-80)]
    Attached Thumbnails Attached Thumbnails solving for t?-daylightsine.png  
    Thanks from topsquark
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  15. #15
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    Re: solving for t?

    Oh, I see!!
    Last edited by nyago; November 25th 2012 at 07:46 PM.
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