trigonometry question

Discussion in 'General Math' started by Jason, Nov 17, 2005.

  1. Jason

    Jason Guest

    Hello, I am having a problem with a trig question maybe someone can
    give me a clue where I am going wrong!

    The height h in metres of water in a harbour above low tide is given
    by h = 14 - 10cos(30t), where t is hours since midday. A ship can enter
    the harbour when the height of the water is > 20 metres. Between what
    times is it safe for boats to enter the harbour?

    Okay, so this is what I tried:

    I think I need to find two solutions where the value of t makes h 20,
    and then format the answer as two times?

    so I can rewrite it as:

    20 = 14 - 10cos(30t) - solve when h = 20
    20 - 14 = 10cos(30t) - move the 14 over
    (20 - 14) / 10 = cos(30t) - move the coefficient over
    30t = cos^-1((20 - 14) / 10) - use arccosine to get t out
    t = cos^-1((20 - 14) / 10) / 30 - move the coefficient over

    so two possible solutions to t would be +/- 1.77?

    so midday - 1.77 hours, and midday + 1.77 hours

    boat can pass between 10:14am and 1:46pm?
    but apparently this answer is wrong and I have to do it again, but I'm
    very stuck!

    please point me in the right direction!

    thankyou for any suggestions.
     
    Jason, Nov 17, 2005
    #1
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  2. If arccos is correct, then at those two times
    the h is 20 and the boat can't pass because
    it's necessary h > 20, not h = 20.
    What can you say about h before and after arccos 6/10 ?
    You want to know when f(t) = h - 20 is positive,
    ie all values of t when f(t) is positive.
     
    William Elliot, Nov 17, 2005
    #2
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  3. Jason

    a1jrj Guest

    Stop right there - you have dropped a sign.

    Looking at the equation, h can only be bigger than 14 if cos(30t) is
    negative.
    Where is cos negative? hint: cos(-x) is cos(x) not as you seem to think
    -cos(x).

    Should be plain sailing from there on ;-)
    Incidentally, drawing a graph to check trig answers is beneficial at
    least initially.

    HTH
    JJ
     
    a1jrj, Nov 17, 2005
    #3
  4. Just curious, why do you carry (20 - 14) / 10) all the way through
    your work without simplifying it? If nothing else, you could save
    pencil lead.

    --Lynn
     
    [Mr.] Lynn Kurtz, Nov 17, 2005
    #4
  5. Jason

    Proginoskes Guest

    He probably got yelled at for using his calculator to find e^2, and
    he's overcompensating for that mistake.

    --- Christopher Heckman
     
    Proginoskes, Nov 18, 2005
    #5
  6. Jason

    Jason Guest

    I find it easier to backtrack later if i make a mistake, i would
    simplify if it got more complicated than this.

    Thanks to everyone who responded, I think I've got the right answer
    now!

    Thanks, Jason.
     
    Jason, Nov 18, 2005
    #6
  7. ^ sign missing here!

    20 - 14 = -10cos(30t), not +10 etc.

    you missed the minus sign! Try it again after this correction and you should
    be OK. Best wishes,

    gwh
     
    Grover Hughes, Nov 19, 2005
    #7
  8. Jason

    ticbol Guest

    Jason, Nov 17, :
    by h = 14 - 10cos(30t), where t is hours since midday. A ship can enter

    the harbour when the height of the water is > 20 metres. Between what
    times is it safe for boats to enter the harbour?
    and then format the answer as two times?
    You transposed the 14 to the lefthand side, allright, but why was the
    "10cos(30t)" changed from negative to positive at the righthand side?
    It should have remained as negative. So,

    6 = -10cos(30t)
    cos(30t) = -6/10 = -0.60 -----negative cosine, meaning the angle 30t
    is in the 2nd or 3rd quadrants.

    30t = arccos(-0.6)
    30t = 126.87 ------in the 2nd quadrant
    30t = (180 -126.87) +180 = 233.13 ------in the 3rd quadrant
    Hence,
    t = 126.87 /30 = 4.23 -------4 +0.23(60) = 4hr +13.8min = 4:13.8 PM
    t = 233.13 /30 = 7.77 -------7 +0.77(60) = 7 +46.2 = 7:46.2 PM

    That means, the ships can safely enter the harbor from 4:14 PM to 7:46
    PM ------answer.

    ---------
    If you want, if you know how to do it, you can draw the graph of
    h = 14 -10cos(30t)
    on t,h cartesian axes---t being horizontal axis, in hrs; h being
    vertical axis, in meters; (0,0) being (12noon,0m).
    It is a sinusoidal curve, a cosine curve, with a period of 360/30 = 12
    hrs, an amplitude of 10m, and it was shifted vertically up at 14m.
    Max h = 14 +10 = 24m
    Min h = 14 -10 = 4m.
    h at 12:00Noon = 14 -10cos(30*0) = 14 -10 = 4m.

    Then draw a horizontal line at h=20m.
    You'd see that this h=20m line cuts the graph at two points---one point
    after the (1/4)period or 3:00PM, and the other point before the
    (3/4)period or 9:00PM.

    ------
    Your "The height h ?in metres of water in a harbour above low tide is
    given
    by h = 14 - 10cos(30t),..." is misleading.

    The "...above low tide..." should have not been there----because it was
    found out that at low tide, h=4m.
    At low tide does not mean there is no water at the harbor.
     
    ticbol, Nov 20, 2005
    #8
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