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Math Help - Application of trig

  1. #1
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    Application of trig

    The depth, D(t) metres, of water at the entrance to a harbour at t hours after midnight on a
    particular day is given by D(t) = 10 + 3 sin(π t/6), 0 ≤ t ≤ 24.
    a) Find the value of t for which D(t) ≥ 8.5.
    b) Boats which need a depth of w metres are permitted to enter the harbour only if the
    depth of the water at the entrance is at least w metres for a continuous period of 1 hour.
    Find, correct to 1 decimal place, the largest value of w which satisfies this condition.

    i was able to sketch the graph, just having trouble with how you figure out D(t) ≥ 8.5

    the answer in the book is {t : D(t) ≥ 8.5} = {t : 0 ≤ t ≤ 7} ∪
    {t : 11 ≤ t ≤ 19} ∪ {t : 23 ≤ t ≤ 24}
    this makes sense, i just dont know how to figure it out algebraically
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  2. #2
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    Hello needhelpplease
    Quote Originally Posted by needhelpplease View Post
    The depth, D(t) metres, of water at the entrance to a harbour at t hours after midnight on a
    particular day is given by D(t) = 10 + 3 sin(π t/6), 0 ≤ t ≤ 24.
    a) Find the value of t for which D(t) ≥ 8.5.
    b) Boats which need a depth of w metres are permitted to enter the harbour only if the
    depth of the water at the entrance is at least w metres for a continuous period of 1 hour.
    Find, correct to 1 decimal place, the largest value of w which satisfies this condition.

    i was able to sketch the graph, just having trouble with how you figure out D(t) ≥ 8.5

    the answer in the book is {t : D(t) ≥ 8.5} = {t : 0 ≤ t ≤ 7} ∪
    {t : 11 ≤ t ≤ 19} ∪ {t : 23 ≤ t ≤ 24}
    this makes sense, i just dont know how to figure it out algebraically
    Here's my quick sketch of the graph (see attachment). Clearly when t = 0, d>8.5, and we can find the values for which t = 8.5 by solving the equation 8.5 = 10+3\sin\Big(\frac{\pi t}{6}\Big).

    So: 8.5 = 10+3\sin\Big(\frac{\pi t}{6}\Big)

    \Rightarrow 8.5-10 = 3\sin\Big(\frac{\pi t}{6}\Big)

    \Rightarrow 3\sin\Big(\frac{\pi t}{6}\Big)=-1.5

    \Rightarrow \sin\Big(\frac{\pi t}{6}\Big)=-0.5

    \Rightarrow \frac{\pi t}{6}= \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{19\pi}{6},\frac{23\pi}{6} ...

    (Is this where you were getting stuck? Look at the point moving around the unit circle, noting that \sin\Big(\frac{\pi}{6}\Big)=0.5
    . So the first value of \frac{\pi t}{6} for which \sin\Big(\frac{\pi t}{6}\Big) = -0.5 is \frac{\pi t}{6}=\frac{7\pi}{6,}, etc)

    \Rightarrow t = 7, 11, 19,23

    ... and using the sketch-graph, the answers then follow.

    Grandad
    Attached Thumbnails Attached Thumbnails Application of trig-untitled.jpg  
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  3. #3
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    thanks that really helps
    i was getting to here sin(π t/6)=-0.5 and then solving for t, didn't realise you had to use the unit circle.
    makes sense now, thanks
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