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Math Help - Trig Application

  1. #1
    Junior Member
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    Trig Application

    Just an exercise I came across in my revision in which I have no idea how to do

    The variation in blood pressure for a typical person can be approximately modelled by the funtion:

    P(t) = 20sin\Big(\frac{5\pi}{2}(t-\frac{2}{5})\Big)+100, where t is the number of seconds over which the persons heart beat is recorded, and P(t) is the blood pressure measured in mmHg

    1. At what time t, will the blood pressure first reach its maximum?

    I take it the maximum will be 120 given the amplitude and 100 - y intercept. How do I find the first t? Or is it just 120= 20sin\Big(\frac{5\pi}{2}(t-\frac{2}{5})\Big)+100

    2. The time between two successive peak pressure readings is called the period of the pulse. Calculate this period.

    I have no idea how to do this one. Please help
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  2. #2
    MHF Contributor alexmahone's Avatar
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    1) The blood pressure will reach its first maximum when sin \Big(\frac{5 \pi}{2}(t-\frac{2}{5})\Big)=1 ie the argument of the sine function attains \frac{\pi}{2}

    \frac{5 \pi}{2}(t-\frac{2}{5})=\frac{\pi}{2}

    5(t-\frac{2}{5})=1

    t-\frac{2}{5}=\frac{1}{5}

    t=\frac{3}{5} sec

    2) The blood pressure will reach its second maximum when the argument of the sine function attains \frac{\pi}{2}+2\pi

    \frac{5 \pi}{2}(t'-\frac{2}{5})=\frac{5\pi}{2}

    (t'-\frac{2}{5})=1

    t'=\frac{7}{5} sec

    Period of the pulse is t'-t=\frac{7}{5}-\frac{3}{5}=\frac{4}{5} sec
    Last edited by alexmahone; June 17th 2009 at 12:56 AM.
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