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Math Help - Question on trigo?

  1. #1
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    Question on trigo?

    Hi I really have no idea which catergory this question falls into, especially so when I don't even know how to start off the question..

    If Person A is 20Km north of Person B at 5pm. Person A walks south at a rate of 7Km/h and Person B walks west at a rate of 9Km/h. What time do these two peopole stop approaching each other and instead, drift further apart?

    How do I start and what formula should I use? I'm really clueless. Is it under bearings..?


    Thank you so much, really!
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  2. #2
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    Re: Question on trigo?

    let person B start at the origin, (0,0)

    person A's position as a function of time, t, in hrs ... (0, 20-7t)

    person B's position ... (-9t , 0)

    distance, r ,between the two at any time t in hrs ...

    r = \sqrt{[0 - (-9t)]^2 + [(20-7t) - 0]^2}

    simplify, then determine when \frac{dr}{dt} changes sign from negative to positive.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Question on trigo?

    This is how I would set it up:

    Let north be the positive y-axis and west be the positive x-axis. Distances are in km and time is in hrs.

    Person A's position at time t is:

    A(t)=(0,20-7t)

    Person B's position at time t is:

    B(t)=(9t,0)

    Let D(t) be the distance separating the two. Hence:

    D^2(t)=(20-7t)^2+(9t)^2=130t^2-280t+400

    We see this is a parabola opening upward, so all you need to do is find the axis of symmetry to find the time when their distance is a minimum.

    Recall the axis of symmetry for the parabola y=ax^2+bx+c is x=-\frac{b}{2a}.
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