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Thread: 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, $\displaystyle (0,0)$

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

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

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

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

    simplify, then determine when $\displaystyle \frac{dr}{dt}$ 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:

    $\displaystyle A(t)=(0,20-7t)$

    Person B's position at time t is:

    $\displaystyle B(t)=(9t,0)$

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

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

    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 $\displaystyle y=ax^2+bx+c$ is $\displaystyle x=-\frac{b}{2a}$.
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