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Thread: Optimization Question - General Coordinates?

  1. #1
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    Optimization Question - General Coordinates?

    The question is this:

    A yacht is travelling due WEST at 6km/h. 3km NORTH WEST of the boats location, there is a boat travelling 4km/h, SOUTH WEST. How close to eachother do the boats get.

    From how I understand these questions, I needed to sub something into a distance forumla (distance between 2 coordinates)... I could sub in some form of general coordinates for each boat's path.... or I could sub in their position equations....

    I don't know how to begin this question!!! I'm drowning in this stuff..
    Last edited by mike_302; Mar 31st 2009 at 12:20 PM.
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  2. #2
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    Quote Originally Posted by mike_302 View Post
    The question is this:

    A yacht is travelling due WEST at 6km/h. 3km NORTH WEST of the boats location, there is a boat travelling 4km/h, SOUTH WEST. How close to eachother do the boats get.

    From how I understand these questions, I needed to sub something into a distance forumla (distance between 2 coordinates)... I could sub in some form of general coordinates for each boat's path.... or I could sub in their position equations....

    I don't know how to begin this question!!! I'm drowning in this stuff..
    Use a coordinate system. Place the yacht at the time t = 0 at the origin. Draw a sketch of this situation! Then the way of the yacht is described by:

    $\displaystyle Y:\left\{\begin{array}{l}x = -6t\\y=0\end{array}\right.$

    The boat at t = 0 is at $\displaystyle B\left(-\frac32 \sqrt{2}\ ,\ \frac32 \sqrt{2}\right)$ (Use Pythagorean theorem on an isosceles triangle). The way of the boat is described by:

    $\displaystyle B:\left\{\begin{array}{l}x= -\frac32 \sqrt{2} - t \\y = \frac32 \sqrt{2} - t\end{array}\right.$

    Now calculate the distance betwee the yacht and the boat:

    $\displaystyle d(Y,B)=\sqrt{(-\frac32 \sqrt{2}-t-(-6t))^2+(\frac32 \sqrt{2}-t-0)^2} = \sqrt{26t^2-18\sqrt{2} \cdot t +9}$

    If$\displaystyle d(Y,B)$ has a minimum then $\displaystyle (d(Y,B))^2$ has a minimum too. Calculate the first derivation of $\displaystyle (d(Y,B))^2$ and solve the equation

    $\displaystyle \left((d(Y,B))^2\right)' = 0$

    for t. I've got $\displaystyle t \approx 0.4895\ h$

    Plug in this value into the equation of the distance to answer your question.
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