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Calculus no clue

  1. Jun 12, 2005 #1
    calculus...no clue :(

    Here is this question, I have no clue how to do it, I have tried it several different ways but cannot manage to get the right answer. Any help will be appreciated.

    There is a 4 km long straight beach. At one end, there is a boat anchored at A, 3 km off the beach. At the other end, there is another boat anchored at B, 5 km off the beach. A sailor from the first boat is to bring a passenger to the beach and proceed to the second boat to pick up another passenger. At what point, C, along the beach should the first passenger be dropped in order to minimize the distance? Fill in the blank to answer this question. The boat whould land _______ km down the beach from a point across from A.

    The answer is 1.50
  2. jcsd
  3. Jun 12, 2005 #2


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    So you are looking for the C point on the beach that satisfies:
    AC + CB is minimum. C is the common point of the 2 lines AC, and BC. Does that make you think of anything?
    What AC equals to? HINT : Draw a line goes through A and is perpendicular to the beach. And let's see if you can get the answer.
    Viet Dao,
  4. Jun 13, 2005 #3
    So you end up with two triangles...do you use similar triangles ?

    5/(4-x) = 3/x
    you get the answer

    if this is right how are the two triangles similar ?
  5. Jun 13, 2005 #4


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    You should not assume similar triangles. Actually, if you just draw the triangle ABC you are trying to minimize something about that triangle. What is it, and how do you find a minimum?
  6. Jun 13, 2005 #5


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    Method of "reflections"- imagine that the second boat, instead of being 5 km off shore is 5 km inland. In other words imagine that the shore line is a mirror and B' is "in the mirror", a reflection of B It is easy to see that the shortestest path from A to B' is the straight line from A to B'. But because "reflection" preserves distances, the point C where the straight line from A to B' crosses the beach also gives the shortest path from A to B that touches the beach at C. NOW you can use the geometry of the situation ("vertical angles") to show that the triangles are similar.
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