數學難題 - 高中級(derivative)
2008-04-25 6:01 pm
Consider a room that has a floor with dimensions 4m by 5m, and walls that are 3m high. A spider is on one of the shorter walls, 1m from the floor and 2m from either adjacent walls. It wants to get to a point on the opposite wall 1m from the ceiling and 0.2m from an adjacent wall. What is the minimum distance that the spider must walk?
回答 (3)
2008-05-02 7:05 pm
✔ 最佳答案
[IMG]http://i92.photobucket.com/albums/l14/AhThir/Solution.jpg[/IMG]From the question, there are 2 possible target Point,
but since they having same distance to Spider, so i only use one of it.
Since the Spider can not "Fly", so it must walk stick to the wall.
The image show how is the Spider and the target Point locate in the room.
If the image can not be display, use this :
http://i92.photobucket.com/albums/l14/AhThir/Solution.jpg
2008-05-02 10:54 pm
The shortest distance at the adjacent 5mx 3m wall is the horizontal distance of 5m at the height of 1 m from ceiling.
The shortest distance at the opposite wall is 0.2 m at the height of 1 m from ceiling.
The shortest distance from the starting point to a point at the edge of the adjacent wall at a height 1 m from ceiling is the hypotenuse made by triangle of horizontal distance 2m and height 1m(the level difference between starting point and end point)=square root 5 m=2.236m
The minimum distance from start to end point =(2.236 + 5 + 0.2)m=7.436m
The shortest distance at the opposite wall is 0.2 m at the height of 1 m from ceiling.
The shortest distance from the starting point to a point at the edge of the adjacent wall at a height 1 m from ceiling is the hypotenuse made by triangle of horizontal distance 2m and height 1m(the level difference between starting point and end point)=square root 5 m=2.236m
The minimum distance from start to end point =(2.236 + 5 + 0.2)m=7.436m
2008-04-25 6:26 pm
Horizonatl distance to travel = 2+5+0.2 = 7.2m
Vertical distance to travel = 1m
Shortest distance = sqrt(7.2^2+ 1^2) = sqrt(52.84) = 7.27m
Vertical distance to travel = 1m
Shortest distance = sqrt(7.2^2+ 1^2) = sqrt(52.84) = 7.27m
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