Calc mistake in #25
editedIn a rectangular room (a cuboid) with dimensions 30x12x12 , a spider is located in the middle of one 12x12 wall one foot away from the ceiling. A fly is in the middle of the opposite wall one foot away from the floor. If the fly remains stationary, what is the shortest total distance (i.e., the geodesic) the spider must crawl along the walls, ceiling, and floor in order to capture the fly?
sorry...the picture was also given...this can help..
source wolfram....
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29 Answers
FOR MORE DETAILS REGARDING THIS PROBLEM.....
(TO VIEW THE SPIDERS PATH ALONG THE CEILING,WALL,FLOOR,WALL.....IN ANIMATION...)
PLEASE SEE
http://mathworld.wolfram.com/SpiderandFlyProblem.html
yup 40 is the correct answer.....[1]
[1][1][1][1]!!!!![1][1][1][1]
There are two other possibilities of squares at right
Numbered them
The first one gives a length of √1658 feet
and the second one gives 40 feet
So, the smaller, 40 feet is the answer right!!!!
hope, I made it
[339]
the question asks shortest path... the path you have chosen is not the shortest one....
there is even a shorter path possible...
the flaw lies in chosing your diagram...
theres another way to construct the cube....
you are veryy close!!! [1][1]
This is the figure U R sPkin about..............
Well, actly there R two short paths, wen U convert it to 2D, U can take the spider to move along the base or along the wall............
wen it moves along the wall, the distance covered is √(42)2+102 = √1864 feet
and wen it moves, along the base(ground/roof), the distance travelled is 42feet = √1764
The shorter of the two is the answer=42 feet
U can make the same room, using many kinds of card boards, it just depends which one u choose
[339]
your figure is wrong....
think you are outside the room......
also think that you are given a cardboard...you are
to make a cuboid..... you don have glue... how would
you cut the cardboard??? in the same way open up the
given cardboard...and solve... you will get your answer...
Here z my method
This gives 42, and 42 is less than √1864
So, √1864, is surely not the answer[12][12]
[339]
@sb... try with #7 (govind's method..)
lets wait and see what other's have to say
ok should i give the final hint.... or wait for some more answers??
@govind: you are going wrong somewhere.....
see for my first answer i posted my method...for the second one..the spider will move 1m abv and then 30m horizontal and 11m down towards the fly..or do it the other way that the spider moves 11m down first then 30m and then 1m...both will give the same answer 42m...
no...... you are all too close...
can any 1 of you post ur method...
@sb : may be.... so whats the smallest...
yup.. the sum was done by somewhat this method only.... so what answer you are getting.. by this method
@ govind...very close...not exact answer
well the way of doing this question is this..just assume the conditions given here..in the room u r sitting..(provided u r sitting in a rectangular room)..then just draw the conditions given on a paper..i mean convert the 3d structure into one..by making all the walls lie in the same plane and then just join the points between which u want to find the distance..then simple use of pythagoras theorem and it's done...