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|dw:1347549611645:dw| By SIMILARITY of Triangles [ three similar , right-angle triangle] we obtain (X+a)/(y+b) = x/b = a/y So ab = xy and bx + ba = yx + bx
y = ab/ x Substitrute int Length^2 = (y+b)^2 + (x+a)^2 AND Make derivation , then equate to 0 and solve for x - here is your SOLUTION !!
(ab/x + b)^2 + (x+a)^2 = a^2b^2/x^2 + b^2 + 2ab^2/x + x^2 + 2xa + a^2
well in the position you mentioned,,the rod cant move,so wont exit out.... |dw:1347549761719:dw|
(a^b^2/x^3) * (-2) - 2ab^2/x^2 + 2x +2a = Derivative
Now Deivative = o means the following should be solved
you're concept it wrong buddy,,see my last comment again,,and correct me if am wrong..
|dw:1347549859821:dw| Well - sorry to be so direct. But you ARE wrong. Rotation is NO PROBLEM
To be precise Rotatio + Simultaneous sliding of the ends
You see - the interval (lets call it "stick") that is the solution is THE SHORTEST POSSIBLE OF ALL ANGLES. Thus, a little motion provides it with MORE SPACE , NOT LESS !
Imagine a stick that fits into this space at "the most restrictive" angle. Well, it will obviously fit into positions at LES restrictive angles, wouldn't it ?
|dw:1347550109862:dw| for up ward movement,,downward movement should also be there,,which is not possible..
sorry didnt get this figure..
Firstly the LEFT END RISES, and the right end SLIDES into positions 3 and 3
it cant slide like that.. |dw:1347550476551:dw|
Look forget - drawing for a second. If Some position is the SHORTEST interval - all its small motions will create SPACE AT BOTH ENDS or AT ONE END at least !
Come on - after rotation the OLD SHORTEST STICK IS NOT FILLING THE NEW - LONGER SPACE
More SPACE - NO PROBLEM FITTING SHORTEST STICK !
You need a stick that will move inside the SHORTEST place. Sooo just take the SHORTEST LENGTH - some infintisimal "epsilon"
you mean.. |dw:1347551005335:dw| ?
This sign "- " above was MINUS
Call the minimal diagonal by M. Now take stick of length\[M - \epsilon \]
and since this dx is there,,it can rotate,,which will create dx upwards also ? right ?
You can take epsilon as small as you wish, BUT in some sense you are right. The reaching of the worst position and getting out of this tightest of all positions is strictly speaking impossible without some nominal "epsilon" reduction"
well,,if you include that epsilon or dx there,,after all the calculations,,we set dx->0 hmm ?