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Integral Calculus
25 Oct 2007 22:37:13 IST
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Nadeem
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Joined: 25 Aug 2007
Posts: 521
25 Oct 2007 22:57:55 IST
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26 Oct 2007 03:57:41 IST
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Given points are
A(1,1) , B(-1,1), C(-1,-1) and D(1,-1)
Take an arbitrary point P(x,y) inside the square.
Now according to the given conditions,
Distance of (x,y) < A(1,1)
x2 +y2 < x-1x2 +y2 <(x-1)2
y2 < -2(x-1/2) -1
this represents the region bound by the parabola y2 =-2(x-1/2) {left-opening with vertex at (1/2,0) } and the given sqaure, in this case.
Similarly using other conditions, we get,
x2 +y2 < x+1y2 < 2(x+1/2) -2
x2 +y2 < y-1x2< -2(y-1/2) -3
x2 +y2 < y+1x2 < 2(y+1/2) -4
Now all these parabolas enclose a region which is symmetric in all the four quadrants, and hence u can find the area in the first quadrant and multiply by 4.
Now in the first quadrant, to find the area, we first find the point of intersection of the parabolas,
y2 = 1-2x and x2 = 1-2y
which comes out to be (2 -1, 2 -1)
2 -1, 2 -1)i.e. it lies on y=x.
Now to find the total area, we can first find the area bounded by line y=x, curve y2= 1-2x and the x-axs, and then multiply by 2. This gives the total area in the first quadrant.
Hence A = 2{1/2 *base *height + [ 2-1]
[1/2] 1-2x dx }
2-1][1/2] 1-2x dx } = 2 { 1/2 (2-1) (2-1) + [ 2-1][1/2] 1-2x dx }
2-1) (2-1) + [ 2-1][1/2] 1-2x dx }on computing its value, u'll get
A = 1/3 (42 -5)
2 -5)Hence S = 4A
= 4/3 (42 -5)
2 -5)Hope u got it!!
Cheerrrsssss!!!
27 Oct 2007 14:22:44 IST
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hey aditya
hav a luk at this................
its given in de ques tat P (x, y) shud be nearer to de origin
now de pt can't exceed beyond de squ
tat is why v r takin de arbitary pt 2 b neare 2 origin
the dist of de pt p has to be less than 1-x ( luk in de diag)
hence x2 + y2 < |1 - x |
x2 + y2 < |1 - x |{For pts along -ve x n y axis take mod}
i guess this was ur doubt
hope its cleared now
4 de rest of soln u cn luk at Ramya's post n nudges
27 Oct 2007 16:51:44 IST
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sorry my parabolas are not correct cos i didnt solve it but . u get the general idea... ramya there has solved the parabolas.. it won't be a square. cos if it was a square then ant the vertex of the square the point is closer to the side of the bigger square than the origin.
27 Oct 2007 20:46:32 IST
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Well mona, the graph is wrong, because u have to draw the graphs of the parabolas to get the correct figure.
Now,
y2 < -2(x-1/2)
this is the parabola we are getting on applying the condition.
So first draw the graph of y2 = -2(x-1/2) which has vertex at (1/2,0) and passes thru. (0,1) and (0,-1)
Similarly, y2 < 2(x+1/2) has its vertex at (-1/2,0) and this is the exact image of the first parabola in the y-axis.
So, when u draw the other two parabolas also, u'll get a curved figure (similar to a square which has curved sides). Try to draw it once, and tell me if its clear.
27 Oct 2007 22:05:54 IST
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WELL YA i know ...................
the grafh is gona come as 4 overlappin parabolas
i drew tat garph just to xplain aditya
tat P is an arbitary pt inside de squ
n those red lines represent its x n y coordinates
just to give an idea of hw those conditions came coz he didn't understand tat part
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