IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

mona840

have a look at this Ques

a square's havin vertices at (1,1), (-1,1), (-1,-1) n (1,-1) let S be a region consistin of all d pts in d squ which r nearer 2 d origin than 2 any edge.

find d area of d region S

nadeemoidu

spideyunlimited

dude it doesnt require equidistant from vertices, it mentions edges!

watch this space for more, im on it.

ok ramyadiamond down there is correct :D

ramyadiamond

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-1

x² +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+1

y² < 2(x+1/2) -2

x² +y² < y-1

x²< -2(y-1/2) -3

x² +y² < y+1

x² < 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)

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 { 1/2 (2-1) (2-1) + _{[ 2-1]}^[1/2] 1-2x dx }

on computing its value, u'll get

A = 1/3 (42 -5)

Hence S = 4A

= 4/3 (42 -5)

Hope u got it!!

Cheerrrsssss!!!

priyesh

excellent work
vaise ramya & gaurav when do you sleep lol

mona840

ramyadiamond

Hey thanks priyesh!!

mona840

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 |

{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

spideyunlimited

NO MONA YAAR! the locus of p will be like a flower...the intersection of y^2 =x, y^2 = -x, x^2 = y, x^2 = -y

spideyunlimited

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.

ramyadiamond

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.

mona840

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

mona840

well

aint tat hw de graf comes

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