- #1
Perrault
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Okay, so I've found the answer to this question but I'd like to know if it's right. Here's what I've got. Check the attachment in my second post (the first one has a small mistake) if you're more of the visual type.
What percentage of the area of a square is closer to the center of this square than to one of it's sides.
Not really equations but information :
-A parabola is the locus of the points that are at equal distance from a point (focus) and a line (directrix).
I gave the square a side of 1 unit.
I split the square in 4 with 2 lines passing through the center of the square.
I have chosen to study the lower fourth.
I have set the origin (0,0) to the center of the bottom line of the square (i.e. the longest line of the part of the square we are studying)
Consequently the focus is at (0, [tex]\frac{1}{2}[/tex]).
The directrix is y=0.
The equation of the parabola is y= x2 + 0.25. This is true because if you let p be the distance between the focus and the directrix, then y= x2/2p if the vertex of the parabola is set at (0,0) (which it isn't because placing the origin on the directrix will help us integrate). We add 0.25 to get the real equation of the parabola.
We find the limit of the parabola in our part of square to be at [tex]\frac{1 - \sqrt{2}}{2}[/tex] and -([tex]\frac{1-\sqrt{2}}{2}[/tex]) (don't forget that minus in front of the fraction! Also, the latter value is the greatest) by finding for which values of x is one of the lines that cut up the square (ex. y= -x - 0.5) equal to our parabola. We find the value of y for the right-most value of x (arbitrary, but we only need one) by plugging the values of x in the equation of the parabola (y= x2 + 0.25). We get ((-2) + sqrroot(2)/2) (sorry but this forum only shows [tex]\frac{1 - \sqrt{2}}{2}[/tex] whenever I put a fraction).
Now, if you integrate the parabola from [tex]\frac{1 - \sqrt{2}}{2}[/tex] to -([tex]\frac{1-\sqrt{2}}{2}[/tex]) (don't forget that minus in front of the fraction!) we get the area under the parabola.
Then we find both identical triangles left out by the integration (but still part of the fourth of square), by multiplying their base by their height and dividing by two (then multiply by two because they're two of them!)
This answer (integration plus triangles) is one fourth of the area further to the center of the square than to one of its side. Divide this by 0.25 (the area of one fourth of the square of side = 1 unit) and you get about 60.947571%. But since we want the area CLOSER to the center we do 100% - 60.947571% and we get 39.052429 %.
Is this right? Thank-you, I'm graded on this.
EDIT : I've corrected a mistake in this post but not on the image. My dad pointed out that the base of each triangle is not right-most the value of x where the line crosses the parabola, but rather 0.5 minus this value of x. Check my next post for an updated, corrected image.
Homework Statement
What percentage of the area of a square is closer to the center of this square than to one of it's sides.
Homework Equations
Not really equations but information :
-A parabola is the locus of the points that are at equal distance from a point (focus) and a line (directrix).
The Attempt at a Solution
I gave the square a side of 1 unit.
I split the square in 4 with 2 lines passing through the center of the square.
I have chosen to study the lower fourth.
I have set the origin (0,0) to the center of the bottom line of the square (i.e. the longest line of the part of the square we are studying)
Consequently the focus is at (0, [tex]\frac{1}{2}[/tex]).
The directrix is y=0.
The equation of the parabola is y= x2 + 0.25. This is true because if you let p be the distance between the focus and the directrix, then y= x2/2p if the vertex of the parabola is set at (0,0) (which it isn't because placing the origin on the directrix will help us integrate). We add 0.25 to get the real equation of the parabola.
We find the limit of the parabola in our part of square to be at [tex]\frac{1 - \sqrt{2}}{2}[/tex] and -([tex]\frac{1-\sqrt{2}}{2}[/tex]) (don't forget that minus in front of the fraction! Also, the latter value is the greatest) by finding for which values of x is one of the lines that cut up the square (ex. y= -x - 0.5) equal to our parabola. We find the value of y for the right-most value of x (arbitrary, but we only need one) by plugging the values of x in the equation of the parabola (y= x2 + 0.25). We get ((-2) + sqrroot(2)/2) (sorry but this forum only shows [tex]\frac{1 - \sqrt{2}}{2}[/tex] whenever I put a fraction).
Now, if you integrate the parabola from [tex]\frac{1 - \sqrt{2}}{2}[/tex] to -([tex]\frac{1-\sqrt{2}}{2}[/tex]) (don't forget that minus in front of the fraction!) we get the area under the parabola.
Then we find both identical triangles left out by the integration (but still part of the fourth of square), by multiplying their base by their height and dividing by two (then multiply by two because they're two of them!)
This answer (integration plus triangles) is one fourth of the area further to the center of the square than to one of its side. Divide this by 0.25 (the area of one fourth of the square of side = 1 unit) and you get about 60.947571%. But since we want the area CLOSER to the center we do 100% - 60.947571% and we get 39.052429 %.
Is this right? Thank-you, I'm graded on this.
EDIT : I've corrected a mistake in this post but not on the image. My dad pointed out that the base of each triangle is not right-most the value of x where the line crosses the parabola, but rather 0.5 minus this value of x. Check my next post for an updated, corrected image.
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