*sigh* no, I'm not trying to get anybody to do my homework, I just have a maths exam tomorrow... and I was doing some questions for practice, and have a really weird answer for one, so I wanted to know if you get the same answer or if I have a mistake (if so, where?)
I'm talking about the square with the circles in it (the circles are all touching other circles/the wall on 4 sides): http://img311.imageshack.us/my.php?image=matheoe1qj.gif
The length of one side of the square is 1. You are supposed to say what the area covered by the circles will be for n² circles.
So, here's what I did:
r(1 circle)=1/(2n)
A(1circle)=π*r²=π/(2n)²=π/(4n²ðŸ˜‰
so, A(n² circles)=(π*n²ðŸ˜‰/(4n²ðŸ˜‰, right?
But then, that would be the same as π/4, which would be a constant number (~0.785). But shouldn't the area be increasing as the number of circles increases, meaning shouldn't it still have a variable of some sort in the equation???
Thanks for any help I can get...
k angie, as far as I can see the answer won't change if you have 1 circle or 100000 circles. Since each circle can have a square cut around it, and a certain percentage of that will be covered in circle, and a certain won't be, no matter how many circles you fill it with.
Your only issue in work may be: r(1 circle)=1/(2n)
If you want to put the number of circles in for n here, it won't give you the correct radius. 4 circles in the square would have a radius of .25, and this gives .125. I was trying to find an equation that would do it, but it doesn't look needed.
Edit: okay I guess I don't exactly follow how you got the n² in your final equation. Nevertheless the work is probably right, though you could have solved it just by finding the answer with 1 circle in the square.
Originally posted by angie88Maths? Pah! Forget the exam... you've got some top tunes on their way to you!
*sigh* no, I'm not trying to get anybody to do my homework, I just have a maths exam tomorrow... and I was doing some questions for practice, and have a really weird answer for one, so I wanted to know if you get the same answer or if I have a mistake (if so, where?)
I'm talking about the square with the circles in it (the circles are all touching other c ...[text shortened]... dn't it still have a variable of some sort in the equation???
Thanks for any help I can get...
Posted today. 🙂
Originally posted by sasquatch672oh yeah you could squeeze them down to one if you aren't following the model in the picture. the model in the picture says that there are the same number across and up and down, and that they line up. Therefore no matter how small they get there is going to be the space.
Okay Angie, here we go. Try to appreciate this intuitively first and then we'll go to the math. As the number of circles approaches infinity, the radius of an individual circle will go to zero, and therefore, the areas of the cirlces can reduce to points - where there is no area - space - between them. So the two areas will converge.
Okay, here w ...[text shortened]... pi*(L^2/n^2)*n*n^2/2
The n^2 terms cancel
= L^2 - pi*((L^2*(n/2))
Somebody check me?
Originally posted by angie88No, the answer will be the same as if it was just one circle
But shouldn't the area be increasing as the number of circles increases, meaning shouldn't it still have a variable of some sort in the equation???
Thanks for any help I can get...
in the square, because of the relationship between the side
of the square and the diameter of the circle(s).
Originally posted by CoconutNice editing Coconut .
k angie, as far as I can see the answer won't change if you have 1 circle or 100000 circles. Since each circle can have a square cut around it, and a certain percentage of that will be covered in circle, and a certain won't be, no matter how many circles you fill it with.
Your only issue in work may be: r(1 circle)=1/(2n)
If you want to put the numb ...[text shortened]... y right, though you could have solved it just by finding the answer with 1 circle in the square.
Originally posted by sasquatch672What?
Yeah but that area goes to zero. As n--> infinity, r-->0, and as r approaches zero, you get an infinite number of infinitely small circles. Therefore, the space between the circles becomes infinitely small. This is the first day of calc II.
As n--> infinity r--> 0 is correct however n never reaches infinity and therefore r never reaches zero.
As a circle inside a square that it touches on all sides takes up a proportion of the squares area independant of scale the answer is constant.
Area covered by circles is pi*0.5^2 or 0.785 (3 sf).
Originally posted by XanthosNZHoly crap that sounds complicated, hence why i never took math!
What?
As n--> infinity r--> 0 is correct however n never reaches infinity and therefore r never reaches zero.
As a circle inside a square that it touches on all sides takes up a proportion of the squares area independant of scale the answer is constant.
Area covered by circles is pi*0.5^2 or 0.785 (3 sf).
Angie is right, XanthosNZ is right except for a little ambiguous language. Angie's question involved a specific fixed n, and the area for any specific fixed n is constant -- pi/4.
Since constant sequences converge to the same constant, the limit as n --> infinity of the area is simply pi/4 (you could even prove this from the definition of a limit: the nth term of the sequence is pi/4. Select any positive real p. Then if n >=1, abs(pi/4-pi/4) < p, so the sequence of areas converges to pi/4 by definition of convergence).
If this doesn't seem intuitively clear (ie if it seems more intuitive that the square would 'fill up' with circles), consider the following 'proof' that Pythagoras' theorem is false. If you can spot the flaw in this, you can spot the flaw in accepting the last parenthetical instead of the penultimate one:
Suppose we have an isoceles right triangle ABC with AB and BC having length n. If we start at A, travel to B and the to C along the edges of the triangle, we travel a distance 2n, obviously. Suppose we approximate the journey from A to C directly by travelling from A parallel to BC for a distance n/2, then parallel to AB for n/2, then parallel to BC for n/2, and then parallel to AB again to C -- 2n units in total.
If we do the same zigzagging, except we replace a journey of n/2 units with two perpendicular ones of n/4, and then do the same with n/8 etc., we find ourselves travelling along a 'staircase'. Oneach attempt, we have more stairs, and each stair is closer to the line AC. However, each time we travel a distance of 2n. Thus, since the staircase tends to the line as the numer of stairs goes to infinity, the diagonal of the triangle has length 2n, so Pythagoras' theorem is b0110cks.
In the above, the italicised line contains the flaw in the argument, which is analogous to the notion that Angie's circles 'fill the square' in the limit.