Limiting Area (1 Viewer)

[Carrotsticks]

So I know most people here have drawn this thing at SOME point in their lives out of pure boredom.

The other day I was just thinking about it and I thought 'Suppose I were to make the interval K approach 0, and thus have an infinitely large number of lines and intervals, what smooth continuous curve does it form?'

My instincts point towards an astroid (http://en.wikipedia.org/wiki/Astroid) but I can't think of a way to prove this.

Also, suppose that the square in the diagram had side length a. Assuming it is an astroid, if I wanted to find the ratio of Red:White, I'm guessing I would integrate the function parametrically, and put it over a^2?

Any ideas on how to find the equation of a curve. defined by the point of intersections of infinitely large number of lines?

The only idea I have so far is to utilise the Method of Exhaustion (http://en.wikipedia.org/wiki/Method_of_exhaustion) and to take a limiting case as k -> 0, but I have no idea how to set up the initial equation.

 

[seanieg89]

I for one haven't drawn that thing before , how is it defined?

Edit: Oh wait, I think I see what you mean. Going to try to figure it out

 

[Iruka]

'Cause I am feeling lazy, I shall link you to the appropriate wikipedia site.

http://en.wikipedia.org/wiki/Envelope_(mathematics)

It even has a gif annimation

 

[seanieg89]

Got R=1/6 by just taking the limit as n->inf of the area enclosed by the envelope with K=1/n (it is fairly straightforward to calculate).

My method should also yield the equation of the curve by using the FTOC (we can calculate partial areas in a similar way and hence obtain a differential equation). A little tedious though.

 

[sdrow]

I just did it by hand (differently to above) and got y = a - 2sqrt(ax) + x as the equation for the curve. This will also give R = 1/6.

 

[seanieg89]

 

[Carrotsticks]

Ahhh yes I understand how this works. Very nice proof!. Although MATH1906,7,2 do not cover the max function (well... not really a function), I have come across various problems involving it in Olympiads, so I decided to read up on it. It was a bit of a coincidence when Question 8 of 2011 HSC involved it!

I just woke up at 2am and decided to post because I came up with a potential solution! I guess the excitement woke me up. Perhaps a more elementary and less elegant solution than yours seanieg, but a solution nonetheless.

I'm just going to plant the idea here for myself to try in another couple of hours, incase I forget about it after I go back to sleep. Apologies in advance if I seem like I'm rambling or being nonsensical (cmon its too early, cut me some slack)

My idea involves taking an arbitrary point on the diagram, namely x, then defining the next consecutive point in terms of k and x. Similarly, define the previous consecutive point.

Afterwards, we should have 3 points, one of which is bounded between the other two. We can now use a principle similar to Differentiation by First Principles, whilst utilising the Squeeze Law, then taking lim k->0 to create a First Order ODE that defines the tangent function. Then integrate as appropriate in order to produce the required function.

It's just an idea that I woke up with. Don't know it it will work though, I shall give it a try once I wake up again.

Many thank to seanieg, sdrow and Iruka. I have long been looking for 1. A wikipedia page for this and 2. A solution, by various methods of course.

I shall post up another proof, if possible, by the end of today.

 

[Carrotsticks]

Okay I'm working on it now, and this is some of the ugliest algebra I have ever dealt with. However, it does simplify a lot to become something quite nice, so I'm guessing I'm on the right track.

Similarly to you seanieg, I saw that the curve is defined by the highest point of intersection. However, I also realised that the highest point of intersection (ie: the points that constitute the curve) come from the point of intersections of the Kth interval and the K+1th or K-1th interval. As shown in the diagram in OP, there will be infinitely many points of intersections, but the only ones to 'appear' on the curve are the consecutive ones.

I have just found the point of intersection of the Kth and K-1th interval, and once I get back home from work, I'll try to find the POI of the K+1th and Kth interval, and then take the gradient of that. Hopefully, that gradient can be manipulated to become the gradient function of the first quadrant of the astroid.

 

[lolcakes52]

Okay, here is how I see the problem. I am sure of very little of what I am saying but, instinctively, it makes sense... to me. Basically, imagine a triangle ABC, with a right angle at C. The AC+CB is equal to a constant, which is greater than zero. The point on the curve is given by the point P, which divides the line AB in the ratio of BC:AC, as the value of AC varies. If you dont understand what I am saying then try drawing a picture with a few triangles labeled ABC with different lengths of AC and CB. If that doesn't help then I can draw a pretty picture for you!

 

[lolcakes52]

Bam. I did some algebra and got something which I think is what you want. y=(x^(1/2)-1)^2 It behaves how you expect the funny shape thingo would between x=0 and x=1, but it changes after that. I used the method in the post above this one.

 

[sdrow]

Bam. I did some algebra and got something which I think is what you want. y=(x^(1/2)-1)^2 It behaves how you expect the funny shape thingo would between x=0 and x=1, but it changes after that. I used the method in the post above this one.

Yeah, if you replace the 1 in your formula with sqrt(a) (for variable a) you would get what I thought it was. It looks like we've used the same method.

 

[seanieg89]

Okay, here is how I see the problem. I am sure of very little of what I am saying but, instinctively, it makes sense... to me. Basically, imagine a triangle ABC, with a right angle at C. The AC+CB is equal to a constant, which is greater than zero. The point on the curve is given by the point P, which divides the line AB in the ratio of BC:AC, as the value of AC varies. If you dont understand what I am saying then try drawing a picture with a few triangles labeled ABC with different lengths of AC and CB. If that doesn't help then I can draw a pretty picture for you!

OK, it seems the different answers stem from the fact that you two used a different definition for the family of defining lines to me. I took it to be the set of line segments from the x-axis to the y-axis of length 1. (Think of the side-on view of a ladder sliding down from a wall with its bottom sliding away from the wall). If we take the family of lines to be the set of lines with "AC+CB" constant, then my method leads to your answer.

 

[lolcakes52]

I used ratios and similar triangles(to some extent). How did you do it? Also I am not sure that each segment has the same length, but it is possible. I havent thought about it. It would be possible to define the point parametrically using an angle and then finding the sides of the triangle using trig. That might be interesting and shed some light on the curve in general.

 

[seanieg89]

Well it is not entirely clear how Carrotsticks defines the family of lines in his diagram. All that is indicated is that the x-intercepts are equally spaced.

If the lines are such that the y-intercepts are also equally spaced, then the curve you wrote down first is correct.

If the lines are such that they all have length 1, then the curve i wrote down first (the astroid) is correct.

My method is worked through a few posts above, using elementary calculus.

 

[lolcakes52]

The line was in the 2010 extension two paper, question 4.