## Wednesday, March 30, 2011

### Integration trick

I was working on this problem, long story short it didn't amount to much. But I thought the technique I used trying to solve it was interesting in itself. If you have a certain graph in a cube like so:
x^2 in 3 dimensions
But you don't want to find the volume of the whole rectangular area looking form the top, you just want it to look like this when you look down on it from the top (you just want between the black lines):
topview cutout
SO that an artists poor rendering might be this:
You want the volume of this triangular wedge thingie
So the technique I use is to make two functions, one that goes from 4 to 10  and one that goes from 0 to -8 in the time it takes y to go from 0 to 1. Like so:
b = 4*(1-y)+10*y
a = 0*(1-y)-8*y
If you mess around with these you can see b = 4 and a=0 when y=1 and a=-8 and b = 10 when y=1 and they go in lines between the points in between.

So I set up my integral like this:

Then integrate with y going from 0 to 1 and x going from 0 to 10...
That gives the wedge shaped thing's volume to be:
V=148

## Friday, March 25, 2011

### SO it's an improvement

Well, I started thinking about the relationship of my formula to the formula for radioactive decay. The formula they use currently says that if you start with 2 grams of something, and it decays by half every second, that the formula for how much you have after t seconds is:
2*e^(-ln(2)*t)

But my problem with this is the integral of the above from 0 to infinity is 2.88, the original function is supposed to measure grams so the integral will be a certain number of grams*seconds.   It's a little hard to explain but it's like they're saying they went through 2.88 grams when they actually only had 2 to start out with.
*Edit I think I found a better way to explain it, over all those seconds the total number of grams*seconds they had was 2.88, but what if you condensed that all to just one second, that would say they had 2.88 grams for one second, but they only started off with 2 so the most they could have for 1 second is 2.
It's kind of like it used 2.88 grams of space-time when it was only supposed to have a maximum of 2.
So I like my approximation to exponential decay better because it adds up to 2 as expected. I wouldn't end up with more when you add up how much I had over time.

*Further edit I don't know if this is going to help or hurt but a thought experiment I'm having... Say if you are able to get a certain amount of work out of a certain amount of mass proportional to the mass. Like it could be pressing down a lever or something. With a 2 gram mass you are able to get 2 units of work done for one second on the lever. What their formula is saying is if the 2 grams were radiosactive and decaying and you kept removing the decayed part, you could somehow do a total of 2.88 units of work over an infinite number of seconds. But this doesn't really make sense you shouldn't be able to do more work just because the mass is decaying, right? Right? I don't know I'm confusing myself.

## Thursday, March 24, 2011

### The smoothness

So I was thinking about this idea that you can add 1 + a half + a quarter + an eighth to infinity and it adds up to 2 over the infinite number of terms. That's all good but what I was wondering is what would be a continuous function that sort of does the same thing? Anyway the first thing I noticed was this:
See it looks like the rectangles take up about half the area between the red and blue lines.

So next I thought maybe the proportion between the red and blue line that is also under a rectangle when compared to the total area between the red and blue line stays roughly constant? So I measured that proportion with the integral between 0 and 1 and used that instead of the (1/2) in the previous formula...
And that weird function down at the bottom of this picture is what you end up with, it follows the area of the rectangles continuously and has the same amount of area under it as you go to infinity as the rectangles do.

*EDIT
It doesn't quite look like it in this drawing I did but in the actual graph without the drawing areas there is a cool thing to notice:

The green areas are equal to the blue areas, meaning the part of the rectangle that gets cut off above the red line exactly matches the hollow part in between the line and the rectangles. That's an easy way to see that the curve has the same area underneath it as all the rectangles added together.