## Friday, May 01, 2015

### Eine Kleine Nacht Integralrechnung (A Little Night Integral Calculus)

I had asked my middle-schooler if he could spare me some time after dinner to talk about some mathematical ideas and he had agreed. So I rushed to put the plates away, and we gathered with papers and a marker.

After a preliminary conversation on the nature of acceleration in cars, we decided to focus our discussion on the topic: how far does a dropped object fall in a given period of time? I illustrated it by dropping the marker from a small height. (Assuming no air resistance, I added.)

We started with the basic notion of acceleration. When we say an object is experiencing constant acceleration, what do we mean? I took up the example of the constant acceleration due to gravity, which we call g. What is the value of it, I asked him? 9.8 he answered, remembering this constant from our previous discussion.

We talked about the units of acceleration, meters per second-squared, which is telling us the additional speed increase every second. So in this case, the speed of the falling object increases every second by 9.8 m/s, I said to him, and he nodded.

I raised and dropped the marker again. We watched it fall with a thud.  What was the speed at the very moment when it started its fall? I asked. 0, he said.

How about a second later? 9.8 m/s, he said.

How about in two seconds? 9.8 + 9.8, which is 19.6 he said. I agreed, right, two times the acceleration g. So then, he asked, to clarify, is the speed 2g's after 2 seconds?

I clarified that we normally refer to g's when talking about acceleration, which in this case will always be 1g, but numerically, yes, the speed at this point, measured in m/s, would be 2 times the value of g. And we extrapolated from there how it would 3 times g after 3 seconds, and so on.

I wrote down the formula: speed = acceleration x time, which corresponded to our extrapolation. I drew points on a plot with the x-axis labelled time and y-axis labelled speed: (0,0), (1, 9.8), (2, 19.6), (3, 29.4), and the line that goes through those points. We talked briefly about how this fits the form of the equation of a line (y = m.x + c) which he saw recently.

Now, I asked, how much distance did the object travel in the first second? He said 9.8 meters. Ah, I responded. That would be true if the object was going 9.8 m/s for the entire first second. But in fact it was starting at 0 m/s and accelerated to 9.8 m/s only at the end of that second. So you would expect it was something less than 9.8 meters, no? Ah yes, he said, smacking his head, I forgot!

Let's see if we can figure it out, I said. I drew a horizontal line from 0 s to 1 s at 9.8 m/s and then a step up to another line from 1 s to 2 s at 19.6 m/s then another line from 2 s to 3 s at 29.4 m/s. Imagine this object, I said. It moves at 9.8 m/s for the first second, then at a constant speed of 19.6 m/s the second second, and then a constant speed of 29.4 m/s for the third second. How far would it travel in these three seconds? We worked it out: 9.8 + 19.6 + 29.4 = 58.8 m.

Then I showed him how the distance traveled in this case could be viewed as the sum of the areas of the corresponding rectangles. I then asked him, do you agree this object is always traveling faster than our free-falling object except at the end of each second when it's equal? He nodded. So this object travels farther than ours. He nodded again.
Now consider another object, I said, which travels 4.9 m/s from 0 s to 0.5 s, 9.8 m/s from 0.5 s to 1 s, and so on. This object's distance traveled would also be the sum of the corresponding rectangles, I pointed out, and would also be higher than the distance traveled by the free-falling object, but a little better approximation. His eyes lit up as I pointed out that we could make better and better approximations, and he could see how the distance traveled by the free-falling object would come to be the area under the line we had drawn showing the speed versus time function for it.

Now, I said, recall that we agreed the speed of the object could be expressed as acceleration x time. He agreed. I continued, the area in question (i.e. under the line) is that of a triangle with this height of "acceleration x time", and a base of length "time". So by the equation for the area of a triangle (half of base x height), we can conclude the total distance traveled would be 1/2 x acceleration x time ^2. I pushed on. This is pretty nifty, to have such a clean answer to the question how far does the object fall in a given time. But often we don't get a nice equation like this. Consider the kind of speed versus time that is typical of when I drive the car. It speeds up, stays constant for a while perhaps, then slows down again, maybe stops for a bit at a light, then picks up again, and so on... We get a strange, arbitrary shape. What would be the distance traveled in this case? By analogy to what we just did, it is just the area under this arbitrary curve. He agreed. And how can we figure out this area? I asked him. By drawing those rectangles, he replied...

His mother, who was listening to us by now as well, chimed in, isn't this Calculus? Yes, I nodded. This is what's called Integration. We talked briefly about Newton and Leibniz and their competing claims to having invented Calculus.

My son's eyes lit up with a question. He picked up the pen from me and drew a blob on the paper. What about the area of this blob? he asked me. Can you calculate it? I said, sure, using this method of rectangles we could calculate the area approximately. No, but exactly, he asked. Well, not sure I could give you an infinite precision answer, I said. Aha, thought so! he exclaimed, feeling satisfied at encountering a fundamental limit to our understanding of things.

I mentioned then a couple of other ways to estimate the area of that blob. We could take a picture of a rectangle around it and count how many pixels are inside the blob and how many are outside, then use that ratio and the sides of that rectangle to get the area. We could ask a computer to generate a sequence of random points inside the square and check for each point if it is inside the blob or not, and the average ratio in this case would also converge over large numbers of samples to the right answer (this is called the Monte Carlo method, and was invented by Ulam while working on the Manhattan project). Or we could carve out that shape with uniform depth in some material and measure how much liquid it holds to estimate the area...

There was a bit of a digression here while we talked about the computability of real numbers briefly and I mentioned Chaitin's number Omega, which is a real number with the property that no algorithm can compute its digits (unlike pi).

My wife wondered if you could not also figure out the area from the perimeter of the blob which would be easy to measure with a thread. Ah, I said, but you can't infer the area from the perimeter. We talked about how a circle and a square having the same perimeter would have different areas. My son agreed though he said he couldn't prove it. I thought we would save that proof for a later conversation, and we ended our discussion for the night...

***
ps: The title I chose for this article is of course a weak attempt to evoke Mozart's famous piece, which I will never forget listening to at a live performance when I was 19 and everything seemed possible...

#### 1 comment:

steve said...

الرائد افضل شركة نظافة بالمدينة المنورة
تقدم خدمات التنظيف الشاملة للمنازل و البيوت و الفلل و الشقق اتصل بنا الان وايضا تقدم خدمات مكافحة الحشرات فهى افضل افضل شركة مكافحة حشرات بالمدينة المنورة
تقضي نهائيا على جميع انواع الحشرات المنزلية