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Wednesday, October 22, 2008

October 22. The Definite Integral

I guess I'm scribe again since I've been caught by Mr. K. That's alright though, no big deal. At least he's still around. :D

Today we started off my watching a youtube video entitled "Lobachevsky - Tom Lehrer" The video is about a mathematician named Lobachevsky who supposedly plagiarized from a fellow mathematician named Gauss.

Well back to the subject. We started out by graphing some values.


The red bars represent the velocity or speed at the given time intervals as it changes gradually. The orange and green bars represent pretty much another graph, where the change in velocity was instant. Whether the change in speed was gradual or instant, we can't determine so. The curve in between the orange and green is the distance travelled, and this is what we will be trying to find.

We then try to find the lower limit by finding area of the lower limits (found in red bars).

Lower limit = sum of area of least possible #'s.

We multiply each value by the time interval (1).
Lower limit = 1.4(1) + 2.7(1) + 3.5(1) + 4.5(1) + 5(1) = 17.1 ft.
Don't add the last value, because this is the upper limit and it can't get any higher than this (t=5).
Now we calculate the upper limit, which includes the orange and green bars.
Upper limit = 2.7(1) + 3.5(1) + 4.5(1) + 5(1) + 5.7(1) = 21.4 ft.
Now estimate the distance by taking the average of these two areas. (17.1 + 21.4)/2 = 19.3 ft.
If we're talking about limits, the range would be 17.1 ft. to 21.4 ft.
At this point we find the ambivalent area (area of uncertainty) which is pretty much the orange-green bars. Ambivalent area = 1.3 + 0.8 + 1 + 0.5+ 0.2 = 4.3 ft.
We have an estimated area of 4.3 ft. under the curve which doesn't exactly match the previous results we estimated.
Now we decrease the time intervals to be more precise, and to see what happens.

Let's find the lower limit, or left edge of this graph. Left edge = 0.7 + 1.35 + 1.5 + 1.75 + 2.05 + 2.25 + 2.4 + 2.5 + 2.7 = 18.2 ft.

Now we find the upper edge which is pretty much the same, but we exclude the number of least value and include the number of most value. It should equal 20.35 ft.

The distance will be between 18.2 ft. and 20.35 ft.
The mean distance is (18.2 + 20.35) / 2 = 19.3 ft.

The average distance remains the same as our last graph.

Therefore by decreasing the size of the time intervals we have brought both the upper and lower limits' ranges lower.
At 1 sec. interval range = 17.1 ft. and 21.4 ft.
At 0.5 sec. interval range = 18.2 ft. and 20.35 ft.
The mean stayed the same at 19.3 ft.

That was all we did in class, our homework is Exercise 3.1 questions 1, 3, 6, 7 and 9. Hope you enjoyed, if any issues, be sure to tell me and I'll be sure to correct them. The next scribe will be well, not Richard, but Not Paul.

4 comments:

dkuropatwa said...

Now THAT'S what I'm talkin' about!

Anonymous said...

Very nice job

Rence said...

Word.

.:. J + ME .:. said...

the hip to his jive. tee hee