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Thursday, February 12, 2009

Omgosh fast scribelists and approximations.

Okay guys, I'm here again, scribing, which makes no sense to me cause like, I scribed only a few weeks ago it feels like. Anyways todays topic, beside omgosh lightning fast scribe posts, is Approximation, or more specifically, how approximate is approximate?

Now the real location of that campground is somewhere around the big purple dot and letters, NOT near the big purple x and the lies that google tells us. According to google, approximate locations means somewhere within about 40 miles which I can safely say is a bit much.

So the main thing to take away from this, was that when dealing with things as approximations you need to know how good the approximations are. Moving on.

You should recognize these as the questions homework from the other day. Mr. K kindly went through the solution of one lightning fast, while simultaneously commenting on how the other should also work out to be an arctan. The actual solution process shown there is somewhat straightforward. Bring the 100 out of the brackets so you get 1 + something, then make the x2/100 a square by turning it into (x/10)2. From there apply substitution, and voila. The second requires some completing the square to solve, nothing really too complicated, just good practice.

K so the next topic, is shown here on the slide. How do you find an integral when you can't find an anti derivative? On the slide you see us trying some methods we know to integrate ex2. We find substitution doesn’t work because, say you choose x squared as your u, you can get rid of the 2 once you find du but you cant get rid of the x. Also, integration by parts doesn’t work because your f’ is e to the x squared which we're trying to integrate in the first place, and conversely, don’t know the derivative of.

Here's some notes about the above slide that I took in class.

- e to the x squared is u shaped, but not a parabola. The parent function of this is similar to x cubed, but is not x cubed because things happen much faster. Everything is very abrupt.

- Finding the definite integral (I think?) of this function on your calculator gives an approx value of 1.4627

Next slide(s)!




Okay, so all this might be a bit overwhelming, and confusing, but all we were really doing here, was comparing the accuracy of these various different approximation methods. As you can see, we compared normal old boring left and right hand sums, moderately exciting trapezoidal sums, and superb midpoint or tangent sums. Overall there were a couple main things to take away from the comparisons.

When trying to figure out if your estimate is and underestimate or an overestimate what it is depends on;
-The tonicity of the function
-If it is concave up or concave down
-If you are taking a right hand sum or a left hand sum.

But what’s the difference between a value of 3 and a value of 300?
-How concave the graph is.


The first derivative test deals with increase and decrease, while the second derivative test deals with concavity.

The main thing to take away from this scribe (and therefore this lesson) can be seen in this quote taken from a very knowledgeable person.

"When you do an approximation its important to say how approximate is approximate. Being off by 40 miles is not cool. In other words, you need to know how accurate it is. "

-Mr. K

Haha. I think he said that anyways. I might have translated it into my own words or something D:

J + Me Is gonna be the next scribe, and your homework is the homework from yesterday if you didnt do it, and the stuff on the last slide. To do the stuff with Trapezoid sums, use the riesum program on your calculator to figure out the right and left hand sums, then add those two, and divide by two. For the midpoint sum, Set your X Choice value in the Riesum program to .5 for the middle :]

Hopefully that wasnt too fast, and hopefully it all made sense.

Justus saying ciao, and goodnight.


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