Wednesday, February 28, 2007

9.1 Modeling with Differential Equations

9.1 - Modeling with Differential Equations

Differential equations discuss rates of change, like a population that changes according to what the limits on the population are or what the population already is.

First order: Here, we have a derivative that is equal to the original function times a constant. Thus, a likely equation for P would be Second order: y’’+y’=x+y
As the number of derivatives increases (and by that I mean first derivative to a second to a third...) , the order does, too. So, because there are both a first derivative and a second derivative in the above equation, it is a second order derivative. To solve it, you must not only find the defintion of the first derivative in terms of x for substituion, but also that of the second derviative (I do not think we will have to be able to do this, though).

Logistic growth:

As you might remember from Precalculus, many populations in the real world begin increasing rapidly, and then gradually level off until it appears not to be increasing at all. That is the curve of a logistic function. One can see that if the population P exceeds the carrying capacity C, then what is in the parentheses will be negative, so dP/dt will be less than zero, and the population will decrease. What happens when we see the graph leveling off is that P is approaching C, so what is inside the parentheses will almost equal zero, thus the dP/dt is almost zero, meaning the population is hardly changing.

There are two types of problems…

First, we will be provided with a potential solution to test.

This will be tested by finding y’ and plugging it into the equation to, hopefully, find that there are infinite solutions.
Then, we will have initial value problems (IVP) which is a bit like implicit differentiation in reverse. The problem provides us with an equation for which we find the general solution with a C-value, and another equation so we have the capability for solving for y – solve the differential equation for the general solution, then use an initial value to solve for the specific solution.

Provided: 1. Separate the values, isolating x's and y's on different sides of the equal sign 2. Find the antiderivatives of both sides. (You only have to use one C) And, now you have the general equation.3. Solve for C.4. Write the specific equation.

Population dynamics and practice

Paul’s (really detailed) math notes

Laurie, you bum, you are next. Unless sum (hah :( ) elaborate switching process has thrown everything off.

Blind Date - a math love story


Tuesday, February 27, 2007

Thursday's Quiz Topics

Here’s a list of topics that will be covered on this Thursday’s Quiz.

Quiz – Sections 8.1, 9.1
Arc length – given curve and interval (8.1, #1,3,5,9,11,29)
Arc length – determine setup (8.1, #1,3,5,9,11,29)
Arc length – determine setup (8.1, #1,3,5,9,11,29)
Arc length – given curve and interval (8.1, #1,3,5,9,11,29)
Arc length – given curve, determine interval (8.1, #1,3,5,9,11,29)
Differential equation – analysis and interpretation (9.1, #11)
Differential equation – verification of solution (9.1, #1,5)

That’s it for now! I’ll be around after school on WEdnesday and online Wednesday evening/night.

In any collection of data, the figure most obviously correct,
beyond all need of checking, is the mistake

(1) Nobody whom you ask for help will see it.
(2) The first person who stops by, whose advice you really
don't want to hear, will see it immediately.

And on another note, look for the simple solution:


Mr. French, for the past 45 minutes, I have been desperately trying to make my blog, but photobucket is "down for maintenance" and I can't post all of my pictures!!! I also tried directly posting them from paint, but the image appeared blank. When I tried to save it as a web page and then convert them to images it said "error: equation.html must support jpeg or gif." I feel like I've done everything in my power to try and fix this, Mr. French!


Ok, this is an update of 10 minutes later. I finally figured out how to upload them as regular images, Mr. French. But in the denominator, the squares aren't showing up because internet explorer doesn't support them or something. So I don't want to post it yet, because it's incorrect information that will confuse people!

Ok, Mr. French it's now 2:37 AM, and I figured out how to get everything to post correctly...except for the graphs. I could do it with photobucket, but that's still down. And I don't want to post it yet, because I don't want to without it being absolutely complete! (I even used winplot like you suggested!)


Ok, Mr. French. It's 3:50 am, and I finally figured out how to get the graphs to come out ok. The numbers are kind of small though. I'm not sure if I should post it. Maybe I should wait for photobucket to come back? Well, I'm sleeping. Night!


Monday, February 26, 2007

8.1 Arc Length


Ok, well, today we're going to figure out how to find the the length of an arc. It really is a beautiful thing. Now, remember that we always like to do things in terms of x and y. So, in order to express a constantly curving line, we will simply use an infinite number of hypotenuses (or is it hypoteni?) of right triangles. In order to express the hypotenuse in terms of y's and x's , we use pythagorean theorem:

*IMPORTANT REMINDER*: The legs must always be in terms of dy and dx, or the change in y and the change in x, for the given interval of the hypotenuse. (So you would start at one end of the hypotenuse and go to the other end, measuring how much the y changed and how much the x changed.)

But of course, we need the sum of all these hypotenuses in a given domain. So we use the integral sign and a dx. that we don't change the nature of the function, we also DIVIDE by the dx (dx/dx =1, and thus the value remains unchanged.)

Note the dividing by dx on the bottom.

So now, we want to try and simplify that devilish little formula. So....we put the bottom dx inside the square root. And to do that, we need to square it, so that we don't change the value of the function. (The square root of dx squared is of course dx, what we had in the first place)

Alright. Excellent. Now that we've done that, we can simplify what's inside those parentheses, since dxsquared over dxsquared is 1:
Unfortunately, Gianna, there is no +C when we have it definite like this. But that's ok. You still get it, right?
If you notice, in parentheses, we have (dy/dx), WHICH IS THE SAME THING AS THE DERIVATIVE OF THE FUNCTION!!! ISN'T THAT WILD???? So it looks like this:
*IMPORTANT REMINDER*: f(x) MUST be continuous on [a,b] and x must fall between or equal to a and b.

It's very true that we could have skipped all that and cut to the chase by going right to this formula. But I think that Mr. French and I would both agree that we all should value process over product. PLUS IT'S JUST PLAIN COOL TO SEE HOW WE DERIVED THAT FORMULA!!!

Ok. So let's test this theory.

Let's use this math problem:

Find the length of the following curve.

To sort of illustrate, this is what the graph looks like (Fortunately, unlike vons, my images actually came out pretty decently the first time round):

Photobucket - Video and Image Hosting

And this is the specific portion that we have to find the length of.

Photobucket - Video and Image Hosting

Let's go ahead and take the derivative. Wait! Just kidding. First we should put the function in more recognizable terms (After all, we don't want it to be so scary-looking that a certain someone would just start frantically trying to change it into slope-intercept form.)

Note that I simply made the coefficeints more recognizable. I changed the divide by 6 and instead multiplied x to the 5th by 1/6 (the same thing). Then I took the xcubed out from the denominator of the second term and multiplied (1/10) by x to the -3 (same thing).

Now take the derivative for real this time:

Let's plug it into our equation:

*IMPORTANT REMINDER*: Dont' forget to add 1 inside the integral!!!! (After all, we know that some people in our class seem to have a bit of a problem with adding.......Go Kristin!)
And plug it into our TI-83's and 84's (Or in John Cynn's case TI-89 Platinum Addition. Man, he thinks he's so cool with it, doesn't he?) with the good ol' fnInt(Y1, X, 1, 2,).

Of course I'm sure you could do the complicated way and do it by hand, taking the antiderivative of that and then plugging in 1 and subtracting it from what you get when you plug in two. But I'm pretty sure the only people arrogant enough to do that would be Mark Chodas. (No offense, Mark. You know we love you.)

This is a great site:
There's obviously not much to explain abou this topic. So I found a site that gives a lot of example problems and shows how to solve them!! :)

So as many know, I am a big fan of Desperate Housewives, and I have no shame in telling people that.....


a) Gabrielle Soliz

b) Edie Britt

c) Lynette Scavo

d) Susan Meyer

e) Bree VandeKamp

(Feel free to leave a comment about your choice.)


Whoever can guess where this is from first will get a prize.....!!!!!! (And it'll be good.)'re up!!

Wednesday, February 21, 2007

Friday's Quiz Topics

Here’s a list of topics that will be covered on this Friday’s Quiz.

Quiz – Sections 7.1,7
Integration by Parts – basic (7.1,#3,7)
Integration by Parts – definite integral
Integration by Parts – f and g won’t go away (7.1,#15 - not assigned, but good practice!)
Integration by Parts – tabular method (7.1, #61)
Trapezoidal Rule (7.7, #1,3,7,29)
Midpoint Rule (7.7, #1,3,7,29)

That’s it for now! I’ll be around after school on Thursday (after 3:30) and in early on Friday. Donut holes and OJ...

At New York's Kennedy Airport today, an individual, later discovered to be a public school teacher, was arrested trying to board a flight while in possession of a ruler, a protractor, a set square, and a calculator. Attorney General John Ashcroft believes the man is a member of the notorious Al-Gebra movement. He is being charged with carrying weapons of math instruction.
Al-Gebra is a very fearsome cult, indeed.They desire average solutions by means and extremes, and sometimes go off on a tangent in a search of absolute value. They consist of quite shadowy figures, with names like "x" and "y", and, although they are frequently referred to as "unknowns", we know they really belong to a common denominator and are part of the axis of medieval with coordinates in every country. As the great Greek philanderer Isosceles used to say, there are 3 sides to every angle, and if God had wanted us to have better weapons of math instruction, He would have given us more fingers and toes.
Therefore, I'm extremely grateful that our government has given us a sine that it is intent on protracting us from these math-dogs who are so willing to disintegrate us with calculus disregard.
These statistic bastards love to inflict plane on every sphere of influence. Under the circumferences, it's time we differentiated their root, made our point, and drew the line. These weapons of math instruction have the potential to decimal everything in their math on a scalar never before seen unless we become exponents of a Higher Power and begin to appreciate the random facts of vertex.
As our Great Leader would say, "Read my ellipse". Here is one principle he is uncertainty of---though they continue to multiply, their days are numbered and sooner or later the hypotenuse will tighten around their necks.

7.7 Approximate Integration

Hey everyone, ready for some fun math? K, well remember Reimann sums? it's backkkkk. but this time it's even more exciting because we use TRAPEZOIDS instead of RECTANGLES!

Let's take a look at a graph:

As you can see, the blue shaded regions are trapezoids, NOT rectangles. Therefore when we find the area, we will need to use the area of the trapezoids. In this example, there are five trapezoids. Their heights are B1, B2, B3, B4, B5 and B6. So now that we know the height, the width and the number of trapzeoids, we can find the total area.

Just in case we forgot geometry, the area of a trapzoid is:

* in this case, our "a" and "b" are B1, B2, B3...and so forth*

Finally we get the general equation which is called the trapezoid rule!

The trapezoid rule is just another way to approximate the area under a curve, but it is more accurate than using midpoints or Reimann sums.

But wait...don't you hate having to add numbers and then mulptify them? Don't you wish there was a simpler way? Well Simpson has the answer! Then using the same graph as above, the Simpson rule states:

*** Don't forget that the Simpson's Rule ONLY applies to even number of intervals.***

Example numero uno:
Use the trapezoid rule and the Simpson's rule to approximate: and also, n=4.

When n=4, Δx = π/4. Therefe when we use the trapezoid rule:

But wait! n=4, which is an even number! that means we can use the Simpson rule!

yay! now we are all gonna ace the quiz on friday right? of courseeee. but if you need more help, i found some really great sites:

  1. Unfortunately we couldn't see the visuals that Mr. French was going to show us, but this site has animations on creating trapezoids to approximate the area undert he curve.

  2. I don't know who the guy who wrote these Calc notes are, but he is AMAZING! This site offers detailed directions and explanations on how to use the two equations!



Tuesday, February 20, 2007

7.1 - Integration by Parts

Alright, folks, the Lord Protector of Math is back for some more Calculus action! Huzzah!

This lesson will be about integration by parts. This is somewhat like the substitution rule, but it deals with the Product Rule instead of the Chain Rule.

The basics of integration by parts are pretty easy. Let's start with the basic formula.

If we were to use u and v in place of the functions, the formula would simplify to this.

This is taken from the product rule for derivatives, and it allows us to integrate when two functions are multiplied.

Let's get an example. Evaluate the integral.

First, we go about setting values. There will be u, v, du and dv. It's generally a good idea to set the simplest function as the u value and the more complicated function as the dv value.
So, let's set x as u, and sin2x as dv.
This gives us:Since dx is the derivative of x and -(1/2)cos2x is the antiderivative of sin2xdx.
Are you with me? We're almost done.

Now we plug these equations into the equation.

See? We just plugged in the equations for u, du, v, and dv.
Now we evaluate the indefinite integral.

Don't forget your + C, or I will laugh at you pretentiously.
Anyway, see where we're going with this? Let's finish up.

Done! All evaluated.

There are some other things you need to remember:

1.You may have to repeat the process more than once to get rid of all of the integrals.
2. If the functions keep repeating, continue until you have reached an equation similar to the starting problem. Then, add/subtract it to one side, and solve.

Finally, the ultimate time-saver!


This is pretty straightforward and easy. If you have an equation with a relatively simple u and dv, it might be better to use this method. Let's try it with this function.

Alright, let me explain this tricked-out diagram. Basically, one column shows u and its derivatives and the other shows dv and its anti-derivatives. Drawing those diagonal lines helps you keep track of which u multiplies with the dv, and putting the alternating signs above the lines tells you whether those will be positive or negative. So, x^2 will go with e^x and be positive, 2x will go with e^x and be negative, etc.

So, the final answer turns out like this.

Don't forget the + C.

If you want more practice, this cool little web page lets you solve a bunch of problems step by step in Flash.

REMINDER: Lisa, you're up next.

And now I finish my post with this Photoshop edit:

Hope this was helpful, folks. Until next time.