AP Calculus BC 2007-2008

Integration by Parts

2008-01-31T16:55:00.000-07:00

Ok here it is! This is what we've all been waiting so long for! The blog post for how to do integration by parts! I know its a bit late but only by 3 weeks.

Integration by parts was basically created with the idea of the product rule in mind. If you have two things being multiplied, you can take the derivative of each separate piece and it would be the same as taking the derivative of the whole thing. Guess what! It works the same way with integrals too! Fancy that, its funny how the world works.

So basically we start out with the equation:

And by the magic of math it is simplified to:

So lets say your trying to integrate

xcosxdx

The first thing you would do is pick a u and a dv.
u=x dv=cosxdx
Which means that
du=dx v=sinx

Then plug it into the equation so you will have

This you can then easily simplify to xsinx-cosx+C

The thing to remember is not every u and v combination will work so if it appears that one just makes the new equation more complicated, stop and start over with a different set of u's and v's. Try and make u something that simplifies when its differentiated, and try to make dv something that remains manageable when its integrated. One easy way to remember what to look for in integrals is by using the phrase:

L n

I nverse Trig

P olynomials

E xponentials

T rig

Sometimes during your mathematical travels you will come across some integrals in which you would have to use integration by parts twice. Basically all you do is do integration by parts once, and then do integration by parts again, but only for the part of the answer that still has to be integrated. Then you can just put that answer back in.

Lastly, we learned one more form of integration. Its called Tabular Integration and its basically the cheating way of integration by parts. In your integral, if you have one piece that if differentiated enough times will end up at 0, and the other piece is easy to integrate over and over and over, then you can just put them in tabular integration. Lets say you have the integral

So obviously the x squared part is the part that is easy to differentaite and the e to the x is the part that can be integrated many times over. So you make a list where in the write column you begin with the x squared and differentiate it until it is zero. Then on the other side integrate the e to the x until you have it lined up with the zero. Then do a diagonal line from the first x squared to the integral of e to the x and keep making those lines until you run out of e to the x's. Then make the first one positive, the second one negative, the third one positive, and so on. That there is a mediocre account of tabular integration since I was unable to use any diagrams, so if you have questions (which I doubt) ask me in person and I will show you.

Logistic Growth and Partial Fractions 1/22/08

2008-01-24T17:13:00.000-07:00

Hi!
We have a test on integration Fri. 1/25 (which is also a food day!)
To review - pg. 358 #'s 1-28
Partial Fractions HW due Tues - pg.452 #'s 7-29 odd

Area Under A Curve

2007-12-04T14:48:00.000-07:00

Sorry this is late, but we had this lesson on Wed, Nov 27, 2007

For some practice problems, go to page 264 and do Exploration 1
Homework: pg 254 #5-13 odd, 20; pg 267 #7-41 odd

Integrals

2007-11-28T17:08:00.000-07:00

Hi Everyone!
This Friday (11/30) is food day! Please bring whatever you decided you were going to because otherwise we'll all be sad (and hungry).
Homework: pg. 274 #'s 1-33 odd
We took lots of notes today - here are scanned copies (to enlarge click on them).

Newton's Method , Differentials, Related Rates

2007-11-05T20:07:00.000-07:00

Hi guys
we have Quiz on Nov 6th, Tuesday.

Newton's Method
The general method

More generally, we can try to generate approximate solutions to the equation using the same idea. Suppose that is some point which we suspect is near a solution. We can form the linear approximation at and solve the linear equation instead.
That is, we will call the solution to . In other words,

If our first guess was a good one, the approximate solution should be an even better approximation to the solution of . Once we have , we can repeat the process to obtain , the solution to the linear equation

Solving in the same way, we see that

Maybe now you see that we can repeat this process indefinitely: from , we generate and so on. If, after n steps, we have an approximate solution , then the next step is

Provided we have started with a good value for , this will produce approximate solutions to any degree of accuracy.

Dfferentials

Given a function we call dy and dx differentials and the relationship between them is given by

Note that if we are just given then the differentials are the df and dx and we compute them the same manner

Related Rates

There is 13ft of string beteen two people. The people holding the string.

One person(x) is moving left and another person (Y)is moving forward. (keeping the distance between people 13ft)

speed of Y person is 4ft/s

We can know that the length of x side is 5ft and length of y side is 12ft to use pythagorean theaorem.

2 2 2

x + y = 13

2x (dx/dt) + 2y (dy/dt) = 0

x(dx/dt) + y(dy/dt) = 0

x(dx/dt) + y*4 = 0

5(dx/dt) + 12*4 = 0

dx/dt = - 9.6

Mean Value Theorem Applications, First Derivative Test

2007-10-20T19:43:00.002-06:00

So this is Jake. Basically, Mr. Marchetti extrapolated on the Mean Value Theorem (grrr theorem). Anyways, the homework he gave us was page 192 # 1-13 odd, 19-37 odd, 42, and 43. Also, he postponed the quiz until Tuesday the 23rd, so cram for that while you're at it.

Now, the MVT (Mean Value Theorem) has 3 basic premises that apply.
1. The Increasing/Decreasing Rule
So this rule basically states that, and I quote, "If, over the course of the interval, the secant line's slope is positive, then the tangent line's slope will also be positive."
In literal terms, this means that when the function is increasing, the secant line, and therefore the tangent line's slopes will be positive.

What this means to us is that functions increase when f'(x) > 0 and they decrease when f'(x) <0

The second rule is that a line with a derivative of zero is constant, so x=5 is just a horizontal line.

The third rule of the MVT is that "3. Functions with the same derivative only differ by a constant." Therefore, if f'(x) = g'(x), then f(x) = g(x) + constant
Here is an example for you: we have both f'(x) and g'(x) equaling 2x
You know that f(x) must = x^2 and that g(x) must also = x^2. However, g(x) could also equal x^2 + 121. That would still result in a derivative of 2x, the same as for f'(x), proving that functions with identical derivatives only differ by a constant.

Now that we have the MVT explained, we can move onto the First Derivative Test. What this rule states is that "When the derivative goes from positive to negative, you have a max. When the derivative goes from negative to positive, you have a min. And finally, if the derivative doesn't change signs, then there is no max or min." Pretty self-explanatory.

In this graph right here, you can see that, at -2, the derivative is negative and then switches to positive. This shows that there is a relative minimum at that point. At -1, the derivative is positive and goes to negative, leading there to be a maximum at that point.

To put all of this together, there is a good example that Mr. Marchetti showed to us on the board. You have the function y= x^3 - 4x. You need to find the maximum(s) and minimum(s). Ready go. So to solve this, you need to first find the equation's derivative. That is y' = 3x^2 - 4. You make that equation equal to 0 in order to determine where the critical points are in this specific function.
0 = 3x^ - 4
4 = 3x^2
4/3 = x^2
+/- (4/3)^.5 = x (this would be a lot less cluttered if I could insert stuff...)
+/- 2 / (3)^.5 = x

Now that you have the critical points of the function (where the derivative changes signs, which results in a max or a min as stated in the First Derivative Test), you can figure out how the function's derivative behaves in each interval. The intervals that you will have in this instance will be (- infinity, -2/(3)^.5), (-2/(3)^.5, 2/(3)^.5), and (2/(3)^.5, infinity). What you do is pick any value within each interval and plug that into the derivative equation that you found earlier (3x^2 - 4) to find how the derivative in that interval behaves. Remember, the endpoints of each interval are 0's, so the derivative crosses the x axis at those points and changes signs.
(- infinity, -2/(3)^.5)...: 3(-5)^2 - 4 = 71, positive in this interval
(-2/(3)^.5, 2/(3)^.5)...: 3(0)^2 - 4 = -4, negative in this interval
(2/(3)^.5, infinity)...: 3(5)^2 - 4 = 71, positive in this interval
now, to find which of these critical points is the maximum and which is the minimum, you refer again to the First Derivative Test. It states that when the derivative goes from positive to negative, you have a maximum. Therefore, at the point -2/(3)^.5), you have a maximum. Also, because the First Derivative Test states that when the derivative goes from negative to positive, you have a minimum, you know that your minimum is at 2/(3)^.5.

That is all that was talked about in class on Wednesday. Again, we'll see if the inserting pictures problem resolves itself or not.

SCRIBE POST - OCTOBER 10, 2007

2007-10-12T12:09:00.000-06:00

Hi guys, it's Devon. I was the scribe last Wednesday (October 10). It was a late start day, so all we covered was something called The Mean Value Theorem. We recieved a review packet, and we had to start working on that in preperation for our quiz Friday, October 19th.

Ok. Here we go.

The Mean Value Theorem connects the average rate of change and instantaneous rate of change of a function. The theorem states that all continuous and differentiable points between A and B on a differentiable curve, have at least one tangent line parallel to AB.

Between points A and B, we find a point C. Point C is the point of a parallel tangent line, and can be found by the previous equation.

A graph that represents The Mean Value Theorum would look like this:

An example of this would be driving. If you went on a 3 hour trip, averaging about 30 miles/hour, The Mean Value Theorem states that you would have to be going exactly that speed at least one time in the duration of your drive. That's basically it, it's pretty simple.

Derivatives of Logs and Exponents + Logarithmic Differentiation

2007-10-07T21:57:00.000-06:00

I do not have powerpoint so everything is done on word.

The information is on the pictures.

Implicit Differentiation

2007-09-26T20:36:00.000-06:00

Implicit Differentiation

An example of a function that is explicitly defined would be y=f(x).

But unfortunately not all functions are explicitly defined, such as the equation of a circle x^2 + y^2 = r^2. This is an example of a function that is implicitly defined. Often with implicitly defined functions it is not possible to solve for y.
             
            To equation of a circle can be divided into two pieces: the top half of the circle and the bottom half of the circle, with         
            each piece a separate function 

The Implicit Differentiation Process
      1. Differentiate both sides of the equation with respect to x
      2. Collect the terms with dy/dx on one side of the equation
      3. Factor out dy/dx
      4. Solve for dy/dx

example 1:  

               x^3 + y^3 - 9xy = 0   

The graph of x^3 + y^3 - 9xy = 0, a folium


                                                       
-This function has well defined tangent lines at every point on the graph. 
-Even though it is not a function in the classic sense, it is still differentiable
-The derivative from the right = the derivative from the left

x^3 + y^3 -9xy = 0 

3x^2 + 3y^2 * dy/dx - 9x*dy/dx -9y = 0           Since y is a function of x, when taking the derivative of any term with y,              
                                                                           multiply by dy/dx   
3y^2*dy/dx - 9x*dy/dx = 9y - 3x^2

y^2*dy/dx - 3x*dy/dx = 3y - x^2

dy/dx( y^2 - 3x ) = 3y -x^2

dy/dx = (3y - x^2)/(y^2 -3x)



*can use notation y' or dy/dx


example 2: 
                         2y = x^2 + siny

2*y' = 2x + cosy*y'

2*y' - cosy*y' = 2x

y'( 2 - cosy) = 2x

y' = 2x / (2 - cosy)


Finding the tangent and normal lines to a curve:
      -to find the slope of the tangent at a point, plug the coordinates of the point into the equation for dy/dx


example 1:
                         x^2 - xy + y^2 = 7  at ( -1,2 )



2x - x*dy/dx - y + 2y*dy/dx = 0

-x*dy/dx + 2y*dy/dx = y - 2x

dy/dx ( 2y-x ) = y -2x

dy/dx = ( y- 2x )/( 2y-x )

dy/dx(-1,2) = (2 - 2* -1)/(2*2 + 1) = 4/5

tangent: y-2 = 4/5(x+1)
normal: y-2 = -5/4(x+1)        
*To find slope of normal line, use opposite inverse of slope of tangent line


In class, we took our first homework quiz over the derivatives thus far, with the exception of the exponential and logarithmic derivatives

Homework: page 155# 3-28, multiples of 3

More Derivation Rules

2007-09-18T16:27:00.000-06:00

I hope everyone knew (and still know) rules for Trig Functions, because we took a quiz on the last week or so today, and if you didn't you missed a good 30-40% of the problems. Whoops. Anyway, we continued Rules today with Inverse Trig Functions and Logarithms/Exponential Functions. They are as follows:

Inverse Trig Functions:

Logarithms/Exponential Functions:

On that note:

Marchetti in 9 Months:
A mathematician went insane and believed that he was the differentiation operator. His friends had him placed in a mental hospital until he got better. All day he would go around frightening the other patients by staring at them and saying "I differentiate you!"
One day he met a new patient; and true to form he stared at him and said "I differentiate you!", but for once, his victim's expression didn't change.
Surprised, the mathematician marshalled his energies, stared fiercely at the new patient and said loudly "I differentiate you!", but still the other man had no reaction. Finally, in frustration, the mathematician screamed out "I DIFFERENTIATE YOU!"
The new patient calmly looked up and said, "You can differentiate me all you like: I'm e to the x."
http://www.onlinemathlearning.com/math-jokes-calculus.html

Yeah, I'm a nerd.

Homework: Page 172, #1-11

I didn't want to draw a picture. Sorry.

Chain Rule, Motion along a line, and R.O.C. of the Area of a circle with respect to the Radius.

2007-09-16T16:28:00.000-06:00

R.O.C. of the area of a circle with respect to the radius:
-The formula for the are of a circle is A= πr^2
-Using the power rule, take the derivative of A= πr^2. to get dA/dr= 2πr.

Example:
When R=5
dA/dr=2π5= 10π

Motion Along a Line:
-This is the relationship between position, velocity, and acceleration.
-Position: Where an object is with respect to time. S(t)
-Velocity: Is measured by both the magnitude and direction. V(t)=S'(t)
-Acceleration: Change in velocity with respect to time. A(t)=V'(t)=S"(t)

Example:
A rock (named ethan) is shot straight up into the air with a velocity of 160ft/sec. Using the equation S=160t-16t^2. A)Find how high ethan goes. B)How long it takes ethan to reach the vertex. C) Find the time when ethan is at 256 feet above the ground. D) Find the speed of ethan when it is 256 feet above the ground. E) Find the acceleration of ethan.

A) S=160t-16t^2
S=160• 5-16•5^2
S=400 feet.

B) First take the derivative of original equation which= ds/dt=160-32t. Then simply set the equation equal to zero.
160-32t=0
t=5 seconds

C)256=160t-16t^2
16t^2-160t+256=0
t^2-10t+16=0
(t-8)(t-2)=0
At time=8 seconds and time=2 seconds ethan will reach 256 feet.

D) To find the speed at those previous points, use the velocity equation.
V(8)= 160-32•8
V(8)=-96 ft/s
V(2)=160-32•2
V(2)=96ft/s

E) To find the acceleration take the derivative of the velocity equation.
a(t)=V'(t)= -32 ft/sec

Chain Rule
This rule is used to take the derivative of a function when that function, according to Mr. Marchetti, "is very ugly."

Chain Rule: dy/du=(dy/du)•(du/dx)

Example:
f(x)=(3x+5)^10

-First take the derivative of the "inside equation" which is u=3x+5, the derivative of that then equals du/dx=3
-Then take the derivative of the "outside equation" which is y=u^10 the derivative of that then equals dy/du=10u^9
-Then simply place those derivatives into the chain rule equation to get 10(3x+5)•3

Homework Due Tuesday September 18: Pg 129, #2,4,9,12,14,16,25 and pg 146 #1-19 odd

graphing nDeriv(, IVT, RoC

2007-09-13T17:24:00.000-06:00

Greetings Kids!

So, it appears as if I chose a good day to have the most siblings/step-siblings of all of you, and if any of you were hiding anything, it's your loss.

As it turns out, we didnt go over much in class, mostly just how to graph a numerical derivative using our calculator, the IVT (Intermediate Value Theorem), and Rates of Change. Based on the fact that I spaced making this post until the night before class, I won't have time to download the software and do the whole screenshot deal, but it's a nice thought. Use your imaginations.

GRAPHING THE NUMERICAL DERIVATIVE:

1. Input the funtion of which you wish to see the derivative into the Y1 entry line.
2. In the Y2 entry line, follow the following (teehee!) progression of buttons:
MATH
8
VARS
→Y-VARS
1
1
,
X
,
X
)
3. Now that you have punched this in, hit the ENTER button. On the resultant graph, you now have the original function and its derivative.
4. Note: When you see in the Y2 entry line “nDeriv(Y1,X,X)” I suppose you may want to know what this all means. This means you are finding the numerical derivative of the Y1 function for the variable X in terms of X. Instead of using nDeriv( to find an individual slope at a point, we are instead using it to graph the slopes of all tangent lines to the curve, THE DERIVATIVE!

INTERMEDIATE VALUE THEOREM

The Intermediate Value Theorem is stated as follows:

If a and b are any two values on a function, f, and f is differentiable then f’ takes on all values between f’(a) and f’(b).

As Marchetti said, that’s a lot of mathspeak, which is something most of us don’t understand. Basically, if a function is differentiable, then a tangent line exists at all points. If I’m wrong on that, then correct me. Basically, it says that if the function can have a derivative between a and b, then the function will.

Rates of Change:

There are two methods for finding Instantaneous RoC, finding the RoC for any point on the function, and finding the RoC at a specific point. Here they are:

Inst. RoC for a Function (Defn. of Derivative):

f'(x)=(f(x+h) - f(x))/h

Inst. RoC at a Point, “a”

f'(a)=(f(a+h) - f(a))/h

I think that’s about it, thank you all for your time and I apologize dearly for my lack of punctuality. The homework, in case you’re looking desperately at this, is pg 140 # 29, 28, 33, 31, and 32 (yes, in that order).

2007-09-12T20:13:00.000-06:00

Ok, I'm encoding this post in HTML so let me know if there are any problems. I figure I can use some coding to slightly alter the color of the math portions, therefore making them easier to read. Alright, so here we go:

Graph of Y' Stuff

Graphs are pretty important in Calculus as they form an easy allabi when you can't solve a problem. Anyways, there are alot of sweet things to know about a graph for example, if we take the function F(x)=X² who's graph is

then you take the Derivative of the function, which happens to be Y' = 2X¹ and you get this

Now you'll notice several things right away. First, is that the degree of the function is one lower. Second, you'll also notice, and the zero of the derivative is at the minimum of the original function. This can clearly be seen when you graph both of them at the same time

and zoomed in

Example Graphs

I figure I'll use a list instead of a table, the blog site probably won't respond to the coding. The function is in black, whilst the derivative is in red. The original function will come first, then both of then together.

F(x)=sinX & Y'=cosX

F(x)=X³ + 2X² & Y'=3X² + 4X

So if we can take two rules from all of this:

If F(x) is a polynomial, than F'(x) will be one degree lower.

If F(x) is continuous and at a maximum or a minimum at A than F'(x) will be 0 at A.

Trig Tricks

Alright, even though Trigonometry is a total pain in the ~~ass~~, we still have to do it. That being said, it's time consuming, stressful, and slightly annoying when you're attempting to find the F'(x) of a trig function. So, there are a couple tricks that make your life a whole lot easier when applied with the other calculus rules. On the left is a function, and on the right in green is the Y' of that function

F(x)=sinX F'(x)=cosX

F(x)=cosX F'(x)=-sinX

F(x)=secX F'(x)=secXtanX

F(x)=tanX F'(x)=sec²X

F(x)=cscX F'(x)=-cscXcotX

F(x)=cotX F'(x)=-csc²X

Below is an example problem of how to use these tricks along with the Quotient Rule:
Your starting function
Y=cosX / (1-sinX)
Apply the Quotient Rule
Y'= (1-sinX)(-sinX) - cosX(-cosX) / (1-sinX)²
Simplify by distributing
Y'=-sinX + sin²X + cos²X / (1 - sinX)²
Next use identies to turn the Sin²X into (1-cos²X) and subtract the cos²X
Y'= -sinX + 1 + - ~~cos²X~~ + ~~cos²X~~ / (1 - sinX)²
Now cancel out a (1 - sinX)
Y'= ~~(1 + sinX)~~ / (1 - sinX)²

Y'= 1 / (1 - sinX)

Alright, that's it: Dalton King

2007-09-08T15:36:00.000-06:00

9/7/07

Sometimes Limits Don't Exist
Provided the limit exists, the definition of a derivative is

However, the limit might fail to exist in four conditions:
1. Corner – one-sided derivatives are different
2. Cusp – the slopes of the secant lines approach ∞ from one side and –∞ from the other
3. Vertical tangent – the slopes of the secant lines approach either ∞ or –∞ from both sides
4. Discontinuity

For examples of what each of these conditions look like, see pages 105-106 in the textbook.

Don’t worry about this until it comes up in problems – you can still take the derivative of the equations!

Differentiability
Differentiability implies two things, which will become more important later:
1. Local Linearity
Graph a function, say y=(x-2)² + 4. Zoom in a lot on one area of this parabola and it looks linear. 2. Local continuity

Data
If you are given a table of values instead of an equation, you can still estimate the derivative of the midpoint between two data points by finding the average rate of change using (y2-y1)/(x2-x1) between the two points.

To see an example of this, refer to pages 99-100 in the textbook. First, they made a scatter plot of Table 3.1 (Figure 3.6) Next, they found the average rate of change and midpoint between the first two points, the 2nd and 3rd points… This is in Table 3.2. The midpoint became the x value and the average rate of change became the y value when they made a graph of the derivative (see page 100).

Numerical Derivatives
This is a calculator trick so that you don’t have to go through all the steps to find the slope of a tangent line.

On the calculator, press MATH then 8:nDeriv(. Fill in the equation, the variable that you want to take the derivative with respect to, and the value given in the problem.Ex. Given y=x³-3x²+4, find the tangent line at x=1. You can press nDeriv(x³-3x²+4,x,1) and get -2.999 which you would round to -3.To do this without a calculator, find the derivative of the equation which in the above example is y’=3x²-6x. Plug the x value into this equation to find the slope y’(1)=3(1)²-6(1)=–3.

If you want to find the y intercept of the equation of the tangent line, you would first plug 1 in the original equation to find f(x) which is f(1)=2. Then use this point (1,2) which you know is on the tangent line, along with the slope you found previously to solve for b.2=–3(1)+b → 5=b Therefore, the equation of the tangent line is y=–3x+5.

Homework Due Tuesday
Worksheet - 1-19 odd (front page) - just apply the rule, don't simplify.
Pg. 101 - exercises 7-12, 15, 17-19, 26
Pg. 111 - 1-10, 11-21 odd
Pg. 120 - 27, 29, 31

Investigating Rules of Differentiation from Friday, Aug 31

2007-09-03T21:33:00.000-06:00

Hey BC Class, first scribe of the year here-AKA Christine or Steener
Basically on Friday our class did a worksheet on Investigating Rules of Differentiation. The worksheet allowed us to do a couple of problems to see if we noticed any patterns. Afterwards we learned some rules that can be very useful for the “Definition of the Derivative”. The rules are as follows:

After we learned the three basic rules, we went deeper into the pool of derivatives and investigated more problems. Notice how each rule was used to solve other problems (multiple rules in one problem). Next we then applied the three basic rules already learned to the next three rules. More Rules:
So those are all the rules we’ve learned so far. I wasn't sure how to put put images on, i couldn't find paint but I hope these explanations help…don't let those carrots confuse you or Laura's engergizer bunny will eat them. That made no sense.

Hold on, I'm not done yet, we are going to have a test on Tuesday, September 9 on the Summer Packet. The homework was page 120 # 1-23 odd. Enjoy and happy break.

Scribe Post

2007-08-28T06:53:00.000-06:00

I hope the beginning of the year has started smoothly. Once we have ended our review portion of the year we will begin having a scribe for each class. The scribe will be in charge of putting all of the notes, announcements, important dates, and homework in a post here. One thing that may be difficult to do is to post graphs and such. There are online tools that can help. More on that later. I will upload electronic copies of as many handouts and other things as possible and post them on our wiki. After the scribe has posted the class can comment on the post, looking for errors and omissions. You will be graded on your posts and any comments you make (that are relevant) will get you participation credit. I look forward to seeing how you do with this. Good luck!

Countdown to the 2007 - 2008 School Year

2007-08-14T08:49:00.001-06:00

Well the countdown is on. Only 7 days left until the first day of school. I wanted to remind you that we will be going over the review packet during the first few days of school, along with some other material and class expectations. Also, there will be a quiz on the content of the packet and some other stuff after about a week. Make sure you are comfortable with a majority of the material in the packet before we begin class. If I were you I would get together with other members of the class and work together. Remember that you can post questions on the blog also, and get answers from your classmates.

That's all for now. See you all in a week. If you have any questions please comment on this post.

Welcome!

2007-07-03T11:11:00.000-06:00

I hope everyone is having a great summer. Mine has pretty much just begun. I went through the moving process during late May and early June, attending the Technology In Education conference where I was a presenter, and attending IMP training at LHS and finally I am able to relax and take some time to blog, and hopefully get you involved in this class and blog.

Remember your summer assignment? Well if you lost it or never got one, email me and I will send you a copy. Also, I would like you to post any questions that you have about the summer assignment here and get some help from your classmates. This is going to act as a precursor to how we use the blog during the school year. I would like you to get used to this thing early.

One more thing, post a comment to the blog just so I know you are getting this. Thanks!