Precal Blog

Today we worked on the data that was given to us before break. We graphed this on our calculators and our homework was to find the equation for this graph.

Next we recieved a paper plate and a piece of reciept tape and were to measure the tape so that it was the length of the circumference. We found that the tape was 3 times the diameter and a little more. We did the same for the radius and it was 6 radius’ and a little more. 

Next we marked the side of the plate to show where the radius’ were on the plate. Since we measured one radius along the side the angle from zero to the end of the radius was the same as the length of the radius.

Hey everyone,

I don’t even know if this qualifies as a legitimate blog post because we had a half day and didn’t really get in-depth into any topic, but for anyone who was absent or forgot what we did due to amnesia, I’ll give you a summary.

First, Mr. Bieniek blew everyone’s minds by showing us an analytics map.   An analytics map is a map which shows statistics of where people are viewing a site.

This was no ordinary analytics map, though, it was that of our very own blog. We found out that even though the map may not have been very specific (Hales Corners had 0 hits), we have an idea of how many people, and from where, are accessing our blog. They may find our site through a search engine, so make sure to keep the titles of your posts relevant to what we are learning, so people from other parts of the country and the world can access our posts.

Next, Mr. B hacked into a WE Engergies online account.  Don’t be alarmed, it was his account.  He showed us his spending records for the last two years.  If you didn’t receive this list, here it is.

January ‘09 ——>  $140.69

December ‘08 ——>$201.09

November ‘08—–>$143.78

October ‘08——>$127.58

September ‘08—–>$83.83

August ‘08——>$37.36

July ‘08——>$25.26

June ‘08—–>$21.13

May ‘08——->18.74

April ‘08——>19.91

March ‘08——->$30.23

February ‘08—–>$42.39

January ‘08—–>$62.90

December ‘07 —–>$101.03

November ‘07——>$115.62

October ‘07 ——>$143.07

September ‘07 ——>121.55

August ‘07 ——->$92.45

July ‘07 ——->$ 56.04

June ‘07 ——>$40.27

May ‘07—–>$35.97

April ‘07 —–>$33.71

March ‘07——>$29.05

February ‘07——>$28.97

We took these values and plugged them into our calculators.  Obviously, the gas costs during the winter will be greater than the costs during the summer.  Could these numbers have a sinusoidal relationship? Perhaps.  We will see on Monday, when we are sure to work with these numbers more closely.

Hope everyone had a great Thanksgiving weekend.

The next scribe, you ask?  The next scribe will be none other than SARA.

We’ve already learned all about the graphs of sine and cosine. We also learned how to transform them using different translations and dilations. Now, we’re learning what types of real world problems would use these transformations.

This graph shows the changes in Mr. Bieniek’s gas bill from December of 2007 to November of 2009. (2 years) If you were to draw a line of best fit for this graph, it would look like a sine graph. Using what we’ve learned about transforming sine graphs we can come up with an equation for a line of best fit for this graph. The equation I came  up with is y=80+65sin30(?).

The line would look something like this and can be used to represent the actual graph.

The next scribe is Pablo.

Today in class we learned about how to transform graphs of tangent and cotangent. We know that tangent = sine/cosine and that cotangent = cosine/sine. Here’s the graph of tangent
And since I can’t find a graph of cotangent using degrees I will attempt to explain it. It is the reverse of the tangent graph with the s shape starting up first going to zero and then going down. The period is still 180° but the first “s” is 0 at 90°. Transforming tangent functions is very similar to transforming cosine and sine. To transform the tangent function y=1+2tan30( -2) You would first find the period. Since the 30 is inside the parenthesis it would be a horizontal translation of 1/30 making the period 6. Also it would be horizontally translated 2 to the right, again because the -2 is in the parenthesis. The center point would be at (2, 1) because of the vertical translation up 1. The asymptotes are at the edge of the period: three to each side because the period is six. The asymptotes are vertical lines at -1 and 5. Finally the points which are usually one on the original tangent graph are one between the center point and right asymptote and negative one between the center and left asymptote. In this example they would be at three and negative one because of the vertical translation by two. The right point would be at (3.5, 3) and the left point (.5,-1). Transforming a cotangent function is exactly the same except the high point is on the left and the low one on the right and the zero is originally at 90° so you must transform it from there.

Today in class we continued to work with some of the trigonometric functions. We mainly focused on tangent and cotangent. We were given different equations and we had to graph them. There was some confusion between Hour 3 and Mr. B, but we got things straightened out eventually. Sorry for the sketchy graphs, winplot was being very uncooperative so I had to draw them all on paint (not as easy as is my seem).

y = -1 + 3cot2(x-30)

 So by looking at the equation we know…

1) the midline is at -1

2) period is 90 (cotangent’s period is originally 180 and this is half of that)

3) the center point is 75,-1) because the center point for cot is originally at 90 so in this case it would be at 45 and then it is translated 30 to the right

4) the asymptotes are at 30 and 120 ( 90 degrees between each asymptote)

5) cot starts high and ends low

6) Quarter points are (52.5, 2) and (67.5, -4)

So put this all together and there you have it. There was quite a bit of confusion on this problem so Hour 2 please thank hour 3 for straightening it out.

y = -cot 45(x)

So looking at the equation we know…

1) the midline is at 0

2) -cotangent makes the graph tangent, starts low and ends high

3) the period is 4 (the period is normally 180, divided by 45 = 4)

4) center point is at (2,0)

5) asymptotes are at 0 and 4

6) Quarter points are (1,-1) and (3,1)

y = 2 + 3tan36(x-2)

So looking at the equation we know…

1) tangent starts low and ends high

2) the period is 5 (180/36= 5)

3) the midline is at 2

4) the center point is (2,2)

5) the asymptotes are at -0.5 and 4.5

6) Quarter points are (0.75,-1) and (3.23, 5)

The last thing we did in class was write the equation for a given graph, I unfortunately don’t have that graph, but it is just like doing the opposite of what we did above.

The next scribe is Elise.

Today, while Mr. Bieniek was gone, we worked on Exploration 3-3b. The first problem on this sheet gives us an equation and we must find the horizontal dilation, period, horizontal translation, vertical dilation, and vertical translation. We can find each of these by looking at the equation. y= 3 + 1/2 tan 5(x – 7). This follows our y= A + B tan C(x – D). We’ve learned that A is the vertical translation. In this case, 3. Now we know B is the vertical translation, or the amplitude, or in tan, how far the quarter points are from the midpoint (vertically). In this problem, it’s 1/2. To find our horizontal dilation, we take 1/C. Using this, we can find the period by doing 180*1/C (in tan graphs, the normal period is 180). In this problem, the horizontal dilation is 1/5 and the period is 36. To find our horizontal translation, we take the opposite of D. D in the problem is -7, so the horizontal dilation is +7. Now we can graph our equation. Our midpoint of the tan graph is going to be at 7. To get the asymptotes, we’re going to add 18 to 7 and subtract 18 from 7. 18 is half of 36 so this is why we add and subtract. The asymptotes are at 25 and -11 From our equation, we find our midline to be at 3. The midpoint will be at (7,3). Since we have 18 degrees on both sides of the midpoint, our quarter points are going to be 9 from the midpoint. Our quarter points are going to be at (-2, 2.5) and (16, 3.5). Now we connect our points and follow the asymptotes to finalize our graph. The next problem has to do with cot which our class hasn’t gone over yet. Then the next pronlems are review of our csc and sec.

The next scribe will be……….. Greg

November 20-

First of all, I apologize for my graphs; I am not good at sketching them, but they should give you a general idea.

We began by adding upon what we did with tangent previously.

We were given the equation y=3+2tan2(?+30).  It can also be seen as y=3+2[(sin2(?+30))/(Cos2(?+30)).

To interpret the equation, we must first consider that the original center point for tan is (0,0).  The period is 180; not 360 like the other graphs.  This is because it is division between the the sin and cos graphs. (For more information on tangent, look in the side bar on the homepage for topics about tangent).

3 tells us the vertical translation would mean we would be up 3.

2 is the vertical stretch.

2 in front of theta helps us determine the period.  1/2 of 180 means the period would be 90.

30 means that the center point is moved 30 to the left.

Using the vertical translation and the horizontal translation, it can be understood that the new center point would be (-30, 3). 

To find the asymptotes, we must go from the center point.  We know that the period is 90 so half from the center point is 45.  The asmyptotes would be at -75 (-30-45) and 15 (-30+45).

The graph would appear as follows:

Cotnagent  has some differences from tangent. Sin’s asymptotes are Cot’s asymptotes. Its zeros are Cos’s zeros.  The centerpoint is at (90,0).  A sketch of the parent functions would look like:

Here are some very general differences between tangent and cotangent:

I’m not sure if we needed a scribe today and since they wouldn’t find out until they read this post, I’ll just tell everyone that we had a substitue and worked on a worksheet about graphs beyond sine and cosine.

The next scribe is Dana.

So I think 2nd hour is a little behind 3rd so you have probably seen tan graphs before this post, so if you need visuals you can just look at past blogs.

We all know that tangent is defined as opposite over adjacent or v/u. Another way we can define tangent is by using sin and cos. By solving for v in the equation sinX=v/r (X means theta) and solving for u in the equation cosX=u/r, you can put the values into the tanX=v/u equation. Giving you tanX=(r*sinX)/(r*cosX). The r’s cancel out and your left with tanX=sinX/cosX.

Now that you have tangent defined as sin/cos, it makes it a little easier to graph. Whenever sinX=0 (ex. X=0, +-180, +-360…), tanX=0 because 0/cosX will equal 0

You can easily find the vertical asymptotes because whenever cosX=0 (ex. X=+-90, +-270, +-450…) , tanX will be undefined (sinX/o).

When sinX and cosX are opposites, tanX=-1. When sinX=cosX, tan=1.

So that is the basic tangent graph. The next step would be to learn how to transform tangent graphs! I know it sounds scary and if you don’t quite get how to do it right away, don’t worry because Mr. Bieniek said tangent graphs are one of the hardest lessons.

The next scribe will be Stefan V.

Hey everyone, its Emily. I’m blogging for George since he wasn’t here yesterday due to his involvement in the fall play. Thankfully, we didn’t do too much during class.  The majority of the hour was spent watching the teaser for the fall play. For the rest of class, we had a substitute because Mr. B was gone. During this time we started working on our claim problems for section 3.3 in the book. This section deals largely with the new graphs (Csc, sec, tan) we’ve been learning about in class. No word yet on when the claims are due, but you can probably bet on having presentations sometime before Thanksgiving break. That’s about it for our pretty uneventful Thursday. The next scribe will be Kate!

So today in class we were given a worksheet to do.  The first three problems on the worksheet dealt with csc and sec graphing and the last problem dealt with tan which we will be learning tommorow.  One of the three problems we were supposed to do in class today was graph y=3+csc6( theta-30). To do this we first need to find the asymptotes and the high and low points of the sine graph since it is csc.  We know the vertical midline of the sine graph is 3 since it is 3+csc6( theta-30). Then the amplitude is one and since the midline is three the graph’s vertical high point would be 4 and its vertical low point would be 2.  Then we need to determine the period by dividing 360 by 6 which gives us a period of 60.  Since the equation has theta minus 30 in it the graph starts at 30 on the x axis.   Then we can plot the sine graph (or you can just do this next step in your head thus taking off the training wheels) and where ever the graph crosses the midline which in this case is 3 we illustrate a veritcal asymptote. Knowing this we can now plot the sec graph which looks like this(without the sine graph and the asymptotes):

After we completed the worksheet Mr. Bieniek threw an unexpected curve ball at us as we were asked to find the equation from the graph this time. He gave us  this graph (link to graph click here and then put in pg 110 for the page on the lower left corner and it is #37.)  For this graph the first thing i did was find out whether it was sec or csc.  To do this i looked at where the high point would have been if we had plotted sine or cosine and realized it was not the starting point of the period but instead it is at 3 on the x axis which is a quarter of the way through the period which is 12.  This meant it had to be csc (the reciprocal of sine)  because cosine graphs always start at their high point.  Then i noticed the amplitude was 1 and since the high and low points were at 1 and -1 the midline was at 0.  Since i figured out that the period was 12 i divided 360 by 12 and got 30.  With all this imformation i got the equation y=0+1csc(30theta).

The next scribe will be AJ.