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.