Precal Blog

Hey guys…  Sorry about not getting the blog up, but my internet connection was disconnecting me every couple minutes, so I couldn’t get one up.  I  just finished fixing it, and so I’m gonna be quick because I don’t know how long it’s gonna last for.  And I have 2 days to blog for.

 Yesterday was half claims, half exploration.  I don’t remember who did what claims, so I’m sorry about that, but here’s a short explaination of all the problems. 

 Problem 10 was the first claim, and it asked to find the exact radian measure of 1080`degrees`.  This can be accomplished by dividing 1080 by 360, since we know that 360 is equal to 2`pi`.  It ends up being 3, so 1080 is equal to 6`pi`.

Number 41 asks you to find the exact value of `tan“pi/6`.  According to our plates, `pi/6` is equal to 30`degrees`.  We know the sine of 30`degrees` is `1/2`, and the cosine is `1/sqrt2`, and tangent is `sin/cos`.  This means that the tangent of 30`degrees`, and of `pi/6`, is `2/sqrt2`. 

46 wants us to find the value of `csc pi/6 sin pi/6`.  This can be done by realizing that the cosecant of `pi/6` is the reciprocal of the sine of `pi/6`, and a reciprocal of something times the original equals 1.  So the final answer is 1.

For number 49, you simply have to pull out the required elements from the graph to come up with an equation.  The vertical displacement is 5, the amplitude is 7, the frequency is 30 (with a period of 12), and a horizontal displacement of 2.  This gives the equation of `y=5 +7cos 30` It’s cosine because the graph starts at the top and goes down.

 For the next half we did an exploration activity about a bug travelling around a square track.  We got coordinates of points, and we ended up graphing the x and y coordinates independently.  They ended up making squarish cosine and sine graphs.  This tied in with today’s activity, which was a bug on a circular track.  I related it to what it would look like if you attached a light to the side of a tire, turned the light off, and rolled it.  It would produce a sinusoid.  Apparently 1 full rotation of a point on a circle can produce 1 period of a sine graph or a cosine graph, depending on if you’re graphing x values or y values.  I have a feeling that’s what we’re gonna be concentrating on in class…..  Just a guess…  And that’s all we did today.

 Well, I didn’t get disconnected, so that’s good.  Hope that was a halfway decent relation of what happened in class.  I’m gonna try to fix this connection for good…  Hopefully I can.  That’d be good. 

Tomorrow’s scribe will be Emily.  See you guys ^_~

~~~~~~Andy~~~~~~

Today we were in the computer lab doing another packet. This packet was called “Bug on a Circular Track. First you had to follow the instructions and create this in sketchpad…

Then when we dragged point E around the circle, we noticed that F is tracking the y-coordinate value of E. Doing some playing around with the Points you also can make the guess that Moving AD, and E moves the x value of F. You can then predict that the graph will become a sinusoid when point E moves around the graph. It will make this…

As you can see when E mobes around the circle it makes a sinusoid graph and to make the bug make a full lap around the circle you must make the circumference 1/3. That is how far we got in the packet today. The SCRIBE for tomorrow is Bryan.

Having used the blog for one entire quarter now we need to focus on making better comments. I blame myself for not providing good examples of comments to begin with so you sort of did what I did and used your comments to judge (nice job! looks great!) rather than enhance the post or take it further than the original author did.

So, let’s change that. As you make comments from here on out (which will be included in your 2nd quarter journal) use your comments to enhance the post. For example, your comments should include things like the following:

• I understood what you said about…..but am still not clear on….
• I thought about….differently; my thought was….
• I still have a question….
• I was wondering…

You could also include links and graphs in your comments that would further the ideas put forth in the post.

So, you get the idea? Don’t limit yourself to my suggestions but show us by your comments that you processed the post you are commenting on; that you thought about it enough to comment intelligently.

I hope everyone had a nice break. It was a super long Thanksgiving break, which I am sure everyone was thankful for. But now, school is back in session and we have four weeks until winter break. 

 Today’s math class was a nice way to ease back into things.  We didn’t actually learn anything new, we just went continued our work on section 3-4 and the claim problems are due tomorrow.

We also got our first Extended Take Home Problem (besides the summer ones of course).  Our assignment is to find a naturally occuring sinusoid and write an equation about it.  Sinusoids appear everywhere in our world. we just need to look for them.  Writing the equation won’t be too bad because we’ve worked on these things already.  Just a note: make sure to use radians unless your object is measured in degrees!   To get the much sought after ‘A’ on the project, you should take the picture yourself and explain when and where you took it.  Mr. B has cameras available for those who need to use one.

Another important bit of information we got today was the paper regarding Semester Portfolios.  Mr. B passed around some examples (my group was given one that was incorrect and we were a little thrown off!).  The basics of the semster portfolio are: a table of contents, a letter to the reader, problems that focus on the broad goals, and problems for specific goals.  Make sure to include explainations about everything.  I believe that it is imparitive here to include as much information as possible, so that your reader (i.e, Mr. B) will not get confused.  The broad goals and the rubric are listed on the sheet.  We were not given an exact due date for this yet, but it seems that we should keep working on it all semester long.  The portfolios that we looked at in class were very long and had tremendous detail as well as graphs.  I guess its time we all became very comfortable with Winplot!

The claimable problems for section 3-4 are :  10, 41, 46, 49 and remember to do your yellow sheet if you haven’t already.  The scribe for tomorrow will be Andy and he will  tell you all about the lovely presentations!

Today in math class we were greeted with a smile from the substitute teacher, Mrs. Portia. When the bell rang, the class was off to the LMC to work in sketchpad. I of course got stuck with the joy of explaining what the class was suppose to get out of the computer lab activity.

The first part of the lab was rather intriging to the class. Most of the students got a kick out of the pattern that point A made when you hit animate. What some of the class may have learned was how interesting it is to stare at the screen watching the pattern go round and round. Some other stidents learned how to change the colors of the petals so that the pattern looked really cool as it went around. While still some other kids discovered that the pattern ended up covering the entire page with the color red if you let the pattern keep animating. Even though some of this information is valuable, the class was suppose to learn more than that from this activity.

We were suppose to learn how radians can be used to estimate the value of pi. (Okay stupid me cannot figure out how to find the symbol for pi on the computer, so if anyone decides to read this over the vacation I’m sorry. I am computer challenged and you cannot paste on the blog page. This leaves me with a problem with the easiest solution beingto write pi out.) When you animate the points, you see in one rotation around the circle that three petals form. When only one petal is formed the arc length is pi divided by two which was what we learned last week in the plate activity. So the lengths of radius, r, moves around the circumference two times pi. This makes sense because circumference equals 2 pi r. Wow everything makes sense now. Finally, you were suppose to determine the number of petals formed in the circle. This was 22 petals in 7 rotations thus meaning that 1 radii= pi circumferences.

The second part of the lab was figuring out what periodic functions are. They repeat themselves over time and the specific type that we were looking at was a circular function. However, Julie and I were only able to complete the second part of the lab up to the back page because we got confused. Then we just gave up. From observing the learning atmosphere in the library, it looked as if we got the furthest in the class, so we could not ask anyone to answer our questions. I do not remember enough of what we did to figure out the rest of the lab based off of memory. My guess as to what we were suppose to learn was how the radians and the revolutions are related.

I hope everyone has a Thanksgiving! What an amazing holiday. A holiday based off of food. That’s crazy. You know else is crazy that its Liz and Rose’s birthday tomorrow, and on Friday Motion City Soundtrack plays at the rave! Have a good vacation. Shelby nominated herself to be the scribe on Monday because everything is alright.

Well its late sunday night, and the Pats just killed the Bills…Who are the Bills?

Lets get down to it, not much went on in math class on friday, we marked some more points on our plate, and discussed arc length a little. Not too in depth though, I’m not worried about it right now.

On a brighter note though, Saturday was opening day for the gun-deer season, boy was i excited… me, pops, and gramps woke up at 5 and headed out into the woods from sun up till sun down, no luck either day, whatever, there is always next year (and dec. 6-9 for doe season)

Scribe for the thanksgiving break-Liza, because i said so.

To start out class we took out our paper plates marked in radi and reviewed what we did yesterday for homework. We went over how we did or should have marked the circle in increments of ‘pi/6′, ‘pi/3′, and ‘pi/4′. Doing this is a color coded fashion really helped. We were also told that we may need to reproduce this…or in otherwords given a hint that this circle marked in radi may be on our next test. Remember ‘2*pi = 360 degrees= a circle’

Next, we learned how to solve for arc length using the known radius measure of a circle and angle measured in Radians. This is a confusing concept at first but once you understand the concept it is pretty simple.

Example 1 : You are given a circle with the radius of 1 and an angle measure of .63 radians. To find the arc length you must multiply.63 (the angle measure) by 1 (the radius measure) to get .63 as the arc length. Now I am sure you are thinking..well why do I do that? You do this becasue one radian= one radius. In other words, x or the arc length is .63 radi therefore you multiply .63 by 1 to get the arc length of .63.

Even though Mr.B said not to get to attached to an equation, one equation you can use for this is ‘Angle measure * radius measure= arc length’. However, remeber on our test it is not equations we will need to know but the concepts. Think of the radians as a percent almost of radi. So if you have 5 radians (this would be your angle measure) than that is equal to 5 times the measure of the radi.

Example 2: Now given the arc length and angle we are going to find the measure of the radius. You are given the arc length to be 16 and the angle to be 2.2. You can either you the equation i stated before or think of it this way:

16 is the value of 2.2radi …..16=2.2(radius)…solve for radi and you get 7.3= the radius measure

In additon, don’t forget that the units of radians or the arc length is the same as that of the radius or vice versa. For example, if the radius is 1 cm then the arc length and angle measure are also in radians

I hoped that this cleared up any questions people had about what we learned in class today. And don’t forget to read section 3-4 for monday and to do the yellow sheet. Mondays scribe will be Rosey who will have 6 whole days to scribe about class! (If I am not mistaken, Rose is the last person on our class rotation)

Hey guys, sorry that this is so late, but my computer’s power cord decided to stop working last night, and so I had to wait until I came to school to do this, and so I have to keep it short. Apparently I’m the scribe instead of Liza, because she’s already been it for this cycle.

When we came to class, we were instructed to partner up with our 3 o’clock partners and we all discovered that we had a piece of adding machine tape and a paper plate. We flattened out our plates into circles (easier said than done) and cut our adding machine tape to the exact (or close to it) circumference of our plate. We then folded our plates into quarters, giving us a radius of our circle. We marked those units on our adding machine tape, giving us a sort of “radius unit” ruler. When we re-wrapped our circles with our “ruler,” we marked off our radius units and were told they are called “radians.” Also, apparently a full circle is 2`pi` radians, a semi-circle is `pi`radians, a quarter circle is `pi/2`radians, and a three-quarter circle is `3pi/2` radians.

Again, sorry for the length and probably boringness of this blog, but when today’s technology decides to act up, what can ya do?

Tomorrow’s scribe is… Ooh, hey, I’m the last scribe? Ok, how about NickK?

~~~~~~Andy~~~~~~

We started today with the instructions to flatten our paper plate. Now we all may be juniors and honors strudents, but it is no walk in the park trying to flatten a paper plate with out destroing it in frustration. But once we had tackeled that obstical were we instructed to wrap our adding machine tape around the circumfrence of the plate and cut it to fit sungly and accuratly around the plate. Can I just say that at this point I was like where is he going with this? Next Mr. B askd us to fold our carefully flattened plate in half and lay out our adding machine tape (Ok pause, I am so sick of typing adding machine tape so from now on it’s going to be AMT, back to the blog). He asked us to measure the flat edge of the plate, or the diameter against the AMT and mark off the exact dimater on the AMT. We found that three full diameters fit on to the AMT and there was a little tape left over. Next we were asked to find the ralationship between the circumfrence and diameter of a circle. Well pulling info from the good old days in Mrs. Hojnacki’s class we all knew that `C=pi*d` so we knew the litttle bit left over on the AMT is the decimal change on the end of `pi`. Now we were instructed to fold our late in half again, so the radius was the flat edge, and mark each radius on the AMT. Again there was a little extra room at the end of the AMT after six radii. Next Mr. B asked us to think of the relationship between the circumfrence and the radius which is `c=2pi*r` so the circumfrence of a circle is equal to `2pi` in relation to the radius. Finally Mr. B told us that that concept of the circumfrence of a circle equaling `2pi` is called radians. The true definition of a radian is the rotation of the radius around the arc of a circle starting at zero on the uv-coodinate plane. When you form an angle with the mark made by the radius and the u-axis, the measure of that angle is a radian. Radians are always expressed in trems of `pi`. If you think of radians in relation to rotation then `360degrees=2pi` and using that concept we can find the values of `90degrees`, `180degrees`, and `270degrees` as `pi/2`, `pi`, and `3pi/2`, respectively. Finally at the tail end of class we were instructed to find the position of `pi/3`, `pi/4`, and `pi/6` on our plate. These measurements create regular intervals around the plate and we were told to mark all the intervals for homework. So that is radians in a nut shell. I hope is was helpful. Tomorrow’s scribe is Natalie. Sorry my dear but at least you don’t have to do it as your eyes are closing after murder a.k.a. practice.

Blog for November 14, 2007.

Once Mr. Bieniek arrived to class, we all entered the room, and decided which problems we were going to claim from last night’s assignment. Then, we wrote down the numbers that we were going to clam, and gave the piece of paper to Mr. Bieniek. He then wrote down our claims and chose people to present the problems. While he decided which people were to present the problems, we all talked about many things that were very irrvalent. My group had an interesting discussion on the fish in Mr. Bienieks classroom. Then we started our presentations.

Stephanie started us out with presenting number 15 from section 3-3(Graphs of Tangent, Cotangent, Secant, and cosecant Functions.)

For those of you that don’t remember, number 15 was the problem regarding the rotating light house beacon problem. I wanted to include a link to this question from the online book, but it wasnt working for me.

Question:

The problem stated that a lighthouse was 500 m from the shore,. See your book, or the book online for the rest of the opening to this question, and for the diaagram, because i can not figure out how to include that in here.

1. Part A stated: Plot the graphs of D and L as functions of `theta`. Use a `theta`-range of o-360 and a y range of -2000 to 2000. Sketch your results.
2. B. When `theta` = `55 degrees`, where does the spot of light hit the shore? How long is the beam of light?
3. C. What is the first positive value of `theta` for which the light beam will be 200 m long?
4. D. Explain why D is negative for 90<theta<180
   and for 270< `theta` <360 and why L is negative for 90 < `theta` < 270.
5. E. Explain the physical significance of the asymptote in each graph at `theta` =90.

Solution:

1. A. To solve for the equations of D and L, you must look at the picture in the book for this question. By looking at it you can see that `tan theta = D/500`. Because tangent= opposite over adjacent. SO, by using this equation, you can solve for d.

`tan x = d/500` simply multiply by 500 on both sides to solve for d.

So, `D=500 times tan(x)`

Then, you do the same to solve for L. By looking at the picture, you see that to find L you must use `cos theta = 500/L` because cos equals adjacent over hypotenuse. So, you then must use this equation to solve for L.

`cos theta = 500/L` multiply both sides by L, so `L cos theta = 500` and then divided both sides by the `cos theta`. So, `L= 500/(cos theta)` Now we have solved for L and D. Next you can simply plug these equations into your calculators grapher.

For D: Y1 = , `D=500*tan(x)`

For L: Y2 = `500/(cos(x))`

B. tan (55) = D/500 So D=714 m along the shore

Then, because you have a right angle, you can plug this into the Pythagorean theorem to solve for L

714^2 +500^2 = L^2

L= 872 M

C. Cos theta = 500/2000

`Theta = 75`

D.D is negative because the light is in the opposite direction on the beach.

L is negative because the distance of L points in the opposite direction.E. When the graph is 90degrees, the light is parallel to the shore, so it would be undefined because it is not a triangle.

After that, Rachel presented section eight, # R3 A and B.

Question:

A.Sketch the graph of `y=tan theta`.

B.Explain why the period of tangent is 180 degrees, rather than 360 degrees like sine and cosine.

Solution:

1. Rachel started by making this problem easier to see by simply graphing cosx, sinx, and tangent x on the TI smartview calculator on the smartbord, instead of physically drawing them out, so that they would be easier to see. She then explained how you would find the tangent graph, by using sine and cosine, because in order to find the tangent graph, you must use the sine and cosine graphs.

She graphed:

\y1=cos (x)

\y2= sin (x)

\y3=tangent(x)

**(x is used in the place of theta when graphing on your calculator)

This looked like:

Once you have plotted the sin (red part of graph) and cosine graphs(green graph), it is actually pretty easy to find the tangent graph (blue graph). You simply look at four specific things on the two graphs.

1). Whenever the cos graph(green) is at zero (cosx=0), then there is an asymptote on the tangent graph, because tangent= sinex/cosx, so if cosx is zero this is undefined, because you cant divide by zero. So when cos=0, I plot a dotted line vertically, meaning there will be an asymptote in the tangent graph.

2). When sin(red graph) = zero, tan is also zero, because a fraction with a zero on top is always zero, so whenever the sin graph is at zero, plot a point at zero for the tangent graph.

3). When cos equals sin (where the graphs of cos (green graph) and sine (red graph) intersect, in positive y), plot a point for tangent at one, because any number divided by itself is one. And, 4).When cos and sin are equal but opposite, plot a point at -1.

B. The period of tangent is 180 degrees because if you look at the graph, the values/graph start repeating themselves after 180degrees, where as in sin and cos, they dontstart repeating until 360 degrees. In tangent, the period ends at each asymptote, which is at every 180 degrees.

Finally, Rose presented number 13 from section three.

Question:

13. What are the dilation and translation caused by the constants in the equation? Plot the graph on your grapher and show that these transformations are correct.

`Y=4+6sec.5(theta +50)`.

Solution:

First, Rose showed us another way to write this same equation.

Because secant equals 1/cosx, you can write this same equation as `4+6(1/cos((.5(x+50)))`

Then, she graphed this equation on the smart board calculator.

She made her window -720 < x <20, with a scale of 90, and -45 < y < 45, with a scale of 5.

agraph
xmin=-15; xmax=15; ymin=-10; ymax=20; xscl=2; axes();
plot(4+6(sec(.5x+25)));
endagraph

`Y=4+6sec.5(theta +50)`.

The values in the secant equation mean the same as the values in the sine and cos equations, so:

• 4 is the midline an the vertical translation
• 6 in the amplitude , because the secant range is not ‘controled’, so this value is how much the graph is stretched by.
• .5 is the frequency, also meaning that and the horizontal dilation is two, because .5 is technically in the parenthesis when you would multiply it in, and we know that everything in the parentheses is in opposite world, or bizzaro land meaning that it is opposite/reciprocal in a way for dialation, so the horizontal dialation is 2.
• 50 is the horizontal translation to the left, meaning that the x values move 50 to the left. It is to the left, not towards the right because +50 is in ( ), so it is opposite, moving to the left instead of the right.

Finally, Rose gave us some further explanations on amplitudes. She said that because you don’t technically have an amplitude (because the midline isn’t in your domain in your graph), the # for the amplitude, in this case 6, would just be how much the graph has stretched vertically.

Then, Mr. Bieniek came up to the board to add on to Rose’s explanation. He asked us, “How then would you create an equation, if you were only given this final graph, instead of the equation originally?” We thought for a moment, and then Mr. Bieniek showed us that by looking at the critical points, you can see a midline between these points. The middling between -2 and 10 is 4. 4 is 6 away from 2, and 6 away from 10, so 4 and 2 is 6, so sec does have an amplitude. If you connect the critical points you would git a translated sin/cos graph.

In sec, where is cos? You see it by connecting the max and min critical points.

Sec is translated the same as cos. 4 + 6 sec.5(theta +50)

The numbers aply to cos also

4=5cos (.5x=50) You can get the cos graph, for this secant equation, and then find sec from the cosine graph. This is easier for translating sec.

If given secant, you can plot the graph as a cos graph, and the 0’s make asymptotes. You can then use that to plot sec.

Sorry guys, this is really hard to explain without being able to draw out these graphs as I explain it.

You can always use cosine top graph the secant

Mr. Beiniek then reminded us that we are not responsible for these types of transformations yet, we are only responsible for the parent graphs of sin, cos, tan, cot, sec, and cos.

The next scribe is Rachel. Sorry!..