Precal Blog

Today in precalculus, we had a substitue teacher.  We got a yellow sheet to do during class that had problems with a sinusoidal function, and we gad to find the values of X on the given graph given Y.  To do this first, you had to find the equation of the sinusoid.  The vertical translation of the graph was 9, the vertical dilation was 7, the period was 13, and the horizontile translation was 4.  So, that made the equation of `Y=9+cos((2pi/13)(X-4))`.  Next you had to make the line Y=5 to calculate the value of X when Y=5.  To calculate the intersection of the two lines all you had to do was use the intersect feature on your calculator and find the location of each intersection.

Today was another great day in precal!  We went over the claims for section 3-4 today with great success! 

This chapter was mostly about Radian-Degree Conversions.  To convert degrees to radians you use the equation pi/180*degree measure.  Converting radians to degrees is basically the opposite which is, 180/pi*radian measure.  For the most part, the class understood this concept.  The tricky part of today was leaning how to do these conversions while including the six trig functions.  One way we did it was to realize that the inverse of tan(5)=x is actually tan(x)=5.  This shows that to undo an inverse, you just swich your given value and the value of x.

Also something epic happened today, the Curtis method was created today.  He said that for any instance, you can replace pi with 180 degrees.  How cool is that?!?!  It made this chapter a lot easier to understand.  Thanks Curtis!

Speaking of Curtis, he is our scribe for tomorrow!  =]

In addition to the Curtis Method and the Ranta Conjecture we had a couple of really good questions come up. One in particular deserves some more attention.

We were trying to graph `y=-4+5sin((2*pi)/3(x+1))`

The first attempt was to try and avoid radians and graph `y=-4+5sin(120(x+1))`. Then the question was asked – “If we are converting `(2*pi)/3` to degrees, then doesn’t the 1 (radian) have to be converted also? So, going with that idea we have: `y=-4+5sin(120(x+57.295…))` or exactly: `y=-4+5sin(120(x+180/pi)` That seemed very reasonable. Before we had time to discuss it any further time ran out.

Looking at a table of values though, it is clear that the two equations above are not equivalent. Why not? After all `(2*pi)/3` radians is equal to 120 degrees, and 1 radian is equal to `180/pi` degrees.

I won’t tell you the answer (I know that surprises you) but I will give you the two equations that are equivalent and you can think about why these two are but the orignal two are not.

`y=-4+5sin((2*pi)/3(x+1))` in radians is equivalent to `y=-4+5sin((2*pi)/3(x+180/pi))` in degrees.

What’s going on?

When I made that last post, all of the pi symbols were turned into ?. I thought I remembered Mr. B talking about how to add the pi symbol, but I am not sure. If he did, can someone remind me how.

Much Appreciated.

First, I just want to say sorry for the time it took to get a post up. As some of you know, the site was getting weird again.

Anyways, the past couple of days, we have been be looking at radians. A radian is the measure of a central angle subtending an arc equal in length to the radius. Radians deal with distance, which is different from our previous way of using degrees. Although, they are two different measures, they can mathematically be interchanged. To get radians, multiply the degree measure by ?/180. To get degrees, multiply the radians by 180/?.

Much like specific sin and cos values, I think that certain radian values should be known. First, 0 degrees is obviously 0 radians. Second, 360 degrees is 2? radians. That is why to switch from radians to degrees (and vice-versa) you use ? with 180 (a.k.a. 2? with 360) If you think about it this is because the diameter will go around a circle 2? times (or 6 radii with about .28th of a radius left). Then, if 360 degrees is 2?, 180 degrees will be ?. Others are 30 degrees=?/6, 60 degrees=?/3, and 45 degrees=?/4.

We also touched a little on arc lengths and other circular things using radians. This was only brief, but we learned that if the radius of a circle is one, the arc length is equal to the measure of radians for the section it subtends. If the radius is not one, then you can solve to find the arc length. If the radius is 10 and there is the radian measure of 1.4, the arc measure would be 14. You can look at this like a simple equation r*m=l where r=the radius, m=the radian measure, and l=the arc length.

This can be used to solve for things other then the arc length too, i.e. the arc length is 12.6 and the radian measure is 4.2. All you have to do is divide the arc length by the radian measure to get the radius. In this case, 12.6/4.2=3, so the radius is three.

Again, sorry for the delay. Hope this helps if anyone is confused. Also, remember to do your claims ^_^

The next scribe will be [((___Bobbie_))]>

P.S. I guess the posts have something against foreign pi so ummm… yah the ?’s are pi.

Next, we did an activity relating to radians.  We each got a paper plate and tried to flatten it as best as we could.  Next, we took a strip of paper, which would turn into our ruler, and cut it to be the length of the circumference of the plate.  We then made marks on the ruler that were the distance of the diameter of the plate.  There were 3 marks and a little extra (`pi`).  Then, we did the same with the radius.  There were 6 marks with a little extra again (2`pi`).  We folded our plate into fourths and on the fold on the right we made it zero.  Then, we took our tape and placed it from zero to the first radius mark, and made a line on our plate.  This is one radian.  Radians are always in the same spot of a circle no matter the size.  We came up for a definition of radians: a distance measured in the radius-units along the circumference of the circle.  Our radian was equal to one, so two radians would be equal to two.  If the radius of a circle was five, two radians would be equal to ten, which is determined by going around the arc length of the circle.

We determined that the whole tape was equal to 2`pi` from our 6+ radians.  If the whole tape was 2`pi`, then half the tape would have to be equal to `pi`.  We folded our tape in half at `pi` and marked it by going around the outside of our plate, and from the zero fold `pi` went halfway around the plate.  We then folded our tape into fourths, making marks at `pi/2` on the top of the plate and `3pi/2` on the bottom.  If you are confused about any of these fractions, you can get them by multiplying the fraction you’re folding by 2`pi` (`3/4 * 2pi = 6pi/4 = 3pi/2`).  We then folded our paper into 3rds, 6ths, and 12ths, and plotted them on our plate.  The final numbers should be: `pi/6`, `pi/3`, `pi/2`, `2pi/3`, `5pi/6`, `pi`, `7pi/6`, `4pi/3`, `3pi/2`, `5pi/3`, `11pi/6`, and finally `2pi`.

I know this is a new concept, so if any part of my entry was confusing to you please leave comments and I’ll try to clarify.  I tried to leave a picture of my plate into the post but it wouldn’t let me upload.  The next scribe will be Brandon!

Today was a half day, so we only had 21 minutes to complete the following problem!

Create a graph sketch of the following situation. Be as accurate as possible and scale your graph as accurately as you can. Write down any assumptions you are making as you go.

*The height off the ground of a buffalo chip stuck to a wagone wheel as a function of distance (over at least 3 revolutions of the wheel).

My group (Rosie, Allison, Zak, and myself) first started off with the completely wrong idea.  Instead of finding the circumference of the wheel, we all decided that 360 degrees was what our periods would be in.  After discussing this with Mr. Bieniek, we soon realized that circumference of course is not a measure of degree, but a measure of distance. In order to find the circumference, you would need to make up some measurements for your wheel. We decided to make the radius of our wheel as 1 foot.  We then found the circumference (2`pi`r) or in our case just 2`pi`. We figured out then that our period would be 2`pi` and that our graph would have to start at 6`pi` because it says there are three revolutions prior to the buffalo chip being stuck. Well this is basically how far we got in such a short class period. We will continue our work tomorrow and will hopefully be successful!

p.s. todays date is 11/11! thats good luck everyone make a wish!!

Also, when i previewed my post all the pi symbols i put in are showing up as question marks and i have no idea how to fix that…. so    ? = pi

The next scribe will be Samantha Betlej =]

Today in class we went over the Exploration 3-3b sheet, which we did in class on friday.  We went over how to properly determine the horizontal dilation which is covered in Haley’s previous post.  We then got another sheet that practiced more transformations.

The first was a transformation of a Tangent graph: y=2+5tan3(`theta`-5°)

As Lauryn and I worked through this transformation we thought we were right on track.  After we compared with Zak and Matt, we found that we had made a huge mistake, instead of having the middle point start at x=5 (point 5,2) we had started the period at x=5.  This threw our graph off completely.  The lesson was learned that the horizontal dilation factor effects where the middle point will be located, not the asymptotes.

The second transformation was for a cotangent graph: y=-1+3cot2(`theta`-30°)

To make this graph we used our knowledge of tangent graphs and essentially flipped it, so that the graph started high with an asymptote and ended with a low one.  We knew we could do this because cotangent is the inverse of tangent which is sin?/cos?, making cotangent cos?/sin?.

The third transformation was for a secant graph:y=4+6sec.5(`theta`+50°)

I think that most of the people in our class find it easier to draw a cosine graph first, then draw a secant graph based off of that.

The fourth transformation was for a cosecant graph: y=3+2csc4(`theta`+10°)

Like the third transformation, I found it easier to draw this graph by first drawing a sine graph.

Hopefully everyone feels a little more confident about transformations of tangent, cotangent, secant, and co secant graphs.

Tomorrows Scribe is Lauryn.

Class today was focused on the topic of (you guessed it) graphing.

We began by reviewing the graphing packet that we looked at last week, just correcting it and addressing a few small problems.  One aspect of the packet that confused a large number of people was the part that dealt with graphing cotangent functions.  Unlike every other graph we’ve been working with over the past three weeks or so, the unchanged cotangent graph does not have a point when `x=0`.  The cotangent graph has an asymptote on the y-axis.

After that, we were given a worksheet with four equations and four blank graphs. The equations were:

`y=2+5tan3(?-5)

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

y=4+6sec1/2(?+50)

y=3+2csc4(?-10)`

I will walk you through the first equation, a tangent graph.  This graph has a horizontal dilation of 1/3, because of the 3 following tangent.  As a result, the period of this equation is 60 (Recall that the period of a normal tangent graph is 180 instead of 360).  There is a horizontal translation of 5 due to the 5 in the parenthesis.  The vertical dilation factor is 5 because of the 5 next to tangent.  The vertical translation is 2.

If you wish to graph this, be sure to be in degree mode.  I suggest using a window of `-100<=x<=100` and `-20<=y<=20`.  Examine the cycle that is in the center of your window.  Important points in this cycle are (-10, -3); (5, 2); and (20, 7); with asymptotes at -25 and 35.

Tomorrow’s scribe will be Andrew.