A timeline for Mr. Cribby's 08-09 Pre-calculus class. It contains announcements, notes, hw assignments, etc.
Created by stcribby on Aug 24, 2008
Last updated: 04/04/11 at 02:44 PM
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The phenomena that i descided to base my graph on was the predator versus prey relationship between the Canadian Lynx and the Snowshoe Rabbit. Here is the original graph that i based mine on:
[caption align="alignleft" width="300" caption="Original Lynx versus Rabbit"][/caption]
Based from this graph you can sort of see the already occurring pattern in the graph. From this i had to average the heights of the Rabbits population and the Lynx's population. After i did this i created two equations. One for the rabbits and the other for the Lynx.
The Canadian Lynx's equation that i came up with was:
The Snowshoe Rabbit equation was: y= -42cos((π/6)x)+42
Both of these are modeled from the original graph. Then once I had my equations I graphed them..and my graph turned otu looking like this:
[caption align="alignright" width="300" caption="My graph"][/caption]
The orange, smaller, graph is the Canadian Lynx's model. The Blue, taller graph, is the Snowshoe rabbit model.
The x-axis is in years from 1838 to 1922. The y-axis is Population in thousands. So when it looks like the species almost goes extinct, they actually don't.
The Lynx's eat the rabbits which causes a rise in the lynx population, but a decrease in the rabbit population. Then, since there are too many lynx's, and not enough rabbits; the lynx's die off. Snowshoe Rabbits then increase there population because their predator is gone. Then this whole cycle starts over again.
But there are some other variables that you need to think of...like hunters, harsh winters, and other things that affected these animals also. Because somewhere, something made the two cycles get a little of sync with each other.
hope ya learned something. =P
As I informed all of you however long ago, there is a relationship between the movement of electrons around the nucleus of an atom and sin. The circular motion of the electron around the nucleus is what creates this relationship. In terms of the the basic sin equation, A sin B (x-h) +k electrons are an issue because there isn't one set equation for their movement.
A is equal to the distance of the electron from the nucleus. This distance is also known as a shell which is a sphere a certain distance around the nucleus in which the electron travels. This can vary between different elements, how many electrons are currently present in the atom, and time, which eventually leads to an electron being caught or thrown out.
B is equal to the time it takes for the electron to go around the nucleus. The problem with this is that electrons move at almost the speed of light (I'm pretty sure they due anyway) and you really can't see them.
h is just equal to where you start your graph at. It really doesn't make much of a difference becuase you can start it anywhere without any ill effects except that shifting it left or right by pi/3 (Or shifting it at all for that matter) will just make graphing it more difficult.
k is equal to the position of the nucleus which is highly impractical to move up or down because the nucleus is a point in space that can't really be represented by a vertical shift. Just make it easy on yourself and leave it at the origin.
That pretty much sums up the relationship of electron movement to sine. My advice would be to leave this to people who know what to do with it because they're the only ones that can determine its relevance.
Oh, just as a clarification, electrons do not move on set paths around the nucleus. They can go wherever they want inside their respective shells as long as it's in a circular path that is equal to the shell's circumference. This erratic movement can lead to electron collisions. This bumps the atoms to a different shell closer or farther away from the nucleus. If an electron collision occurs in the outermost shell, the electrons can be flung from their atom entirely and be picked up by a new one.
Yay for electrons. Fast little dudes that do head-on collisions for a living.
Hi everyone. I did my proect on tidal movements.
There are 4 tides in a day 2 in the moring and 2 at night. They are coused by the moon pulling the water on the Earth as it rotates around the Earth. Because the waater on earth is constantly moving the moon has a force over our oceans. When you graph the tides for a month you will notice that the 4 tides of each day form what looks like a sin graph. The power point
Hello. As you all know (Because everyone was in today) sin, cos, and tan graphs are much easier than we thought they actually were. Instead of having to copy down numerous points and guess on what the coordinates of the graph peaks are, we have now been enlightened. Now, we could probably do this stuff in our head (Unless the equation is large and overly complicated).
This, as you all should hopefully know, is a basic sin graph. In terms of translating this graph, there are six crucial aspects of this graph that you need to know in order to graph one period. The period in this graph is 0°-360°. This label system is very confusing to those of us that use the way that gives us better answers in terms of pi. Hence this convenient list:
0°-360° = 2pi
0°-180° = pi
0°-90° = pi/2
0°-60° = pi/3
0°-45° = pi/4
0°-30° = pi/6
0°-22.5° = pi/8
Anyway, the standard sin/cos/tan version of an equation is: y=A sinB(x-h)+k.
A, represents vertical stretches or shrinks.
B, represents horizontal stretches or shrinks.
h, represents horizontal translations.
k, represents vertical translations.
Just as a helpful note, the standard period for sin and cos graphs, along with their reciprocals, is 2pi. For tan and cot, it's just pi.
Now for the six crucial elements. Period (per), amplitude (amp), center line (c.l.), maximum (max), minimum (min), and phase shift (p.s.), are the six crucial elements. With these six pieces of information, you can easily create any sin, cos, or tan graph, no matter how warped it is.
per = 2pi/B (pi/B for tan and cot graphs)
amp = A
c.l. = k
max = k+A
min = k-A
p.s. = h
Here's an example. Let's say you have y=15sin3(x-(pi/6))+1.
per = 2pi/3
amp = 15
c.l. = 1
max = 16 (15+1)
min = -14 (1-15)
p.s. = pi/6
amp, c.l., max, and min are all vertical changes, while p.s. and per, are horizontal changes.
In order to make the graph you just put everything together. The original coordinate of the max was ( pi/6, 15) and the min was ( pi/2, -15). With the translations, these become a max of ( pi/3, 16) and a min of ( 5pi/6, -14). The final points on either side of the c.l. are ( pi/6, 1) and ( pi/1). The centerpoint of the period was originally ( pi/3) but when you add in the phase shift, it becomes ( pi/2, 1). Using all of these points, you can create the graph.
In order to make an equation, such as y=5-3sin((2x)+(4pi)), easier to use, just change it to standard form.
Then you figure out everything else.
Have at it Cribby. Fix any problems you find.
Without any changes, this is what sin(x) looks like:
[caption align="alignnone" width="300" caption="a graph without any translations or add ons"][/caption]
Now lets see what the graph of 2sin(x) will look like...what is your prediction? well, since you are multiplying sin(x) by two the translation will be a vertical stretch.
[caption align="alignleft" width="300" caption="2sin(x)"][/caption]
another way to express this is to do y/2=sin(x)
So you know that if sin(x) is being multiplied by a number (meaning the number is in front) then the translation is vertical. No lets talk about what happens if the number is in with the x.
[caption align="alignnone" width="300" caption="sin(2x)"][/caption]
This is the graph of sin(x) but the x is multiplied by 2. can you see the difference between this and the very first graph? now i will show a few graphs of what happens when you add and subtract inside the parantheses.
[caption align="alignleft" width="300" caption="sin(x+1)"][/caption]
This is the graph of sin(x+1) so it is a horizontal translation to the right by one.
Now here is a graph were you subtract:
[caption align="alignright" width="300" caption="sin(x-1)"][/caption]
This is the graph of sin(x-1) and so it is also a horizontal translation to the left. So you can conclude from your observations that if there is a number inside the parentheses that is being added or subtracted, then it is a horizontal translation.
But what do you do when there are both multiplication and addition or subtraction??
[caption align="alignleft" width="300" caption="2sin(2x+1)"][/caption]
This is the graph, 2sin(2x+1) so in this case, there is a vertical stretch of 2, and a horizontal translation to the right by 1, and it is halved by 2.
So just remember that is the number is infront of the sin the it is a vertical translation, but if the number is in the parentheses then something is happening horizontally. :-)
Today we learned a new way of looking at sine, cosine, and tangent. This is just a little diagram that may save you some time:
Cosecant, Secant, and Cotangent (recipricals of sine, cosine, and tangent)
The reciprocated trig ratios are:
csc(angle) = 1/sin(angle) = H/O = 1/y
sec(angle) = 1/cos(angle) = H/A = 1/x
cot(angle) = 1/tan(angle) = A/O = x/y
Like before, you can use a unit circle to come up with a table of values for special triangles (notice how they're all recipricals of sine, cosine, or tangent. For example, sin30= 1/2, where csc30= 2/1):
And finally, here are the graphs of cosecant (csc), secant (sec), and cotangent (cot)
(Mr. Cribby, can we scan and enter my homework graphs tomorrow at school?)
In we graphed the trig ratios from the packet or from angle sin/cos/tan table scribed easier.
The first graph we did was the sinθ vs. angle. In this graph we noticed that the point form a wave that starts a 0° goes up to 1°, back down to 180° then to -1° coming back to 360°.
The next graph we did was of the trig ratio cosθ vs. angle. this graph looked more like a V shape. at 0° the cos was 1 at 180° it was -1 and at 360° it was back at 1.
The last graph we did was the trig ratio tanθ vs. angle. This graph looked more like three individual lines then one flowing one.This was because there is no tanθ for angles 90 or 270.
Just by graphing the data we learnd that there is more of diference then what appears on the table of sinθ/cosθand tanθ.We learned what the data truly looked like.
In class we graphed the trig ratios form the paket also shone in the angle/sin/cos/tan/ table scribes earlyier.
In Trapper's Post we learned about the unit cirlce and how it's used. Today we're going to take the unit cirlce one step further:
All of these angles have something in common... Yep, you guessed it, They are all either 30 degrees, 45 degrees, or 60 degrees from the x-axis. That means they can all form a special triangle:
Remember sin, cos, and tan? We can use those and what we know about these special triangles to come up with the trig ratios for the angles in the top diagram:
If you notice, sometimes the ratios are positve and sometimes they're negative. That is a LOT of memorizing if you want to know all these ratios without always having to do the math. Here's an easy way to remember if the particular ratio you're using is positive or negative. First, look at the following graphs:
If you remember from Trapper's Post, sin, cos, and tan have specific ratios in a unit circle:
You're probably thinking, "Great, what am I going to do with a bunch of random letters?"
"A" means that all ratios in that quadrant are positve. "S" means that only sin is positive in that quadrant. "T" means that only tan is positve in that quadrant. "C" means that only cos is positive in that quadrant.
To help you out, here's a fun acronym: All Students Take Calculus. You can use it whenever you're dealing with trig ratios in a unit circle.
The line where a rotation starts (0°) is known as the "initial side". The line that ends the rotation is called the "terminal side." If the rotation goes past 359°, the resulting angle is called a "coterminal". This can be represented as any angle below 359° and above 0° once the ratation is done. The coterminal can be solved quite easily.
Ex: 823° → ? (coterminal)
823° / 360° = 2.28611 Only .28611 is relevant anymore because that's what's left over.
360° • .28611 = 102.9996 Round up to get 103°. 103° is the coterminal.
This is a unit cirle.
The unit circle makes life much easier. Observe. sinΘ = opp. / 1 -or- sinΘ = opp.
cosΘ = adj. / 1 -or- cosΘ = adj.
tanΘ = opp. / adj. -or- tanΘ = y / x -or- tanΘ = sinΘ / cosΘ
Since this is a 30-60-90 triangle, the x = .5 because the hypotenuse is 1. It also makes y = √3 and Θ = 30°
This means that: sin30° = 1 / 2
cos30° = √3 / 2
tan30° = 1 / √3 -or- tan30° = √3 / 3
If you continue on to 150°, you make a line connecting to the hypotenuse perpendicular with the x axis. This makes another 30-60-90 triangle to play with. The only differences are the signs used to describe the side lengths. √3 becomes -√3. This changes the ratios to:
sin150° = 1 / 2
cos150° = -√3 / 2
tan150° = ( 1 / 2 ) / ( -√3 / 2 ) -or- tan150° = -√3 / 3
In order to divide fractions you: ( 1 / 2 ) • ( -2 / √3 ) multiply by the reciprocal.
1 / -√3 The twos cancel out.
( 1 • -√3 ) / ( -√3 • -√3 ) Multiply by -√3.
-√3 / 3 Finish.
What's true for a rotation of 150° is also true for a rotation of 240°. Draw the connecting line perpendicular to the x axis to form a new 30-60-90 triangle (Again.) The difference is that 1 is also negative along with -√3.
Oh by the way: sin0° = 0 / 1 -or- sin0° = 0
cos0° = 1 / 1 -or- cos0° = 1
tan0° = 0 / 1 -or- tan0° = 0
Consider yourself enlightened. :D :D
A 45-45-90 triangle when using sines, cosines, and tangents becomes:
sin45=1/√2 cos45=2/√2 tan45=1
sin could also be expressed as √2/2=sin45
Then there is the 30-60-90 triangle:
in this case, it would be the following:
In cases where you don't know if the triangle has a right angle you must use the law of sines:
Also to help you to make a right triangle when there doesn't appear to be one is to cut the triangle down the center as shown in the picture to the left. Then you have two right triangles to deal with. :-)
But in the example to the right it is undefined due to the 37 being an impossible number on the side that it is on due to the rest of the triangle.
Now lets look at some examples using the law of sines. :-D
So with this example on the right you would do the following:
Law of cosines:
The law of cosines is:
Now that you know this lets look at some examples:
AND REMEMBER DON'T COMBINE TERMS THAT YOU THINK ARE ALIKE, BUT REALLY
Next is another example:
But there is also another way to write the law of cosines:
but the thing with this equation is that you can only write it this way to solve for angles. You MUST use the other equation for the law of cosines to find sides. :-)
As the person says, "YAY!" now I hope that you've got it. =) hope these notes were good enough.
Start typing bold maybe
start something new wysiwyg
Hi pre-calculites! This is your first scribe, done by Mr. Cribby, as an example. One thing to remember: over the course of the year this website will be our textbook. It's created for you... and by you... It also has a global audience. Take it serious and write good posts so we can be proud of our "online textbook" when we are done. So... here goes:
We started right-triangle trig today by "reviewing" the first three trigonometric ratios. We defined the ratios as the following:
sin θ = O/H
cos θ = A/H
tan θ = O/A
If you struggle remembering the ratio think SOH CAH TOA!
You can use these ratios to solve for side lengths like:
sin θ = O/H
sin (72) = 25.54/x
25.54/sin (72) = x or x = 26.854
To solve for angles, you use the inverse operations.
If: tan θ = 24/13 (notice we already have the ratio, we need to "undo" the process using inverse tangent)
tan^(-1) (24/13) = θ
or θ =61.557
That covers how to use trig to solve for missing sides or angles of a right triangle. Someone noted in that the adjacent side is not always "on the bottom." It's "next to" or touching the angle in question.
chs_precalc: #2 - Trig ratio worksheet (1.1 1-3,29-32; 1.2 1-4,15,16,20,22,23)
chs_precalc: #1 Read the "Welcome" article on the website. Register, log in, and post a comment on the article answering its questions. (10 pts)
Hello Pre-calculus students... Welcome to an exciting new year at CHS! This is YOUR website. This website is a blog that everyone in the must create an account and participate in various ways. Anything that goes on during will be posted here. You will be contributing notes and completing certain assignments here. You have the ability to post articles and scribe notes (which will be explained to you in comment, and ask Mr. Cribby or your for help. It is a place for discussing pre-calculus and its topics.
Become familiar with this site. There is already general information available about the course. Read through the links in the menu on the left. Here is your 1st assignment (10 points):
Read the Course Description and the Syllabus.
Even though there's not a lot on it... Check out the Dipity timeline.
Create an account on this website following the directions given you in>
Log in and post a comment on this article.
Your comment should answer the following questions:
Have your past math been a positive experience? Explain why or why not.
Describe what you think math is. In other words, what do you think math is all about?
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What is your least favorite thing about math?
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chs_precalc: Getting ready for school to start!
In we graphed the trig ratios form the paket also shone in the angle/sin/cos/tan/ table scribes earlyier.
Start typing bold maybe
start something new wysiwyg