BQ #7: Unit V- Difference Quotient

The Difference Quotient is something we've become very familiar with this year. But what we might not know is what it means and where is it derived from. Simply enough, the difference quotient formula is just the slope formula mixed around, more specifically the slope of all tangent lines.

http://cis.stvincent.edu/carlsond/ma109/diffquot.html

 

 

The slope formula is what? (y2-y1)/(x2-x1). If we refer to the points of f(x+h), F(x), and such, "h" representing the distance between the first point and the second. Our ordered pairs will be seen as (x, f(x)) and (x+h, f(x+h)). By plugging in both ordered pairs to the formula, the result would be f(x+h)-f(x)/x+h-x. When we cancel the x's we would get: f(x+h)-f(x)/h and VOILA the difference quotient!

BQ #6: Unit U

1. What is a continuity? What is a discontinuity?

Up to this point, we've all become familiar with continuities. A continuity can be described or classified by many different aspects, one being it is predictable. By the graph being predictable, we always know where its headed on either sides because it continues on forever and ever. Because the function reaches infinity on both sides, we must know why; because it has no jumps, no breaks, and no holes. Why? Because it's continuous! A break is when the graph stops at a simple value, a jump os when the graph breaks into two separately placed functions that land on different left and rights; a hole is the lack of a value. These all three entail that at some point, the function stops which means it is no longer continuous. A continuity can be drawn without ever lifting a pencil.

http://faculty.wlc.edu/buelow/calc/nt2-4.html

2. What is a limit? When does it exist? What is the difference between a limit and a value?

To start, a limit is an indented height of a function. A function has many, many limits because they can be found basically anywhere on the graph. However, this does not mean that EVERY function has a limit, no because there are several types of functions. As we've gone over the difference between continuities and discontinuities, we know that discontinuities are grouoed into two categories: removable (a point discontinuity where the function has a limit but also has the value in a different y-value) and un-removable(a jump where they have different left and right positions, an oscilating which means it has no set x-value, and an infinite which is classified by an asymptote that causes unbounded behavior).

http://en.wikipedia.org/wiki/Classification_of_discontinuities

 This example shows two different un-removable discontinuities, an oscillating and an infinite. Because the oscilatiing entails that there is no set value, then we know that the limit does not exist. Same goes for the infinite; because it reaches towards infintity by the vertical asymptote, the value is virtually never reached because infinity is not a true value. This means that the value is never reached ergo the limit does not exist.

http://en.wikipedia.org/wiki/Classification_of_discontinuities

There is one more where the limit does not exist; in jump functions. A limit is where both the left and the right mesh to the same place when going towards the center. In a jump, we cannot trace to the same position by the left and right so we say that it has different left-and-right. Ultimately, this means that the limit does not exist.

From this, we can assume that limits do not exist in unremovable discontinuities. The same cannot be said for removable which is a point discontinuity. If we refer to the picture above again and avoid one of the functions, we will see that it has one function with a hole and a strange dot that signifies the actual value. This is a point discontinuity where the limit and value both exist which means that the limit also exists.

3. How do we evaluate limits?

When we evaluate limits numerically, we basically put them in a table. This table is used to show where the function goes when approaching a value.

x	-5.1	-5.01	-5.001	-5.0001	-5	-4.9999	-4.999	-4.99	-4.9
g(x)	23.2	23.1	23.001	23.0001	24	249999.9	24999.9	249.9	24.9

http://people.hofstra.edu/stefan_waner/realworld/tutorials/frames2_6a.html


Usually, this is done by moving by a tenth each time. as seen in the photo above, we see that the graph gradually gets closer and closer to the value in the middle.

When we evaluate limits graphically, we do just that! By graphing the function.

http://people.hofstra.edu/stefan_waner/realworld/tutorials/frames2_6b.html

 

To evaluate graphically just means that we need to graph the function and see the graph and the way it actually looks. This generally is better way of evaluating because we can see the values and limits all in one photo.

 

Algebraicallly, a limit can be evaluated one of three ways: through substitution, factoring, and rationalizing.

10. https://www.youtube.com/watch?v=CvB4080WC48
Although we always want to start with direct substitution. This means that we substitute the number being approached for every x value. If the value is 0/0 then we have to change methods. The factoring method is next because it allows us to cancel out to then simplify. If neither work, we refer to rationalzing and multiplying with conjugates. Usually this allows us to break it down into an easier solution. Then after conjugating we once again directly substitute for all the x variables.

BQ #4: Unit T Concept 3

Why is a “normal” tangent graph uphill, but a “normal” Cotangent graph downhill? Use unit circle ratios to explain.
We know that the ratio for tangents are y/x which is sine/cosine. We also know that sine and cosine have all values so if cosine were to be zero than that would make the ration undefined! Hmm, what was that thing about undefined again? OH, right! Undefined means there is an asymptote! Here is where we look at the graph and decide where tangent will be negative or positive. According to the unit circle, we already know that tangent is positive in the first quadrant which but negative in the second! SO when looking at a graph, we see the tangent line high above the x-axis in range of 0 to pi/2 but as soon as we go pi/2 to pi, what happens? it becomes negative and goes downhill but because it starts off as negative (by the asymptote) and then reverses to positive, a tangent graph is normally uphill!. This is because it is just another presentation of the Unit circle!
Cotangent is much like its reciprocal but has its own qualifications. Its ratio is x/y- or cosine/sine. To find our asymptotes now we want SINE to be equal to zero. According to the Unit Circle sine has asymptotes at zero and pi! Then we continue with the same steps as for tangent and decide in which quadrants is cotangent positive and negative in: positive in one and three but negative in two and four! The Asymptote really gives us where the graph really begins. And because we can see that the cotangent graph starts out as a positive above the x-axis but then transcends into the negative, it is called a downhill! 

(Mrs. K's unit T: Exploration Packet)

BQ #3: Unit T Concept 1-3

How do graphs of sine and cosine relate to the others? Emphasize asymptotes.
For the rest of our lives we are going to refer to sine and cosine as being our "favorite cousins." Why, you might ask? Because they relate to our other trig functions through their ratios! By knowing this we already know that they will have some grand effect on the other four trigs.

Tangent
We know that Tangent's ratio is... y/x! Which, when referring to the unit circle values (sin = y/1 & cos = x/1), is equivalent to Sin/Cos. If we imagine the unit circle being unfolded to lay as the graph, we can picture the four sections that represent four quadrants; at the same time, we would place both sine ans cosine in the graph segment as well.  The area between zero and pi/2 would be known as Quadrant one. if we focus specifically on where sine and cosine are, we see that both are abooooove the x-axis! Therefore, because of tangents ratio, we know that tangent is also positive n the first quadrant. Curious about where the we get the placements of the asymptotes? Easy! Where there is an undefined value, there is an asymptote and we'd only get undefined by dividing by ZERO! Ergo, asymptotes will be wherever cosine is at zero.

Cotangent
Cotangent is very similar to tangent because they are what? Reciprocals! This means that the ratio for cotangent would be x/y because sine is the value of y and cosine the value of x. Again, we know that cosine, sine, and cotangent will intersect on the graph because of that one ratio. Now, we have to figure out the asymptotes. Because y is the denominator, we noe have to figure out where sine is at zero because it would create an undefined value therefore an asymptote! The only places where y equals zero is at (1,0) and (-1,0). Here, the radians are at zero and pi.

Secant
Secant is a little more different than tangent and cotangent because its period does not end at pi, but at two-pi which makes a complete revolution. Secants ratio is 1/cos. Although it takes longer, we can tell the positive or negative sign through the same method. If cosine is positive then of course secant will be positive as well! Our asymptotes for secant will always be the same as the asymptotes for tangent because they both have cosine as their denominator! However, because secant only has cosine in its ratio, instead of having a normal curve for tangent, their graph becomes a parabola facing the direction of positive or negative, depending on the quadrant.

Cosecant
With cosecant, we are going to see very similar things as we did with secant. The only difference is the value in its ratio. the ration for Cosecant is 1/sine. This means that the asymptotes for cosecant and cotangent will be equivalent because they too have the same denominator. When drawing a cosecant or secant graph, we always start with drawing its reciprocal FIRST because it is the general foundation of this graph, After drawing the reciprocal, we draw the asymptotes. For cosecant our asymptotes will be wherever sine is zero! From there, we draw the parabolas starting at the maximums and minimums of each hill on the reciprocal.

BQ #5: Unit T Concept 1-3

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.

 We can explain why sine and cosine do not have asymptotes simply by referring to their trig ratios: sin= y/r & cos= x/r. Because we are using the Unit Circle, the radius will always be the absolute value of one. If the radius will always be one then we know that the ration will never be undefined because the value cannot be zero. Now here is where we can see the connection: if there is no undefined value, automatically we know that there are no asymptotes.
This cannot be applied to their reciprocals nor tangent or cotangent because these trig function rations can have any value as the denominator which includes zero that would make it undefined. This is because both the y and the x values have no restrictions.  

BQ #2: Unit T Concept Intro

How do trig graphs relate to the Unit Circle?
Ultimately, we can say that a trig graph, in its first period, is just a different presentational form of the Unit Circle that has been unwrapped on a graph. In referring to the original form of the Unit Circle, we know that each trig function has two quadrants where they are positive and two where they are negative. For example, sin is positive in both the first and second quadrants but negative in the third and fourth. By knowing that they are positive in the first and second, we know that the distance from the beginning of Quad 1 and Quad 2 is 0 to pi; the distance from Quad3 to Quad 4 is 3pi/2 to 2pi where we know sine is negative. What a trig graph generally represents is the sign change for all trig function values by stating if the values are positive (above the x-axis) and if they are negative (below the x-axis). This forms wave like motions on the graph.

 

The trig graphs will go on forever repeating the same period for each trig function.

 



Why is the period for sine and cosine 2pi, whereas the period for tangent and cotangent is pi?
Sine and Cosine both have a period that end at 2pii because that is when the pattern ends. A period is the end to a pattern on the graph. Because Tangent has a pattern of positive-negative in the first two quadrants and repeats for the third and fourth, we know that the period for tangent and cotangent are cut down to half which is just pi.

How does the fact that sine and cosine have amplitudes of one (and the other trig functions don’t have amplitudes) relate to what we know about the Unit Circle?
So why is it that sine and cosine have an amplitude of one? Maybe the Unit Circle can answer that. When we look at the trig functions sin = y/r and cos = x/r; because it is the unit circle, we automatically know that r is the value of one. So it is because we may only divide by one that we must assume that the move must be a value of one or negative one. This also explains why the trig of sin and cosine may not equal any value that does not lie between one and negative one. However, these do not apply to any other function. Why not, you ask? Well let's look at tangent; the ration for tangent is y/x. There are no restriction values for either y or x which means that there is a possibility for the number to exceed the distance of one in either direction. the same applies for the other trigs.

Reflection #1 - Unit Q: Verifying Trig Identities

1. Reflect on your learning this unit, specifically with concept 1 and 5.

Although this Unit made me pull my hair out, I did learn much more about trig functions! :)
  1. What does it actually mean to verify a trig identity?
Well, my dear, all it means is just that: to verify the answer. When verifying, you are given an "equation" that has two separate portions equal to each other. To "solve" we would have to prove how the problem on the left can equal what is on the other side. But there is one thing to always remember: never touch the right side! In other words we cannot do anything such as divide, multiply, square or square-root because those would have to apply to BOTH sides which we cannot do of course!
  2. What tips and tricks have you found helpful?
It is always better to right your identities on a flashcard and memoriiiiiiiiiiiiiiiiize  all of them! This will cut the time of having to keep referring to the beginning of your packet and just make these stubborn trig functions easier in general. Also, practice-practice-and-more-practice! You might feel comfortable with a few problems but just remember that there are so many different ways an equation can be set up that we want to be as familiar as we can with any method.
  3. Explain your thought process and steps you take in verifying a trig function.
When verifying, i always want to see if i can change them all to functions of sin and cosine simply because they're easier to work with. If we are given fractions, most of the time it would be helpful to also separate in portions to cancel and/or substitute with better identities. Also remember that our main focus is to get the same answer as on the right without ever ever EVER E-V-E-R touching it! Just keep practicing and it'll come to you eventually.  

SP #7: Unit Q Concept 2 (Finding All Trig Functions When Given One Trig Function and Quadrant)

Solving with SOH CAH TOA:

1.

2.Solving with Identities:

When finding actual trig functions, we can use two methods: SOH-CAH-TOA or Identities! As you can see from the first photo, we refer back to our Unit circle and can use the Pythagorean Theorem to get our missing side with the two we already have. In our case, we go to the sin where we know the fraction is y/r. If we place it into the formula, our "x" will be one! The second method is through identities! When using our identities, we are always given two trigs which cuts down a portion of our work! With these two, we first figure out in which quadrant the answers will be either negative or positive in (please refer to the second picture). After that is when we either substitute or solve for the proper trig that we are looking for.

BQ #1: Unit P Concepts 6-7 (Law of Sines and Law of Cosines)

Law of Sines

1. Why do we need it?

• First and foremost, we have to understand that not all triangles are right triangles. This means that not all triangles have a ninety degree angle, meaning we cannot use the Pythagorean theorem to solve for more. Then how will we solve  for irregular triangles you ask? Fear not, my beloved friend because this is where the Law of Sines come in to play.

-The Law of Sines, however, may only be used when given SSA (side-side-angle)

2. How do we derive from what we already know?

• From what we already know, the triangle does not have a right triangle. this is where we draw a Line straight down the middle to make two right triangles. This length is labeled as "h."

 

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

 

• If we look at this from a graph perspective, we see that the sin<A will equal h/c because it's opposite over adjacent. If we multiply both sides by "c" then we would get cSin<A=h. Sin<C will equal h/a and multiply by "a" then aSin<C will also equal "h" which will make all angles of the same value.
• This can also be said for angle B therefore we plug Sin<A/value = Sin<B/x and cross multiply! This is how we derive the Law of Sine.

 

I/D #2: Unit O Concepts 7-8

Inquiry Summary Activity:

A. 30-60-90 triangle
 We can derive a special right triangle of 30-60-90 angles from an equilateral triangle. An equilateral triangle is a triangle that has all three angles of the same degree to add up to 360 degrees which would make them 60 degrees. Along with having all angles of same degree, all sides are of equal length as well. In this case one. We derive this special right triangle by cutting the equilateral. By doing this we create two triangles of 30-60-90 degrees, however, we just need to use one.

Because one side was unaffected by the split, we know that it has a side length of one while we know that the other original side was cut in half giving us 1/2. With these two side lengths, we can use the Pythagorean Theorem to get the missing length of b- theorem being a^2 + b^2 = c^2. By solving, we get the value of radical3/2.

 

After every value has been found, we place them in their sides such as the photo below. In looking at the picture, we can see that fractions can be bothersome. A method to remove fractions would be to multiply all sides by two to become:

 

 

The original values can be labeled as "n" or really any variable because they represent the basis of the Special Right Triangle (SRT). Here is when we begin to see the pattern completely because no matter what the length values will be, n will always be connected to the 30-60-90.

 

 

B. 45-45-90 Triangle

 



The 45-45-90 triangle is much like the one in prior concept of 30-60-90. The difference is, we derive these from a square. Our lengths are already given to us to have an value of one all the way around. Again we are going use a line to cut the shape differently. Please refer to the photo below.

 

 

This would have cut two of the 90 degree angles to make four forty-five degree angles, giving us once again two new angles. This time, our degrees would be a special right triangle of 45-45-90 degrees. Because the outer lengths, also known as the legs (or a and b) were unaffected by the cut, we can once again use the Pythagorean theorem to get the missing leg of the hypotenuse. Your work should be plugged in and solved similarly to the photo below.

 

 

 

Because our leg lengths are once again the value of one, we want to replace it with variable "n." This "n" now represents the basic pattern value of a 45-45-90 triangle because it applies to any length of this type of triangle.  This will give us two legs with the value of n and our hypotenuse with n times radical two. Keep in mind that if the value of the triangle is expanded, it must be multiplied by the original values of n and so on.

 

 

 

Inquiry Activity Reflection:

1. “Something I never noticed before about special right triangles is…” That there is always a pattern in finding values of sides.

2. “Being able to derive these patterns myself aids in my learning because…” they ensure that I will remember the values or the proper method of getting any of my missing values.



 

 

I/D #1: Unit N Concept 7 (identifying degrees, radians, and ordered pairs of a unit circle)

All special right triangles (SRT) may be found not only in the first quadrant of a Unit Circle but all four. Special right triangles either have degrees of 30, 60, 90, or 45, 45, 90. The hypotenuse of either triangle is and always will be 1.

Inquiry Activity Summary:

1. The 30 degree triangle:

The thirty degree triangle is one of the Special right triangles. Assuming we don't know the values of the adjacent and opposite sides (we know the hypotenuse us always one), we can use the one to find value of x. The formula for the Hypotenuse is 2x, for the side where 30 degrees is opening up is x, for the side that 60 degrees opens toward is x-rad3. R is the length of the hypotenuse and as we know, the length of a Unit Circle's Hypotenuse is always one. Since we know 2x = 1, we can find the value of x (and the length of one of our sides) and get 1/2. For the last length, we just evaluate the x to get rad3/2. In the picture below shows the ordered pairs.

2. The 45 degree triangle:

The forty five degree angle has different standards. Although the hypotenuse value of x-rad2 is still equal to 1, the other two legs have a value equal to x. Because both sides have the same value, we only need to find the value once. first, we set x-rad2 equal to 1 to get the value of x. At first we would get 1/rad2 but because the denominator must never have a radical we have to multiply both the bottom and the top by rad2. This will give us an x value of rad2/2 and therefore give us both opposite and adjacent values. In the bottom picture below shows the ordered pairs.



*Correction: In the blue box of the horizontal value it says says rad2/rad/2, I meant to write rad2/2.

3.  The 60 degree triangle:

The 60 degree triangle is almost identical to the 30 degree angle accept they lie differently on the Unit circle. In the 30 degree triangle, the 30 degree angle lies on the x-axis, whereas in the 60 degree triangle, the 60 degree angle lies on the x-axis.  This just means that the x and the x-rad3 values are switched, the hypotenuse stays the same regardless. This also means that when writing the ordered pairs, the x and y values are just switched as well. In the picture below shows the ordered pairs.



4. This activity helped me derive the Unit Circle because not only does it apply to one quadrant but also to the entire circle as well. Through this I discovered the points and ordered pairs to come more automatically to mind. Knowing the special right triangles allow me to understand the bases of a Unit Circle in its entirety.  

5. The Unit Circle is basically made of five degrees known as the "magic five." Three of which we've went through in this activity, the other two being 0 degrees and 90 degrees.  This five angles are repetitive, separating each quadrant in the same angles. This is where conterminals and reference angles come into play because with them we can find many other degrees. As seen in the photo below, there is a 60 degree angle, a 30 degree angle, and a 40 degree angle that can be found also in quadrant II, III, and IV.

Inquiry Activity Reflection:

The coolest thing I learned from this activity is that Unit Circles is really just made up of one portion repeated throughout the circle.

This activity will definitely help me with this unit because there are many things that seem more complicated then they actually seem.

Something I never realized about the special right triangles is how they are all connected in their degrees, plots, and positions.

 

Refer to this website for the picture above:

http://the-crafty-crayon.blogspot.com/2012/09/blog-post_22.html



RWA #1: Unit M Concept 5 (graphing and identifying all parts of an Ellipse)

1. The mathematical definition of an ellipse is "the set of all points such that the sum of the distance from two points is a constant."
-Crystal Kirch
2. Algebraically, the equation of an ellipse may look like this:

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=DPKcpdDOYy4itM&tbnid=w0dDhRwnk-AOBM:&ved=0CAUQjRw&url=http%3A%2F%2Fwww.mathwarehouse.com%2Fellipse%2Fequation-of-ellipse.php&ei=vdX6UvLvKs2xqwGvhoAI&bvm=bv.61190604,d.b2I&psig=AFQjCNGfO6poWHy-sP11pQ6Htlr3jjQt6w&ust=1392256786266063
With this equation alone, the center can always be found. The "h and k" are always with the  X and Y terms. If the X and Y squared terms are alone without being in parenthesis it simply means that they are both zero, for example "x^2/4 + (y-4)^2/16," the X term is being simplified or otherwise may be re-written as "(x + 0)^2/4." To find the vertices we would need to look at the bottom portions of the term fractions, generally these would be referred to as "a^2" and "b^2." We can easily tell which is the a because with an ellipse, the A is always bigger than B. If the A is under the X then the Y of the vertices would not change. This would also be associated as the major axis line. If the A was under the Y then it would be the X of the vertices that wouldn't change and then the X be associated (in linear form) as the major axis. The vertices, major axis, and foci will always have this one digit in common, depending on the placement of the A. Along with finding A, we've also found B- together they can be plugged into the a^2 -b^2 = c^2 to get C. With the C we can finish the other half of the foci.

Another way to describe an ellipse is graphically. By having the equation in standard form, which has both terms squared and equal to one, we can infer whether the graph will be "skinny" by having the a under the Y or "fat." by having the a under the X. In other words, if the longer distance is along the x-axis then it would be a horizontal ellipse therefore can be called "fat," now if the longer square root distance goes along the Y-axis then it would be vertical, therefore can be called "skinny."

3. A real life example of an ellipse I found was a race/running track with "designs that help designers take into account top speeds and such depending on shape." According to this website: http://www.barrington220.org/cms/lib2/IL01001296/Centricity/Domain/112/conics%20review%20word%20problemsP2.pdf, the extended circular form of the track. Inside the ellipse are the two  points that are like the foci on the track.

Another thing I found out is that you can use a runner running along the track to find some of its portions. The runner acts as a point on the ellipse because they touch the actual track. If the runner's circling is further away, we can infer that they are currently long the major axis, which as we already know, the longer constant distance that decides the shape of the ellipse.

Please refer to the video below for any further questions:
http://youtu.be/6pDh42E2bbA

4. - http://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=DPKcpdDOYy4itM&tbnid=w0dDhRwnk-AOBM:&ved=0CAUQjRw&url=http%3A%2F%2Fwww.mathwarehouse.com%2Fellipse%2Fequation-of-ellipse.php&ei=vdX6UvLvKs2xqwGvhoAI&bvm=bv.61190604,d.b2I&psig=AFQjCNGfO6poWHy-sP11pQ6Htlr3jjQt6w&ust=1392256786266063
-http://www.barrington220.org/cms/lib2/IL01001296/Centricity/Domain/112/conics%20review%20word%20problemsP2.pdf
-http://youtu.be/6pDh42E2bbA

WPP #10: Unit L Concepts 9-14 (basic probability)

Create your own Playlist on LessonPaths!

WPP #9: Unit L Concepts 4-8 (FCP, Combinations, Multiple Events, Permutations etc)

Create your own Playlist on LessonPaths!