Tuesday, June 3, 2014

BQ #7: Unit V: Derivatives and the Area Problem

1.  Explain in detail where the formula for the difference quotient comes from now that you know! Include all appropriate terminology (secant line, tangent line, h/delta x, etc).


To find our derivative, which is the slope of all tangent lines, we use the difference quotient. How is the difference quotient derived? The first point is (x,f(x)). The second point is (x+h), f(x+h). The distance across is h. We then plug it into our slope formula and get (f(x+h)-f(x)/h). The h cancels out when we simplify so it concludes to be the difference quotient, which is f(x+h)-f(x)/h. H is also known as delta x. The difference quotient is used to find the possible slopes on any graph. A tangent line touches the graph once. On the other hand, a secant line touches the graph twice. To find the slope, we find the limit as h approaches 0. We do this by simply plugging in 0 for h. We then get a derivative, which we then can use to find values and the slope.


Here is an application of my explanation. 



Sunday, May 18, 2014

BQ #6: Unit U

1. What is a continuity? What is a discontinuity?
A continuous function is predictable. It contains no breaks, no jumps, and no holes. It can be drawn without lifting a pencil. The intended height and the actual height of the graph are equal. On the other hand, a discontinuity has breaks, jumps, and holes. The intended height (limit) and actual height (value) are not equal to each other. The two families of discontinuity are removable and non-removable. The removable discontinuity is point discontinuity, in which there is a hole. The non-removable discontinuities are known as jump discontinuity, oscillating behavior, and infinite discontinuity. The limits don't exist in non-removable discontinuities. For jump discontinuity, the left and right limits are different, thus they don't exist. For oscillating behavior, it is very wiggly, so a limit cannot be determined because it does not approach a value. For a infinite asymptote, there is a vertical asymptote which means unbounded behavior.

Removable and non-removable discontinuities
(Please forgive my coffee stain, I think it's a sign that I need to cut back on coffee...)

2. What is a limit? When does a limit exist? When does a limit does not exist? What is the difference between a limit and a value?

A limit is the intended height of a function. It exists at point discontinuities. Why? Because limits exists from when the left and right side matches, and meet, therefore they have the same intended height. An actual height is the value, while the intended height is it limit. A limit does not exist, however, at the three non-removable discontinuities. The right and left limits aren't the same. In an oscillating is wiggle and does not reach a single value, therefore it does not have a limit. An infinite discontinuity does not have a limit because it has unbounded behavior due to a vertical asymptote.

Infinite discontinuity; unbounded behavior because of vertical asymptote
Limit DNE
Oscillating behavior; no limit because does not go to single value
Limit DNE 
Jump discontinuty, limits from left and right don't match

3. How do we evaluate limits numerically, graphically, and algebraically?


For evaluating limits numerically, we use a table. The number that X is approaching in the middle is what we are using to find the limit. First, we need to subtract .1 from the given number and that would be in the first end on the table (beginning). We add .1 to the given number to get the end on the table (last). Then, basically the numbers will get smaller as we go towards the middle. Soon, it will be clear the numbers approach a certain value that will be our limit. The limit may or not be reached depending on the function though.

To find a limit graphically, we basically use our fingers to see if the right and left side meet. If they don't meet, the limit does not exist. Below is a video that will show my explanation with a real graph for a visual. He also explains more into depth!



To find limits algebraically, there are three different ways. First is direct substitution. In direct substitution, you plug in the given number and see what you get. If you get 0, a numerical answer, undefined (limit DNE), then you are done. If you get 0/0, that is an indeterminate form, and you have to use another method! The next method is factoring method. You have to basically factor the numerator and denominator. After, cancel common terms, then plug in the given into the equation that is left. The next method is conjugate method, where you rationalize depending where the radical is in the numerator or denominator. You take the conjugate of what has a radical- either in the numerator or denominator. Then you multiply the entire fraction by the conjugate. Remember, you want to try direct substitution first because you don't want to go through a huge hassle if you can already get an answer using it. That is just a cheese bucket move, and no one wants to be called a cheese bucket because of this silly mistake!


SSS packet

Saturday, April 19, 2014

BQ #4: Unit T Concept 3

Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill?

Basically, a normal tangent graph is uphill because it is based on its asymptotes. They are located at pi/2 and 3pi/2. Furthermore, the quadrants are shown as positive, negative, positive, and negative (based on sin/cos). Cotangent is downhill because its asymptotes are located at 0 and pi. Because of where the asymptotes are located, it is downhill. The quadrants are positive, negative, positive, negative. The asymptotes are where sine and cosine are equal to 0 (making the ratio undefined).

                          Visual of Tangent 

As you can see, the colors represent the different quadrants (I,II,III,IV)

Asymptotes at pi over 2 and 3pi over 2

  Visual of Cotangent

Asymptotes at 0 radians and pi

BQ#3 – Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each of the others?  Emphasize asymptotes in your response.


Sine and cosine is related to tangent because the ratio identity for tangent is (sin/cos). The quadrants in the Unit Circle is shown in the tangent graph. Using our previous knowledge of the unit circle, we know where sine and cos is positive or negative in the quadrants.  If sine or cosine is negative, then tangent would be negative. If sine or cosine was positive or negative, tangent would be positive. In the first quadrant, sine and cosine are positive, so tangent is positive. In the second quadrant, cosine is negative, but sine is positive. But, overall tan is negative because cosine is negative. In the third quadrant, both sine and cosine are negative, making tangent positive (the negatives cancel out). The fourth quadrant for tan is negative because cosine is positive, but sine is negative, making tan negative. Tangent has asymptotes when cosine is equal to 0. Therefore, at pi/2 and 3pi/2, there are asymptotes. The asmymptotes repeat itself over and over again, as the domain goes on.

Visual of Tangent
In cotangent, the ratio identity is cosine/sine. It is very similar to tangent. Cosine and sine are positive in the first quadrant, so cotangent is positive. In the second section, the sine is positive and cosine is negative, so cotangent is negative. For the third quadrant, cosine and sine are negative, making cotangent positive (cancels out).  For the fourth quadrant, cos is positive and sine is negative, making cotangent negative. Sine must equal to 0 to find the asymptotes. Remember, asymptotes are undefined values. Therefore, we know that sine is equal to 0 are (1,0) and (-1,0). That is at 0 and pi. Cotangent is different from tangent because of the asymptotes. 
Visual of Cotangent


The reciprocal of secant is cosine. So the graph is just cosine flipped. So since cosine is positive in the first and last quadrant, and negative in the second and third, so is secant.  If cosine is positive, so is secant. Sine has no effect on the graph because it is not part of the ratio. The asymptotes are (0,1) which is pi over 2 and (0,-1) and 3pi over 2. This is where cosine is 0, making the ratio undefined. This is similar to the tangent graph, which had cosine as the denominator. 

Visual of Secant


Cosecant is just secant reversed. The ratio for cosecant is 1/sin. So basically, since sin is positive. so is cosecant in the first quadrant. In the second quadrant, cosecant is positive as well. In the third and fourth, cosecant is negative. The asympytotes are where sine is equal to 0 which is at (1,0) and (-1,0). Therefore, it is at 0 and pi. Remember, the graphs don't EVER touch the asymptote. 

Visual of Cosecant 

Thursday, April 17, 2014

BQ#5 – Unit T Concepts 1-3

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.
          Sine and cosine do not have asymptotes because they are never undefined. Remember, sine means y/r and cosine means x/r. R stands for the hypotenuse, and in the unit circle the radius is always 1, which is the hypotenuse. The other four trig graphs do have asmpytotes because they do not have 1 (r) as their denominator. Instead, their denominators can be any numbers, according to their trig ratio. For example, tangent's ratio y/x. X can be any number; there are no restrictions.
        The denominator, however, can be 0 (from the coordinates (0,1) or (-1,0), which makes the ratio undefined once in the according trig function. In cotangent, the ratio x/y also can assume asymptotes. If the coordinate is at (-1,0) or (1,0), when plugged into the ratio, it will be either positive or negative 1 over 0, which is undefined. Also, this applies to secant. Secant is r/y. Y is 0 at (1,0) or (-1,0). This results to be 1 over 0, which makes it undefined. In cosecant, the ratio is r/x. X is 0 as (0,1) and (0,-1). Thus, when plugged in, the ratio is 1/0, making is undefined as well. An asymptote is basically when the ratio is divided by 0, meaning its undefined. 

Wednesday, April 16, 2014

BQ#2 – Unit T Concept Intro

How do the trig graphs relate to the unit circle?
So, you may ask, how do trig graphs relate to the unit circle? Well, it uses the positive and negative signs of the quadrants of the trig function! The trig graphs are basically the unit circle unwrapped or uncoiled to a line. The points on the graph are relevant to the points in the unit circle (radians). FOR EXAMPLE, in the unit circle, sine is positive in the first two quadrants, and negative in the third and fourth concepts (we know this from our previous units).  Thus, on a trig graph, the graph will show a pattern of going uphill, uphill, downhill, downhill. This represents the quadrants. It is positive in the first two quadrants and negative in the other two. Lets use another example. For cosine, in the unit circle it is positive in the first quadrant, negative in the second, negative in the third, and positive in the fourth. So on the trig graph, the period is positive and then goes negative, and eventually goes back positive according to the marks. For tangent and cotangent, it will have positive values first, then negative, then positive, then negative. The trig graph has a shorter period and this is because in the first two quadrants, it goes positive then negative, and then it is basically repeated in the other two quadrants. 

Beautiful visuals of the trig graphs and the patterns that I talked about in my earlier paragraph.

Period? - Why is the period for sine and cosine 2pi, whereas the period for tangent and cotangent is pi?
The period for sine and cosine is 2pi because basically when you look at the unit circle, sine is positive, positive, negative, negative on the quadrants. In cosine, it is positive, negative. negative, positive. The period is 2pi and it takes all the quadrants (one whole revolution) to finish the pattern. But, for tangent and cotangent the pattern is positive negative. If it is positive and negative in the first two quadrants, and it repeats after that, then basically it only takes the period of pi in a revolution, in which it repeats itself twice.

Amplitude? – 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?
Sine and cosine have amplitudes of one. Why? Well think about what sine and cosine consists of. Sine is y/r and cosine is x/r. R is the hypotenuse, which is 1. Sine and cosine cannot be greater than one or less than one. On the unit circle, y and x are either 1 or -1. When you divide using the given ratios, it still equals to 1 or -1. So that clarifies that the amplitude is always one or negative one because of this rule. But, for other trig functions it is different. There are boundaries to what values it must equal to because x and y can vary in different numbers. For example, cosecant, it is r/y. Your y CAN be 1 or -1, OR it can be any other number like 2 or 3. Same goes to the other trig functions. So there are no amplitudes. 

Thursday, April 3, 2014

Reflection #1: Unit Q-Verifying Trig Identities

1. What does it mean to verify a trig identity?
To me, verifying an identity basically means to prove that an equation is true by showing that both sides are equal. I think of it as taking steps to simplify one side of the equation to make it equal to the other. Verifying a trig identity takes a gradual process of manipulation, alterations, and logical thinking! It may be difficult at first, but all you have to do is to make both sides equal and there are multiple steps so it is not as difficult as other math units.
2. What tips and tricks have you found helpful? 
I found that remembering the identities (ratio, reciprocal, Pythagorean) make verifying a trig identity much faster. Also, I learned that it's a little different than other units we have went through, because there are many ways to verify an identity and it's very ambiguous. Usually, we are usually used to using a formula and getting our answer right away, but with this we basically we have trial and error. Another tip is to not touch the other side! You simply cannot, no matter how tempting it is. 
3. Explain your thought process and steps you take in verifying a trig identity.
First, I look at the problem and I try to convert anything to a trig identity of sine or cosine.  Next, I usually look to see if I can take out a LCD (least common denominator). I factor if I need it. Then, I usually check if I can use any identities, especially Pythagorean identities. I also find that powering up can lead me to the Pythagorean identity. I sometimes multiply by conjugate if there is any fraction. If I get confused at all, I usually go to three steps back and review what I did. If I can't make sense of it, I usually start from scratch because I don't want to over complicate the equation and continue if I accidentally made an error. 

Wednesday, March 26, 2014

SP#7: Unit Q Concept 2: Finding All Trig Function Values Using Identities

Please see my SP#7, made in collaboration with Vivian Pham, by visiting her blog here Also be sure to check out the other awesome posts on her blog.

Wednesday, March 19, 2014

I/D #3: Unit Q Concept 1: Using Fundamental Identities to Simplify or Verify Expressions

Inquiry Activity Summary

Where does sin²x + cos²x  = 1 come from to begin with? 
First off, in order to understand the concept, we must acknowledge that an identity is a proven fact and formula that is always true. Therefore, the Pythagorean Theorem is an identity because it is a proven formula that is always true. The Pythagorean Theorem is usually demonstrated through the formula a^2+b^2=c^2. However, in reference to the unit circle, the Pythagorean Theorem is x^2+y^2=c^2. To get the theorem equal to 1, we divide both of the sides by r^2. The result is x^2/r^2+y^2/r^2=1. These can also be written as (x/r)^2+(y/r)^2=1. Now, it is in complete reference to the unit circle! Cosine for the unit circle is x/r and sine is y/r. Therefore, if you substitute the equation with cos^2θ and sin^2θ. The new equation, cos^2x+sin^x=1 is now referred as an Pythagorean Identity. We can prove this by plugging in a pair from the unit circle such as 60 degrees. The pair for 60 degrees is (1,2, radical 3 over 2). When plugging it into the formula, after simplifying, you get 1=1. IT IS TRUE. :D 

How to derive the two remaining Pythagorean Identities from sin2x+cos2x=1.
This shows how to get tan^2x+1=sec^2x. First, you have to divide both sides by cos^2x to get tangent and secant. cos^2x/cos^2x cancels out to become 1. After, using your identities, you simplify the equation. Thus, with our memory of identities, we get the identity tan^2x+1=sec^2x.

This Pythagorean Identity consists of cotangent and cosecant. To get there, we divide by sin^2x for both sides. We know that sin^2x/sin^2x cancels out to become 1. We then use the identity for cos^2x/sin^2x which is cot^2x. Then, we know that 1/sin^2x is csc^2x. The final answer is 1+cot^2x=csc^2x

Inquiry Activity Reflection: 
1. “The connections that I see between Units N, O, P, and Q so far are that it uses the unit circle and trig functions. Also, it uses the Pythagorean Theorem as we have been using in triangles.
2. “If I had to describe trigonometry in THREE words, they would be crazy, mind-blowing, interesting.

WPP #13 and 14: Unit P Concept 6 and 7: Law of Sines and Law of Cosines

Please see my WPP13-14, made in collaboration with Christine Nguyen, by visiting her blog here Also be sure to check out the other awesome posts on her blog.

Sunday, March 16, 2014

BQ# 1: Unit P Concept 1-5: Law of Sines and Area of an Oblique Triangle

Law of Sines:
Why do we need it? 
The law of sines is needed when solving for triangles. To find angles and lengths in right triangles,we use the Pythagorean Theorem and normal trig functions. However, when we are faced with a triangles that are not RIGHT triangles, we result to using the law of sines, which we can find using ASA or AAS.

How is it derived from what we already know? 
We start off with a triangle that is not a right triangle.

We draw a perpendicular line down from angle A, forming two right triangles. Label the line h (for height).

Since we have two right triangles, we can use trig functions. We have to find h, and we know the equations are sinA=h/c and sinC=h/a in right triangles. To get h by itself, we multiply the denominators by itself  for both equations. So the result is csinA=h and asinC=h. Since they are both equation to h, you can make them equal to each other. So it will become csinA=asinC. Next, we divide the equation by the coefficients together, ac. The result is sinA/a=sinC/c.

Using the same concept, we can now find angle B. Let's draw a perpendicular line through C.

With the two new right triangles, the trig functions are asinB=h and bsinA=h. Set them next to each other by the transitive property. Next, we again divide the denominators multiplied together, ab, and we get sinB/b=sinA=a.

The result: Law of Sines 

Area of an Oblique Triangle:
How is the "area of an oblique triangle" derived? 
The area of a triangle is a=1/2bh. When we want to know the area of an triangle, but we don't know the height, this is when when use an oblique triangle. In order to, we have to have two sides and one angle. To find our height, we will use the trig functions. For example we know sinA=h/c. Multiply both sides by c and you get csinA=h. After, you can plug this into the area of a triangle formula, which results to a=1/2b(csinA). This is our new formula for the area of the oblique triangle. We can also find the other two angles as well using the same method.

Our final formula for area of an oblique triangle


How does it relate to the area formula that you are familiar with? 
We used the original area formula a=1/2bh to find the area of the oblique triangle. The only difference is that we had to find the height using trig function sine, instead of normally being given it. The rest of the area formula is the same.

(and beautiful triangles drawn by me!!! :D) 

Wednesday, March 5, 2014

WPP#12: Unit O Concept 10: Solving Angles of Elevation and Depression Word Problems

Bulgaria: RGI's Europa Capital Completes Acquisition of 'Mall of Sofia'
The Problem: 

Part A: Carrie is going to go to the store. She is 110 feet away from the store. The angle of elevation is 30 degrees. She wants to know the height of the building because she is very curious, what is it?

Part B: Carrie decided to climb the store because she is crazy. She is on top of the store. She is 63.5 feet higher than a car she wants to jump on. She estimates the angle of depression from where she is now to the car to be 20 degrees. How big will her fall be if she decided to jump? 

The Solution:
 Part A:

Part B:

Tuesday, March 4, 2014

I/D2: Unit O - How can we derive the patterns for our special right triangles?

Inquiry Activity Summary

1. In order to derive a 30-60-90  triangle from an equilateral triangle, we start off with the given information that since it is an equilateral triangle, all sides are the same, and in this case they are 1. Also, since the sides are all the same, this also means that the angles are the same as well. Therefore, having a triangle add up to 180, we can conclude that each angle is 60 degrees if you divide 180 degrees by the amount of sides, which is three. To form the 30-60-90 triangle, we simply split the triangle in half. Once we do that, there is now a 90 degree angle on the interior. Since we split the triangle in half, we split the 60 degrees angle in half as well, thus making it 30 degrees on each side. We now have two 30-60-90 triangles.

 To find the sides, we know that the equilateral triangle has sides of 1. But, since we split the split the triangle in half, the side across the 30 degrees is also split in half. We know the angle across 90 degrees is still 1 because the splitting of the triangle didn't affect the side (now the hypotenuse). To find the angle across the 60 degrees with our newly made triangle, we use Pythagorean Theorem. Pythagorean theorem is a^2+b^2=c^2. The hypotenuse is c, and the angle across the 30 is a. We are looking for b, which  turns out to be radical 3 over 2. 

Lastly, we multiply each side by 2 because we want to get rid of the fractions. After multiplying by 2, our sides conclude to be a=1, b= radical 3, and c equaling to 2. The importance is that a pattern begins to form. Since a=1, we see that side c is twice of it. This creates the notion that a is equal to n, and side b is 2n. The third side concludes to be n radical 3 since 1 times radical 3 stays the same. We place n to show that there is a pattern, that the numbers from the sides don't just magically appear by memorization. There is a relationship between each sides in relation to their angles and the n shows that, and you can solve it by using n.

2. In order to derive a 45-45-90 triangle from a square, we know that a square adds up to 360 degrees and all sides are equal. We also know that a square has equal angles of 90 degrees. 
To create a 45-45-90 triangle, we draw a diagonal through the square. Because each side was 90 degrees in the square, we see that two of the angles are split because of the diagonal. Therefore, each of the angles split is 45 degrees, forming a 45-45-90 triangle. 

We know that each side from the 45 degrees remains 1 because the diagonal cut did not affect them. To find the hypotenuse, the diagonal cut, we use Pythagorean Theorem. We know that leg a is 1 and so is leg b, so we plug it in to 1^2+1^2=c^2. We then result to 2=c^2. To get our c we square root both sides and we get radical 2=c. Thus, we see that our sides is a=1, b=1, and c=radical 2.

 This is when n comes to play. We see that n is equal to 1, therefore n takes place for a and b. Since the hypotenuse is radical 2 times 1, and we know that n is equal to 1 (which both sides a and b have), we can conclude that our legs are going to be n radical 2. N represents the relationship between the sides in the 45-45-90 triangle and no matter what numbers we are given, we can find the other sides using the relationship of N. N IS POWERFUL! :) 


  1. “Something I never noticed before about special right triangles is…” there is an actual reasoning behind why 30-60-90 and 45-45-90 have the sides that they do. Before I just assumed to know from memorization!
  2. “Being able to derive these patterns myself aids in my learning because…” IT GOES BACK TO THE UNIT CIRCLE AND IT HELPS ME UNDERSTAND THE UNIT CIRCLE SO MUCH BETTER NOW :)

Friday, February 21, 2014

I/D #1: Unit N Concept 7: How do SRTs and the UC relate?

Inquiry Activity Summary 
The activity that we did in class demonstrated to us how special triangles are related to the unit circle. It was review of the special right triangles and the side lengths of the triangle depending on the special right triangles of 30º, 60º, 45º.

30º Triangle! 

A 30º triangle has special qualities. The shortest side opposite of 30º is X. The hypotenuse is two times x and is across the 90º angle. The side across the 60º angle is x times radical three. There are rules to the 30º triangle. To simplify the three sides, the hypotenuse is set to 1. In order to get to 1, we must divide the hypotenuse by itself, making it 2x/2. Thus, r is equal to 1. Since we did it to one side, we must do it to the others. The opposite side (x) is radical 3 over two because we have x over 2x and the x's cancel out, leaving you with 1/2.  With the adjacent side, you do x radical 3 over 2x. The x's cancel out leaving you with radical 3/2. When you draw it on a coordinate plane, you think of your adjacent as your x axis and opposite as your y axis.

45º Triangle! 

A 45º triangle is special because there are two angles that are equivalent. That means that both of their opposite sides are the same as well. The hypotenuse is x radical 2. For a 45º, we divide the hypotenuse by x times radical 2, so the hypotenuse becomes 1. We divide x by x radical 2 to the other two sides. The x's cancel out and when you rationalize, it simplifies to radical two over two. Therefore, the y value is radical 2 over 2 because you are going up radical 2 over 2.

60º Triangle!

A 60º triangle has the same characteristics as 30º, as in the sides. Thus, we divide the hypotenuse by itself to get it to 1. Though it is similar to the 30º,  the triangle's visual is different, as in the orientation. This changes the x coordinate in the adjacent side to 1/2. The opposite side length (vertical) coordinate will become radical 3 over 2 for the y coordinate. 

Here is a video that clarifies how I find the sides of each triangle. 

How does this activity help you derive the Unit Circle? 
Once you draw a plane for the triangles, it is visible that the new triangles are in the first quadrant. This shows how the ordered pairs work in the Unit Circle. Since the hypotenuse is equal to 1, that makes the radis of the Unit Circle 1. Thus, with the origin being at (0,0), we soon see the side lengths as coordinates that we use in the Unit Circle. We also see that every quadrant in the plane looks the same, just the difference is the negative values in either x or y in II, III, and IV. But, with the triangle drawn in quadrant 1, we see that it reflects onto the other quadrants, which creates the same reference angles, which is key to completing the Unit Circle.  The point at 0 degrees is (1,0), 90 degrees is (0,1), 180 degrees is (-1,0), and 270 degrees is (0, -1) because the radius is 

What quadrant does the triangle drawn in this activity lie in?  How do the values change if you draw the triangles in Quadrant II, III, or IV?
The triangle in this activity lies in the first quadrant. The values change from quadrant to quadrant because they share a common origin and its just a reflection of the triangle in each quadrant.


If you reflect a 30º triangle to the 4th quadrant, you see that the for the side across the 30º angle changes. Instead of being a positive on both the x and y coordinates as seen in the first quadrant, the y value is -1/2. This is because in the fourth quadrant, the y value is always negative. Therefore, the coordinate is (radical 3 over 2, -1/2). The adjacent angle (across 60º) stays the same because it is not being affected as it reflects. 


If you reflect a 60º triangle in the third quadrant, both values are negative. As you can see, everything stays the same as in value, but it becomes negative because it is in the third quadrant. The fourth quadrant is similar, the x value is positive and the y value is negative. 



If you reflect a 45º triangle to the second quadrant, the coordinates remain the same. The only difference is that the x value is negative. This is because the second quadrant always contains a negative x value. 


  1. The coolest thing I learned from this activity was that the Unit Circle is not just memorization, but there is a meaning behind it all which makes everything so much simpler.
  2. This activity will help me in this unit because last year in Algebra II I didn't understand the unit circle and I struggled a lot, mainly because I didn't understand the triangles. Now, I can solve the unit circle because I know why it works the way it does!
  3. Something I never realized before about special right triangles and the unit circle is that THEY ARE RELATED! It helps so much because now I don't have to convert degrees to radians!

(p.s.) Sorry for the white background shadowing for the words, I don't know how to fix it! If you could teach me during class one day it would be great!)

1. (http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_34.gif)
2. (http://02.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_21.gif)
3. (http://01.edu-cdn.com/files/static/learningexpressllc/9781576855966/The_Unit_Circle_45.gif)

Sunday, February 9, 2014

RWA# 1: Unit M Concept 5: Graphing Ellipses Given Equation

1. The mathematical definition of an ellipse is "the set of all points such that the sum of the distance of two points, known as the foci, is a constant" (Kirch).
2.     The standard formula of an ellipse is (x-h)^2/a + (y-k)^2/b = 1. On a graph, an ellipse is depicted as an oval, or a squashed circle. In other words, it is stretched out. The key features of an ellipse are the standard form, the center, vertices, co-vertices, two foci, major axis, minor axis, a, b, and c, eccentricity,  and if it is seen as "skinny" or "fat". You can find these graphically or algebraically.
         Lets start off with algebraically. By looking at the standard form of an ellipse, you can easily find the center. The x value stands for "h" and the y value stand for "k". After you find your center, you can find your major axis by looking at the denominator. If the denominator is bigger under the first term (the x), that is your major axis. This also means that your ellipse will stretch horizontally, being "fat". If the bigger denominator is under the y, then then your graph will stretch vertically, being "skinny". Once you have your major axis, you can find your minor axis because it would be opposite. The smaller number x is your minor axis and horizontal. If the smaller number is under y then it is vertical.  For major and minor, if one is vertical, you know the equation is x= h is. If it is horizontal, y=k.You can find your a and b from the standard formula too. The bigger denominator is the a^2 term so you have to square root it to find your a. The smaller denominator is your b^2 term and you also have to square root it to find your b. After that, you can find your c term by plugging your "a" and "b" into the formula, a^2+b^2=c^2. To find the vertices, you add and subtract "a" from your center. The major axis stays the same throughout. To find the co-vertices, you add and subtract "b" from your center, the minor axis staying the same. To find the foci, you add "c" to the value that's changing. It's on the major axis. If major axis is x=, you add "c" to your "k" (y value of center). It is plus or minus because there are two vertices. If the foci is closer to the center, the ellipse is more circular. If it is farther, it is more stretched out. Eccentricity is  a measure of how much the conic section deviates from being circular. If it is closer to 1, is it more stretched out. If it is closer to 0, it is more in the circular form. The eccentricity is between 0 and 1. To find the eccentricity, you use the formula c/a and round to the thousand's place.
             To identify an ellipse graphically, it is simple as well. The major axis is the longer diameter of the ellipse, and is seen with a straight line (no dashes). The minor axis is the axis intersecting the major axis, it is shorter, and seen with a dashed line. The vertices are the points that lie on the major axis. The co-vertice are the points that connect the minor axis, at the end. With the vertices and co-vertices, the ellipse can be drawn. The center of the ellipse is where the major and minor axis intersect. If the ellipse is seen more horizontally, then it is "fat". If it is seen as more vertically stretched, it is "skinny" You can determine a by counting how many points are away from the center from each vertices on the major axis. Same goes with how to find b, you can count how many points are away from the center of the co-vertices on the minor axis. The foci is within the vertices of the major axis. The closer the foci to the center, the more it is in circular form. The farther the foci, the more it deviates from being a circle. If the foci is close to the center, the eccentricity is close to 0. If foci is far from center, eccentricity is big.

Here is a website that shows how to graph an ellipse!

Here is an ideal graph of an ellipse with some of the points I talked about. 

This is a cool video on how to find the foci! 

3.       A real world application of an ellipse is that it is used in lithotripsy. "One important property of the ellipse is its reflective property. If you think of an ellipse as being made from a reflective material then a light ray emitted from one focus will reflect off the ellipse and pass through the second focus. This is also true not only for light rays, but also for other forms of energy, including shockwaves. Shockwaves generated at one focus will reflect off the ellipse and pass through the second focus."(http://mathcentral.uregina.ca/beyond/articles/Lithotripsy/lithotripsy1.html). 
           Therefore, it has been used my medical experts to create a device that helps kidney stones and gallstones. A lithtripter uses shockwaves to SHATTER a kidney stone or gallstone. Patients with kidney stones or gallstones can be saved because of the shape of an ellipse because it reflects shockwaves. HOW COOL IS THAT?! MATH SAVES LIVES AND IT WILL CONTINUE TO. :D 

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