Unit V BQ #7

Where does the difference quotient come from?

PC: Kelsea; graph

In the quadratic graph above, there is a specific point labeled (x, f(x)). This specific point creates a tangent line. A Tangent line is a line that only touches the graph once. The second point, (x+h, f(x+h)) creates a secant line. A secant line is a line that touches two points on the graph. Now, you're probably wondering where we go the two points. Well, it's kind of simple even though it looks complicated on the picture right now.

So, let's begin by observing the point (x, f(x)). This point is the original one given to us. Since on the graph, the x value is stated, all we need to do to find the y value is plug it into the function, that is why it is f(x). Now, to the solid point on the x-axis, we call it h since we do not know the exact value of it. Now, how do we find the point on the graph above h? Well, for the x value of it, we know that if we add x+h, it will give us the x-value. Now, to find the y-value, all we do is plug in the x to the function, making it f(x+h). Yay! Now we have the two exact points for our secant line! Now, how do we find the equation for the tangent line of (x+h),f(x+h))? Well, in the picture below, we use the slope equation to help with that. It is worked out below. Now, we can see that this is where the difference quotient comes from! Yay! So, in short, the difference quotient comes from slope equation, that is solved but just didn't use actual numbers.

PC: Kelsea: equation

References:

Mrs. Kirch's videos

Photo's: Kelsea

Unit U BQ#6

1. What is continuity? What is discontinuity?

Continuity is when we are able to draw the graph without lifting the pencil from the paper. So that means the graph cannot have any jumps, holes, or break. It is also predictable. This means that we now it is approaching infinity on both sides, either positive or negative. Now what is discontinuity? Look above, and picture the exact opposite of what I just explained. Discontinuities can be draw by lifting the pencil. So there are jumps, holes, and breaks. It is also unpredictable, meaning we do not know where it is headed from the left and from the right at any given point. Now, there are four discontinuities. There are two families of discontinuity: removable and no-removable. Removable contains point discontinuity. Non-removable contains point discontinuity, oscillating behavior, and infinite discontinuity. 

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

A limit is the intended height of the graph. That means the limit can be a hole on the graph, because it was in fact intended to go to that point on the graph. The limit exists only at removable discontinuities, which is specifically point discontinuity. It exists here only because the intended height is specific from both the left and the right of the graph. However, the limit does not exist at non-removable discontinuities, which are point discontinuities, oscillating behavior, and infinite discontinuities. It doesn't exist at point discontinuities because there are two separate intended heights coming from both the left and the right of the graph. It doesn't exist at oscillating behavior because the graph is so wiggly, that there is no specific point on the graph and in general, cannot make up it's mind on where it wants to go. Lastly, it doesn't exist at infinite discontinuity because it occurs when there is a vertical asymptote, which leads to unbounded behavior which means two separate directions from both the right and the left.  The limit is the intended height however the value is the actual height, which only includes shaded in holes and lines on the graph, not open circles. 

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

We evaluate limits numerically by using tables. We put numbers that are approaching a certain number from the left and the right. That way we can see what is the intended height. For graphically we observe the graph and use onr finger on each side and see if they approach at the same point. If its a hole but they both meet then its a limit. If there are two seperate points, then there is no limit. For algebraically, there are three methods. The first method you should alwats try first is substitution. Thats when we directly substitute x and solce to find the answer. If the answer is 0/0 we use factoring to see if we can cross out any equation which means there will also be a hole. After factoring and crossing out, we then substitute the x and solve. If we cant factor though, we use conjugate if there are any Radicals to find anything that can cancel. And that's how we solve (:

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.

Before we begin, it is important to note that the equation for sine is (y/r) and cosine is (x/r), which means they can never be undefined (on the unit circle) and therefore not be an asymptote.

Now, let's begin with tangent. Tangent, in terms of identities, has the equation of (sin/cos). In terms of graphing, we know that if cos is 0, it will be undefined and give the equation asymptotes, restrictions. The asymptotes help restrict the graph as well as how it will look. We know that tangent has a pattern, in terms of unit circle, of +-+-. So asymptotes help determine where the graph will lie. It will be easier to show you so look below for reference.

©Mrs. Kirch's Desmos

As we can see, the unit circle is divided into four and the tangent (orange line) pattern in the unit circle helps determine the pattern that the tangent has as well as what it will look like. Now, this relates to sine and cosine because In the first quadrant, all are positive and are above the x-axis. In the second quadrant, only sine is positive, so cosine and tangent are negative. In the third quadrant, only tangent is positive and is the only one above the x-axis. Lastly, tangent and sine are negative so sine is the only one above the x-axis. There are similarities between the two every from quadrant to quadrant.

Now as for cotangent? Same rules apply like tangent! Look below and you can see that the only difference is the direction because cotangent goes left to right and tangent goes right to left.

©Mrs. Kirck's Demos

Now as for secant? Well, lets look at the picture below. (Secant is the blue line.)

©Mrs. Kirch's Demos

Now we can see that there are asymptotes restricting the secant graph and making multiple ones. However, when we look, we can see a similarity between secant and another function. Can you see it? Whenever cosine is positive, the secant graph goes up. When the cosine graph is negative, the graph goes down. I know, I know, you've just been mind blown right now. But hold on, there's one more.

Look at cosecant below. (Cosecant is the blue line.)

©Mrs. Kirch's Desmos

Do you see a similarity despite the asymptotes restricting the cosecant graph? BY JOLLY YOU'VE GOT IT! Whenever the sine graph is negative, the cosecant graph is negative and whenever the sine graph is positive, the cosecant graph is positive.

BQ#5: Unit T Concepts 1-3

Why do sine and cosine NOT have asymptopes, but the other four trig graphs do? Use Unit Circle ratios to explain. 

First, let's find out how to get an asymptote in the first place. We get an asymptote when the equation's answer becomes undefined. Now, let's look at the equations for the trig functions. Now, below are the equations. We know that r will always be 1 because we are referring to the unit circle. Sine and Cosine have a denominator of r so it will be impossible to obtain an undefined answer when the denominator is 1. Cosecant and Cotangent have a denominator of y so if y is 0, then the answer would be undefined and would result in an asymptote. Actually, this would only work for degrees of 90 and 270.  For Secant and Tangent, the denominator is x so if x is 0, that will lead to an answer of undefined and an asymptote. This would only work for degrees of 0 and 180. 

©Kelsea DC

©Kelsea DC

References:

Mrs. Kirch's SSS Packet

Desmos Packet

BQ#2: Unit T Intro

How do the trig graphs relate to the unit circle?

A trig graph is actually a unit circle. Well, not the whole thing since the graph goes on forever, but from 0 to 2π, which is essentially the length around a whole unit circle.
To test this, we can use our trig functions to help. For example, lets use sine. For sine, it is positive in the first two quadrants and negative in the last two. Now, how is this related at all? Well, when we uncoil the unit circle and make a trig graph for sign, essentially what is quadrant one and quadrant two will be above the x-axis and will be positive. What would be the third and fourth quadrant would be below the x-axis since it is negative.
Make sense yet? Somewhat? Let's try another one! Let's use cosine.Cosine is positive in quadrant one and quadrant four and negative in quadrant 2 and 3. Now, let us once again imagine the unit circle unwinding and placing it on a graph What would have been quadrant one and quadrant four would be above the x-axis because it is positive for cosine in those quadrant. Now, the second and third quadrant will be below the x-axis since cosine is negative in those quadrants.
Making more sense? Need just one more example? Let's use tangent! Tangent, is positive in the first and third graph and negative in the second and fourth graph. Now, for the last time, uncoil that unit circle and think, how would our graph look like? ... Quadrant one and three would be above the x-axis because it's positive and quadrant 2 and 4 would be below the x-axis since it is negative.

©Kelsea DC

Period?-Why is the period for sine and cosine 2π, whereas the period for tangent and cotangent is π?

Lets refer back to what we reviewed above to help us. Also, note that a period is when a part of the graph reaches both above the x-axis and below the x-axis. As we know, sine, cosine, and tangent are positive in certain area. Sin: ++--; Cos: +--+; Tan:+-+-. These are the patterns for these. Now a period is how much it takes for something to repeat. When we look at sine, how long does it take to obtain a period of ++--? Well, it takes 2π and also because there are no repeating factors in the format and that how long it would take for it to repeat again. . For cosine, what kind of period would it take for +--+? Well, 2π again since there are no repeating factors and that's how long it takes in order for it to repeat again.  Now, tangent is +-+-. What's it's period? 2π? WRONG! It's π! Why? because in one period, the graph reaches both above the x-axis and below the x-axis. Need a visual? NO WORRY! Look Below!

©Kelsea DC
Green=Quadrant 1 Orange=Quadrant 2 Blue =Quadrant 4 Yellow=Quadrant 4

Amplitude?-How does the fact that sine and cosine have amplitudes of one (and other trig functions don't have amplitudes) relate to what we know about the unit circle?

Sine and Cosine can only have amplitudes of 1/-1 because on the unit circle, they are restricted to it. Their denominator in their equation is 1. If we try to use a number larger or smaller than it and plug it into the calculator, it becomes an error. However, tangent has an equation of y/x so it isn't restricted because y and x aren't really a specified number like r is. Same applies to cotangent, only the equation is flipped. 

Reference:

Mrs. Kirch's SSS Packet 

Hand Drawn Photos By Moi 

Reflection#1: Unit Q: Verifying Trig Functions

1. To verify a trig function means to make sure that the two equations are equal to each other. You are unable to touch both sides at the same time so you have to use one side and from that, use identities to get the exact answer as it is on the other side of the equal sign.
2. I have found that remembering and at least knowing the unit circle is a big help because it can be used not only to check answers on one of the concepts, but it is very useful when finding degrees because there are many right triangles referenced in unit Q. Also, to remember which equation have sin and cos first because those are the most vital identities to being successful in this unit.
3. When looking at an identity, whether I need to verify or simplify, I look to see if all the trig functions have seomthing in common. Then I try to use reciprocal and ratio identities if there is nothing squared in the equation. However, if there is something squared in the equation, I reference the Pythagorean Identities to check if anything can be substituted. I often look for the least common denominator, taking out GCF, and factoring. Also, reciprocal and conjugates are my best friend in this unit!

Lastly, as a side note, I would like to mention that the easiest and best way to pass this unit is to simply become familiar with the identities. How you ask? CONSTANT PRACTICING! Becoming familiar with the identities with equations is one of the best ways to master it. And do what feels right. Do not think too hard about the problem. If you do think to hard and find it too difficult, you need to relax and take a breather. This unit is easy. It just takes some time to process all the information.

SP 7: Unit Q: Concept 2: Finding all trig functions when given one trig function and quadrant

Please see my SP7 made in collaboration with Eriq, by visiting their blog post here.  Also, be sure to check out our other collaborative blog post as well as his awesome individual ones! (:

I/D #3: Unit Q: Concept 1: Pythagorean Identities

http://www.softmath.com/tutorials-3/relations/articles_imgs/5385/fundam5.gif

There are the reciprocal identities that I will constantly be referring to.

http://calculustricks.com/wp-content/uploads/2011/03/Trigonometric-Identities-3.jpg

These are the ratio identities that I will constantly be referring to.


1. Identities are proven facts or formulas that are always true. The Pythagorean Identity sin²x+cos²x=1 comes from using the Pythagorean theorem.  The Pythagorean theorem, instead of using a²+b²=c², we will replace the letters with x²+y²=r², which is the same exact thing when placed on a coordinate plane. Then we want the right side to equal 1 so we divide both sides by r² and we get(x/r)²+(y/r)²=1. When looking at what we now have, we can make to changes which we learned previously. First, we can change (x/r)² to cos² (cosine) because the equation for cosine on the unit circle is x/r. Also, we can change (y/r)² to sin² (sine) since that is the equation for sine on the unit circle. Now, to make sure that this is really identity, we must apply the unit circle to this to get the Let's plug in an angle of 45* to this identity. First, we find cos of 45 and sin of 45 which turn out to be √2/2. So we plug it into the Pythagorean Identity that we found. So it is going to be (√2/2)²+(√2/2)²=1. When we distribute the power and add both of them together, our answer becomes 4/4 which equal 1. Remember to put x or theta because that means that it can be any angle. Now we have one of our Pythagorean Identity which is cos²x+sin²x=q.

©Kelsea

©Kelsea

2. There are two remaining Pythagorean Identities, 1+tan²x=sec²x and 1+cot²x=csc²x. To obtain the first equation I mentioned, first divide both sides by cos²x. By doing so we get cos²x/cos²x+sin²x/cos²x=1/cos²x. Now when we look at our ratio identities and our reciprocal identities, we can see that some of the fractions we have created are equal to some sort of single identity: sin²x/cos²x=tan²x and 1/cos²x=sec²x. So we can replace the fractions with these single identities to get the equation 1=tan²x=sec²x.  We now have one last Pythagorean Identity to find. To do so, we divide by sin²x. Now our identity will be cos²x/sin²x+sin²x/sin²x=1/sin²x. Again, the fractions correspond with some of the ratio and reciprocal identities. cos²x/sin²x=cot²x and 1/sin²x=csc²x. Now replacing the fraction identity with a single identity, we get the last Pythagorean Identity which is cot²x+1=csc²x.

©Kelsea

References:

http://www.softmath.com/tutorials-3/relations/articles_imgs/5385/fundam5.gif

http://calculustricks.com/wp-content/uploads/2011/03/Trigonometric-Identities-3.jpg

WPP#13-14: Unit P: Concept 6 and 7

This WPP 13-14 was made in collaboration with Eriq on his blogspot! Please visit his other awesome posts by going here.

http://www.kevinandamanda.com/whatsnew/travel/sweet-tooth-fairy-bakery.html

The Problem: Bakery Meet Up

While on her way walking around town, Kelsea notices her friend Eriq across the street, about 20 feet away. He notices her and waves back. They both text each other, to avoid walking, and look ahead and see a bakery in which they both decide to meet up. Kelsea is N30*E and Eriq is N25*W. 

a) what is the distance from Kelsea to the bakeshop?

b) what is the distance from Eriq to the bakeshop?

The Solution:

Copyright: Kelsea and Eriq

The Problem: Leaving

After having a few sweets and catching up on each other's lives, their rides arrive. They bid each other goodbye and leave at the same time, traveling on a course with an angle of 150* between them. If Kelsea, on the right, travels 30 mph and Eriq, on the left, travels 40 mph, how far apart are they after two hours have passed?

The Solution:

Copyright: Kelsea and Eriq

BQ#1:UnitP: Concept 1-4:Non-Right Tiangle Equations

1. Law Of Sines
For non-right triangles we need an equation that will help us find the angles and sides. We can't use the Pythagorean theorem because that only applies to right triangles. So observing a non-right triangle, we can try and find an equation to help us. First of all, let's try and make that non-right triangle into two right triangles by taking angle B and putting a straight line through it.

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

Above, we can see how the line that cuts in between angle B has created two right triangles. Also, the line is now being called h. Now, we use Sine to find and equation with h in it.
In order to do so, we use Sin A=h/c and for the other side we use Sin C=h/a. Then we cross multiply to get cSinA=h and aSinc=h. Now, here's where it gets a little tricky.

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

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

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

In the above picture, we can see how we already did the first two steps. Now to put it simply, all we had to do was equal both equations to each other since they both equal h. Then we divide both by the side by the side that it has. So we divide by ac. Then one we do that we get the equation SinA/a=SinC/c. Now this is the equation for sines. Since we have this, we can safely assume that SinB/b also equals the others. And if you are skeptical, we can do the exact same thing but instead just cut a straight line through angle A

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

We did the same thing we did at the beginning but the line cuts through angle A. Now we solve the exact same way but for angle B. We did the same thing by cutting a straight line but only to a different angle. Then let's call it k. It could be h but we already used it.

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

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

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

 we solved the exact same way but for angle B and got the same and similar answer. So now we know that not only do they all equal each other but also that this works for all triangles

4. Now let's try and find the area. As we know, for right triangles, the area equation is A=1/2bh. Now of course' this is for right triangles. Now, to find one for non-right triangles, we simply replace h with

aSinC. Now you're probably wondering how that is so. It's simple. Remember when we were finding the origins of the law of Sines each one was equal to h? Well instead of using the equation area for right triangles we simply replace the h with the sin that we got when fixing the Law of Sine.

(http://pcb2s11.blogspot.com/2011/06/chapter-6-summary.html)

References:
http://pcb2s11.blogspot.com/2011/06/chapter-6-summary.htmlhttp://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

WPP#12: Unit O Concept 10

http://cdn.snowboardermag.com/files/2009/10/IMG_4157.jpg

The Problem:Lauren has a week off from work and decides to go snowboarding at the mountains and stay at a resort. When she arrives, she is admiring the snow white mountain at ground level at the snowboarding resort "Mountain High." 

a) Lauren measures the angle of elevation to the top of the mountain to be 60*. If, at this point, she is 60 feet away from the mountain, what is the height of the mountain?

b) Lauren reaches the top of the mountain and see's her friends waving at her and encouraging her to ride down the mountain. She now see's the flag where she is supposed to snowboard down and she measures the angle of elevation to the flag to be 20*. How long is the path to the flag? [keep in mind to use the height of the mountain]

The Solution: a) Draw out the mountain and draw a person some distance away from it. Then draw a line from the top of the mountain to the persons feet and then draw a line connecting the person to the bottom of the mountain. That line will be 60. The angle we just made will be 60*. Now label the height of the mountain x and that is the one we will be solving for. Now we are going to be using tangent to find the height of the mountain. The equation should be tan(60)=x/60. multiply both sides by 60 and the answer should be x=103.9 ft. 

b) Draw the mountain again but this time put a person on top of the mountain and a flag some distance from the mountain. Connect the flag to the person and that will be our x. Make a flat line across the top of the mountain and then connect the flag to that line. That will create a height which will be the height of the mountain, being 103.9. Now, using sin, we will find the distance from the person to the flag. The equation should be sin(20)=103.9/x. Multiply both sides by x and then divide both sides by sin(20). The answer should be 303.8 ft. 

*anything highlighted are important to solving the problem*

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

Inquiry Activity Summary

[Property of Kelsea]

1. A 30-60-90 triangle is derived from an equilateral which is cut in half. An equilateral is a triangle where all sides are equal as well as all angles. All the angles in the equilateral triangle are 60* and all the sides are 1. When we cut the triangle in half, we get different angles and sides. half of the triangle is now made up of angles which are 90*, 60* and 30*. The 60* is the one that was not affected by the cut. The 30* is the one at the top where the angle was cut in half, which makes sense since half of 60* is 30*. The 90* comes from the bottom where the side is perfectly perpendicular to the bottom side. The sides are now different as well. The hypotenuse, the slanted on, is not affected by the cut so is remains 1. The bottom one, which is cut in half becomes 1/2. However, there is no given value for the side that has been cut. In order to find that side, we use the Pythagorean Theorem [a^2 + b^2 = c^2]. The hypotenuse is c, the bottom side is a, and w solve for b. Once we plug it all in and solve for b, the answer should be b=√3/2. Now the sides should all be as follows: a=1/2, b=√3/2, and c=1. Now, since nobody is very fond of fractions, we can multiply all the sides by 2 to eliminate the fractions. Now the sides should be a=1, b=√3, and c=2. Now we are almost done! Now we must add an n to each side number because although the angles will remain, the sides are subject to change. That means n can be any number and the angles will remain the same. So finally, one last time, lets list what the angles are: a= n, b=n√3 [n should not be behind it], and c=2n. Our 30-60-90 triangle is now derived and conquered!

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

2. A 45-45-90 triangle is derived when a square is cut diagonally. A square is a shape where all four sides are equal [in this case we will set them equal to 1] and all sides are 90*. When cut diagonally, the two angles change, and one side changes. Since it is cut diagonally, the two angles affected will become 45* which makes sense because half of 90 is 45. The other angle remains 90* since it was not affected by the cut and it is perfectly perpendicular to the bottom side. Two of the sides were not affected, the ones that are not slanted, so they remain 1. However, one was, the hypotenuse. In order to find the hypotenuse, we must once again use the Pythagorean Theorem [a^2 + b^2 = c^2] to find the hypotenuse. Hypotenuse will be c, the bottom side will be a, and the other side will be b. Now we plug our numbers into the equation and solve for c. The answer should b √2. Now all our angles should be as follows: a=1, b=1, c=√2. Yay! Wait ... we are forgetting something that we did in the 30-60-90 triangle. What is it? Oh Right! Our lovely buddy n. Add n to all the sides because, as stated before, the angles can remain the same but the sides are subject to change! So now, our sides should be a=n, b=n [It is 1n but since any number multiplied by 1 is that number, we do not need the 1], and c=n√2 [remember, no n behind the radical!]. Congratulations! The angles are now derived and we now know how to derive the 45-45-90 triangle! Good Job!

[Property of Kelsea]

[Property of Kelsea]

[Property of Kelsea]

Inquiry Activity Reflection

1. "Something I never noticed before about special right triangles is ..."  that the sides are found using the Pythagorean Theorem equation. [Hypotenuse for 45-45-90 and the b for the 30-60-90]
2. "Being able to derive these patterns myself aids in my learning because..." if I ever forget what the side is, I can derive it using the what I learned from this activity, which is to cut a square diagonally for a 45-45-90 or an equilateral triangle for a 30-60-90 and then use the Pythagorean Theorem to find the angle that is missing.

I/D #1: Unit N Concept 7: How Do SRT And UC Relate?

Inquiry Activity Summary

[Kelsea Del Campo; worksheet]

1. The 30* triangle is one of the special right triangles that is taught to us in Geometry. This means that each side has a special equation that goes with it. The hypotenuse is 2x, the horizontal angle is x, and the vertical angle is x radical 3. Now, we have to identify which is the x and y. The horizontal is x and the vertical is y. The hypotenuse is r, which refers to being a reference angle, the angle from the terminal side to the closest x axis. Now if we set the reference angle equal to one, we have to do that to all of the other sides by dividing by 2x. This means that we have to divide by 2x on the horizontal and vertical side. This means our hypotenuse will equal 1, our vertical angle equal to 1/2, and our horizontal angle equal to radical 3 over 2. Next I drew a coordinate plane where the triangle was in the first quadrant. Then I labeled the vertices into ordered pairs. The edge is (0,0) since it is at the middle of the graph. The one to its right is ( radical3/2, 0) because the x is radical3/2.  The one above it is (radical3/2, 1/2). The last ordered pair is a point on the unit circle and its angle is 30*. This can help us identify the ordered pair on the unit circle with a reference angle of 30*, depending on which quadrant one or both will be negative. 

[Kelsea Del Campo: worksheet]

2.  The 45* angle is also a special right triangle. The hypotenuse is x radical2, the vertical side is x, and the vertical side is also x. Now we also have to identify x and y. X will be the vertical side and y will be the horizontal side. The hypotenuse is, again, r. We have to set the hypotenuse equal to 1 again so we divide all three sides by x radical2. After doing so, our r should equal 1, our y should equal radical2/2, and our x should equal radical2/2. Next draw a coordinate plane with the triangle in the first quadrant. Label the verices as ordered pairs. The edge is (0,0), the one beside it is (radical2/2, 0), and the one above it is (radical2/2, radical2/2). The last ordered pair is a point on the unit circle with an angle of 45* and can help us find the ordered pairs for an angle with a reference angle of 45*, depending on which quadrant one or both will be negative. 

[Kelsea Del Campo: worksheet]

3. The 60* angle is very similar to the 30* except that everything is switched. This means that the vertical side is x radical3 but remains y. The horizontal is now x but remains the x. The hypotenuse remains the same. Set the hypotenuse equal to 1 by dividing by 2x and divide the rest by 2x as well. Now the hypotenuse should equal 1, the y should equal radical3/2, and the x should equal 1/2.Next, draw a coordinate plane where the triangle is on the first quadrant. Then identify the vertices as ordered pairs. The edge should be (0,0), the one beside it should be (1/2,0) and the one above it should be (1/2, radical3/2). As before, the last ordered pair is a point on the unit circle and is the same ordered pair for that with a reference angle of 60*, depending on which quadrant one or both will be negative.

4. This activity helps us derive from the unit circle because when we look at the unit circle and look at all reference angles of 30*, it has the same ordered pair, depending on the quadrant a negative will be present. Any angle with a reference angle with 45* has the same ordered pair, depending on the angle there will be a negative on the x or y, or both. Lastly, for any angle with a reference angle of 60*, it will have the same ordered pair, and depending on the quadrant, the x or y or both will be negative.

[Kelsea Del Campo: phone]

5. The triangle drawn in this activity all lie in the first quadrant. If we draw the triangle in the second quadrant, it will be a mirrored image of the first quadrant, the third quadrant will be a mirrored image of the second quadrant, and the fourth quadrant will be a mirrored image of the third quadrant. The closest angle to the x-axis is going to be an angle with a reference angle of 30*, the middle will that with a reference angle of 45*, and the one closest to the y-axis will e the one with the reference angle of 60*. The x's in the second quadrant will be negative, x and y will be negative in the third quadrant, and the y in the fourth quadrant will be negative. The 45* angle is shown in the second quadrant and is like a mirror image of the one in the first quadrant. It has the same ordered pair but its x is negative. The 30* angle is in the third quadrant and is upside down and the x and y of the ordered pair is negative. Lastly, the 60* angle us also upside down but only the y is negative.

[Kelsea Del Campo: phone]

[Kelsea Del Campo: phone]

Inquiry Activity Reflection

1. "The coolest thing I learned from this activity..." was how the special right triangle corresponds are are very similar in each quadrant. I didn't learn in Algebra II where these numbers came from so that was also interesting.

2. "This activity will help me in this unit because..." we are currently memorizing the unit circle to help us the sin, cos, and tan so it will help in finding the answer. 

3. "Something I never realized before about special right triangles and the unit circle is ..." that we get the ordered pairs from the special unit circle when we set the hypotenuse equal to 1 and that the triangle can be found in the unit circle itself. 

RWA1: Unit M: Concept 4-6: Conic Sections In Real Life

[http://www.youtube.com/watch?v=7mqUYWtRL5Y]

1. "The set of all points such that the sum of the distance from two points is a constant. " (Mrs. Kirch/SSS Packet)

2. The equation for this conic section, the ellipse, is (x-h)²/a²+(y-k)²/b²=1 or (x-h)²/b²+(y-k)²/a²=1. The center will always be the h and k. H will be paired with the x and the k will be paired with the y. The a will always be bigger than the b. Where the a is placed will determine whether the graph is skinny or fat. If the a² is under the (x-h)², it will be fat/wide. If the a² is under the (y-k)², it will be skinny/long. The equation will always be equal to 1 and the equation will always be added.The ellipse on a graph looks like an elongated circle. It can look fat/wide or it can look skinny/tall. 
The key features on the ellipse are the center, two vertices, two co-vertices, two foci, major axis, minor axis, and the position/look of the ellipse. The key features in the equation that will be needed to find the key features on the graph are the standard form, center, a, b, c, and the eccentricity. The easiest way to graph an ellipse is to obtain the standard form first. To find the equation, you complete the square. After doing so, divide everything by the number that is on the right side so the equation will be equal to one. After that, identify the circle by looking at both numerators. The h will be associated with the x and the k will be associated with the y. Since the numerators are  (x-h)² and (y-k)². That means the center is just (h, k). However, if your standard form has a (x+h)², then the h will be a negative; the same concept applies to the y and the center may be (-h, -k) depending on which numerator has addition. Then plot the center. Now go back to the equation and determine if the graph is going to be fat/wide or skinny/tall. The graph will be fat/wide if the a² is the denominator for the (x-h)². The graph will be skinny if the a² is the denominator for (y-k)². After identifying the shape of the graph, find your a and b. Since in the equation is is squared, you need to square root it to find the distance from the center to one endpoint of the major/minor axis and the other endpoint of the major/minor axis.Now that a and b have been found, to find the vertices and co-vertices (a is associated with vertices and b is a associated with co-vertices) we need to observe the shape of the graph again. If it is wide /fat, the y (meaning the y from the center) in the ordered pair for the vertices will stay the same and the x (meaning the x from the center) in the ordered pair so you have to add and subtract the y from the center to get the other number for the ordered pair. If it is wide/fat, the x in the co-vertices will remain the same so we have to add and subtract the y from the center to get the other numbers for the ordered pair.The same thing goes for skinny/tall but vice-versa meaning the a in the vertices stays the same and the y in the co-vertices stays the same. Plot the vertices and connect the two endpoint with a solid line and plot the co-vertices and connect the two endpoint with a dashed line, that way it will be easier to identify which line is which. Now just draw you ellipse and you are done! Just kidding! But you're almost there! Now we have to find c, which will help us in finding the foci. To find c, we use the equation a²-b²=c². After plugging in  the equation and finding the c, we need to find the ordered pair of it so we can graph it. Look at the vertices and whatever number stays, the x or the y, then the same will go for the foci since the foci will be on the major axis like the vertices. Now just add or subtract from the other number, x or y depending on the shape, and plot. The purpose of the foci is to know the how fat/wide or how skinny/tall it is since the sum of two distance from the foci is a constant. The farther from the center, the more elongated it will be and if closer to the center it will be less elongated. Now we need one more thing! The eccentricity! to find the eccentricity, we need to use the equation c/a. The number should be between 0 to 1. Since it is in between 0 and 1, then that means an ellipse deviates from being a circle a little bit. 

3. Conics of ellipses can be seen where very few men have gone: space. The rotation of the planets around the sun form an ellipse. Another space example would be Haley's Comet. It has a rotation that is shaped like an ellipse. An ellipse that isn't so out of this world is a glass of water. When tilted slightly, the surface of the water can be viewed to look like an ellipse. Ellipses can be found in common places as well: buildings, statues, car logos, and even more places! Can you find all of them?

4. Works Cited:
http://britton.disted.camosun.bc.ca/jbconics.htm
http://www.youtube.com/watch?v=7mqUYWtRL5Y
http://intmstat.com/plane-analytic-geometry/earth.jpg