Thursday, February 27, 2014
Saturday, February 22, 2014
I/D#1: Unit N Concept 7:The Unit Circle and the Magic Five
Inquiry Activity Summary:
Angles on the first quadrant
30 degree angle
Figuring out the 30 degree angle triangle was very simple, though at first it was a bit difficult. The reason being because I didn't know the measurements of each side. But once I logged onto legoogle.com, I was able to find all the sides. Side R = 2x, side Y = x and side X = xradical3 (as seen in the picture.) Our hypotenuse/sideR had to equal one, therefore, we divided out 2x from the 2x and we got side R to equal 1. Simple right, are you done yet? Nope. In order to divide out the 2x from sideR we had to do that to all the sides. SideX when we divided and simplified it gave us radical3 over two. For sideY we did the same and the x's cancelled therefore leaving us with 1/2.
The second part of the 30 degree angle was finding the points as if it were on a graph. Taking that initiative we drew out the x and y axis. Our first point on the graph was the center point and that was simply just (0,0). The second point on the the x-axis, on the right, would turn out to be (radical3/2, 0) because of the information we had solved earlier. According to rise over run, our last point would have to be (radical3/2, 1/2). Why? Because we went over radical3/2 and rose 1/2. Lastly we filled in the blanks for r, x, and y.
45 degree angle
Figuring out the 45 degree angle was much simpler than the first angle we had to figure out because we already had an idea of how to solve it out. According to our sources, google.com, sideR was xradical2, sideX and sideY were both x. Just as we did for angle 30, sideR had to equal 1. Therefore, we divided out xradical2 from xradical2 to get one. We divided out xradical2 from all the sides. When dividing out xradical2 from x we ended up with 1/radical2, of course we all know, that no radical can be in the denominator, so we multiplied radical2 to both the bottom and top. Our final result was radical2/2. From there we drew the x and y axis and figured out the points, resulting with what is above in the picture. Lastly, we filled in the r, x and y blanks.
60 degree angle
While we all solved this angle we slowly realized that everything was the exact same as the 30 degree angle except everything was reversed. SideR was the same. But the X for the 60 degree angle was 1/2 instead of being radical3/2 and sideY was radical3/2 instead of 1/2. As for the points as well! Instead of going through all the work, since we already knew how to do it from the previous 30 degree angle, we just reversed everything and finished our 60 degree angle in seconds.
How did the activity help us derive the Unit Circle?
Well, glad you asked. These three angles are key to finding the rest of the Unit Circle. All we need to know is each and every angle (referring to 30, 45, 60 degree angle) and we can find the rest of the unit circle.
http://fac-web.spsu.edu/math/edwards/1113/unitcircle.htm
As seen from the Unit Circle above, the only points that are repeatedly seen are 1/2, radical3/2 and radical2/2. That's because all angles have the same common thing, they all have reference angles of 30, 45, and 60. 30 and 150 reflect off one another and have the same point except for the negative x, but that makes sense since it is on the negative side of the x-axis. This method goes to the rest of the angles as well they all reflect one another and tie back to either a 30, 45 or 60 degree angle. The only thing that changes throughout the Unit Circle is the degrees and radians but that's simply because as we go around the circle the angle gets bigger and bigger.
Quadrant II (30 degree angle)
This angle above in fact is the 150 degree angle. But it has the reference angle of 30 degrees. The only thing that really changes compared to the original 30 degree angle is that the 150 degree angle has a negative radical3/2 and of course the degree and radian are different as well.
Quadrant III (45 degree angle)
The angle above is actually 225 degrees. This angle has a reference angle of 45 degrees. The only thing that really changes compared to the original 45 degree angle is that the 225 degree angle's the x and y points are negative. To conclude, the degree and radian are one of the other factors that are different.
Quadrant IV (60 degree angle)
The angle above is actually 300 degrees. This angle has a reference angle of 60 degrees. The only thing that really changes compared to the original 60 degree angle is that the 300 degree angle's x and y points are negative. Lastly, the degree and radian are one of the other factors that are different.
Inquiry Activity Reflection:
The coolest thing I learned from this activity was that I finally understood how to solve the Unit Circle! Last year, in Algebra II, when we were learning this, I was so lost. I wanted to love the Unit Circle but the Unit Circle didn't love me. It was a very depressing time for me. But now me and the Unit Circle are on the same terms. I get it and the circle gets me. It's a happy world isn't it? Yes, very.
Monday, February 10, 2014
RWA#1-Unit M: Ellipses
Ellipses
Definition: The set of all points such that the sum of the distance from two points is a constant. (http://www.lessonpaths.com/learn/i/unit-m-conic-section-applets/ellipse-drawn-from-definition-geogebra-dynamic-worksheet)
Definition: The set of all points such that the sum of the distance from two points is a constant. (http://www.lessonpaths.com/learn/i/unit-m-conic-section-applets/ellipse-drawn-from-definition-geogebra-dynamic-worksheet)
Algebraically:
Ellipses can be algebraically seen as the pictures above. An ellipse can be seen in two ways, "fat" and "skinny". How do we determine which is which? Well if the larger denominator, a^2, is under the x it will be a horizontal ellipse or "fat" ellipse. If the larger denominator, a^2, is under the x it will be a vertical ellipse or "skinny" ellipse. "A" give us the indication of the length of the major axis and the "b" gives us the indication of the length of the minor axis. With our "h" and "k" we can find our center. Do keep in mind that the "h" goes with "x" and "k" will always go with "y". With our equation we can find about everything needed to graph our ellipse. To find our a, b and c, we use "a^2-b^2=c^2".
By looking at the equation, we can find the "a" and "b". "A" pertains to our vertices, "b" to our co-vertices and "c" to our foci. Depending whether its horizontal or vertical, our vertices and foci have something in common, either the "x" or "y" for both vertices and foci will be alike. In finding our vertices if the bigger denominator(a^2) is under the y (vertical/"skinny" ellipse), our vertices and foci would have the similar x's. So what would we do with "a"? You add and subtract that to the "y" of the center two get your two vertices. This would be the same process for the "c" and foci. Finding our co-vertices would mean that the "y" would stay the same for both co-vertices. So in this case we would add and subtract "b" to our center to find our two co-vertices. This whole process would be the same for horizontal/"fat" ellipses except vice-versa.
By looking at the equation, we can find the "a" and "b". "A" pertains to our vertices, "b" to our co-vertices and "c" to our foci. Depending whether its horizontal or vertical, our vertices and foci have something in common, either the "x" or "y" for both vertices and foci will be alike. In finding our vertices if the bigger denominator(a^2) is under the y (vertical/"skinny" ellipse), our vertices and foci would have the similar x's. So what would we do with "a"? You add and subtract that to the "y" of the center two get your two vertices. This would be the same process for the "c" and foci. Finding our co-vertices would mean that the "y" would stay the same for both co-vertices. So in this case we would add and subtract "b" to our center to find our two co-vertices. This whole process would be the same for horizontal/"fat" ellipses except vice-versa.
Graphically:
Graphically, ellipses look like the picture above. The vertical/"skinny" ellipses consists of the same parts as the horizontal/"fat" ellipses. The center of our ellipse is found in our equation, as mentioned above. With our "h" and "k" we can find our center. Do keep in mind that the "h" goes with "x" and "k" will always go with "y". Our graph consists of a center, vertices, co-vertices, foci, a minor axis, a major axis and eccentricity. To determine the major axis and minor axis would only mean looking at the vertices and co-vertices. Looking at the vertices, whichever, x or y, is similar that would be the major axis. For the co-vertices, whichever, x or y, is similar that would be the minor axis. The only factor that has not been explained is eccentricity. Eccentricity is "c" divided by "a". This determines how much the ellipse deviates. If it's closer to one, it means that it looks more like a circle and vice versa if its closer to zero (0 < e < 1). Eccentricity is not necessarily seen on the graph but it helps the student to get an estimate of how the graph might look like when they draw it.
Here's more help with ellipses:
Real World Applications (RWA)-
Ellipses can be represented in the orbit of the planets in outer space. How do you say so? Looking at the picture demonstrates it perfectly and why. The sun would always serve as one of the foci. The other focus would be empty but still be there (http://oneminuteastronomer.com/8626/keplers-laws/). As described in the Kepler Law, "their orbits are similar ellipses with the commonbarycenter being one of the foci of each ellipse. The other focus of either ellipse has no known physical significance. Interestingly, the orbit of either body in the reference frame of the other is also an ellipse, with the other body at one focus."(http://en.wikipedia.org/wiki/Ellipse#Ellipses_in_optimization_theory)
The things remaining around the ellipse have no physical "significance" but are still there. The basic real world of application of the ellipse is centered around the foci. One thing that is obvious from the picture above is the foci and the planet. It goes along with the concept of d1+d2= a constant. Which is what ellipses our all about! The picture above perfectly describes why an ellipse can be applied in the real world with the orbit of planets from outer-space.
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