I/D#1 Concept 7: Deriving Unit Circle

In this activity, we are able to derive the unit circle from where it came from. The unit circle is formed by the 30,60,90 degrees and the 45,45,90 degrees triangle. We derive this by using the special right triangle formula.

Inquiry Activity Summary

  The hypotenuse is always labeled r. The verticle value is y and horizontal x. The value for the hypotenuse is 2, x value is radical 3 and the y value is 1. To make the hypotenuse equals to , divide everything by 2.

This triangle is the same, however the angle is in a different position. The x,y,r is label the same with the values. However, the ordered pairs will be in different.

The special right triangle of the 45,45,90 will have a different value. The label of the x,y and r will be the same. To derive it, we need to find the value of each angle. Since 90 degrees value is radical 2 and we want it equal to 1, we must divide everything by radical 2. Since the other angle is being divided by radical 2 we must rationalize it because the bottom can't have a radical number.

This activity help me derive the unit circle because the unit circle is basically made up of the special right triangles including the 45,45,90 degrees and 30,60,90 degrees.

The value of these triangle changed when it is drawn in the second, third, and fourth quadrant by the negative and positive signs

In the second quadrant it is still in the first half, which is positive but it is in the left side. Therefore the x value will be negative while the y value is positive.

The third quadrant is in the bottom half. Because it is on the bottom and to the left, both the x and y value will be negative

Finally, the value of the x and y in the fourth quadrant will be positive and negative. The x is positive and the y is negative because you are starting from the right side of the unit circle and going down.

Conclusion

The coolest thing I learn from this activity is that the unit circle is derive from the special right triangle

This activity will help me in this unit because I did not have to memorize a lot of thugs because I can work out the problems.

Something that I never realize about the special right triangle and the unit circle is that they are both related to each other in many ways

RWA#1: Unit M Concept 4-6 Conic Sections in real life

Hyperbola 

Definition: a set of all point such that the difference in the distance from two point is constant.  

Algebraically: The equation is (x-h)^2 over a^2 minus (y-k)^2 over b^2 equals to 1.You can tell if the hyperbola is open left/right if x comes first. It will open up/down if y comes first in the equation. Also, the needs to be in standard form to make it easier to classify the hyperbola. The equation needs to be subtracted for it to be a hyperbola. Otherwise, it will be an ellipse.

Graphically: By looking at the hyperbola the x and the y affects how the graph will shape. It will shape the graph left/right, up/down. There will be a focus on one side and the vertex on the other side of the graph. The "C" will go through the focus while the transverse axis connect the two vertices together.The "a" and the "b" will affect where the two vertices and co vertices will be placed/ how much it will move on the graph.

Key: To start the equation needs to be in standard from which is (x-h)^2 over a^2 minus (y-k)^2 over b^2 equals to 1. The center could be find by taking the opposite of h and k from the standard form. From there, you can determine if it is up/down if y comes first and left/right if x comes first. To find "a" you take the square root of a^2. Similarly, you can find "b" by taking the square roots of b^2.  Plot the center and count number of "a" from the center and "b" from the center to find the vertices and co vertices. The numbers that dominates in the ordered pairs will be the transverse axis and conjugate axis.You can use a^2+b^2=C^2 to find c. As for the eccentricity, you take C divided by A to get the answer. The two foci will be (x,y+-eccentricity). "The 'foci' of an hyperbola are 'inside of each branch and each focuses is located some fixed distance c from the center."  (credit) Finally, to find the asymptote, you find the m slope by taking a/b or b/a depending on the direction of the branches. After finding the slope, plug in to y=mx+b .

Real World Application:

Hyperbola can be applied to the real world. For example, it could be the shape of your lamp, where it will cast a hyperbolic shape on your wall. It can also be found in many things such as Dulles Airports, sonic boom, cooling tower of nuclear reactors, stone in lake, etc. 

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