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Wednesday, March 5, 2014
Tuesday, March 4, 2014
1/D: #2 Unit O Concept 7-8: Derive the SRTs
Inquiry Summary Activity:
To derive the Special Right Triangles, I started with a square with the side lengths of one and a equilateral triangle with the side lengths of 1. Theses two shapes help us derive the 45-45-90 Special Triangle and the 30-60-90 Special Triangle.The Pythagorean Theorem was also used to find missing parts of the triangles.
1) 30-60-90 Triangle:
To derive the 30-60-90 triangle, I used a equilateral triangle with the side lengths of 1. An equilateral triangle has angles that are 60 degrees and all 3 angles are the same. Since a equilateral triangle has the same angles and the same sides, I sliced it down the middle. By slicing it down the middle a height and a 90 degree angle were created. A 30 degree angle was also created by splitting the 60 degree angle in half. Since the triangle was sliced now the base is 1/2. Since we know that on side is 1 and the base is now 1/2, using the Pythagorean Theorem we can figure what the height is. We use the a^2+b^2=c^2. a=One squared is a=1 and b=1/2 squared is b= 1/4. You add them and that gives radical 3 over 2.We multiply everything by 2 to get radical 3, 2 and 1. This translates into the normal n radical3, 2n, and n. The n is used as variable meaning that any number can be substituted in. N is used to expand the problem as needed.
2) 45-45-90 Triangle:
To derive the 45-45-90 triangle, I used a square with the sides of 1. The 4 angles of a square are 90 degrees.I sliced the square down its diagonal. By cutting it down its diagonal created a hypotenuse and two 45 degree angles. Since we know that two sides are 1 we must now find the hypotenuse of the triangle created. I used the Pythagorean Theorem to find the hypotenuse. The equation is a^2+b^2=c^2. A=1 and B=1, which means c stays the same. 1 squared is 1 so that means a+b equals one. To get c we must get rid of the square root by squaring c and 2. That means c equals radical 2. The sides of the triangle are 1,1, and radical 2. That translates to the original pattern of a 45-45-90 triangle which is: n,n, n radical 2. The N is used a variable, which means any number can be substituted and that means that N is also used to expand the problem as needed.
Inquiry Activity Reflection:
1) Something I never noticed before about special right triangles is that they were created form other shapes.
2)Being able to derive these patterns myself aids my learning because if I need to do a problem that involves this and the triangles are not given I can do it on my own.
To derive the Special Right Triangles, I started with a square with the side lengths of one and a equilateral triangle with the side lengths of 1. Theses two shapes help us derive the 45-45-90 Special Triangle and the 30-60-90 Special Triangle.The Pythagorean Theorem was also used to find missing parts of the triangles.
1) 30-60-90 Triangle:
To derive the 30-60-90 triangle, I used a equilateral triangle with the side lengths of 1. An equilateral triangle has angles that are 60 degrees and all 3 angles are the same. Since a equilateral triangle has the same angles and the same sides, I sliced it down the middle. By slicing it down the middle a height and a 90 degree angle were created. A 30 degree angle was also created by splitting the 60 degree angle in half. Since the triangle was sliced now the base is 1/2. Since we know that on side is 1 and the base is now 1/2, using the Pythagorean Theorem we can figure what the height is. We use the a^2+b^2=c^2. a=One squared is a=1 and b=1/2 squared is b= 1/4. You add them and that gives radical 3 over 2.We multiply everything by 2 to get radical 3, 2 and 1. This translates into the normal n radical3, 2n, and n. The n is used as variable meaning that any number can be substituted in. N is used to expand the problem as needed.
2) 45-45-90 Triangle:
To derive the 45-45-90 triangle, I used a square with the sides of 1. The 4 angles of a square are 90 degrees.I sliced the square down its diagonal. By cutting it down its diagonal created a hypotenuse and two 45 degree angles. Since we know that two sides are 1 we must now find the hypotenuse of the triangle created. I used the Pythagorean Theorem to find the hypotenuse. The equation is a^2+b^2=c^2. A=1 and B=1, which means c stays the same. 1 squared is 1 so that means a+b equals one. To get c we must get rid of the square root by squaring c and 2. That means c equals radical 2. The sides of the triangle are 1,1, and radical 2. That translates to the original pattern of a 45-45-90 triangle which is: n,n, n radical 2. The N is used a variable, which means any number can be substituted and that means that N is also used to expand the problem as needed.
Inquiry Activity Reflection:
1) Something I never noticed before about special right triangles is that they were created form other shapes.
2)Being able to derive these patterns myself aids my learning because if I need to do a problem that involves this and the triangles are not given I can do it on my own.
Friday, February 21, 2014
I/D#1:Unit N Concept 7: Derive the Unit Circle Activity
Inquiry Activity Summary:
In this activity we were given 3 triangles which had the measurements of: 30, 60,90, 45,45,90 and 60,30, and 90 degrees. These measurements are of Special Right Triangles. We had to label each triangle according to the rules of Special Triangles. Which are:
The hypotenuse of each triangle had to equal 1. The first triangle was a 30 degree which meant that the sides where labeled as 2x, x radical 3, and x. To make the hypotenuse 1, we had to dive all sides by the hypotenuse and simplify. After that we labeled the hypotenuse "r", horizontal value"x", and vertical value "y". The next step is to draw a coordinate(this has to be done to very triangle given) with the origin at the labeled measure, which for the first one was 30 degrees. The vertices had to be labeled as ordered pairs for each triangle. For the first triangle, which is 30 degrees, The hypotenuse equal 2x(r), and the sides were x(vertical value), and x radical 3(horizontal value). To simplify I had to divide all sides by 2x, which gave me 1 for the r value, radical 3 over 2 for x, and 1/2 for y. I later drew the coordinate and labeled the order pairs which were (0,0), (radical 3/2,0), and (radical 3/2, 1/2). That is how the ordered pairs that are in the Unit Circle came to be for any reference angle of 30. The pairs are the same for any reference angle of 30.
1)30 Degree Triangle:
2) 45 Degree Triangle:
For the 45 degree angle, the hypotenuse was x radical 2(R), vertical side was x(x), and horizontal side was x(Y). For this triangle we had to divide by radical 2, which gave R=1, X= radical 2/2, and Y= radical2/2. After drawing the coordinate plane, the vertices were: (0,0), (radical 2 over 2, 0), and (radical 2 over 2, radical 2 over 2). That is how the ordered pairs for the quadrants in the unit circle came from, for angles that were reference angles of 45 degrees.
3) 60 Degree Triangle:
For the 60 degree triangle, the hypotenuse was 2x(R), horizontal side was x(x), and vertical side was x radical 3(y). We had to divide by 2x which gave r=1, x=1/2, and y= radical 3 over 2. After drawing the coordinate plane, the vertices were (0,0), (1/2,0), and (1/2, radical 3 over 2). That is how the pairs for any reference angle of 60 came to be. For any reference angle of 60 in the unit circle the pairs will be the same.
4) This activity helps us obtain the unit circle because the ordered pairs that we got for the tree triangles are the same in each quadrant of the unit circle, which means that if you know the pairs for the 30, 45, and 60 degree angles you will know the complete unit circle. As each of the tree angles have reference angles in each quadrant.
5)The triangles drawn lie in the first quadrant, which makes the ordered pairs positive. If the triangles were drawn in different quadrants the pairs would change to negative depending in what quadrant they are. After re-drawing the triangles, for the 30 degree triangle in the second quadrant, the x values of the ordered pair became negative. For the 45 degree triangle the x and y values both became negative in the third quadrant. For the 60 degree triangle the y values became negative when drawn in the fourth quadrant.
Inquiry Activity Reflection:
1. "The coolest thing I learned from this activity was" where the unit circle came from.
2. "This activity will help me in this unit because" it will help be get reference angles faster and the ordered pairs.
3. "Something I never realized about special right triangles and the unit circle was" that they had so much in common or that the triangles were used to make the unit circle.
In this activity we were given 3 triangles which had the measurements of: 30, 60,90, 45,45,90 and 60,30, and 90 degrees. These measurements are of Special Right Triangles. We had to label each triangle according to the rules of Special Triangles. Which are:
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| http://www.math.hmc.edu/calculus/tutorials/reviewtriglogexp/Add caption |
The hypotenuse of each triangle had to equal 1. The first triangle was a 30 degree which meant that the sides where labeled as 2x, x radical 3, and x. To make the hypotenuse 1, we had to dive all sides by the hypotenuse and simplify. After that we labeled the hypotenuse "r", horizontal value"x", and vertical value "y". The next step is to draw a coordinate(this has to be done to very triangle given) with the origin at the labeled measure, which for the first one was 30 degrees. The vertices had to be labeled as ordered pairs for each triangle. For the first triangle, which is 30 degrees, The hypotenuse equal 2x(r), and the sides were x(vertical value), and x radical 3(horizontal value). To simplify I had to divide all sides by 2x, which gave me 1 for the r value, radical 3 over 2 for x, and 1/2 for y. I later drew the coordinate and labeled the order pairs which were (0,0), (radical 3/2,0), and (radical 3/2, 1/2). That is how the ordered pairs that are in the Unit Circle came to be for any reference angle of 30. The pairs are the same for any reference angle of 30.
1)30 Degree Triangle:
2) 45 Degree Triangle:
For the 45 degree angle, the hypotenuse was x radical 2(R), vertical side was x(x), and horizontal side was x(Y). For this triangle we had to divide by radical 2, which gave R=1, X= radical 2/2, and Y= radical2/2. After drawing the coordinate plane, the vertices were: (0,0), (radical 2 over 2, 0), and (radical 2 over 2, radical 2 over 2). That is how the ordered pairs for the quadrants in the unit circle came from, for angles that were reference angles of 45 degrees.
For the 60 degree triangle, the hypotenuse was 2x(R), horizontal side was x(x), and vertical side was x radical 3(y). We had to divide by 2x which gave r=1, x=1/2, and y= radical 3 over 2. After drawing the coordinate plane, the vertices were (0,0), (1/2,0), and (1/2, radical 3 over 2). That is how the pairs for any reference angle of 60 came to be. For any reference angle of 60 in the unit circle the pairs will be the same.
5)The triangles drawn lie in the first quadrant, which makes the ordered pairs positive. If the triangles were drawn in different quadrants the pairs would change to negative depending in what quadrant they are. After re-drawing the triangles, for the 30 degree triangle in the second quadrant, the x values of the ordered pair became negative. For the 45 degree triangle the x and y values both became negative in the third quadrant. For the 60 degree triangle the y values became negative when drawn in the fourth quadrant.
Inquiry Activity Reflection:
1. "The coolest thing I learned from this activity was" where the unit circle came from.
2. "This activity will help me in this unit because" it will help be get reference angles faster and the ordered pairs.
3. "Something I never realized about special right triangles and the unit circle was" that they had so much in common or that the triangles were used to make the unit circle.
Monday, February 10, 2014
RWA#1: Unit M Concept 5: Graphing ellipses given equation
Section 1:
- The set off all points, such that the sum of the distance known as the foci, is constant.
- Equation: (x-h)^2 / a^2 + (y-k)^2 / b^2=1 or (x-h)^2 / b^2 + (y-k)^2 / a^2=1 and a^2-b^2=c^2
- The key points of an ellipse are: the center, a, b, c, 2 vertices, the major axis, 2 co-vertices, the minor axis, 2 foci, and the eccentricity. To find A and B is the standard form is given A will always be the bigger number. Depending if x or y come first the graph will be either skinny or fat. If Its skinny the x value will not change from the center. to get the vertices, if its skinny the x values wont change and you will need to add and subtract whatever number a is to find the y values. For the co-vertices the y will be the same as the center and subtract and add whatever number b is to find the x values. The major axis will depend on if its skinny or fat, if its skinny major axis will be x= whatever number x is for the center and the minor axis will equal whatever y is in the center. For a fat graph the numbers will change y will become major and x minor. To find the foci you need to know c and depending if its skinny or fat the x or y values will change from the numbers that make up the center. To find C you use a^2-b^2 =c^2. The eccentricity will be C over A.
Section 3:
A real world application that displays ellipses are earrings. Earrings can be worn on the ears and can be different material and color. Earrings can be in the form of ellipses.Most women over the world use earring. This real world application contains different sizes for ellipses.
Section 4:
http://www.mathamazement.com/Lessons/Pre-Calculus/09_Conic-Sections-and-Analytic-Geometry/ellipse.html
http://official-stardollfashion.blogspot.com/2009/06/8-hottest-trends-for-summer-2009.html
Sunday, January 12, 2014
Tuesday, December 17, 2013
WPPP#9: Unit L Concept 4-8- Calculating combination and permutations
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Sunday, December 8, 2013
SP#6: Unit K Concept 10- Writing with repeating as a rational number
To write a repeated decimal as a rational number the geometric series can be used. This number is infinite. You need to find a sub 1 and r.
Look for things like matching the decimal with its fraction. You must leave it as a fraction and not as a decimal.
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