Part One: |
Proving the Pythagorean Theorem:
We started this project by proving the Pythagorean theorem. To prove this we worked on a worksheet called "Proof by Rugs". In this worksheet we worked with two squares broken up into different right triangles and squares. We used these triangles and our knowledge of angles to prove that squares, A and B combined were the same size as triangle C. We were trying to prove that both rugs had the same amount of shaded space by using the surrounding triangles. Thus proving that a^2+b^2=c^2. |
Using the Pythagorean Theorem to derive the distance formula:
Using the Pythagorean Theorem, we derived the distance formula. The distance formula is used to find the distance between two points on a coordinate plane. We put a right triangle onto a coordinate plane and translated the Pythagorean Theorem to work with the triangle on the coordinate pane. Since we were trying to find distance we couldn't used a,b or c because we didn't know their values. We instead used what we did know, which was the coordinate pints of the triangle. Then we used the coordinate points and plugged them into a new equation. We knew the distance between x2 and x1 was x2-x1, this would give us A in the Pythagorean Theorem. We knew the distance between y2 and y1 was y2-y1, this would be B in the Pythagorean Theorem. Then we used D to be the length of our hypotenuse. |
Using the Distance formula to derive the equation of a circle centered at the origin of a Cartesian coordinate plane:
From the distance formula we moved to the equation of a circle centered at the origin of a Cartesian coordinate plane. We did this by placing a circle onto a coordinate plane, we then made a line to represent the radius or r. We dropped a perpindicular from the end of the radius to the x-axis and now we were back to using a triangle. Again we used our knowledge of the Pythagorean Theorem to find the equation. We knew or coordinate pints, x and y and we needed to find r. If we looked at the in term of a triangle we would see that r is our hypotenuse, meaning that x^2+y^2 would equal r^2. This is because x is our distance from the origin to x and y is our distance from the x-axis to y. These are just the a and b in the Pythagorean theorem. The reason we need this equation, x^2+y^2=r^2 is because it is working with coordinate points and circle, rather than length and triangles. This equation is used to find points in a unit circle. |
Defining a Unit Circle:
A unit circle is simply a circle with a radius of one. We know all circles are similar, since we can center them at an origin and dilate them to become the same size. Because we know this, we can use the unit circle to be a basis for all other circles, simply scaling up or down when needed. Using the unit circle, we can also say that any right triangle based off of it will be similar, given that they share angles. |
Finding a point on the unit circle:
To find a point on a unit circle we can use the same steps we used to find our equation of a circle. If we are solving for a 30 degree unit circle the first thing we do is we drop a perpendicular from the end of our radius to the x-axis. This gives us a right triangle for which we will fill the remaining angles. We know that a triangle must add up to 180, so we know we have a 90 degree angle and 30 degree angle, leaving us with the last angle being 60 degrees. We now reflect this triangle over the x-axis which will give us an equilateral triangle. We know it is an equilateral triangle because all of the angles will now be 60 degrees. Since we are working with a unit circle we know our radius is 1 meaning one length of the triangle is 1. Since we are working with an equilateral triangle, we know all of the sides must be 1. Now we want to divide the entire triangle in half, or un-reflect it. Now we can plug in our knowledge into the equation of a circle, 1/2^2+y^2=1. Solve for y, which would be 3/4. Reduced this would give us the side lengths , square root of 3 over 2 and 1/2. |
Using symmetry to find remaining points on a unit circle:
We can use symmetry to find remaining points on a unit circle by reflecting. If we have the point (3,3) and we want to find the remaining points, we would reflect over the x-axis, which would give us, (3,-3). Then reflect that over the y-axis to get (-3,-3). Then reflect back over the x-axis which would give us (-3,3).
We can use symmetry to find remaining points on a unit circle by reflecting. If we have the point (3,3) and we want to find the remaining points, we would reflect over the x-axis, which would give us, (3,-3). Then reflect that over the y-axis to get (-3,-3). Then reflect back over the x-axis which would give us (-3,3).
Using the unit circle to define sin and cosine (of angle theta):
We used the unit circle to define sine and cosine next. If we made another right triangle using our radius on the unit circle as the hypotenuse and dropping a perpendicular, we can define the cosine and sine of theta. The cosine of theta is is simply the adjacent side of the triangle from theta over the hypotenuse or the radius. This equation would be cos(theta)=a/h. This equation would can used to find a x coordinate on a unit circle. The sin of theta is the opposite side of the triangle from theta over the radius or hypotenuse. This equation would be sin(theta)=o/h. This equation can be used to find a y coordinate on a unit circle. |
Using the Mount Everest problem to discover the "Law of Sines" and deriving the "Law of Sines":
The next thing we did was approach a problem called the Mount Everest problem. In this problem we were asked to find the lengths of the triangle, this was the first problem we had that wasn't involving a right triangle. We approached this problem by dropping a perpendicular from point c, creating two right triangles. By working through this problem we were able to derive the law of sines. The law of sines is, sin(A)/a=sin(B)/b=sin(C)/c. Unlike SOH,CAH,TOA, all you need for this equation is two angles and one side length. |
Deriving the "Law of Cosines":
The last thing we did was derive the "Law of Cosines". We used the help of our Distance formula to find this equation. To use the Law of Cosines, you need two side lengths and one angle. The Law of Cosines is c^2=a^2+b^2-2abcos(theta).
The last thing we did was derive the "Law of Cosines". We used the help of our Distance formula to find this equation. To use the Law of Cosines, you need two side lengths and one angle. The Law of Cosines is c^2=a^2+b^2-2abcos(theta).
Part Two:
The second part of this project was measuring an object and finding the volume of it. We also had to use an area function and a trigonometric function in this part of the project.
My group and I decided that we wanted to measure a watermelon. We chose a watermelon because we thought it would be an easy object to cut into different shapes and if we chose not to cut it, we would be finding the volume of a shape we haven't worked with before, an ellipsoid. After we bought our watermelon we realized it sort of looked like a mix between an ellipsoid and a sphere. So, we decided to find the volume of it as if it was an ellipsoid and as if it was a sphere. We also found the watermelon's surface as an extra challenge, as we have never worked with the surface area of an ellipsoid. The first thing we did is get measurements of the watermelon. In order to do this we found what we needed in order to find the volume. We needed the radius, half the width and half of the height. After we found these measurements we started plugging our numbers into the equations. We ended up getting a total of 113.06 cubic inches for the volume of the watermelon using the volume of a sphere equation. We found that the volume of our watermelon using the volume of a sphere equation was 110.27 cubic inches. We integrated area equations by finding the area of the base of the cross section of a sphere and ellipsoid. The area of the circle was 28.27 square inches, and the area of the ellipse was 27.57 square inches. We integrated trigonometric functions by creating three right triangles using the measurements we got for our ellipsoid. The triangles were triangles ABx, ACy, and BCz. We used the Pythagorean theorem to find our missing side lengths. Then after getting these side lengths we found all of the missing angles. We did this by using the inverse cosine function. As an extra challenge I decided to find the surface area of the ellipsoid. This was a challenge because it was a new equation to me. Not only was the equation new to me, but some of the function inside of it were new. For example before attempting this equation I did not know how to approach decimal exponents, but by trying this equation I was able to learn how to solve decimal exponents. The surface area of the ellipsoid was 109.2 square inches. Reflection: I think that this project went very well. My group and I were able to very well together. My biggest challenge was attempting the surface area problem because of it's complexity. It is a very dense equation and I had to figure out how to do decimal exponents. Even though that was a challenge, I think it was really good that I attempted it because it taught me how to do something new and that's what we were supposed to do. One habit of a mathematician that I used was "Stay Confident and Persistent". I used this one mainly during the surface area problem because I got frustrated often because I working with so many numbers, but I decided to keep trying. This caused me to use another habit, "Stay Organized", I realized that if I wanted to solve the problem I would have to keep the numbers organized and neat. The only thing I would do differently is choose another item that is a more complex shape. |