BY ILONA MILEWSKA & ANNA UCHWAT

1. BURNING MIRRORS PROJECT BY ILONA MILEWSKA & ANNA UCHWAT

2. “Archimedes is considered one of the three greatest mathematicians of all time along with Newton and Gauss. In his own time, he was known as “the wise one,” “ the master“, and “ the great geometer” and his works and inventions brought him fame that lasts to his very day. He was one of the last great Greek mathematicians.” (Paul Golba) ARCHIMEDES

3. PART 1 • In the Second Punic War the Romans were fighting against the people of Syracuse. They had difficulty winning the bottle, because Archimedes used a mathematical principle to burn the Roman ships. He burned up the whole Roman fleet by discovering that on a curved mirror he can concentrate the sun’s rays onto anything burnable with enough intensity to cause the object to burst into flames in seconds.

4. Archimedes would not burn the enemies’ ships if the mirror that he used was a plane mirror. Sun’s rays would not reflect the way that they would cross each other in one point. This point where all the sun rays are crossing is called focus point. PLANE VS. CURVED MIRROR

6. Part 2 1. A To find the vertex using the standard form of the quadratic function . There is a very easy rule to find it: Vertex = (h, k). So, in the given quadratic function the vertex is V = . point of the quadratic To find if the vertex is a maximum or minimum function without graphing, we have to determine in the function. If , the vertex of the parabola would be a minimum point, but if the , then the vertex of the parabola would be a maximum point. In our function is equal to -1 so it is less than 0. We can conclude that the vertex is the maximum point of the function.

7. 1.B , we have to use To find the vertex of the function in the form , the formula . First, we have to determine what is a, b and c, so , and . Now, the founded numbers we can already plug into the vertex coordinates formula. The vertex of the quadratic function is (1, 1). To determine if the vertex of given function is minimum or maximum point of the we need to find out if or . If parabola of the quadratic function as in , then the vertex is a minimum point of the parabola, but when our problem where then the vertex is the maximum point of the parabola of quadratic function.

8. 2.A To graph the quadratic function we have to follow these steps: STEP 1 Find if the graph curves up or down. Because , so 1>0, then we know that it is going to curve up. STEP 2 Find the vertex. As we know already from the standard form of the quadratic function we identify vertex , so the . In our function , and coordinates of the vertex are . STEP 3 Find x and y intercepts. Y-intercept To find y intercept we have to replace x with 0 in the given function, because it is a point where graph crosses the x-axis. The y intercept is (0, 5)

9. X-intercept To find x intercept we have to replace y with 0, because it is a point where graph crosses the x-axis. Now we have to solve this quadratic equation by using this formula , so In our example , , and Because under the square root is a negative number it means there is no real number solution, no x-intercepts.

10. STEP 4 Graph the parabola AXIS OF SYMMETRY To find the axis of symmetry we have to take the x coordinate of the vertex or read it from the graph. For this function

11. 2. B To graph the function we have to follow these steps: STEP 1 Find if the graph curve up or down. Because a = -1, then a < 0 so we know it is going to be curve down. STEP 2 Find the Vertex As we already know that the function is in the form so we have to use the formula . so the coordinates of the vertex are (0, 1)

12. STEP 3Find x and y intercepts. Y-intercept To find y intercept we have to replace x with 0 in the given function, because it is a point where graph crosses the x-axis The y intercept is (0, 1) X-intercept To find x intercept we have to replace y with 0, because it is a point where graph crosses the x-axis. The x intercepts are (-1, 0) and (1, 0).

13. STEP 4Graph the parabola. AXIS OF SYMMETRY To find the axis of symmetry we have to take the x coordinate of the vertex or read it from the graph. In this function x=0.

14. PART 3-1 The plane curve that emerges when you graph the function of the form is called parabola.

16. The graph of the function is the same wide as the function y = ax^2 it means the graph are congruent but their positions are different, because h and k moves them to the right, left, up or down. The coordinates of the parabola. Ax + bx + c = a [ x^2 + b/a x ] + c = a [( x + b/2a )^2 – b^2/4a^2 ] + c = a ( x + b/2a )^2 – b^2/4a + c = a ( x + b/2a ) + 4ac – b^2/4a = a ( x – (- b/2a))^2 + 4ac – b^2/4a Comparing with y = a(x – h)^2 + k gives h = - b/2a and k = 4ac – b^2/4a

17. PART 3-2 Though not so simple as the circle, the ellipse is nevertheless the curve most often "seen" in everyday life. The reason is that every circle, viewed obliquely, appears elliptical.

18. Any cylinder sliced on an angle will reveal an ellipse in cross-section (as seen in the Tycho Brahe Planetarium in Copenhagen).

19. Tilt a glass of water and the surface of the liquid acquires an elliptical outline. Salami is often cut obliquely to obtain elliptical slices which are larger.

20. The early Greek astronomers thought that the planets moved in circular orbits about an unmoving earth, since the circle is the simplest mathematical curve. In the 17th century, Johannes Kepler eventually discovered that each planet travels around the sun in an elliptical orbit with the sun at one of its foci. The orbits of the moon and of artificial satellites of the earth are also elliptical as are the paths of comets in permanent orbit around the sun.

21. On a far smaller scale, the electrons of an atom move in an approximately elliptical orbit with the nucleus at one focus.

22. The ability of the ellipse to rebound an object starting from one focus to the other focus can be demonstrated with an elliptical pool table. When a ball is placed at one focus and is thrust with a cue stick, it will rebound to the other focus. If the pool table is live enough, the ball will continue passing through each focus and rebound to the other.

23. Statuary Hall in the U.S. Capital building is elliptic. It was in this room that John Quincy Adams, while a member of the House of Representatives, discovered this acoustical phenomenon. He situated his desk at a focal point of the elliptical ceiling, easily eavesdropping on the private conversations of other House members located near the other focal point.

24. One of nature's best known approximations to parabolas is the path taken by a body projected upward and obliquely to the pull of gravity, as in the parabolic trajectory of a golf ball. The friction of air and the pull of gravity will change slightly the projectile's path from that of a true parabola, but in many cases the error is insignificant.

25. This discovery by Galileo in the 17th century made it possible for cannoneers to work out the kind of path a cannonball would travel if it were hurtled through the air at a specific When a baseball is hit into the air, it follows angle. a parabolic path; the center of gravity of a leaping porpoise describes a parabola.

26. The opposite principle is used in the giant mirrors in reflecting telescopes and in antennas used to collect light and radio waves from outer space: the beam comes toward the parabolic surface and is brought into focus at the focal point The largest parabolic mirror in existence is in a telescope located in the Caucasus mountains in Russia. It is nearly 20 feet in diameter and was completed in 1967. The easiest way to visualize the path of a projectile is to observe a waterspout. Each molecule of water follows the same path and, therefore, reveals a picture of the curve.

27. Parabolas exhibit unusual and useful reflective properties. If a light is placed at the focus of a parabolic mirror (a curved surface formed by rotating a parabola about its axis), the light will be reflected in rays parallel to said axis. In this way a straight beam of light is formed. It is for this reason that parabolic surfaces are used for headlamp reflectors. The bulb is placed at the focus for the high beam and in front of the focus for the low beam.

28. Two types of images exist in nature: real and virtual. In a real image, the light rays actually come from the image. In a virtual image, they appear to come from the reflected image - but do not. For example, the virtual image of an object in a flat mirror is "inside" the mirror, but light rays do not emanate from there. Real images can form "outside" the Heat waves, as well as light and sound system, where emerging waves, are reflected to the focal point of a light rays cross parabolic surface. If a parabolic reflector is and are caught turned toward the sun, flammable material - as in a Mirage, an placed at the focus may ignite. (The word arrangement of two "focus" comes from the Latin and means concave parabolic fireplace.) A solar furnace produces heat by mirrors. Mirage's 3-D focusing sunlight by means of a parabolic mirror arrangement. Light is sent to it by set illusions are of moveable mirrors computerized to follow similar to the holograms the sun during the day. Solar cooking formed by lasers. involves a similar use of a parabolic mirror.

29. PART 3-3 • There is a perfect relation between the zeros of a quadratic function y=f(x) and the solution of the quadratic equation f(x)=0. But what is the solution of the quadratic equation? It is the value or values of x that makes the function 0. To find the solutions of the quadratic equations we need to find the formula for it , by solving general quadratic equation for x.

30. By using this quadratic formula we can find the solutions of the where . The discriminant which is the expression under the radical tells us how many solutions does the particular quadratic equation has one, two or none. If , the equation has two real solutions If , the equation has one real solution If , the equation has no real solution

31. We found two solutions ½ and 2. We found two solutions -1 and 5.

32. . and We found one solution. We found two solutions

33. is undefined in real numbers, so there is no solution.

34. PART IV Conclusion Once we start studying scenic section we get a new definition of a parabola as the focus of point equidistant to a fixed point called focus and a line called divectrix. From this definition it was constructed the equation of the parabola we get among other 3 standard forms that: = 4 cy was such equation (check construction if you want. Here c is the focal distance (distance from the vertex to the focus of the parabola) So, y = is of the form y = ax (here a > 0) which coincide with the old quadratic function whose graph we use to call a parabola. Now let’s prove it has the focal property we claim it has. Let’s take any point ( x , y ) in the graph of “ such” parabola then Po = (x , y ) = (x ) So if y = To define the angles of incidence and reflection we are using the angle formed with the tangent line to the curve at F.

35. We need to prove that P QF is isosceles and that the tangent line through P is angle bisector of angle FP Q, in other words we need to prove that P CQ = P CF (congruent) which make FP C = CP Q and sine CP Q = incidence angle (vertical angle) we are going to be done. We need to find the equation of tangent line to check that this line intercept segment FQ in the midpoint and than everything will follow to do so; we need the slope of the tangent line so we need to borrow from Calculus I the geometrical interpretation of the derivative of a function at a point f(x) = f(x) = f(x ) = on slope tangent line

36. Belongs to tangent line so, we have 0 = - x P (x , y ) and m = + - - 0 = = =0 y- y = (x - x ) eq of tangent line. Check Midpoint FQ = ( , 0) x y =

37. THE END THANK YOU