A bridge in New York City has to be build. The transportation authority in New York

Rupa 1 Raushan Rupa Shyyam Khan Pre calc Introduction A bridge in New York City has to be build. The transportation authority in New York City have planned to construct a new bridge over the East River in New York. By building the east river bridge it can connect the east with Long Island and west to the New Jersey. Also, the East River is a salty river and keep changing the direction of the flows. That s why, it is not reliable to travel by boats or ships. It s also very busy river and by building the bridge the traffic can be avoided in a vast number. A strong structural constructed bridge can make the transportation easy for both sides of the river. In order to build the bridge, the space is between the supports needs to be 1050 feet; the height at the center of the arch needs to be 350 feet. Also an tanker needs a 80 foot clearance to pass beneath the bridge. The suggested shapes to build the bridge are whether a parabolic or a semi elliptical. The equation for both parabola and semi ellipse can be find out by the provided informations. The channel width is going to vary because of the shape of a parabola is different than a semi ellipse. The width of the channels for both conic section can be used to compare which plan will be best for building the bridge on the East river. The plan (parabola or semi ellipse) with bigger channel width will be accepted by the transportation authority of New York City. Parabola is one of the conic section with a special curve which looks like an arch. Standard equation of a parabola is, y = ax + bx + c. The parabolic equation changes when the center is not in the origin but in a point called (h,k) and that equation is Vertical equation of

Rupa parabola. The vertical equation is, y = a (x h) + k, where (h,k) is the vertex, y is the y intercept. For parabola, there is always just one focus and focus is a fixed point in the axis of symmetry. Directrix is a fixed line which is perpendicular (Two nonvertical lines are perpendicular if and only if the product of their slopes is 1 and Vertical lines are perpendicular to horizontal lines) of the axis of symmetry and parallel ( Two lines are parallel when the product of their slopes is 0) to the line where vertex is present. A parabola has a axis of symmetry perpendicular to its directrix and that passes through its vertex. The vertex of a parabola is the midpoint of the 90 segment connecting the focus and the directrix. A parabola can be defined as, A parabola is the set of all points P (x,y) in a plane that are an equal distance from both a fixed point, the focus and a fixed line, the directrix. (8.1, Conics) This definition can be explained by the diagram below: In this diagram P is a point in the parabola which is (x,y), S (0,a) is the focus and T (x, a) is a point in the directrix. For any point like P in a parabola will have the same relationship which is PT=PS. Focus and Directrix always have the same distance from the Vertex which is (0,0) in this diagram. Parabolic shape can be seen everywhere in life like bridges, water

Rupa 3 fountain, satellite dish, reflection of flashlights and also as a logo in Mcdonalds. Engineers use the equation and shape of a parabola to build many architecture like bridges. The Golden Gate of Bridge is a good example of a parabolic shaped bridge. The provided information can be used in order to build the East River bridge in a parabolic shape. The height of the bridge which is 350 feet can be described as the distance from the vertex to the x axis. The width of the bridge which is 1050 feet is the space from one point to the other in the parabola of the x axis. In order to give the bridge a proper shape, the parabola will be till the height of 350 feet and width of 1050 feet in total. This bridge will limit the infinity parabolic shape to limited. The mid line which will separate the two sides of the bridge to 55 feet each is called the axis of symmetry. The standard equation of parabola which is y = a (x h) + k, here (h,k) is (0,350) in graph scale for this bridge. The equation can be solved for a by placing the measurement of vertex, y intercept. After figuring out a in the equation, the parabolic channel width can be find out. For the channel width y = 80 feet in y = (x h) + k, because that s the length of the clearance. The result for x will be the width of the channel for parabolic bridge. Ellipse is also another form of conic section. An ellipse is a curve which looks like x y squashed circle. Standard equation of an ellipse is, + = 1 for horizontal and + = a x b b 1 for vertical. For ellipse, the a is always bigger in number than the b. If the center of the ellipse is not in the origin and any other point like (h,k) then the equation will change to, + (y k) (x h) (y k) b a b y a (x h) a = 1 for horizontal and + = 1 for vertical ellipse. Also an ellipse can be

Rupa 4 described as, An ellipse is the set of all points on a plane whose distance from two fixed points remains constant. (8.1, Conics) The definition can be described by the diagram below: In the diagram, the distance from two focus is described as d 1 and d. The length of the major axis is a and minor axis is b. The center is in the origin. According to the definition, d 1 + d = a. The point (x,y) in the ellipse can be changed but the total distance from both focus will always be a, the major axis. In this diagram, the vertex are, (a,0) and ( a,0) and co vertex are (0,b) and (0, b). Elliptical shape can be seen everywhere in life like football, satellite and planet orbits, elliptical pool table, bridges, architecture and also the shape of our eyes. One example of an elliptical shape could be, the Statuary Hall in the United States Capitol Building. For making the East River bridge elliptical, the measurement of the major axis will be 1050 feet and half of minor axis will be 350 feet. In order to make the bridge, the ellipse s shape will be cut half by making it a semi ellipse. The semi ellipse will be horizontal and the center x y a b will be in the origin. That s why the equation for this semi elliptical bridge will be, + =

Rupa 5 1. In this equation, a = (1050 ) = 55 feet and b = 350 feet since the ellipse is semi. In order to figure the width of the elliptical channel, the equation can be solved by, y = 80 feet ( the length of the clearance of tanker). The result for x will be the channel width of the semi elliptical bridge. I predict that, the parabolic design of the bridge will be selected by the transportation authority of New York City instead of the semi elliptical design bridge. Because, the parabolic bridge will be wider than the semi elliptical bridge. Also in the world, there is more parabolic than the elliptical bridge. Ellipse shape can more be seen in architectural side. The parabolic shape gives the bridge a good look and also have more space. Also for bridges like one in the East River bridge, engineers mostly design parabolic shape. Problem Solving and Fluency The semi elliptical bridge is one recommended shape for the East River Bridge. The measurements of the bridge like length, height and width can be used in the standard equation of ellipse. The length of the bridge has to be 1050 feet from one side of the bridge to the other. The height of the bridge from the middle is 350 feet. The vertex for the ellipse standard equation will be (1050 ) = 55 feet since 1050 feet is the major axis which is a. The co vertex for the ellipse equation will be 350 feet = b, because the height of the bridge from middle is 350 feet which is half of the minor axis. The center of this semi ellipse will be in the origin. The equation for semi elliptical bridge will be then, x 55 y + = 1. By drawing this equation in desmos, it will look like a bridge on the positive 350 side of y axis.

Rupa 6 In order to get a table of points for this equation, the equation has to be solved for y, x 55 y + = 1 350 y = 1 x 350 55 y = 350 ( 1 ) x 55 y = ± 350 ), if we put this equation in a scientific calculator, it will draw the ( 1 x 55 ellipse and give a table for the equation. Some points for this equation is solved below in the table:

Rupa 7 Value of x Equation: y = ± 350 ( 1 x 55 ) Value of y 0 y = ± 350 ) ( 1 0 = ± 350 ( 1 0 ) = ± 350 1 = ± 350 1 = ± 350 55 + 350, 350. ± 315 y = (±315) 55 ± 350 ( 1 ) ± 350 ( 1 0.36 ) ± 350 0.64 ± 350 0.8 ± 80 + 80, 80. ± 55 (±55) y = ± 350 ( 1 ) 55 ± 350 ( 1 1 ) ± 350 0 ± 350 0 0 Since it s a semi elliptical bridge, that s why the points on the positive side of y axis is taken in order to build the bridge in the calculator. So, few points of the semi elliptical bridge are, (x,y): (0,350), (315,80), ( 315,80), (55,0) and ( 55,0). In order to find the channel width of the semi elliptical bridge, the equation needs to be solved for x. According to the Transportation Authority of New York City, the bridge has to have a clearance of 80 feet to pass the tankers. 80 feet is the measurement of the bridge s height from the surface of the bridge, that s why in the ellipse s equation y = 80 feet and the channel width is the value of x (value of x is just one side of the bridge from the origin and to find the whole channel width the x needs to be multiplied with ). 0

Rupa 8 x 55 x 55 y + = 1 350 80 350 + = 1 x 55 + 0.64 = 1 x 55 x 55 = 1 0.64 = 0.36 x = 0.36 55 x = 99,5 x = ± 99, 5 x = ± 315, since it s a distance, the positive number is only valid from the value of x not the negative number. After solving the equation for x with y = 80, the value of x is 315. So, the channel width for the semi elliptical bridge is, x = ( 315) = 630 feet. Another recommended shape for the East River bridge is parabola. The equation for the parabolic bridge will be the vertical equation as 350 feet is height of the bridge from middle point and the vertex is, (0,350). On the x axis the parabola will spread till 55 feet for both sides. The x intercepts are, (55,0) and ( 55,0). So, the parabolic equation is, y = a (x h) + k. The value of a is, 0 = a ( 55 0 ) + 350, since the vertex, (h,k) = (0,350) and x intercepts =(55,0) and ( 55,0). 55 a = 350 350 a = 55

Rupa 9 a = 0.0017. The vertical parabolic equation of the East River bridge is, y = 0.0017 (x 0) + 350 y = 0.0017 x + 350. With this equation a parabolic shape bridge can be built in the desmos which will have the same measurements of East River bridge. The parabola will spread till the x axis to make it look like a bridge. y = a (x h) + h, if we put this equation in a scientific calculator, it will draw the parabola and give a table for the equation. Some points for this equation is solved below in the table:

Rupa 10 Value of x Equation: y = 0.0017 (x 0) + 350 Value of y 0 y = 0.0017 (0 0) + 350 = 0.0017 0 + 350 = 350 350 ± 55 y = 0.0017 ( ( ± 55) 0 ) +350 = 0.0017 ( ± 55) + 350 = 0.0017 7565 + 350 = 350 + 350 = 0 0 ± 34.8 ( 34.8,80). y = 0.0017 ( ( ± 34.8) 0) + 350 = 0.0017 ( ± 34.8) + 350 = 0.0017 55131.04 + 350 = 70 + 350 = 80 Few points on the parabolic bridge are (x,y): (0,350), (55,0), ( 55,0), (34.8,80) and In order to find, the channel width of the parabola, the parabolic equation have to solved for x. The approval of the design bridge depends on a clearance of 80 feet tanker s pass. 80 feet is the measurement of the bridge s height from the surface of the bridge, that s why in the parabolic equation y = 80 feet and the channel width is the value of x (value of x is just one side of the bridge from the origin and to find the whole channel width the x needs to be multiplied with ). y = 0.0017 x + 350 80 = 0.0017 x + 350 0.0017 x = 80 350 80

Rupa 11 0.0017 x = 70 x = 70 0.0017 x = 55,118.11 x = ± 55, 1 18.11 x = ± 34.77, since it s a distance, the positive number is only valid from the value of x not the negative number. After solving the equation for x with y = 80, the value of x is 34.77. So, the channel width for the parabolic bridge is, x = ( 34.77) = 469.54 feet. The channel width for both the bridges are 630 feet for semi ellipse and 469.54 feet for parabola. The difference between the two channel widths are = ( 630 469.77) feet = 160.3 feet. So, the channel width is better for elliptical bridge than parabolic bridge by 160.3 feet. It will be a better to have an elliptical bridge than a parabolic bridge because the channel width is wider and the tanker can pass through it better. Reasoning and Proof In the diagram below, the blue curve is the semi elliptical bridge and the red curve is the parabolic bridge. The small straight lines inside the curve is the channel widths for both suggested shaped bridges. The diagram drawn in desmos easily showing that the semi elliptical bridge has wider channel width than parabolic bridge. Though they have the same height and length, the channel width is different because of their shapes. Even if, there is any flood and water level goes up by 10 feet, the semi elliptical bridge will be wider and is more acceptable shape.

Rupa 1 The graphs for both bridges also shows the points of the length of channel width intercepted in both positive and negative sides of x axis. For parabolic bridge the points are, (34.8,0) and ( 34.8,0). For semi elliptical bridge the points are, (315,0) and ( 315,0). By using the distance formula between two points, the length of the channel width can be found for both suggested shape bridges. The distance formula is, d = ( x x 1 ) + (y y ) 1. So, for parabola if we have graph and points from the data table, than the length of the channel width is, d = ( x x 1 ) + (y y ) 1 d = ( x x 1 ) + (y y ) 1 Connection Elliptical shape bridge is quite popular and can be seen a lot also. The tunnels are mostly semi elliptical shape and the engineers can find the equation for the tunnel or the bridge by just measuring the lengths and width of the place. In my math class for a project, I researched on

Rupa 13 conic shaped things in the world. My fellow classmates come up with eyes having a shape of ellipse, foods having a shape of parabola, fried egg having a shape of circle and many more. I worked on Golden Gate bridge which is a parabolic shaped bridge. Semi ellipse and parabola can not only be seen in architectural parts but also in things around us. connect conic section formulas to other topics in math and show how they are related. In the Biochemistry class in order to design an experiment one should understand the question they are looking for, have a hypothesis, plan a procedure, solve the problem with fluency and coming with a calculative and relevant conclusion. All these steps are part of designing a bridge in a place. Conclusion At last, the semi elliptical bridge is better than parabolic bridge because of its wider space and also for its beautiful shape. Also most of the tunnels are semi elliptical because the shape gives more space inside than other conic sections. My hypothesis was wrong since I thought parabolic bridge will have more channel width than elliptical bridge. After calculation it is clear that semi elliptical bridge has a channel width of 630 feet and parabolic bridge has 469.54 feet for the given measurements. An elliptical bridge will look nice at the East River also.

Rupa 14

Rupa 15 "Parabola." Parabola. N.p., n.d. Web. 03 Mar. 015. < https://www.mathsisfun.com/geometry/parabola.html >. "8.1 Conics." 8.1 Conics. N.p., n.d. Web. 03 Mar. 015. < https://people.richland.edu/james/lecture/m116/conics/conics.html >. http://jwilson.coe.uga.edu/emat6680/dunbar/assignment/parabola_kd.htm https://www.desmos.com/calculator