FC: Sydney Harbor Bridge | By: Alec Shumway
1: Question 11: The main arch of the Sydney Harbor Bridge is approx. parabolic in shape. It is 503 meters between two pylons. If x is the number of meters measured horizontally from one pylons and H(x) is the height of the arch in meters above water level at the point x, it has been that: H(x)=x(503-x) 475
2: Step 1: Draw it out | Draw half of a | Then draw a long line underneath it to connect the two points which is 503 m. | Draw a line down the middle for the height, make where it touches the X-axis equal x. This is your diagram.
3: Step 2: Solve for H(x) when x = 90 | In the equation H(x)=x(503-x) 475 Set x to equal 90 and you will get H(90)=90(503-90) 475 which will get you 0.87
4: Step 3: Solve for H(x) when x = 280 | Its the exact same as step 2 except you set x to equal 280 so you will get an answer of .47
5: Step 4: Where is the arch 90 meters above the water? | To find this, you type in your original equation for H(x) in Y1 in your calculator, then in Y2 you type in 90 and graph it. You find the intersection and what you get for X is your answer which is 108.31m
6: Step 5: What is the height of the arch at it's highest point? | To find it's highest point,you type in H(x) into Y1 and hit graph. Then you hit Second->Trace->4->enter. Then move your marker to the right of the parabola and then hit enter->enter->enter. X=133.16m is your answer.
7: Step 6: The road across the bridge is 45m. Find the two points where the arch crosses over the bridge. | To do this you keep Y1 the same and you set Y2 as 45. Then you hit graph->second->5->enter(3x) and x=46.86m is one of the points. Then you hit second->5->arrow key right until you are near the other point of intersection->enter(3x) and x=456.14m is the other point.
8: Step 7: What is the length of the road between the two points where the arch crosses the road? | The length of the road is simply the second point previously found minus the first point so x=456.14-46.86=409.28m.
9: Step 8: What would the equation be if for a bridge as wide as Sydney but half as tall? | To find this, you divide 475 by 2 and make that your new denominator so it would be H(x)=x(503-x) 237.5 then you plug in for x
10: Step 9: What is the equation for a bridge as wide as Sydney but double the height? | You do the same as you did for step 8 but instead of dividing the denominator by 2, you double it. So the equation would become H(x)=x(503-x) 950
11: The End