# Thread: Applying graphical methods in solving problems

1. I need help guys D:

Part One:
A symmetrical bridge is being designed to cross a river.
The maximum height of the bridge is 12 metres and the total length is 60 metres. The shape of this section of the bridge could be modelled by any of these functions: a parabola, square root or logarithm.
* Draw graphs of the three different functions that could be used to model the half bridge.
* Give the equation of the function used in each model.
* Discuss the features or properties for each of the functions you have developed.

Part Two:
The symmetrical bridge will be 60 metres long in total and have its highest point 12 metres above the river. Model the complete bridge using your designs from Part One. You may use a piecewise function if needed.
For each design:
* Give the equation of any function(s) used in your model.
* Discuss the features or properties for each of the functions used in your model.

Part Three:
The complete bridge design may be needed to cross rivers of varying widths. The highest point will remain 12 metres above the river. The total length of the bridge will be w metres.
Generalise your models in Part Two to meet these requirements.
For each model:
* Give the equation of any function(s) used in your model.
* Discuss the features or properties for each of the functions used in your model.

For part one, I have come up with these equations -

Parabola: y = -0.0133(x-30)(x-30)+12
Features and properties of parabola:
vertex = (30,12)
y-incpt = (0,0)
domain = 0 < or equal to x < or equal to 30
range = 0 < or equal to y < or equal to 12

Square root: y = 2.19sqrt(x)
Features and properties of parabola:
y-incpt = (0,0)
domain = 0 < or equal to x < or equal to 30
range = 0 < or equal to y < or equal to 12

I'm not quite sure how to go about the logarithm equation.
For part two, and part three I am just so confused, and don't know where to start. Any help would be GREATLY appreciated!  2.

3. Using common logarithms, I get y=f(x)=8.05*LOG(x+1) for 0 ≤ x ≤ 30. f(0) = 0, and f(30) =12. Is that what you're looking for?

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