• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Mathematical Methods - Cables.

Extracts from this document...

Introduction

Mathematical Methods Portfolio #2 2003

Type 3 (Modelling): Cables

Everyday, people are faced with problems that can be solved through the use of mathematics, particularly in the fields of design and construction, where geometry, trigonometry and algebra become very important. This is the point at which this portfolio piece becomes very relevant. The task at hand is as follows:

Engineers put forward three plans to provide a telephone link between three towns: A, B and C. An isosceles triangles is formed by these towns with the distance              AB = BC = a km and the distance from A to C = 2 km.

Engineers wish to use the least amount of cable to join the three towns.

image21.pngimage22.pngimage30.pngimage00.pngimage01.png

                   C - Plan                                        V- Plan                                Y - Plan

The objective of the portfolio is to ascertain, for varying positions of town B, and the resultant differing lengths of a, which plan will in the end provide the link using the least amount of cable.

NB. A must be greater than 1 in all circumstances, as all towns must be linked and the plans are all based on triangles.

...read more.

Middle

Now that equations have been obtained for each of the models/plans, the use of technology, to create visual representations of the equations can now be used. Through this a deeper, clearer understanding can now be gained. One can quite obviously observe the differences between models/plans for the amount of cable required for the varying values of a.

In this graph, the y axis represents the total amount of cable used in the selected model/plan and the xaxis represents the varying values of a.

Thus, it can be deduced that the equation that possesses the lowest y value for any value of x will be the most effective model/plan for that specific x value.

image24.png

Diagram 1.

Through a quick observation of the above diagram, it can quite easily stated, according to the previous deductions, that the most effective values of x (those which have the lowest values of y (use the least amount of cable)),for each of the plans, are:

NB.

...read more.

Conclusion

image10.png

Proof using simultaneous equations:

image13.pngimage11.png

Both of these equations cannot be solved algebraically with the techniques we have been taught at this stage of our education. Although, they can be solved through the implementation of graphics calculators where it can be found that in the modified Y-plan vs. the V-plan, a will equal image29.png and that in the modified Y-plan vs. Y-plan, a will equal 2.

The modified Y-plan does not ever seem to intersect with the C-plan, the modified y-plan always being a few units lower. This may not be the case at high values of a.

The modified Y-Plan provides the most effective use of cable for all possible values of a.

To conclude, the modified Y-plan provides the engineers with a telephone link which uses the least amount of cable possible. The diagrams used were drawn in Graphmatica and a majority of testing was done on a Texas Instruments TI-83 graphics calculator.

Mathematical Methods              Portfolio Piece 2-Cables (Type 3 – Modelling)                

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. 2D and 3D Sequences Project Plan of Investigation

    2b + c = 4 12a + 2b = 12 We can simplify this equation to: 6a + b = 6 My next calculation is below: N =3 _27a + 9b + 3c = 26 12a + 6b + 3c = 12 15a + 3b =14 (15a + 3b = 14)

  2. Best shape for gutter and further alegbra - using Excel to solve some mathematical ...

    By using algebra I can find ? at which the optimum capacity exists. Angle to give Optimum capacity; =w2 sin ? = 0 Cos ? = 0 ? = 90� Semi-Circular cross-section l = ? x r l = ?

  1. MEI numerical Methods

    0.633079233 1/3 0.712208165 1/4 0.761616673 There are two relationships which this graph shows: Assuming ? is +ve, as k -> ?, ? -> 0 Assuming ? is -ve , as k -> -?, ? -> 0 To simplify if K is a positive number then it has a negative correlation,

  2. Investigate the relationships between the lengths of the 3 sides of the right angled ...

    4a + 2b + c = 12 -eqn2 - a + b + c = 4 -eqn1 3a + b = 8 -eqn5 Equation 4 - Equation 5 I am doing this to eliminate B and finally work out what A is worth.

  1. Although everyone who gambles at all probably tries to make a quick mental marginal ...

    Payout Difference Squared 10068347520 $2 000 000.00 $1 956 292.86 1910314086.98 228826080 $1 721.80 $1 708.34 181.17 5085024 $62.20 $42.75 378.30 110544 $10.00 $27.33 300.33 Total =1910314946.78 This formula is a good representation of the relationship between Payoff and Odds of winning for the Lotto 649 Lottery.

  2. Math Portfolio Type II - Applications of Sinusoidal Functions

    Assume that the function representing time of sunrise in terms of day number has a period of 365 days. Explain how the value of parameter b could be determined algebraically. The value of parameter could be determined algebraically by determining the period, which is 365 days.

  1. Mathematics portfolio - Translations.

    It moves left 2 units. y = (x - 1)2 is the effect of translation vector of . It moves right 1 unit. 9. Curves move either left or right horizontally. The units they move is according to the number after x in the equation .

  2. Coding and Modelling - The tools used in my spreadsheets.

    The maximum volume of box is achieved at the peak of the curve, i.e. when the gradient of the curve is equal to 0. Applying Calculus and differentiating the formula can find the point where the graph has a gradient of 0.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work