The Nature of Mathematical Modeling - Essay Example

A mathematical model is an abstract model that uses mathematical language to describe the behavior of a system. It is an accurate representation of the relationship between two or more variables relevant to a given situation or problem

The process of developing such models is known as mathematical modeling. Mathematical models are used particularly in natural sciences and engineering disciplines such as physics, biology, and mechanical engineering but also in social sciences such as economics and political science.

Mathematical models can take many forms, including but not limited to dynamical systems, statistical models, differential equations, or game theoretic models. These and other model types can overlap, with a given model involving a variety of abstract structures. There are six primary groups of variables: decision variables, input variables, state variables, exogenous variables, random variables, and output variables. Mathematical modeling problems are often classified as a black box or white-box models, according to how much prior information is available from the system.

The theoretical problem-solving method is much more than just drawing graphs or solving equations. It contains the following steps:

1. understand the problem
2. identify the important features
3. make assumptions and simplifications
4. define variables
5. use sub-models
6. establish relationships between variables
7. solve the equations
8. interpret and validate the model (i.e. question the results of the model)
9. make improvements to the model

• explain the outcome

The above figure illustrates the relationship between the real world, and the mathematical world through formulation and interpretation. 

In this report, the principles of mathematical modeling are used to solve problems in the real world. The model creates an accurate representation of some phenomenon for better understanding. They create matches of observation with symbolic representation Inform Theory and assist in explanations of the problems, mathematically.

QUESTIONS

1. The brochure for the Feel the Fear coaster says that the height of the coaster can be determined by this polynomial model for 12 seconds after the coaster comes out of a loop

For maximum and minimum heights, the derivative of the function is used, because at maximum and minimum points the derivative = 0

For drawing the graph, the following table when t=0 and t=12 are essential

t

0

1

2

3

4

5

6

7

8

9

10

11

12

H(t)

100

36

0

-14

-12

0

16

30

36

28

0

-54

-140

 and the original equation is given below

Let H be a function of time, H (t), then the first derivative = H ′ (t), and the second derivative is H ″ (t)

 Or

H ′ (t) = -3t2 + 34t – 80,

Solving this by quadratic formula,

           Where a=-3, b=34 and c=-80

t =  and t= 8

Now, to get the maximum or minimum height, these values are substituted into the original polynomial.

H= -14.8 when t=sec   and using t=8, H= 36.

At a maximum point, the second derivative is negative

The second derivative occurs at 

= -6t +34

When t = =    

Minimum height at t =

=-14.8

Maximum height of t=8

2. The Giant Coaster is modeled by

For drawing the graph, the following table is used from when t=0 and t=12

t

0

1

2

3

4

5

6

7

8

9

10

11

H(t)

0

30

36

24

0

-30

-60

-84

-96

-90

-60

0

Like the above question, to get the maxima and minima, the derivative of the polynomial is used.

 Or

H ′ (t) = 3t2 + 30t + 44, and solving this by the quadratic formulae,

And substituting the values in the equation

Where a=3, b=30, and c=44, and therefore t =1.79 seconds and 8.2 seconds. 

Substituting these values into the original polynomial

H =36.43 and -96.43.

Proving the above to find the maximum and minimum by the second derivative

= 6t - 30

Maximum height = 36.43 and the minimum height is -96.43

3. The ride has 100 meters of fencing to make a rectangular enclosure as shown. It will use existing walls for two sides of the ring and leave an opening of 2 meters for a gate.

From the basics, the area is given by l*w

The gate allowance is 2m, the width is x m and the total perimeter fenced is 100m, then the l + w= 102m but width=x m then length= 102-xm, and width is x m so the area becomes (102-x)* x m = 102x –x2

Now if the area = 102x –x2  

A (x)’ = 102- x2   , A (X)” = -2

Equating the above first derivative to 0,

102-2x = 0

X= 51

The maximum possible area is when x= 51

The area from the equation above

A(X)’ = 102 (51) - 512 = 2601M2

4. Snacks will be provided in a box with a lid (made by removing squares from each corner of a rectangular piece of card and then folding up the sides)

L*w*h gives volume

Length = (40 – 2x), width = (20 – x), height = x

The volume of the box

V = x (40 – 2x) (20 – x)

V = x (40-2x) (20-x)

V= x (800- 80x2 + 2x2)

V= 2x3-80x2 +800x

The derivative of this gives

 is from the solution of the above equation

   Substituting the values into the equation

Where a= 6, b+ -160 and c= 800

X= 20 and 6

For x=20        =12(20)-160=80

For

Maximum volume is when  

Height

Length:

Width:                                                                             

Maximum volume is:

 Conclusion

Mathematical modeling has been used effectively in the above exercise to create models that have solved the problems given. The answers to those problems and the procedures used in arriving at them are clearly outlined, in the workings of each question.