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

Logans Logo

Extracts from this document...

Introduction

Mathematics

Portfolio Type 2

LOGAN’S LOGO

January 2009

Maria Stormo

image00.png

This diagram shows a rectangle which is divided into three regions by two curves. The shaded region between the two curves is a logo which can be represented by two mathematical functions. The rectangle is 6.5cm by 6.1cm with a 0.5mm thick frame and the lines of the curves are 0.5mm thick. The aim for this portfolio is to manage to find mathematical functions to represent the logo, be able to modify the logo so that it fits other types of dimensions, and in the end to be able to calculate the area fraction of the logo on the business card. What we can ask ourselves is if the area fraction of the business card will be of the same value as the original logo.

1. Developing mathematical functions for the logo

1.1. Finding the coordinates

The problem is to find the two functions that model these two curves. One type of function that can represent these curves might be sine functions. To check this thesis out, it is first necessary to know the coordinates of the curves.

A parameter is a quantity that defines characteristics of functions. There are several parameters relevant to this portfolio:

  • The thickness of the lines: 0.5 mm
  • The length of the logo: 6.5 cm
  • The height of the logo: 6.1 cm
  • The type of function used

A variable is a value that may vary. There are several variables relevant to this portfolio:

  • The x-value
  • The measurements, as others may have measured differently
  • The size of the card
...read more.

Middle

2,7

2,3

3

2,7

3,2

3,1

X

Y

3,5

3,6

3,6

3,8

3,8

4,1

4

4,4

4,2

4,7

4,5

5,1

4,9

5,5

5,2

5,7

5,6

5,7

6,1

5,2

6,4

4,6

6,5

4,4

1.2. Sine regression

After listing these coordinates into the list function on my GDC, a Texas TI-84 Plus, I can use the sine regression to find an appropriate function for the points.

The lower curve (g(x)):

y = a ∙ sin ( bx + c ) + d

a = 1.668240274

b = 0.8100728241

c = –2.216031762

d = 1.96525663

g(x) = 1.668240274 sin ( 0.8100728241 x – 2.216031762 ) + 1.96525663

The upper curve (f(x)):

y = a ∙ sin ( bx + c ) + d

a = 2.378686945

b = 0.7859285241

c = –2.565579771

d = 3.202574671

f(x) = 2.378686945 sin ( 0.7859285241 x – 2.565579771 ) + 3.202574671

image01.png

By comparing the two functions to the points of the logo, it can be noticed that these functions do not represent the two curves in the logo very precisely. This can be explained by that the curves are not sine functions. Evidence for this is that for a function to be a sine function it has to be symmetric, and the curves of the logo do not have that property.

1.3. Cubic regression

Another type of function that may represent the curves can be polynomial function of 3rd order, a cubic function. Using the graphing package, “Graph 4.3” by Ivan Johansen downloaded at http://www.padowan.dk/graph/, I can find the polynomial function that best represent the curves.

The lower curve (g(x)):

y = ax3 + bx2 + cx + d

a = –0.089515466

b = 0.74847837

c = –0.88572438

d = 0.58866081

R2 = 0.9995

g(x) = –0.089515466x3 + 0.74847837x2 – 0.88572438x + 0.58866081

The upper curve (f(x)):

y = ax3 + bx2 + cx + d

a = –0.13020708

b = 1.2745508

c = –2.4169654

...read more.

Conclusion

image03.png

image04.png = image05.png

image06.png = image07.png

An accurate way to calculate this is to use the integral function on the GDC.

This gives the area(A) beneath the two curves to be:

A(g(x))         =15.5419

A(f(x))         = 23.6114

area         =  23.6114 – 15.5419

= 8.0695

Total area of the card: 9∙5 = 45

image08.png        =        0.179322

It is impossible to simplify this fraction, but if we first simplify the answer to fewer decimals it is possibe to get a simplified fraction.

0.18 = image09.png

So the logo occupy image09.pngof the card surface.

This might be an important aspect of a business card because it is important for a logo to be noticed but at the same time not take up to much space.

4. Conclusion

In the introduction I asked if the area of the logo on the business card was the same as the area of the original logo, and to answer this the area of the initial logo has to be calculated.

Using the GDC the area of the initial logo is given to be:

0.18 = image09.png

So the answer is yes, the area of the logo on the business card is the same as the area of the original logo. This means that all the calculations and modifications are correct and that the logo is precicely modified into other dimensions.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths 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 International Baccalaureate Maths essays

  1. Math IA - Logan's Logo

    The amplitude of the function is the distance between the center line (in this case the x-axis) and one of the maximum points. In order to determine the value of variable a, we must first look at the maximum and minimum (highest and lowest)

  2. Derivative of Sine Functions

    a) Consider several different values of c. 1.Consider c=1, c =2 and c =-1 1 when c =1: f(x) =sin(x+1) 2when c =-1 f(x) =sin(x-1) 3when c =2 f(x) =sin(x+2) the graph is shown below: graph of f(x)=sin(x+1),f(x)=sin(x-1), f(x)

  1. Mathematics Higher Level Internal Assessment Investigating the Sin Curve

    Lastly we will look at transforming the last equation into the desired form. The first step, and only step to transform the equation is to factorize by , as this would change the equation to: . Now it can be seen that the value of and lastly, .

  2. Population trends. The aim of this investigation is to find out more about different ...

    The final equation of this model is or as it is represented by the researcher, . The researcher's model with the constants I put in is a good way to represent the population in China between the years 1950 and 1995.

  1. Creating a logistic model

    have now changed to: (60000, 1) (10000, 2.9) To find the linear growth factor, we form the two equations: 2.9 = m(10000) + b 1 = m(60000) + b Solving this on the GDC, we have m = -0.000038 and b = 2.8 so the linear growth factor is: rn = -0.000038 � un + 3.28 and therefore

  2. derivitaive of sine functions

    Therefore a limits proof will be made to determine the derivative of the function. We will first begin with the general formula used to determine the limits or the derivative of a function.

  1. This essay will examine theoretical and experimental probability in relation to the Korean card ...

    however in this essay, there will be 2 people playing the game. There are two ways of dealing; first one is each player receive two cards each and then reveal the cards to see who has the highest hand. Second one is each player receive three cards, create a highest

  2. The investigation given asks for the attempt in finding a rule which allows us ...

    The function provides a unique challenge ? the vertical asymptote present proves to be problematical for the general statement. If a = 0, while b= 2.5, and the number of trapeziums used is 4, the general statement is undefined since the trapezium cannot cross the vertical asymptote.

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