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Logans Logo

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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.

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