This document discusses modeling the winning heights of Olympic gold medalists from 1932 to 2008 using various functions. The data is initially scaled and a quadratic function is fit manually and automatically. While the quadratic function fits the original data well, it does not extrapolate reasonably to future years. When additional data from 1984 to 2008 is added, the quadratic function is no longer a good fit. An arc tangent function is proposed to better model the leveling off of heights over time.

1. Maiko Yoshida
GOLD MEDAL HEIGHTS
The table below gives the height (in centimeters) achived by the gold medalists at various
Olympic Games.
Year 1932 1936 1948 1952 1956 1960 1964 1968 1972 1976 1980
Height 197 203 198 204 212 216 218 224 223 225 236
(cm)
Using technology, plot the data points on a graph. Define all variables used and state
any parameters clearly. Discuss any possible constraints of the task.
Note: The Olympic Games were not held in 1940 and 1944.
Year 8 9 12 13 14 15 16 17 18 19 20
Height 1.97 2.03 1.98 2.04 2.12 2.16 2.18 2.24 2.23 2.25 2.36
(cm)
X=1900+4x
Y=100y

2. Maiko Yoshida
If the numbers were big, a computational error would be big, as well. In order to avoid
this, and also to calculate easily, I scaled the numbers.
Year: X=1900+4x
Ex. 1932=1900+4(8) x=8
Height: Y=100y
Ex. 197=100(1.97) y=1.97
Parameters
y ax 2 bx c
a represents a direction and how the graph opens
For example, if a is positive value, it means the quadratic graph open to
upward.
b
the relationship between a and b is x
2a
c represents y-intersect
Possible constraints are that X can only be 8,9,10,11,12,13,14,15,16,17,18,19,20.
y can only be positive real number.
What type of function models the behaviour of the graph? Explain why you chose
this function. Analytically create an equation to model the data in the above table.
The height once decreases but basically it increases, as years pass so I think quadratic
equation is the best fit. I picked three examples up to make a manual equation to model
the data. I used 1932, 1960 and 1980 because these are at the first year, last year and
middle year of the data so it can be said that these three points are typical of this graph.
For creating the equation, I used matrix.
First year (1932) 1.97=64a+8b+c
Middle year (1960) 2.16=225a+15b+c
Last year (1980) 2.36=400a+20b+c

3. Maiko Yoshida
64 8 1 a 1.97
225 15 1 b = 2.16
400 20 1 c 2.38
y 0.0010714286 x 2 0.0025 x 1.881428571
On a new set of axes, draw your model function and the original graph. Comment
on any differences. Discuss the limitations of your model. Refine your model if
necessary.
My model function almost fits the original dots but some point such as x=12 or 13 do not
suit to the graph. In order to refine the graph, I focused on the turning point. In this graph,
there are 4 turning points. Therefore, I made quintic equation.
y ax 5 bx 4 cx 3 dx 2 ex f

4. Maiko Yoshida
This equation seems much better to fit the graph within x=8 to 20.
Use technology to find another function that models the data. On a new set of axes,
draw both your model functions. Comment on any differences.
I tried several equations by using logger-pro and checked the RMSE, which means Root
Mean Square Error. The smaller this value, the graph is much better fit.
y ax 2 bx c
Quadratic function RMSE=0.03868

5. Maiko Yoshida
y= a bx cx 2 dx 3
Cubic function RMSE=0.03942
y aln(bx)
Natural log function RMSE=0.05681

6. Maiko Yoshida
y alog(bx)
Base-10 logarithm RMSE=0.05681
y mx b
Linear function RMSE=0.04523
Because RMSE of quadratic function is the smallest value, it can be said that quadratic
equation is the best fitted function.

7. Maiko Yoshida
The graph which shown above is the auto fit quadratic function
The graph which shown above is the manual quadratic function.

8. Maiko Yoshida
The graph which two graphs combined is shown below.
The RMSE of manual equation is 0.0480387.
The RMSE of auto fit equation is 0.03868.
Although there are some errors between manual equation and auto fit equation, it seems
almost same because two of RMSE are nearly same, as well.

9. Maiko Yoshida
Had the Games been held in 1940 and 1944, estimate what the winning heights
would have been and justify your answers.
Year: 1940
If I use the manual fit quadratic equation, the formula would be
1940 1900 4 x
x 10
y 0.00107 102 0.0025 10 1.881
y 2.013
Y 100 2.013 201.3
A. 201.3 cm
If I use the auto fit quadratic equation, the formula would be
1940 1900 4 x
x 10
y 0.00181 102 0.0203 10 2.027
y 2.005
Y 100 2.005 200.5
A. 200.5cm
Year: 1944
If I use the manual fit quadratic equation, the formula would be
1944 1900 4 x
x 11
y 0.00107 112 0.0025 10 1.881
y 2.035
Y 100 2.035 203.5
A. 203.5cm

10. Maiko Yoshida
If I use the auto fit quadratic equation, the formula would be
1944 1900 4 x
x 11
y 0.00181 112 0.0203 11 2.027
y 2.02
Y 100 2.022 202.2
A. 202.2cm
The answers are appropriate because the answers which I got are almost same value as the
graph above.
Use your model to predict the winning height in 1984 and in 2016. Comment on
your answers.
Year: 1984
If I use the manual fit quadratic equation, the formula would be
1984 1900 4 x
x 21
y 0.00107 212 0.0025 21 1.881
y 2.405
Y 100 2.405 240.5
A. 240.5cm
If I use the auto fit quadratic equation, the formula would be
1984 1900 4 x
x 21
y 0.00181 212 0.0203 21 2.027
y 2.398
Y 100 2.398 239.8
A. 239.8cm

11. Maiko Yoshida
Year: 2016
If I use the manual fit quadratic equation, the formula would be
2016 1900 4 x
x 29
y 0.00107 292 0.0025 29 1.881
y 2.853
Y 100 2.853 285.3
A. 285.3cm
If I use the auto fit quadratic equation, the formula would be
2016 1900 4 x
x 29
y 0.00181 292 0.0203 29 2.027
y 2.961
Y 100 2.961 296.1
A. 296.1cm
As year passes, the value of height is going to be unbelievable number because the human
will never achieve 285 or 296.1cm. Therefore, the quadratic function gradually cannot be
used as the year passes.
The following table gives the winning heights for all the other Olympic Games since
1986.
Year 1896 1904 1908 1912 1920 1928 1984 1988 1992 1996 2000 2004 2008
Height 190 180 191 193 193 194 235 238 234 239 235 236 236
(cm)
I scaled again.

12. Maiko Yoshida
Year -1 1 2 3 5 7 21 22 23 24 25 26 27
Height 1.90 1.80 1.91 1.93 1.93 1.94 2.35 2.38 2.34 2.39 2.35 2.36 2.36
(cm)
How well does your model fit the additional data?
As graph shows, the quadratic equation does not fit the dots very well. If I use the
quadratic equation, human can achieve 280cm in 2060. Obviously, human cannot jump
280cm so the equation which I used has restriction. Therefore, it can be said that this is
interpolated example.

13. Maiko Yoshida
Discuss the overall trend from 1896 to 2008, with specific references to significant
fluctuations.
At first, the slope of the graph seems constant except in 1904. However, from 1936 to
1980, the sloop becomes steep. Then, the slope seems to be constant again from 1980.

14. Maiko Yoshida
What modifications, if any, need to be made to your model to fit the new data?
The graph looks like arc tangent which function is constant at first, then becomes sharp
and be constant again. Because the ability of human’s jump does not increase forever, it
should be stable in the end. Therefore, I will use a atan x b even though the value of
RMSE is big.