Technology for Chapter 2: The Modeling Process, Proportionality, and Geometric
Similarity.
In chapter 2, we estimate the slope of proportionality line using the two point method.
Given (x1,y1) and (x2,y2) then the slope, k, is
(y  y )
k 2 1 .
( x2  x1 )
We will use the following to illustrate the use of technology.
Proportionality/Geometric Similarity Modeling
1. PURPOSE: To provide the student the opportunity to develop proportionality
arguments and test them using technology.
2. OBJECTIVE: Determine if the hearts of mammals are geometrically similar using
your knowledge of proportionality models and technology. Provide a summary of your
analysis using the following data to support or refute your argument.
Animal
Heart Weight
(in grams)
Mouse
Rat
Rabbit
Dog
Sheep
Ox
Horse
0.13
0.64
5.80
102.00
210.00
2030.00
3900.00
Length of cavity of
left ventricle (in)
0.55
1.00
2.20
4.00
6.50
12.00
16.00
3. PROCEDURE:
a. Develop the proportionality model relating heart weight (HW) to the length (L)
of the cavity of the left ventricle. You should get HW  L3.
b. Test the proportionality model using the data provided. Testing a
proportionality model often involves the following steps:
(1) Enter the raw observed data into your calculator and computer.
(2) Plot the raw data to check for smoothness and potential "outliers", and
to give you a "feel" of the type of proportionality you might find.
(3) Make any necessary transformations to the data for the model.
(4) Plot the transformed data to test the proportionality (It must form a
straight line through the origin).
(5) Estimate the constant of proportionality (eyeball a straight line
through the origin and use slope =rise/run to find the constant).
(6) Obtain an overlay of the actual vs. computed data or find the errors
(yactual-ypredicted) to see if your model properly captures the trend of the data. Comment
about the fit.
Proportionality Template
Enter the raw data as Xdata and Ydata
Enter the transformation to X or Y as needed
Give the point to use in the slope
Interpret the output.
We illustrate with the same lab.
Heart Weight Length of cavity of
Animal
(in grams)
left ventricle (in)
Mouse
Rat
Rabbit
Dog
Sheep
Ox
Horse
0.13
0.64
5.80
102.00
210.00
2030.00
3900.00
0.55
1.00
2.20
4.00
6.50
12.00
16.00
>> weight=[.13,.64,5.8,102,210,2030,3900];
>> length=[.55,1,2.2,4,6.5,12,16];
>> plot(length,weight,'O')
4000
3500
3000
2500
2000
1500
1000
500
0
0
2
4
6
8
The data is increasing and curved.
10
12
14
16
>> length3= [.55^3,1^3,2.2^3,4^3,6.5^3,12^3,16^3];
4000
3500
3000
2500
2000
1500
1000
500
0
0
500
1000
1500
2000
2500
3000
Appears to be a reasonable straight line.
Slope:
>> slope=weight(7)/length(7)
slope =
0.9521
3500
4000
4500
4500
4000
3500
3000
2500
2000
1500
1000
500
0
0
500
1000
1500
2000
2500
3000
3500
4000
Original data plot and curve (W=.9521*L3)
8000
7000
6000
5000
4000
3000
2000
1000
0
0
2
4
Finding the errors:
>> r=weight-.(9521*l^3)
6
8
10
12
14
16
18
20
r=
-0.0284 -0.3121 -4.3380 41.0656 -51.4705 384.7712
0.1984
Note: In order to get the powers correctly I had to use several steps. This is for finding x3.
Step 1: Enter x
Step 2: nx = 3*log(x)
Step 3. nx1=exp(x)
nx1 is the x3 data.