Class 7.2: Graphical Analysis
and Excel
Solving Problems Using
Graphical Analysis
Learning Objectives
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Learn to use tables and graphs as problem
solving tools
Learn and apply different types of graphs and
scales
Prepare graphs in Excel
Be able to edit graphs
Be able to plot data on log scale
Be able to determine the best-fit equations
for linear, exponential and power functions
Exercise
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Enter the following table in Excel
Independent Dependent Dependent
Variable, x Variable, y1 Variable, y2
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1
1
1
2500
5000
10
100
50
100
7500
1000
150
10000
10000
200
You can make your tables look nice by
formatting text and borders
Axis Formats (Scales)
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There are three common axis formats:
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Rectilinear: Two linear axes
Semi-log: one log axis
Log-log: two log axes
1 km
Linear scale:
Length (km)
1 km
10 km
Log scale:
Length (km)
Use of Logarithmic Scales
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A logarithmic scale is normally used to
plot numbers that span many orders of
magnitude
1
10
100
1000
10000
Creating Log Scales in Excel
Exercise (2 min): Create a graph using x and y1
only.
New Graph
12000
10000
8000
y1
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6000
y1 data
4000
2000
0
0
2000
4000
6000
x
8000
10000
12000
Creating Log Scales in Excel
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Now modify the graph so the data is
plotted as semi-log y
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This means that the y-axis is log scale and
the x-axis is linear.
Right click on the axis to be modified
and select “format axis”
Creating Log Scales in Excel
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On the Scale tab,
select logarithmic
“OK”
Next, go to Chart
Options and select
the Gridlines tab.
Turn on (check) the
Minor gridlines for
the y-axis.
“OK”
Result: Graph is straight line.
New Graph
10000
y1
1000
100
y1 data
10
1
0
2000
4000
6000
x
8000
10000
12000
Exercise (8 min)
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Copy and Paste the graph twice.
Modify one of the new graphs to be
semi-log x
Modify the other new graph to be loglog
Note how the scale affects the shape of
the curve.
Result:semi-log x
New Graph
12000
10000
y1
8000
6000
y1
4000
2000
0
1
10
100
x
1000
10000
Result: log-log
New Graph
10000
y1
1000
100
y1
10
1
1
10
100
x
1000
10000
Equations
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The equation that represents a straight line on
each type of scale is:
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Linear (rectilinear): y = mx + b
Exponential (semi-log): y = bemx or y = b10mx
Power (log-log): y = bxm
The values of m and b can be determined if the
coordinates of 2 points on THE BEST-FIT LINE are
known:
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Insert the values of x and y for each point in the
equation (2 equations)
Solve for m and b (2 unknowns)
Equations (CAUTION)
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The values of m and b can be determined if
the coordinates of 2 points on THE BESTFIT LINE are known.
You must select the points FROM THE LINE
to compute m and b. In general, this will not
be a data point from the data set. The
exception - if the data point lies on the bestfit line.
Consider the data set:
X
Y
1
2
3
4
5
6
7
8
9
10
4
8
10
12
11
16
18
19
20
24
Team Exercise (3 minutes)
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Using only the data from the table, determine
the equation of the line that best fits the data.
When your team has completed this exercise,
have one member write it on the board.
How well do the equations agree from each team?
Could you obtain a better “fit” if the data were
graphed?
Which data points should be used to
determine the equation of this best-fit
line?
Example of Be st-Fit Line
30
Y, Dependent Variable (No Units)
25
20
15
10
5
0
0
2
4
6
X, Independent Variable (No Units)
8
10
12
Which data points should be used to
determine the equation of this best-fit
line?
Example of Be st-Fit Line
30
Y, Dependent Variable (No Units)
25
20
15
10
5
0
0
2
4
6
X, Independent Variable (No Units)
8
10
12
Comparing Results
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How does this equation compare with
those written on the board (i.ecomputed without graphing) ?
CONCLUSION: NEVER try to fit a curve
(line) to data without graphing or using
a mathematical solution ( i.e –
regression)
What about semi-log graphs?
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Remember, straight lines on semi-log
graphs are EXPONENTIAL functions.
y b*e
mx
ln( y )  ln(b * e )
mx
ln( y )  ln(b)  ln( e )
ln( y )  ln(b)  mx * ln( e)
ln( y )  ln(b)  mx
mx
What about log-log graphs?
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Remember, straight lines on log-log
graphs are POWER functions.
y b*x
m
ln( y )  ln(b * x )
m
ln( y )  ln(b)  ln( x )
ln( y )  ln(b)  m * ln( x )
m
Example
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Points (0.1, 2) and (6, 20) are taken from a
straight line on a rectilinear graph.
Find the equation of the line, that is use
these two points to solve for m and b.
Solution:
2 = m(0.1) + b
a)
20 = m(6) + b
b)
Solving a) & b) simultaneously:
m = 3.05, b = 1.69
Thus: y = 3.05x + 1.69
Pairs Exercise (10 min)
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FRONT PAIR:
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Points (0.1, 2) and (6, 20) are taken from a straight
line on a log-log graph.
Find the equation of the line, ie - solve for m and b.
BACK PAIR:
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Points (0.1, 2) and (6, 20) are taken from a straight
line on a semi-log graph.
Find the equation(s) of the line, ie - solve for m and
b.
Interpolation
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Interpolation is the process of estimating a value
for a point that lies on a curve between known
data points
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Linear interpolation assumes a straight line between the
known data points
One Method:
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Select the two points with known coordinates
Determine the equation of the line that passes through
the two points
Insert the X value of the desired point in the equation
and calculate the Y value
Individual Exercise (5 min)
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Given the following set of points, find
y2 using linear interpolation.
(x1,y1) = (1,18)
(x2,y2) = (2.4,y2)
(x3,y3) = (4,35)
Assignment #13
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DUE:
TEAM ASSIGNMENT
See Handout