Chapter 11: The Dynamic Geometry of Calculus
In Chapter 6 we introduced you to Sketchpad’s function plotting capabilities. In this
chapter we shall use those capabilities to explore the geometry of calculus, starting with the
relation between secant lines to a curve, tangents to that same curve and the derivative of the
function that generates the curve. We also introduced in Chapter 6 the idea of using sliders
(variable horizontal line segments) to vary the parameters of a function. The following section
leads you though the steps to create your own slider tool that you will use for creating many
functions in this chapter.
Coordinate-based Sliders
It is fairly easy to create sliders (dynamic line segments) to provide values that can be
manipulated simply by moving the end-point of the slider.

In a New Sketch go to the Graph menu and select Define Coordinate System.

Create a free point in the sketch and label it A.

Select point A and the x-axis and construct a line parallel to the x-axis through point A.

Place a free point B on this line.

Hide the line through points A and B.

Construct a segment between A and B.

Select points A and B and measure their abscissa.

Calculate xB-xA and re-label this calculation b.

Hide the values for xB and xA and turn off the label for point A (click on point A with the
text tool).
You are now ready to create a slider tool from this construction. Follow the steps below:

Select the x-axis, the segment and its two endpoints, and the value labeled b.

Click on the last button on the tool bar

Select Create Tool from the sub-menu that appears.

Name the tool slider_tool and check the Show Script View box.
and hold the mouse button down.
A window similar to Figure 11.1 below should appear:
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Figure 11.1: Script View window for the Slider_Tool

Double click on the first given (1. Straight Object x).

Check the box “Automatically Match Sketch Object” in the dialog window that appears.
“Straight Object x” should move into the Assuming portion of the Script View window.

Save your sketch as “Slider_tool” into the Tool Folder.
Your slider_tool will now be available for use whenever you open GSP 4.
Creating a Function using Sliders as Parameters
Use your slider tool to create four sliders in a new sketch. The coordinate axes will be
created automatically. Your Sliders (segments) will have only one end-point labeled. Four
measures will appear. Re-label these measures according to the labels on your sliders. So as not
to confuse which endpoint is the moveable one you can hide the unlabelled endpoint of each
slider. Use the measures of each slider to create and plot a new function: f(x) = ax3+bx2+cx+d.
You should have something like figure 11.2 below.
2
a = 0.42
A
b = -1.30
c = -1.15
d = 2.70
6
B
C
D
4
f x  = a  x 3 +b  x 2 +c  x+d
2
-10
-5
5
10
-2
-4
-6
Figure 11.2: Plot of a cubic function using 4 sliders as parameters.
Explore the roles of the four parameters on the cubic function by adjusting each of the
sliders. Create a new cubic function using three of the sliders as roots of the function: e.g. g(x) =
a(x-b)(x-c)(x-d). Explore how these two cubic functions differ and how they are similar.
Constructing a Secant to a Function
In this activity we shall use our slider tool to create a control point that will be used to measure
small changes in x. Call this control point ∂x. Place a free point on the x-axis of your function
plot for the cubic function f(x) = ax3+bx2+cx+d. Call this free point x. Measure the abscissa of
this point and label this measurement x. In the calculator create the measure of (x - ∂x/2) and (x
+ ∂x/2). Using the GSP calculator and each of these measures and the function f(x), create the
values f(x - ∂x/2) and f(x + ∂x/2) respectively. The following steps will plot these values of f(x)
as points on the graph and the secant between them.
1. Select the measures of (x - ∂x/2) and f(x - ∂x/2) in that order and Plot as (x, y) under the
Graph menu. A point should appear on your function graph.
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2. Do the same for (x + ∂x/2) and f(x + ∂x/2) A second point should appear on your function
graph.
3. Select these two new points on the graph and construct a line through them.
4. Vary the length of ∂x to see what happens to this secant line.
As ∂x becomes very small this secant line appears to become tangent to the curve
representing the cubic function. With ∂x very small (<0.1) move your free x point along the xaxis to see how the secant line travels along the curve (see Figure 11.3).
a
b
c
d
=
=
=
=
0.31
-1.30
-1.15
2.70
A
Žx
6
x x-
B
C
D
f
4
f x  = a x 3 +b x 2 +c x+d
x x = 2.43
Žx = 0.11
-10
f x x  = -3.25

Žx
x x-
2
= 2.38
2
2

= -3.14
Žx
Žx
x x+
x
-5
5
f
-2

x x+
= 2.49
2
10
Žx
2

= -3.35
-4
-6
Figure 11.3
Constructing the Tangent and the Derivative
Lengthen ∂x so that you can see the two points distinctly on the graph of the function.
Plot the point x, f(x) on your graph. This point should appear between the two points that define
your secant line. Select the secant line and the point between the two points defining the line.
Construct a line parallel to the secant line through the point x, f(x). Measure the slope of this
“tangent” line. As you make ∂x very small the secant line becomes the tangent line. If you make
∂x large (∂x > 2.0) and move your free point x, it will be obvious that the line parallel to the
secant line through f(x) is not always tangent to the cubic function. Make ∂x small again. Now
4
plot the point defined by the measure of x and the slope of the line through the point f(x) [Select
the measure of x and the slope of the “tangent” line in that order and Plot as (x, y) from the
Graph menu.] With this new point still selected, construct its locus with respect to the free point
x on the x-axis. You should now have a sketch that looks something like figure 11.4. What shape
is this locus curve? How does this curve relate to the derivative of the cubic function?
a
b
c
d
=
=
=
=
0.31
-1.30
-1.15
2.70
A
Žx
6
x x-
B
C
D
f
4
f x  = a x 3 +b x 2 +c x+d
x x = 3.60
Žx = 0.07
f x x  = -3.58
x x+
x
-5
Slope tangent
Žx
x x-
= 1.74
2

= -3.64
Žx
Žx
-10

2
= 3.56
2
5
f
-2

x x+
= 3.64
2
10
Žx
2

= -3.51
tangent
-4
-6
Figure 11.4: Constructing the tangent to a cubic function and plotting its slope
Create the expression for the derivative of your cubic function by selecting the expression
for f(x) and choosing Derivative from the Graph menu. You should get the following
expression: f’(x) = 3ax2+2bx+c. Use the calculator to obtain the value of this expression for
point x. Make ∂x very small. What do you notice about the slope of the tangent line and the
value of this expression at point x? Move your free point x. Do these values remain the same?
Do they remain equal to each other?
The value of the derivative varies with the position of x, so we can graph this relationship.
Select the expression for the derivative (f’(x) = 3ax2+2bx+c) and choose Plot Function from the
Graph menu. What shape is the graph of the derivative function? Does this curve co-inside with
the locus of the slope of the tangent line through f(x)? Enlarge ∂x until you see some separation
5
between these two curves. Do the shapes of the curves remain similar even when they are
separated? Reduce ∂x until the two curves are again coincident.
Vary your coefficients and then write three things about the relationship between the graphs of
the cubic function and its derivative (see Figure 11.5).
a
b
c
d
=
=
=
=
0.31
-1.30
-1.15
2.70
A
Žx
6
x x-
B
C
D
f
4
f x  = a x 3 +b x 2 +c x+d
x x = 3.60
Žx = 0.07
f x x  = -3.58
x x+
x
-5
Slope tangent
= 1.74

Žx
x x-
= -3.64
2
Žx
Žx
-10

2
= 3.56
2
5
f
-2

x x+
= 3.64
2
10
Žx
2

= -3.51
f' x  = 3 ax 2 +2 b x+c
f' x x  = 1.74
-4
tangent
-6
Figure 11.5: Graphs of a cubic function and its derivative
Investigating the Derivative of the Quadratic
Start a new sketch, create the axes and place a free point on the x-axis. Re-label this point
x. Use your slider tool to create three control points, a, b and c. Follow the method you used for
the cubic function to create the function f(x) = ax2+bx+c and construct its graph. Create a slider
∂x and use this to construct a secant to your quadratic function. Plot the point x, f(x) on your
graph. Construct the line through f(x) parallel to your secant line. Move point x to see the
“tangent” and secant lines move around your parabola. Vary ∂x. Make it as large as you can.
Move x again. Does the “tangent” line ever appear not to be tangent to the curve? Go back to
your cubic sketch and try the same thing. What seems to be special about the quadratic function?
Assignment 11.1: Create the expression and plot the derivative of your quadratic
f’(x)=2ax+b. Does the value of f’(x) = slope of the secant line for any ∂x? Prove (algebraically)
6
that the secant line through the points f(x-∂x/2) and f(x+∂x/2) will always be parallel to the
tangent at f(x) for any quadratic function!
Tangents to sine and cosine functions
Using the construction methods above create a graph of a general sine function: f(x)=
a•sin(b•x+c) and plot the points (x+∂x/2, f(x+∂x/2)) and (x-∂x,/2 f(x-∂x/2)) to construct the secant
to the sine curve. [Note: You will need to change the Preferences for the angle measure under
the Edit menu to Radians.] Make ∂x very small and measure the slope of the secant-tangent line.
Plot x and slope as (x, y). Construct the locus of this new plotted point as x varies. In what ways
is this new curve similar to your sine curve? Vary a and c. How do these coefficients affect the
two curves? Vary b. How does this coefficient affect the two curves? Make b = 1.00. What do
you notice about the two curves? How are they different and how are they alike? Vary a and c
again. What do you notice? Set both a and b equal to 1; set c equal to zero. Make a conjecture
about the function that would produce the curve that graphs the slope of a tangent to sine x.
Assignment 11.2: Create the derivative function, g(x) of a•sin(b•x+c) using the parameter
measures and the New Function option from the Graph menu. Plot g(x). The graph of your
derivative function should coincide with the graph of the slope of the tangent to f(x) for very
small ∂x. If it does not, vary the coefficient b. What role does b play in the slope function?
7
a = 1.69
b = 1.48
c = 0.78
x: (0.30, 0.00)
x x = 0.30
4
Žx = 0.04
x x + Žx = 0.34
x x – Žx = 0.26
a sin( b ( x x + Žx ) + c ) = 1.62
a sin( b ( x x – Žx ) + c ) = 1.56
a sin( b x x + c ) = 1.59
a cos ( b x x + c ) = 0.57
Slope NM = 0.84
2
-8
-6
-4
0
-2
x
1
2
4
6
-2
Figure 11.6
In Figure 11.6 the red curve is the graph of f(x)= a•sin(b•x+c). The blue line is a secanttangent to this curve. The tall, dark blue curve is the graph of the slope of this tangent line as x
varies. The light blue curve is the graph of a•cos(b•x+c). When b=1 the two blue curves
coincide. Why? In Figure 11.6, b > 1. How is b affecting the slope function? Check your
derivative function by asking for the derivative of f(x) under the Graph menu.
Using the same process as above, construct a secant-tangent line to the general cosine
function g(x)= a•cos(b•x+c) (the light blue curve in figure 11.6). Plot the slope of this line as x
varies. Make a conjecture about the derivative of the cosine function.
Reflecting on secants, tangents, rates of change and derivatives
Reflect on your understanding of tangents to function curves and how they relate to “rate
of change” of the function value with respect to the independent variable (x-value). How does the
slope of a secant line relate to this “rate of change”? What is happening as the secant line gets
closer and closer to the tangent at a single point? It might help to think of the speed you are
traveling when driving a car. You can calculate an average speed for one hour of driving time by
finding how far you traveled during that hour. At any point in time during that hour, however you
8
8
will be traveling at a specific speed that may be different from that average speed. You can think
of your speed at any particular time as the change in distance during a very small time period
divided by that time period (e.g. one second). But this speed is again really an average speed:
your average during that one-second of travel time. If you plotted a graph of your distance
traveled against time traveling, then your speed for this one-second period would be the slope of
the secant line connecting the two distances at the beginning and end of that one-second period
(change in distance over change in time). To find your actual speed at any particular moment in
time you would need to reduce the time interval to very close to zero and the slope of the secant
line would become the slope of the tangent to the distance-time curve at that point in time. Did
varying the small change in x (the ∂x-segment) dynamically help you to visualize this limiting
idea of a secant line becoming a tangent? Does it make sense now to refer to the derivative of a
function y=f(x) as dy/dx (the limit of ∂y/∂x as ∂x->0)?
Area Under a Function Curve, Integration and the Anti-Derivative
Constructing Polygons to Approximate the Area
The word Integration carries with it the notion of joining together. If we consider the
situation of driving a car on a journey for two hours, for instance, the total distance traveled
during the two hours would be our average speed multiplied by the time traveled. If we were
able to travel the whole journey at this one speed (say 50 mph) then the speed-time graph would
be a horizontal straight line and the distance traveled would be the area of the rectangle indicated
200
in Figure 11.7. – that is 100 miles (50x2).
150
100
Speed
f  x  = 50
50
-3
-2
-1
1
2
3
Time
-50
Figure 11.7: Distance Traveled is Area of Rectangle
-100
On a real journey, however, it would be very difficult to travel for the first two hours at
-150
exactly one speed, as we have to start from rest and accelerate up to our cruising speed. If we
-200
came to a halt after two hours (without crashing) we would also have to slow down from 50 mph
to zero during the last few minutes of our journey. Our speed-time graph might look more like
Figure 11.8.
9
200
150
100
Speed
50
-3
-2
-1
1
2
3
Time
4
5
-50
Figure 11.8: Speed Variation Over 2-hour Journey
How could we figure out the
-100 total distance we had traveled given such variation in our
speed? Analyzing the three different parts of the journey indicates that for the first 30 minutes we
-150
accelerated smoothly from zero to 50 mph. We then traveled for an hour at 50 mph. The final
half hour we decelerated smoothly from 50 to zero. During the middle portion of the journey
when we were traveling at the constant speed of 50 mph we traveled 50 miles (speed x time). We
could argue that our average speed for the first half hour was 50/2 or 25 mph as we went from
zero to 50 smoothly. Similarly for the last half hour as we went from 50 to zero smoothly. Thus,
we could argue that we traveled 12.5 miles (25x0.5) during the first half hour and 12.5 miles
during the last half hour. Thus the total distance traveled would be 12.5+50+12.5, which equals
75 miles. The area between the graph of our speed function and the time axis is a trapezoid
whose area is 75 square units, thus the total distance traveled is still the area between the function
and the time axis. We have integrated the three different parts of the journey.
The picture of our speed variation
in figure 11.8 is still somewhat idealistic. The graph in
200
Figure 11.9 provides another possible way in which our speed could vary over the two-hour
150
journey. How could we figure out the total distance traveled in this situation?
100
Speed
50
-3
-2
-1
1
Time
2
3
4
5
-50
Figure 11.9: Variation of speed over a 2-hour journey.
-100 situation in figure 11.8 – that is, we could divide our
We could do what we did for the
journey into smaller parts for which we
could assume a fairly constant speed (or assume a
-150
reasonable average speed for that time interval) and then accumulate the distances traveled over
each of these small parts of the journey. We could do this systematically by taking a reading of
10
our speed every minute (say) and assume the average speed for that minute of travel was the
beginning speed, S0 plus the speed at the end of the minute S1 divided by two: (S0+ S1)/2. This is
in fact what we did for the first and last half hours of the journey depicted in figure 11.8. To
calculate the distance traveled during that one-minute interval we would multiply this average
speed by 1/60 of an hour (one minute). To find the total distance, we would add together all of
these small distances for all 120 minutes of the journey. Figure 11.10 shows a geometric
representation of the product (S0+ S1)/2*1/60 on a zoomed image of the graph in Figure 11.9; it
is a trapezoid with parallel sides of height 29.3 (S0) and 30.8 (S1) and width 1/60 of an hour. For
this particular one minute of our journey, the average speed was approximately 30 mph. Thus we
would have traveled approximately 1/2 mile during this one-minute interval. An approximation
for the total distance traveled would be the combined (integrated) areas of all of the trapezoids
for each minute of the journey, and this area would approximate the area of the region between
the curve and the time axis, shown in Figure 11.11.
40
Speed
30
20
10
0.1
0.2
0.3
Time
0.4
0.5
Figure 11.10:
Area of Trapezoid Indicates Distance Traveled During One Minute
-10
-20
-30
11
0.6
0.7
13 0
12 0
11 0
10 0
90
80
70
Spee d
60
50
40
30
20
10
E
0.5
Time
1
1.5
2
G
-10
signed area betw een curve and axis
= 89.44
-20
Figure 11.11: Showing All 120 Trapezoids and the Accumulated Area
According to the calculation in Figure 11.11, the total distance traveled on a trip whose
speed variation is represented by the curve in Figure 11.11 would be approximately 89.5 miles.
This method of calculating the area between a function curve and the horizontal axis
(independent variable axis), between two points on this axis (the lower limit and upper limit), is
called Riemann Integration after the great German mathematician Georg Friedrich Riemann
(1826-1866). The measure of the area is called the Riemann integral or Definite Integral of the
function between the upper and lower limits of the independent variable.
Riemann used rectangles rather than trapezoids for accumulating the area under the curve.
Three different rectangles could be used. With respect to the trapezoid shown in Figure 11.10,
the left side of the trapezoid and the width could define one rectangle. Using this rectangle would
give rise to the Lower-Riemann Sum. If we create a rectangle with height of the right-hand side
of the trapezoid and the width, this would give us the Upper-Riemann Sum. If we replaced the
trapezoid with a rectangle whose height was mid-way between the two sides of the trapezoid we
would generate the Mid-Riemann Sum (assuming local straightness of our curve) and this would
be the same as using our trapezoid (why?). An interactive sketch called Integration.gsp that
dynamically models the Mid-Riemann sum is included in the Calculus folder in the sample
sketches that come with GSP 4. Figure 11.12 is taken from this sketch. I encourage the reader to
explore the sketch, changing the number of rectangles and also changing the parameters for the
cubic function. The sketch also provides a brief explanation of how it was constructed. It is
possible to also change the function in this sketch but the value for the definite integral is
calculated for the cubic function only. Nevertheless, it is interesting to see how the sum of the
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areas of the mid-rectangles changes as the number of rectangles used between the fixed upper
and lower bounds changes for any continuous function.
5
Approximating a Definite Integral
by Accumulating Values
n=4
fx = ax3+bx2+cx+d
Animate n
Change the number of increments ( n) or
the function f(x) by double-clicking them.
Change the coefficients a, b, c and d, or
the upper and lower domain bounds A
and B by dragging them.
Sum of Areas = 13.27340
4
3
a
2
-0.11
b
c
0.63
1
-0.10
d
0.75
-6
-4
-2
A
2
-1.00
4
B
5.00
6
-1
-2
Show Value
Show Explanation
Show Questions
How This Sketch Was Made
-3
Figure 11.12: From the sketch Integration.gsp
Using GSP 4 to Display and Calculate the Riemann Integral for any Curve
Note: This section is for those who feel comfortable using the GSP iteration transformation and
would like the challenge of creating your own sketch. Others may opt to use the ready-made
sketch entitled new_Riemann_area.gsp available on the CD and skip to the next section
after the:
******************************************************************
With the new function, parameter and iteration capabilities of GSP 4 it is possible to
display and calculate the area under any curve formed by a continuous function, f(x) and the xaxis. The method we shall use essentially creates the Mid-Riemann sum described above (see
13
figures 11.10 and 11.11) using the powerful “iterate” transformation option. The plan is as
follows:
Step 1: Create a function, f(x) using the function editor.
Step 2: Create a parameter (∂x) for the small change in the independent variable, x.
Step 3: Graph the function on a square coordinate system
Step 4: Place points LL and UL (for lower and upper limits of the Riemann integral) on
the x-axis. Note that point LL should be to the left of point UL.
Step 5: Obtain the x-coordinates of points LL and UL (xLL and xUL).
Step 6: Calculate f(xLL) and plot the point (xLL, f(xLL)) on the function graph.
Step 7: Construct the vertical segment connecting the point LL to the point (xLL, f(xLL)) on
the graph. This segment is essentially the left vertical side of the trapezoid in figure 11.10.
Let’s assume we started with a quadratic function defined using sliders for the parameters
a, b and c. Your GSP 4 sketch would look something like Figure 11.13 at this stage in the
construction.
2
Žx = 0.02
f  x  = a x 2 +b x+c
x LL = -0.50
1
x UL = 1.51
f x LL = 0.59
LL
UL
-2
2
4
-1
Figure 11.13: Initial Segment for Displaying Area Under a Function Curve
Figure 11.13 shows the initial conditions for creating a sequence of segments under the
quadratic function similar to those shown-2 in figure 11.11. However, we also want to actually
calculate the area under the curve as we create the sequence of segments that define the heights
of the mid-rectangles in the Riemann Sum. Each mid-rectangle will have a width of ∂x (shown
-3
with a value of 0.02 units in Figure 11.13). The height of each mid-rectangle will be the value of
the function at f(xi - ∂x/2), starting with the point x1=(xLL + ∂x), the right-hand side of the midrectangle. We use the right-hand point (rather than xLL) in order to define an iteration image, and
14
6
we use f(xi - ∂x/2) as our height rather than f(xi + ∂x/2) so that the last mid-rectangle will have the
point UL as its right-hand defining point, and a height of f(xUL - ∂x/2). The following steps will
construct the first image points for the iterative mapping:
Step 8: Plot the point (xLL+∂x, 0) [You can obtain the zero value by using YLL]
Step 9: Obtain the x-coordinate of this new point (xi) and calculate the value of f(xi - ∂x/2)
using the GSP 4 calculator.
Step 10: Calculate ∂x* f(xi - ∂x/2) using the GSP 4 calculator.
This last step provides us with the area of the first rectangle in the mid-Riemann sum. We now
need a way to accumulate (or sum) the area of each of the rectangles that will be produced in the
iteration process. We also need a way of specifying how many levels of iteration we need in
order to create the mid-Riemann sum between points LL and UL. This latter problem is easily
solved by dividing the coordinate distance between points LL and UL by ∂x (why?). However,
the iteration level needs to be a positive integer (why?). As the first level of iteration has already
been constructed, we need to truncate the number (XUL-XLL)/∂x to find the appropriate positive
integer.
The goal of accumulating and summing the areas of each mid-rectangle can be
accomplished by actually constructing a rectangle of unit width, and height defined by a
translation vertically of that unit width by each value of the area of each mid-rectangle (∂x* f(xi ∂x/2)). There is one problem with this plan, however. In GSP, translation has to be by a measured
length or distance. The area ∂x* f(xi - ∂x/2) is in terms of coordinate units. We therefore need to
convert this coordinate quantity to a measure by multiplying it by the actual length of a
coordinate unit. We shall then use this measure to translate the unit width of our accumulation
rectangle vertically. Under iteration, the translated images of the end-points of the unit width will
build up the total area of the mid-Riemann sum. The accumulation rectangle will also provide a
visual display of the area under the curve as a rectangle of unit width. If we place our unit width
on the x-axis, this method also has the advantage of displaying the signed area under the curve.
This is critically important for calculating the Riemann integral, as area between the x-axis and
parts of the curve that fall below the x-axis must be subtracted from the accumulating Riemann
15
sum. These areas formed by negative values of f(x) are negative areas1. By using ∂x* f(xi - ∂x/2)
to calculate the area of each mid-rectangle we do obtain negative values when f(x) is negative. If
more area falls below the x-axis than above it, the resulting Riemann sum will be negative and
our accumulation rectangle will appear below the x-axis. The actual value of the Riemann sum
can be obtained by simply measuring the y-coordinate of the last translated image of the unit
width of the accumulation rectangle. The following steps calculate the level of iteration and then
set up the accumulation rectangle ready for the iteration mapping.
Step 11: Calculate the following value using the GSP 4 calculator: trunc((XUL-XLL)/∂x ).
This will be used for the level of iteration.
Step 12: Measure the distance from the origin to the unit point of the coordinate system
and then use the GSP 4 calculator to multiply ∂x* f(xi - ∂x/2) by this distance.
Step 13: Place a free point H on the x-axis and translate this point horizontally by the
coordinate unit distance measured in Step 12. HH’ defines the unit width for the accumulation
rectangle.
Step 12: Translate both H and H’ vertically by the marked distance calculated in step 12
(∂x* f(xi - ∂x/2)*coordinate unit distance).
We are now ready to define the iteration mapping. In order to be able to see the
constructed image points of LL and H it will be necessary to increase ∂x to a larger value, e.g. 0.2
before defining the iteration.
Step 13: Change the value of the ∂x parameter to 0.2. Images of points LL (XLL+∂x) and
H (vertical translation of H) should now be discernable.
Step 14: Select points LL, H and the calculation from Step 11 (for depth of iteration).
Hold down the Shift key and click on the Transform menu. The last item in this menu should say
“Iterate to depth.” Select this item. An iteration mapping dialog box should appear indicating that
points LL and H are to be mapped to image points that you need to select. Select the point
1
The idea of negative area makes sense in the case of the graph representing the speed of a car
versus time traveled, where the area under the curve represents the total distance traveled during
the journey. When the speed becomes negative, the car is traveling backwards (or back towards
its starting point) thus the distance it travels during this time must be subtracted from the distance
traveled towards its end point.
16
(XLL+∂x) for the image of LL and the vertical translation of H for the image of point H and
iterate.
A screen similar to Figure 11.14 should result from the 14 steps above.
2
Žx = 0.20
f  x  = a x 2 +b x+c
x LL = -0.50
1
x UL = 1.67
f x LL = 0.62
x LL +Žx = -0.30
y LL = 0.00
LL
UL
F
D
H
G
x F = -0.30
 
 
Žx
f x F-
= 0.72
2
Žx f x F-
-1
Žx
DG = 2.54
2
= 0.14
cm
Žx
  
 
Žx f x F-
trunc
n
-2
DG = 0.36 cm
2
x UL - x LL
= 10.00
Žx
x LL
f x LL
x LL+Žx
-3
y LL
      
Žx
xF
f x F-
Žx f x F-
2
Žx
Žx f x F-
2
Žx
2
 DG

x U L- x LL
trunc
Žx
0
-0.50
0.62
-0.30
0.00
-0.30
0.72
0.14
0.36 cm
10.00
1
-0.30
0.81
-0.10
0.00
-0.10
0.89
0.18
0.45 cm
9.00
2
-0.10
0.96
0.10
0.10
1.02
0.20
0.52 cm
8.00
3
0.10
1.08
0.30
0.00
-4
0.00
0.30
1.13
0.23
0.57 cm
7.00
4
0.30
1.16
0.50
0.00
0.50
1.19
0.24
0.61 cm
6.00
5
0.50
1.21
0.70
0.00
0.70
1.23
0.25
0.62 cm
5.00
6
0.70
1.23
0.90
0.90
1.23
0.25
0.62 cm
4.00
7
0.90
1.21
1.10
0.00
-5
0.00
1.10
1.19
0.24
0.60 cm
3.00
8
1.10
1.16
1.30
0.00
1.30
1.12
0.22
0.57 cm
2.00
9
1.30
1.07
1.50
0.00
1.50
1.01
0.20
0.52 cm
1.00
10
1.50
0.95
1.70
0.00
1.70
0.88
0.18
0.44 cm
0.00

-6
Figure 11.14: Results of Iterating Points LL and H
The table of values generated by the iteration is optional (it can be deselected in the
iteration dialog box when defining the structure of the iteration). I intentionally generated the
table so that we could check the values generated through the 10 levels of iteration in this
example. The lower limit of the Riemann integral is at -0.50 and the upper limit is at 1.70. The
point F is the initial (constructed) image of point LL that defines the iteration. The value XF is
increased by ∂x (0.2) at each level of iteration, ending with the same value as the upper limit.
Thus we are using the correct depth of iteration to generate the values of the areas of all of the
mid-Riemann rectangles (∂x* f(xF- ∂x/2)). Points D and G are the origin and unit points for the
coordinate system. The iteration generates 11 values for the areas mid-rectangles. Each of these
17
values is multiplied by the measure of DG (the length of the coordinate unit) to produce the 11
vertical translation distances. The two sequences of vertical dots that begin at the point H on the
x-axis are the result of the iteration of points H and H’ from step 13 above under the vertical
translation by each of these successive distances. Thus these two sequences of dots form the
vertical sides of the accumulation rectangle for the area under the curve between points LL and
UL as desired. The following steps construct the interior of the accumulation rectangle using the
terminal points for the two iteration images (the vertical dots).
Step 15: Select the left vertical sequence of dots and choose Terminal point from the
Transform menu. Select the topmost dot on the right-hand vertical sequence (this is important as
there are actually two iteration images in this vertical sequence of dots) and then create the
terminal point for this iteration image. Hide all 3 iteration images (vertical sequences of dots).
You should be left with the two terminal points, the two original points on the x-axis (H and H’)
and two points immediately above H and H’.
Step 16: Hide the two points immediately above H and H’. Select the four remaining
points (in cyclic order) and construct the quadrilateral interior.
Step 17: Select the terminal point above H and measure its y-coordinate. This will be the
value for the signed area under the curve. Change its label to read “area under curve”.
Step 18: Select the iteration images for the points on the curve and the points on the xaxis under the curve (do not select the vertical segments created under the curve) and hide these
images. Also hide the table of values for the iteration.
Step 19: Change the value of ∂x to 0.02. You can hide any of the measures or calculations
that you don’t need visible.
The result of all of the above steps should look something like Figure 11.15.
18
N
2
Žx = 0.02
f  x  = a x 2 +b x+c
x LL = -0.50
1
x UL = 1.67
LL
UL
D

Žx f x F-

Žx
area under curv e

= 2.30
= 0.01
2
x UL - x LL
trunc
H
G
Žx

-1
= 108.00
-2
Figure 11.15: Area
Under the Curve of a Quadratic Function
-3
Move points LL and UL and observe the change in the area under the curve and the
-4 to the right of the function curve. Note that the area “under”
accumulation rectangle. Move UL
the curve now extends below the x-axis. What happened to the accumulation rectangle as you
-5
moved UL to the right of the curve?
Move UL further to the right until the accumulation
rectangle vanishes. What can you say about the area between the x-axis and the curve in this
situation?
-6
Vary the parameters of your quadratic function (by changing your sliders or your GSP 4
parameter values if you used parameters rather than sliders). You can also change the function by
editing it. Try creating a cubic function or a trigonometric function. You may have to reset angle
units to radians to see any of the trig function graphs.
******************************************************************
Varying the coefficient of the x-term (b) for area under a quadratic
Return to a quadratic function controlled by your sliders or parameters for a, b and c.
Construct the reflected image of point LL about the vertical axis. Move point UL to the reflection
of LL using a movement button. Now vary the parameter (b) for the coefficient of the x-term in
your function. What do you notice about the accumulation rectangle or the value for the area
under the curve? Figures 11.16 and 11.17 illustrate the situation for two different values of b.
19
Move UL to LL'
2
N
Žx = 0.02
f  x  = a x 2 +b x+c
x LL = -1.22
1
x UL = 1.22
c = 1.02
LL
UL
D
-2

Žx f x F-

Žx
2
4

H
6
area under curv e
8
= 2.00
= -0.01
2
x UL - x LL
trunc
G
Žx

-1
= 122.00
a = -0.43
-2
b = 0.60
-3
Figure 11.16: Area Under a Quadratic with Lower Bound = -Upper Bound
Move UL to LL'
-4
2
N
Žx = 0.02
f x  = a
x 2 +b
x+c
-5
x LL = -1.22
1
x UL = 1.22
c = 1.04
-6
LL
UL
D
-2

Žx f x F-

Žx
2
Žx
2

H
4
area under curv e
6
= 2.00
= 0.03
x UL - x LL
trunc
G

-1
= 122.00
a = -0.43
-2
b = -0.74
-3
Figure 11.17: Area Under a Quadratic Following Change in Coefficient b of the x-term
While the area under the quadratic
curve may vary very slightly, it basically is unaffected
-4
by changes in the coefficient of the x-term. The very slight variations are due to the error
generated by our method for generating -5the Riemann Integral. The theoretical integral is the limit
of the sum of the areas of the mid-rectangles as ∂x approaches a value of zero. Try making ∂x
even smaller (try 0.01) and see if there is-6 variation as b changes.
20
Assignment 11.3: Write an explanation for this apparent phenomenon. Are there other
functions (non-quadratic) that might exhibit a similar behavior?
Plotting the anti-derivative
Given that the area under the curve does vary as we move the point UL along the x-axis,
it is possible to plot the graph of this variation by first plotting a point with x-coordinate the same
as UL and y-coordinate the value of the area under the curve, and then constructing the locus of
this plotted point as UL varies. The following steps accomplish this:
Step 1: Select the values for XUL and “area under curve” in that order and choose Plot as
(x, y) under the Graph menu. A point should appear above (or below) the point UL. If you cannot
see a new point move LL closer to UL or to a position that reduces the area under the curve until
the new point appears.
Step 2: Select the new plotted point and the point UL in that order and choose Locus
from the Construct menu. A new curve should appear, passing through the new plotted point.
Experiment with moving point LL to see how this new curve changes. You can also
change the coefficients of your quadratic to see how that affects this new curve. Figure 11.18
illustrates one particular situation:
Move UL to LL'
2
Žx = 0.01
f  x  = a x 2 +b x+c
x LL = -2.29
1
x UL = 1.11
LL

Žx f x F-

x UL - x LL
trunc
Žx
Žx

UL
D
-2
G
2

4
6
N
= -0.03
2
H
area under curv e
= -0.26
-1
= 340.00
-2
c = 1.04
a = -0.43
-3
b = 0.93
Figure 11.18: The Graph of the Variation of the Area Under a Quadratic as UL Varies
-4
-5
21
-6
The new curve in Figure 11.18 looks like the graph of a cubic function. Note that the
curve becomes parallel to and almost coincident with the x-axis to the left of point LL. Try
moving pint UL to the left of point LL. What happens to the iterated images of the segment from
LL to the quadratic? What happens to the accumulation rectangle? What happens to the value
that determines the depth of iteration (trunc((XUL-XLL)/∂x))? The parameter that determines the
depth of iteration in GSP 4 must be a positive integer. If it is not, no iteration is performed. Thus
the area under the curve remains the value of the area of the first mid-rectangle calculated during
the initial conditions for the iteration (∂x* f(xF- ∂x/2)).
Assignment 11.4: Using the function editor, create a cubic function whose graph would
contain the curve you have just plotted. How does this cubic function relate to your quadratic
function? Why might it be called the “anti-derivative”? When you have a cubic function that fits
your locus for the variation of the area under the quadratic, ask GSP for the derivative of this
cubic function (select the function expression, not the graph, then go to the Graph menu and
choose Derivative). Compare this derivative function to your original quadratic function.
Plotting the slope of a tangent to the “area-under-the-curve” locus
Hide the graph of your cubic function so that you can see the locus for the “area-underthe-curve” variation. Using a method similar to that described in the section above titled
“Constructing a secant to a function curve” we can construct a secant to this locus curve that
approximates to a tangent as ∂x gets very small. First we shall use the plotted point that defined
the locus for the “area-under-the-curve” curve (step 1 from the previous section). We need one
more point very close to this point that will also be on this curve. We can use the point that is
defined by XUL+∂x. What will be the area of the mid-rectangle with base defined by XUL and
XUL+∂x? It will be ∂x*f(XUL+∂x/2). Why? Calculate these new values using the GSP calculator.
We now need to add this extra area to the area under the curve to calculate the new value for the
area under the curve if UL moved to the point UL + ∂x along the x-axis (why?). We should then
be able to plot a new point on our anti-derivative using the values (XUL+∂x) and (area-under the
curve + ∂x*f(XUL+∂x/2)) as the x and y coordinates. Construct a line through the original point on
the “area-under-the-curve” locus curve and this new point (that should also be on the curve).
22
Measure the slope of this line and then vary the point UL. Your line should move along your
locus curve as if it were a tangent to the curve.
We can now plot a point that will define the locus for the variation of the slope of this
“tangent” line as point UL moves along the x-axis. Select the values XUL and slope (in that order)
and Plot as (x, y).You should find this plotted point on the graph of the quadratic function above
or below point UL. As you move UL does this plotted point stay on your original quadratic
curve? If so, why? Construct the locus of this new point as UL varies. Does its locus coincide
with your quadratic curve?
The important point to reflect on in this latter part of the investigation is that we did not
do any symbolic integration or differentiation. We worked with the numeric quantities generated
by our construction of the Riemann Sum for the area under a curve. What we have demonstrated
is that the rate of change (slope of the tangent line) of the area-under-the curve of a quadratic
function varies as does the quadratic function. In other words the rate of change of the area under
the function (for a given range of x) is the function itself! This is the fundamental theorem of
calculus. Another way of putting this is indicated by the symbolic result we obtained when taking
the derivative of the cubic function that fit our “area-under-the-curve” locus: The derivative of
the ant-derivative of a function is the function itself! The following two figures illustrate the
locus of the slope of the tangent line to the “area-under-the-curve” locus, and the symbolic
verification of the fundamental theorem of calculus.
In figure 11.20, the function g(x) was generated from f(x) by the usual rules of integration
for polynomial expressions. The graph of g(x) passes through the origin as there is no constant
term. Note that the cubic function for the anti-derivative was formed by generating j(x)=g(x)g(XLL). Why did I need to subtract the constant g(XLL) from g(x) to obtain the correct position of
the cubic graph? Note that j’(x) (the derivative of j(x)) is identical to f(x) except for the extra
constant term 0/6, which has value zero.
23
2
Žx = 0.02
locus of area
locus of slope fn
f  x  = a x 2 +b x+c
x LL = -2.17
1
y=f(x)
Slope fn
LL
-4

Žxf x F-

Žx
G
4
area under curve

= -1.80
tangent
-1
x UL = -0.53
= 81.00
Žx
H
2
= -0.06
2
x UL - x LL
trunc

D
UL
-2
Slope tangent
O
P -2
= 0.44
N
c = 1.04
a = -0.43
-3
x UL +Žx = -0.51

b = 0.93
Žxf x UL +
Show Anti-Derivative
 area under curve
-4
Žx
2

= 0.01
 
+ Žxf x UL +
Žx

= -1.79
2
Figure 11.19: Locus of Slope of Tangent to “area-under-the-curve” Locus
3
g x  =
ax 3
3
bx 2
+
2
y=g(x)
+c x
j  x  = g  x -g x LL 
2
Move UL to LL'
j'  x  = a x 2 +b x+c+
f  x  = a x 2 +b x+c
Žx = 0.02
0
y=j(x)
locus of area
y=f(x)
6
1
x LL = -2.17
x UL = -0.53
LL
-4

Žxf x F-

Žx
2
x UL - x LL
trunc
Žx
-2

D
UL
G
2
= -0.06
H
4
area under curve

= -1.80
-1
= 81.00
O -2
N
c = 1.04
a = -0.43
-3
b = 0.93
Show tangent & slope locus
Hide Anti-Derivative
Figure 11.20: Derivative (j’(x)) of Anti-derivative (j(x)) of f(x) Compared to f(x)
24
The whole of the above investigation of area-under-a-curve was illustrated using a
general quadratic function. GSP 4 does not limit us to remaining with the original quadratic
function. The entire construction can be changed by simply editing the original function. For
example, Figure 11.21 shows the change in Figure 11.19 when f(x) is edited to become
a*sin(x2)+b*x+c and the angle unit is changed from degrees to radians.
3
2
Move UL to LL'
f  x  = a sin x 2 +b x+c
y=f(x)
Žx = 0.02
1
locus of slope fn
Slope fn
x LL = -2.17
x UL = -0.53
tang ent
LL
-4

Žx f x F-

Žx
2
x UL - x LL
trunc
Žx

D
UL
-2
G
H
2
= -0.01
4
6
area under curve
O
P

= -0.61
N
-1
= 81.00
x UL +Žx = -0.51
Slope tangent

= 0.44
Žx f x UL +
-2
2

= 0.01
c = 1.04
a = -0.43
 area under curve
-3
Žx
 
+ Žx f x UL +
Žx

= -0.61
2
b = 0.93
Hide tangent & slope locus
Show Anti-Derivative
-4
Figure 11.21: Graph of f(x)= a*sin(x2)+b*x+c and Its Accompanying Constructions
Reflections
Reflect back on all of the activities in this chapter. Did constructing the sketches for
tangents to a function (based on the limit of a secant to the function) provide you with insights
concerning the derivative of a function? Does the geometry of the situation (slope of the tangent
line) help understand the calculus?
Did creating the iterated construction for area under a curve (following the Riemann Sum
approach) provide you with any insights into integral calculus? Did the demonstration of the
25
Fundamental Theorem of Calculus (by constructing a tangent to the anti-derivative curve
generated from the Riemann Sum) provide a rationale for this theorem? How would you adapt
these activities for a pre-calculus or AP calculus class in High School?
26