Still photos – modelling with functions, regression and loci ( May 2012)
Adrian Oldknow aoldknow@yahoo.co.uk www.adrianoldknow.org.uk
Professor Emeritus, University of Chichester, UK; Research Fellow, University of Amsterdam,
Netherlands
( This is an extract from the longer article: GeoGebra - a vehicle for STEM )
Note that the organisers of the competition wish to thank Prof Oldknow for permission to use this
article as a resource.
“I took a photograph of some bridges over the River Cam in Cambridge – where Zsolt works
sometimes! I guessed that the arch of the bridge in the foreground might be well fitted by a
quadratic function – but how to test it? Well – import the picture into a GeoGebra graphics view, fix
it, and make it a bit more transparent. Now you can plot points, draw lines and show graphs over
the picture. By moving the origin of axes to a suitable point, and adjusting the scales by Zooming we
can simplify the task to finding the value of the coefficient c in the equation y = c x2. Using a slider
for c we find a good “by eye” fit when c = 0.04 .
We can approach the problem in a different way if we capture data from a set of points constructed
to lie on the image of the arch of the bridge. We can use the command “A1=x(D)” to make cell A1 of
a spreadsheet hold the x-coordinate of the point D. We can then use the command“B1=y(D)” to
make cell B1 of a spreadsheet hold the y-coordinate of the point D. We can do this for all the points
we placed on the bridge. We can highlight columns A and B and convert the data to a “List of
Points” which I’ve called “Bridge”. In the Spreadsheet view we can perform a “two variable
regression analysis” and fit a polynomial model of order 2 to the captured data. In the “Input” line
we can enter:
“f(x) = FitPoly[Bridge,2]” to create the function f(x) which is a least squares fit to the data – with c =
.038 .
As well as graphing functions, plotting
scattergrams and finding regression
models we can also make geometric
constructions. O is the origin and with
the x- and y-axes passing through it.
Point D is on the y-axis and D’ is its
reflection in O. The line through D
parallel to x-axis is drawn, and E is a
point on it. The parallel to the y-axis
through E is drawn. We seek to
construct the point F on this line so that
FE = FD’. This is the defining property of
a parabola. So D’FE must be an isosceles
triangle, and so F must be its vertex, and
hence it must lie on the perpendicular
bisector of its base D’E through the
midpoint M. So F is constructed as the
intersection of the perpendicular
bisector of D’E with the perpendicular through E to the “directrix”. The locus of F with E is a
parabola, with DE as its directrix and D’ as its focus. We can then slide D on the y-axis until the locus
is a good fit to the arch of the bridge. The set of green lines will envelope the parabola. The normal
to the locus at F bisects the angle D’FH – which is why a parabolic reflector is used in telescopes,
including Hubble, satellite dishes and flashlights!
It looks as if the focus of the parabola might just be on the surface of the River Cam! It would have
been helpful if I had included something of known length or height in the picture so we could get a
sense of scale. Maybe we could measure the length of the bridge in Google Earth, say? We’ll leave
that till later.
Further afield, in Singapore Harbour,
there is the Merlion water-spout. We
can find the height AB by searching on
Merlion on the Web! Does this give us
enough information to find the speed
with which the water-spout hits the
Finally, here is an exercise for you, based on an article I wrote a while ago on “Cubism and Cabri”!
You can find it at https://www.ncetm.org.uk/public/files/8423816/ATM-MT206.pdf.
Quite soon we will be able to model this using the 3D graphics in GeoGebra. For now, here are a
couple of screen shots showing a possible construction for the hexagonal pyramid and its frustum.