Category Archives: construction

Linear transformations, geometrically

 

Following a recent blog post relating a transformation of points on a line to points on another line to the graph of the equation relating the input and output I thought it would be interesting to explore the linear and affine mappings of a plane to itself from a geometrical construction perspective.

It was ! (To me anyway)

These linear mappings  (rigid and not so rigid motions) are usually  approached in descriptive and manipulative  ways, but always very specifically. I wanted to go directly from the transformation as equations directly to the transformation as geometry.

Taking an example, (x,y) maps to (X,Y) with the linear equations

X = x + y + 1 and Y = -0.5x +y

it was necessary to construct a point on the x axis with the value of X, and likewise a point on the y axis with the value of Y. The transformed (x,y) is then the point (X,Y) on the plane.

The construction below shows the points and lines needed to establish the point(X,0), which is G in the picture, starting with the point D as the (x,y)

 

transform of x

The corresponding construction was done for Y, and the resulting point (X,Y) is point J. Point D was then forced to lie on a line, the sloping blue line, and as it is moved along the line the transformed point J moves on another line

gif for lin affine trans1

Now the (x,y) point (B in the picture below, don’t ask why!) is forced to move on the blue circle. What does the transformed point do? It moves on an ellipse, whose size and orientation are determined by the actual transformation. At this point matrix methods become very handy.(though the 2D matrix methods cannot deal with translations)

gif for lin affine trans2

All this was constructed with my geometrical construction program (APP if you like) called GEOSTRUCT and available as a free web based application from

http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

The program produces a listing of all the actions requested, and these are listed below for this application:

Line bb moved to pass through Point A
New line cc created, through points B and C
New Point D
New line dd created, through Point D, at right angles to Line aa
New line ee created, through Point D, at right angles to Line bb
New line ff created, through Point D, parallel to Line cc
New point E created as the intersection of Line ff and Line aa
New line gg created, through Point E, at right angles to Line aa
New line hh created, through Point B, at right angles to Line bb
New point F created as the intersection of Line hh and Line gg
New line ii created, through Point F, parallel to Line cc
New point G created as the intersection of Line ii and Line aa

G is the X coordinate, from X = x + y + 1 (added by me)

New line jj created, through Point G, at right angles to Line aa
New line kk created, through Point D, at right angles to Line cc
New point H created as the intersection of Line kk and Line bb
New point I created, as midpoint of points H and B
New line ll created, through Point I, at right angles to Line bb
New point J created as the intersection of Line ll and Line jj

J is the Y coordinate, from Y = -x/2 + y  (added by me)
and K is the transformed point (X,Y) Point J chosen as the tracking point (added by me)

New Line mm
Point D moved and placed on Line mm

 

1 Comment

Filed under algebra, conics, construction, geometrical, geometry app, geostruct, math, ordered pairs, rigid motion, teaching, transformations, Uncategorized

Halving a triangle, follow-up number two, pursuing the hyperbola

Halving the triangle, any triangle, led to the equation XY = 2 as the condition on the points on two sides of the triangle, distant X and Y from the vertex.
The envelope of this set of lines turned out to be a hyperbola.
But XY = 2 defines a hyperbola – what is the connection ?

I took xy = 1 for the condition, on a standard xy grid, and wrote it as representing a function x —-> y, namely y = 1/x
The two points of interest are then (x,0) on the x-axis and (0,1/x) on the y-axis.
We need the equation of the line joining these two points, so first of all we have to see that our x, above, is telling us which line we are talking about, and so it is a parameter for the line.
We had better give it a different name, say p.
Now we can find the equation of the line in x,y form, using (p,0) and (0,1/p) for the two points:
(y – 0)/(1/p – 0) = (x – p)/(0 – p)
which is easier to read as yp = -x/p + 1, and easier to process as yp2 = -x + p

Now comes the fun bit !
To find the envelope of a set of straight lines we have to find the points of intersection of adjacent lines (really? adjacent?). To do this we have to find the partial derivative (derivative treating almost everything as constant) of the line equation with respect to the parameter p. A later post will reveal all about this mystifying procedure).
So do it and get  2yp = 1

And then eliminate p from the two equations, the line one and the derived one:
From the derived equation we get p = 1/(2y), so substituting in the line equation gives 1 = 2xy
This is the equation of the envelope, and written in functional form it is
y = 1/(2x), or (1/2)(1/x)
Yes ! Another rectangular hyperbola, with the same asymptotes.
(write it as xy = 1/2 if you like)

Now I thought “What will this process do with y = x2 ?”
So off I go, and to cut a long story short I found the following:
For y = x2 the envelope was y = (-1/4)x2, also a multiple of the original, with factor -1/4
parabola by axa track point
parabola by axa track point and line

Some surprise at this point, so I did it for 1/x2 and for x3
Similar results: Same function, with different factors.
Try it yourself ! ! ! ! ! ! !

This was too much ! No stopping ! Must find the general case ! (y = xk)
Skipping the now familiar details (left to the reader, in time honoured fashion) I found the following:

Original equation: y = xk

Equation of envelope: y = xk multiplied by -(k-1)k-1/kk

which I did think was quite neat.
—————————————————————
The next post will be the last follow-up to the triangle halving.

—————————————————————
Constructions made with GEOSTRUCT, an online browser application:

To get geostruct from the net click
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

and to download the .doc instructions file
http://www.mathcomesalive.com/geostruct/geostruct basics.doc

3 Comments

Filed under algebra, calculus, construction, envelope, geometrical

Halving a triangle, follow-up number one, ellipse

The previous post is “Analytic (coordinate) geometry has its good points, but elegance is not one of them”, in which a formula for any line cutting a triangle in half was found. The envelope of the cutting line for one section (of three) of the triangle was found to be always a hyperbola, which got me thinking “How do I get an ellipse?”. Clearly not by cutting a triangle in half, which involved taking two points A’ and B’on adjacent sides of the triangle, and making the product of their distances from the point of intersection of the sides equal to half the product of the lengths of the two sides.

So we cannot take two distances along two lines from the same point, lets try two distances from separate points on two lines, and keep the product of the distances constant. Magic:
envelope ellipse cropped
The base line is AF. Line DL is set parallel to AF. B and G are the two points of interest, where AB and DG are the two distances, and the envelope of the line BG is an ellipse which touches AF and DL. The really interesting thing about this is that the lines do not have to be parallel, and that as the points A and F are placed nearer and nearer to the point of intersection the ellipse becomes more and more hyperbolic at the nearby end.

The next post will be a different follow-up to the triangle halving.

Constructions made with GEOSTRUCT, an online browser application:

To get geostruct from the net click
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

and to download the .doc instructions file
http://www.mathcomesalive.com/geostruct/geostruct basics.doc

 

Leave a comment

Filed under conic sections, construction, geometrical

Vertex of a parabola – language in math again

Here are some definitions of the vertex of a parabola.

One is complete garbage, one is correct  though put rather chattily.

The rest are not definitions, though very popular (this is just a selection). But they are true statements

Mathwarehouse: The vertex of a parabola is the highest or lowest point, also known as the maximum or minimum of a
parabola.
Mathopenref: A parabola is the shape defined by a quadratic equation. The vertex is the peak in the curve as shown on
the right. The peak will be pointing either downwards or upwards depending on the sign of the x2 term.
Virtualnerd: Each quadratic equation has either a maximum or minimum, but did you that this point has a special name?
In a quadratic equation, this point is called the vertex!
Mathwords: Vertex of a Parabola: The point at which a parabola makes its sharpest turn.
Purplemath: The point on this axis which is exactly midway between the focus and the directrix is the “vertex”; the vertex is the point where the parabola changes direction.
Wikibooks: One important point on the parabola itself is called the vertex, which is the point which has the smallest distance between both the focus and the directrix. Parabolas are symmetric, and their lines of symmetry pass through the vertex.
Hotmath: The vertex of a parabola is the point where the parabola crosses its axis of symmetry

Scoring is 10 points for finding the garbage definition and 5 points for the correctish definition !!!! Go for it!

When I studied parabolas, back in 1958 or so (!) the parabola had an apex. So I checked the meaning of vertex, and found that the word was frequently misused.

Here is a good account: https://en.wikipedia.org/wiki/Vertex_(curve)

Basically a vertex of a curve is a point where the curvature is a maximum or a minimum (in non math terms, most or least curved).

Here are two fourth degree polynomials, one has three vertices and the other has five. The maximum curvature points are indicated. The minimum curvature points are at the origin for the first curve, and at the points of inflexion for the second curve (curvature = zero)

01Gquartic201Gquartic0

A hyperbola has two vertices, one on each branch; they are the closest of any two points lying on opposite branches of the hyperbola, and they lie on the principal axis. On a parabola, the sole vertex lies on the axis of symmetry. On an ellipse, two of the four vertices lie on the major axis and two lie on the minor axis.

For a circle, which has constant curvature, every point is a vertex.

The center of curvature at a (nice) point on a curve is the center of the closest matching circle at that point. This circle will usually lie “outside” the curve on one side of the point, and “inside” the curve on the other side. Look carefully at the picture. It is called the osculating or kissing circle (from the Latin).

The center of curvature can be estimated by taking two point close to the point of interest, finding the tangents at these points, and then the lines at right angles to them and through the points. the center of curvature is roughly at the point of intersection of these two lines

01center of curvature

The diagram below shows this estimate, for the blue parabola, at the vertex.

02center of curvature

Finally (this has gone on further than expected!) I found this delightful gif.

01Lissajous-Curve+OsculatingCircle+3vectors_animated

4 Comments

Filed under conic sections, conics, construction, geometry, language in math, teaching

Real problems with conic sections (ellipse, parabola) part two

So suppose we have a parabolic curve and we want to find out stuff about it.

Its equation … Oh, we have no axes.

Its focus … That would be nice, but it is a bit out of reach.

Its axis, in fact its axis of symmetry … Fold it in half? But how?

Try the method of part one, with the ellipse. (previous post)

parabola find the focus1

This looks promising. I even get another axis, for my coordinate system, if I really want the equation.

Now, analysis of the standard equation for a parabola (see later) says that a line at 45 deg to the axis, as shown, cuts the parabola at a point four focal lengths from the axis. In the picture, marked on the “vertical”axis, this is the length DH

parabola find the focus2

So I need a point one quarter of the way from D to H. Easy !

parabola find the focus3

and then the circle center D, with radius DH/4 cuts the axis of the parabola at the focal point (the focus).

Even better, we get the directrix as well …

parabola find the focus4

and now for the mathy bit (well, you do the algebra, I did the picture)

parabola find the focus math

Yes, I know that this one points up and the previous one pointed to the right !

All diagrams were created with my geometrical construction program, GEOSTRUCT

You will find it here:

www.mathcomesalive.com/geostruct/geostructforbrowser1.html

Leave a comment

Filed under conic sections, construction, engineering, geometry, teaching