Category Archives: geometry

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

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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

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Real problems with conic sections (ellipse, parabola)

So there is an oval hole in a metal casting. It’s supposed to be an elliptical hole. Is it ????? How can we find out ?????
A good start would be to find the line which would be the major axis if it was elliptical. This turns out to be an engineering problem, not a mathematical one (I cannot see a way!). If the oval curve has an axis of symmetry then the method below will find it:

Firstly, get a computer picture of the oval.
Take two circles, of different radii, and push them along until each one touches the oval in two places.

ellipse12

The line joining the two centers will be the axis of symmetry if there is one (this can be shown mathematically).

ellipse34

ellipse5

The equation of an ellipse uses the lengths of the major and minor axes. Do it !

The closeness to elliptic can be assessed in various ways. Think of one.

next…..finding the focus of a parabolic shape

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GEOSTRUCT, a program for investigative geometry

I have been developing this computer software / program / application for some years now, and it is now accessible as a web page, to run in your browser.

It provides basic geometric construction facilities, with lines, points and circles, from which endless possibilities follow.

Just try it out, it’s free.

Click on this or copy and paste for later : www.mathcomesalive.com/geostruct/geostructforbrowser1.html

.Here are some of the basic features, and examples of more advanced constructions, almost all based on straightedge and compass, from “make line pass through a point” to “intersection of two circles”, and dynamic constructions with rolling and rotating circles.

help pic 1
Two lines, with points placed on them
help pic 3
Three random lines with two points of intersection generated
help pic 6
Five free points, three generated circles and a center point
help pic 7
Three free points, connected as point pairs, medians generated
help pic 5
Two free circles and three free points, point pairs and centers generated
gif line and circle
GIF showing points of intersection of a line with a circle
hypocycloid locus
Construction for locus of hypocycloid
circle in a segment
gif002
GIF showing a dilation (stretch) in the horizontal direction
gif piston cylinder
Piston and flywheel
gif touching2circles
Construction for circle touching two circles
gif parabola
Construction for the locus of a parabola, focus-directrix definition.

 

 

 

 

 

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Maths(UK) is full of surprises

The radical axis theorem
radical axis theoremradical axis theorem2
How come that after 55 or so years of involment with math this came as a big surprise !!!!!

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Duality, fundamental and profound, but here’s a starter for you.

Duality, how things are connected in unexpected ways. The simplest case is that of the five regular Platonic solids, the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron. They all look rather different, BUT…..

take any one of them and find the mid point of each of the faces, join these points up, and you get one of the five regular Platonic solids. Do it to this new one and you get back to the original one. Calling the operation “Doit” we get

tetrahedron –Doit–> tetrahedron –Doit–> tetrahedron
cube –Doit–> octahedron –Doit–> cube
dodecahedron –Doit–> icosahedron –Doit–> dodecahedron

The sizes may change, but we are only interested in the shapes.

This is called a Duality relationship, in which the tetrahedron is the dual of itself, the cube and octahedron are duals of each other, and the dodecahedron and icosahedron are also duals of each other.

Now we will look at lines and points in the x-y plane.

3x – 2y = 4 and y = (3/2)x + 2 and 3x – 2y – 4 = 0 are different ways of describing the same line, but there are many more. We can multiply every coefficient, including the constant, by any number not 0 and the result describes the same line, for example 6x – 4y = 8, or 0.75x – 0.5y = 1, or -0.75x + 0.5y + 1 = 0

This means that a line can be described entirely by two numbers, the x and the y coefficients found when the line equation is written in the last of the forms given above. Generally this is ax + by + 1 = 0

Now any point in the plane needs two numbers to specify it, the x and the y coordinates, for example (2,3)

So if a line needs two numbers and a point needs two numbers then given two numbers p and q I can choose to use then to describe a point or a line. So the numbers p and q can be the point (p,q) or the line px + qy + 1 = 0

The word “dual” is used in this situation. The point (p,q) is the dual of the line px + qy + 1 = 0, and vice versa.

dual of a rotating line cleaned up1

The line joining the points C and D is dual to the point K, in red.  The line equation is 2x + y = 3, and we rewrite it in the “standard” form as  -0.67x – 0.33y +1 = 0  so we get  (-0.67, -0.33) for the coordinates of the dual point K.

A quick calculation (using the well known formula) shows that the distance of the line from the origin multiplied by the distance of the point from the origin is a constant (in this case 1).

The second picture shows the construction of the dual point.

dual of a rotating line construction1

What happens as we move the line about ? Parallel to itself, the dual point moves out and in.

More interesting is what happens when we rotate the line around a fixed point on the line:

gif duality rotating line

The line passes through the fixed point C.  The dual point traces out a straight line, shown in green.

This can be interpreted as “A point can be seen as a set of concurrent lines”, just as a line can be seen as a set of collinear points (we have fewer problems with the latter).

It gets more interesting when we consider a curve. There are two ways of looking at a curve, one as a (fairly nicely) organized set of points ( a locus), and the other as a set of (fairly nicely) arranged lines (an envelope).

A circle is a set of points equidistant from a central point, but it is also the envelope of a set of lines equidistant from a central point (the tangent lines).

So what happens when we look for the dual of a circle? We can either find the line dual to each point on the circle, or find the point dual to each tangent line to the circle. Here’s both:

dual of a circle4

In this case the circle being dualled is the one with center C, and the result is a hyperbola, shown in green.  The result can be deduced analytically, but it is a pain to do so.

dual of a circle3

The hyperbola again.  It doesn’t look quite perfect, probably due to rounding errors.

The question remains – If I do the dualling operation on the hyperbola, will I get back to the circle ?

Also, why a hyperbola and not an ellipse ? Looking at what is going on suggests that if the circle to be dualled has the origin inside then we will get an ellipse. This argument can be made more believable with a little care !

If you get this far and want more, try this very heavy article:

http://en.wikipedia.org/wiki/Duality_(mathematics)

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What is Algebra really for ?

An example tells a good tale.

Translation of a line in an x-y coordinate system:

Take a line  y = 2x -3, and translate it by 4 up and 5 to the right.

Simple approach : The point P = (2, 1) is on the line (so are some others!). Let us translate the point to get Q = (2+5, 1+4), which is Q = (7, 5), and find the line through Q parallel to the original line.  The only thing that changes is the c value, so the new equation is  y = 2x + c, and it must pass through Q.  So we require  5 = 14 + c, giving the value of c as -9.

Not much algebra there, but a horrible question remains – “What happened to all the other points on the line ?”

We try a more algebraic approach – with any old line  ax + by + c = 0, and any old translation, q up and p to the right.

First thing is to find a point on the line – “What ? We don’t know ANYTHING about the line.”

This is where algebra comes to the rescue. Let us suppose (state) that a point P = (d, e) IS on the line.

Then ad + be + c = 0

Now we can move the point P to Q = (d + p, e + q)  (as with the numbers earlier), and make the new line pass through this point:  This requires a new constant c (call it newc) and we then have  a(d + p) + b(e + q) + ‘newc’ = 0

Expand the parentheses (UK brackets, and it’s shorter) to get  ad + ap + be + bq + ‘newc’ = 0

Some inspired rearranging gives  ‘newc’ = -ap – bq – ad – be, which is equal to -(ap + bq) – (ad + be)

“Why did you do that last step ?” – “Because I looked back a few lines and figured that  (ad + be) = -c, which not only simplifies the expression, it also disposes of the unspecified point  P.

End result is:  Translated line equation is  ax + by + ‘newc’ =0,  that is,  ax + by + c – (ap + bq) = 0

and the job is done for ALL lines, even the vertical ones, and ALL translations. Also we can be sure that we know what has happened to ALL the points on the line.

I am not going to check this with the numerical example, you are !

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Angle between two lines in the plane……Vector product in 3D…….connections???????

So I was in the middle of converting my geometry application Geostruct (we used to call them programs) into javascript

 get it here with the introduction .doc file here

when I decided that the “angle between two lines” routine needed a rewrite. Some surprises ensued !

angle between two lines pic1

Two lines,  ax + by + c  = 0  and  px + qy + r = 0

Their slopes (gradients) are  -a/b = tan(θ)  and  -p/q = tan(φ)

The angle between the lines is  φ – θ,

so it would be nice to know something about  tan(φ – θ)

Back to basics, where  tan(φ – θ) = sin(φ– θ)/cos(φ– θ),

and we have the two expansions

sin(φ– θ) = sin(φ)cos(θ) – cos(φ)sin(θ)   and

cos(φ– θ) = cos(φ)cos(θ) + sin(φ)sin(θ)

So we have  tan(φ – θ) = (sin(φ)cos(θ) – cos(φ)sin(θ))/( cos(φ)cos(θ) + sin(φ)sin(θ))

Dividing top and bottom by  cos(φ)sin(θ)  and skipping some tedious algebra we get

tan(φ – θ)   =  (tan(φ) – tan(θ))/(1 +  tan(φ)tan(θ))

This is where the books stop, which turns out to be a real shame !

Going back to the two lines and their equations, the two lines

ax + by = 0  and  px + qy = 0

have the same angle between them (some things are toooo obvious)

Things are simpler if we look at these two lines through the origin when they both have positive slope.

Take b and q as positive and write the equations as   ax – by = 0  and  px – qy = 0

Then the point whose coordinates are (b,a) lies on the first and (q,p) lies on the second.

angle between two lines pic2

Also, the slopes of the two lines are now  a/b , tan(θ)   and  p/q , tan(φ)

Let us put these into the  tan(φ – θ)   equation above, and once more after tedious algebra

tan(φ – θ)  = (bp – aq)/(ap + bq)

which is a very nice formula for the tan of the angle between two lines.

This is ok if we are interested just in “the angle between the lines”,  but if we are considering rotations, and one of the lines is the “first” one, then the tangent is inadequate. We need both the sine and the cosine of the angle to establish size AND direction (clockwise or anticlockwise).

The formula above can be seen as showing  cos(φ– θ)  as  (ap + bq) divided by something

and  sin(φ– θ)  as  (bp – aq) divided by the same something.

Calling the something  M  it is fairly clear that    (ap + bq)2 + (bp – aq) 2 = M2

and more tedious algebra and some “observation and making use of structure” gives

M= (a2 + b2)(p2 + q2)

and we now have

sin(φ– θ)  = (bp – aq)/M  and  cos(φ– θ) = (bq + ap)/M

and M is the product of the lengths of the two line segments, from the origin to (b,a) and from the origin to (q,p)

It was at this point that I saw M times the sine of the “angle between” as twice the well known formula for the area of a triangle. “half a b sin(C)”, or, if you prefer, the area of the parallelogram defined by the two line segments.

Suddenly I saw all this in 2D vector terms, with bq + ap being the dot product of (b,a) and (q,p) , and bp – aq as being part of the definition of the 3D vector or cross product, in fact the only non zero component (and in the z direction), since in 3D terms our two vectors lie in the xy plane.

Why is the “vector product” not considered in the 2D case ??? It is simpler, and looking at the formula for sine , above, we have a 2D interpretation of the “vector”or cross product as twice the area of the triangle formed by (b,a) and (q,p). (just as in the standard 3D definition, but treated as a scalar).

So “bang goes” the common terms, scalar product for c . d  and vector product for  c X d

Dot product and cross product are much better anyway, and a bit of ingenuity will lead you to the reason for the word “cross”.

This is one of the things implemented using this approach:

gif rolling circle

Anyway, the end result of all this, for rotating points on a circle, was a calculation process which did not require the actual calculation of any angle. No arctan( ) !

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Rigid motions are actually useful !

I am currently reconstructing my geometrical construction application, Geostruct, to run in a web page using javascript.

One of the actions is to find the points of intersection of a straight line with a circle. Here is a gif showing the result:

gif line and circle

The algebra needed to solve the two simultaneous equations is straightforward, but a pain in the butt to get right and code up, so I thought “Why not solve the equations for the very simple case of the circle centered at the origin and the line vertical, at the same distance (a) from the centre of the circle

line vertical circle at origin

Then it is a simple matter of  rotating the two points (a,b) and (a,-b) about the origin, through the angle made by the original line to the vertical, and then translating the circle back to its original position, the translated points are then the desired points of intersection.

The same routine can be used for the intersection of two circles, with a little bit of prior calculation.

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An almost real life geometry problem

I needed to move a point around a circle, in a computer graphics application, using the mouse pointer. It is clearly not sensible to have mouse pointer on the point all the time, so the problem was

“For a point anywhere, where is the point both on the circle and on the radial line?”

point on circle 2

It may help to see the situation without the coordinate grid on show:

point on circle 1

This is a problem with many ways to a solution, some of them incredibly messy !

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