Kristian Lum

MATH499

 

These are notes on the paper On Caustics of Plane Curves by J.W. Bruce, P.J. Giblin, and C.G. Gibson. This paper was published in The American Mathematical Monthly, Vol. 88, No.9 (Nov., 1981). (This is older than me!!)

 

 

 

Claim: SQ makes the same angle with the tangent to W at Q as SP makes with the tangent to M at P.

Proof: We want to show that α=β.

From the above picture, we can deduce the angle made between the reflection off point P and the angle between SP and its reflection. By definition, the angle between SP and the tangent line to M at P and the angle between the tangent and the reflection are the same. Since the tangent is a straight line, we also know that the angle between those is π- 2α.

 

the line PQ intersects the tangent line to M at P forming two sets of vertical angles. The only one that is important to us is the one opposite the angle between the tangent and the reflection. From there, we also know the angle between SQ and its reflection off W. We know that SQ is perpendicular to the tangent to M at P, so that angle is π and the remaining angle is π/2 α.

By vertical angles again, the angle opposite the π/2 α angle is also π/2 α.

By Supplementary angles, we also know all the angles above the line tangent to W at Q.

And again by vertical angles, we know that α=β.

 

Example: For the circle x2+y2=r2, if the light source is within the circle, the negative pedal is an ellipse. In the graph below, the light source (for simplicitys sake) is at the center of the circle, and the circle is the unit circle. The green star denotes the light source, and the red lines are the negative pedal. (The MatLab code is at the end)

 

Here is another example where the point is at (.75, 0):

 

And lastly, one more example from inside the circle where the point is off center with respect to both x and y. This light source is at (-.5, .25).

 

 

When the point source is outside the circle, the negative pedal is a hyperbola. In this case, the point is at (-1.25, 0).

One more example of the light outside the circle, this time off center. This light is at (1.25, 1):

When the point is on the circle, the negative pedal looks like:

 

 

 

Example: Take the curve and the axes so this curve passes through (0,0). Let

f(x)= x6 + 10x2 + x. Then f(x)= 6x5+20x +1; f(x) = 30x4 +20. So, the curvature of f at (0,0) is 20/(1+1)3/2= 5*21/2.

 

Let g(x)=x^2. Then g(x)=2x and g(x)=2. So, the curvature of g at (0,0) is 2/(1)3/2= 2

 

So, the curvature of f > curvature of g at (0,0). f(x)

 

 

g(x)

 

You can see from their graphs that f(x) is much more curved than g(x). We already know this since it has a higher degree, but the curvature actually assigns it a value.

 

f(k)(x)=g(k)(x) for k=1, , n-1 but f(n)≠g(n).

 

Example: We can always orient the axes so that a point P is at (0,0) and the line tangent to it is y=0. So, for the following, let P be (0,0). Use the old f(x) and g(x).

 

f(0)=g(0)=0

f(0)=1; g(0)=0

So, our original f and g have 0 point contact.

Example: Now consider h(x)=x6 + x5 + x4 + x3 + x2 + x and

k(x)= x5 + x4 + x3 + x2 + x.

h(0)=k(0)=0

h(0)=k(0)=1

h(0)=k(0)=2

h3(0)=k3(0)=3!

h4(0)=k4(0)=4!

h5(0)=k5(0)=5!

But.

h6(0)=6! And k6(0)=0

So, h and k have 5 point contact.

 

Claim: K(0)=f(0)

 

Proof: K(x)=f(x)(1+(f(x))2)-3/2

We have set up our axes so that f(0)=f(0)=0, so

K(0)=f(0)(1+02)-3/2 = f(0)

Claim: K(0)=f(0)

Proof: K(x)= f(x) (-3/2) (1+(f(x))2)-5/2(2)f(x) f(x) + (1+(f(x))2)-3/2 (f(x))

= -3(f(x))2 f(x)(1+(f(x))2)-5/2 + f(x)(1+(f(x))2)-3/2

Again, we have set up our axes so that f(0)=f(0)=0, so

K(0)= -3f(0)2 (0) (1+02)-5/2 + f(0)(1+02)-3/2

= f(0)

 

Claim: Two smooth curves having a common tangent at a common point P have at least 3-point contact at P if and only if they have the same curvature at P.

 

Proof: f and g have at least 3 point contact f(xo)=g(xo), f(xo)=g(xo),

f(xo)=g(xo).

 

Kf=f(xo)(1+(f(xo))2)-3/2

But, f(xo)=g(xo) and f(xo)=g(xo), so

Kf=g(xo)(1+(g(xo))2)-3/2 = Kg

 

f and g have the same curvature Kf = Kg

 

Kf=f(xo)(1+(f(xo))2)-3/2

Kg =g(xo)(1+(g(xo))2)-3/2

 

But, we know that they have a common tangent at P= xo, so f(xo)=g(xo).

 

Kf=f(xo)(1+(g(xo))2)-3/2= g(xo)(1+(g(xo))2)-3/2 = Kg

You can cancel off the (1+g(x)2)-3/2, and we are left with

Kf=f(xo)= = g(xo) = Kg

And this is exactly the definition of at least 3 point contact:

 

(xo)=g(xo), f(xo)=g(xo), f(xo)=g(xo).

 

MatLab Code:

This is the program that maps out the negative pedal. To start, type pedal (or whatever you save it as).You have to input an x coordinate and a y coordinate, preferably near the unit circle or else the plot is too zoomed out. If you actually use this, dont worry about all the Warnings about dividing by 0; its more work than its worth trying to avoid those points that would cause this to divide by zero.

 

function pedal

 

close all

 

x = input('x-coordinate?');

y = input('y-coordinate? ');

 

for i=-1:.015:1

y1=sqrt(1-i^2);

y2=-y1;

p1=(x-i)/(y-y1)+y1;

p2=-(x-i)/(y-y1)+y1;

p3=(x-i)/(y+y1)+y2;

p4=-(x-i)/(y+y1)+y2;

plot([i-1 i+1], [p1 p2] , 'r')

plot([i-1 i+1], [p3 p4], 'r')

hold on

axis equal

plot(x,y,'g*');

end

 

for t=0:.01:2*pi

plot([cos(t)], [sin(t)], 'b.')

hold on

end

hold off