Curves in a plane




Direction of a tangent line of a curve.

Say P(x,y) is a variable point of the curve.

Equal directions

 
  The direction defined by (a,b) = the direction defined by (c,d)

<=>  (c,d) is a multiple of (a,b)

<=>  The following determinant is zero

       | a  b |
       | c  d |


<=>  a d - b c = 0

Curve associated to a set of curves

Consider a curves Kp depending on a parameter p. For each p-value, the curve Kp takes another shape or position. If p varies in an interval, we have a set of curves.

With each Kp, we can associate just one point Qp. The position of that associated point Qp depends only on the value of the parameter p.

If p changes smoothly, we have a set of curves Kp and the point Qp describes a curve C. It is the locus of the point Qp. C is a curve associated to the set of curves Kp.

Examples:

Envelope curve

Two curves are tangent to each other if and only if both curves share a common tangent line at a common point.

The envelope curve of a set of curves Kp is an associated curve C such that

  1. The associated point Qp is on Kp
  2. C is tangent to Kp in point Qp.


Contact conditions

Let F(x,y,p) = 0 be the equation of a set curves Kp. The parameter is p.

Suppose that x = f(p) and y = g(p) are unknown parametric equations of the envelope curve C of the set. The associated point Qp has coordinates (f(p),g(p)).

First condition : The associated point Qp is on Kp

 
     F(f(p),g(p),p) = 0   for each p.         (1)
Second condition : C is tangent to Kp in point Qp

The tangent line at curve C in point Qp has the direction (f'(p),g'(p))

The tangent line at curve Kp has the direction (Fy'(x,y,p) , - Fx'(x,y,p)).
In point Qp is that direction (Fy'(f(p),g(p),p), -Fx'(f(p),g(p),p)).

These directions must be the same for each p. Thus,

 
    - f'(p).Fx'(f(p),g(p),p) = g'(p).Fy'(f(p),g(p),p)  for each p.

<=>

    Fx'(f(p),g(p),p) . f'(p) + Fy'(f(p),g(p),p) . g'(p) = 0 for each p.   (2)

(1) and (2) are the contact conditions.

We simplify the second condition :

Since (1) hold for each p, we get a new identity if we calculate the derivative of (1), relative to p, using the chain rule extension.

 
Fx'(f(p),g(p),p). f'(p) + Fy'(f(p),g(p),p).g'(p) + Fp'(f(p),g(p),p)=0    (3)
By virtue of this result (2) becomes
 
    Fp'(f(p),g(p),p) = 0      (4)
(1) and (4) are the contact conditions.

Conclusion :

With all this we can state:
Let F(x,y,p) = 0 be the equation of a set curves Kp. Say x = f(p) and y = g(p) are the parametric equations of an envelope curve C. Then the contact conditions are

 
     F(f(p),g(p),p) = 0

    Fp'(f(p),g(p),p) = 0
This means that the parametric equations x = f(p) and y = g(p) of an envelope curve are a solution of the system
 
     F(x,y,p) = 0

    Fp'(x,y,p) = 0
Conclusion:
If x = f(p) and y = g(p) are the parametric equations of an envelope curve of the set curves F(x,y,p) = 0, these parametric equations are a solution of the system (S)
 
   /   F(x,y,p) = 0
   |                           (S)
   \   Fp'(x,y,p) = 0
Note: the reverse is not necessarily true
In special cases, it is possible that x = f(p) and y = g(p) form a solution of (S) and that they are not enveloping the curves F(x,y,p) = 0.

A set of lines and its envelope

A first example

We start with a set of lines
 
    x cos(p) + 2 y sin(p) - 4 = 0
The parameter is p. If p varies, we have a set of lines. The parametric equations of an envelope curve are a solution of the system (S)
 
    x cos(p) + 2 y sin(p) - 4 = 0

   - x sin(p) + 2 y cos(p)  = 0
If we solve this system for x and y, we find:
 
    x = 4 cos(p)
    y = 2 sin(p)
These are the parametric equations of a curve that is tangent to each line of the given set of lines. It is an ellipse.
 
    
You can see that by eliminating p. We have
 
    cos(p) = x/4  and sin(p) = y/2

So, (x/4)2 + (y/2)2 = 1

<=>  x2/16  + y2/4 = 1
The ellipse is a curve C such that C is tangent to every member of the set of lines.

A second example

A circle has center M(1,0) and the radius is four. Point A is A(-1,0) and P is a variable point of the circle. Point S is the center of segment AP. The perpendicular bisector of this segment is the variable line l. We'll find the envelope of the variable line l.
 
  The circle has equation

  (x-1)2 + y2 = 16

  and parametric equations

   x - 1 = 4 cos(t)
      y  = 4 sin(t)

  The variable point P is

   P ( 4 cos(t) + 1 ,  4 sin(t) )

  Then   S ( 2 cos(t) , 2 sin(t))

  Now, we calculate the variable line l.
  After simplification, one finds

  (2 cos(t) + 1) x  + 2 sin(t) y - 2 cos(t) - 4 = 0

  t is the parameter

  Now, we write the system (S)

  / (2 cos(t) + 1) x  + 2 sin(t) y - 2 cos(t) - 4 = 0
  |
  \    sin(t) x  - cos(t) y  - sin(t) = 0

  To eliminate t, we solve the system for sin(t) and cos(t)

                 y (4-x)
     sin(t) = -------------------
               2  ((x-1)2 + y2)

               (x - 1)(4-x)
     cos(t) = -------------------
              2  ((x-1)2 + y2)


    sin2(t) + cos2(t) = 1

<=>    y2(4-x)2 + (x - 1)2(4-x)2 = 4 ((x-1)2 + y2)2

<=>   (4-x)2 ( y2 + (x-1)2 )  -  4 ((x-1)2 + y2)2 = 0

<=>   ( y2 + (x-1)2 ) . ((4-x)2 - 4 ((x-1)2 + y2)) = 0

   The first factor gives us no curve.
   The equation of the envelope is

   (4-x)2 - 4 ((x-1)2 + y2) = 0

<=> ...

<=>   x2/4 + y2/3 = 1

Exercise

In an orthonormal coordinate system is A(0,1) a fixed point and B(a,0) is a variable point.
The line r, through point B and perpendicular to AB, is variable.
Find the envelope of the lines r.

Singular points and the system (S).

Suppose that each curve Kp : F(x,y,p) = 0 has a singular point. In this point Fx'(x,y,p) = Fy'(x,y,p) = 0. For each p-value, we have a singular point. These points form a curve S with parametric equations say, x = f(p) and y = g(p).

Since the singular point is on the curve Kp we have

 
     F(f(p),g(p),p) = 0   for each p.
We get a new identity if we calculate the derivative, relative to p, using the chain rule extension.
 
Fx'(f(p),g(p),p). f'(p) + Fy'(f(p),g(p),p).g'(p) + Fp'(f(p),g(p),p)=0 (3)
Since Fx'(f(p),g(p),p) = Fy'(f(p),g(p),p) = 0, last equation becomes
 
   Fp'(f(p),g(p),p)=0    for each p.
Therefore x = f(p) and y = g(p) are a solution of the system (S), and this without mentioning tangent lines.

The locus of a singular point of the variable curve Kp, is a solution of the system (S). In general, this locus is not an envelope curve of the set curves Kp.

Example 1.

We take the semi-cubic parabola Kp with equation
 
   (y-p)3 = (x-p)2  <=>  (y-p)3 - (x-p)2 = 0

Here F(x,y,p) = (y-p)3 - (x-p)2

and Fx'(x,y,p) = - 2(x-p)   Fy'(x,y,p) = 3(y-p)2

The point Q(p,p) is a singular point for each p.
The system (S) for this set of curves is
 
    (y-p)3 - (x-p)2 = 0

    3(y-p)2 .(-1) - 2 (x-p).(-1) = 0
We see that x=p ; y=p is a solution of that system. It is the first bisector line of the coordinate system and the locus of all the singular points.

 
    

Example 2.

We take the semi-cubic parabola Kp with equation
 
   (y-p)3 = x2  <=>  (y-p)3 - x2 = 0

Here F(x,y,p) = (y-p)3 - x2

and Fx'(x,y,p) = - 2 x   Fy'(x,y,p) = 3(y-p)2

The point Q(0,p) is a singular point for each p.
The system (S) for this set of curves is
 
    (y-p)3 - x2 = 0

    3(y-p)2 .(-1)  = 0
We see that x=0 ; y=p is a solution of that system. It is the y-axis, the locus of all the singular points.

 
              

Converse theorem

If the set of curves Kp , with equation F(x,y,p)=0, has only regular points, then we can show that a solution x = f(p) and y = g(p) of the system (S) constitute parametric equations of an envelope curve of the set Kp.

Given:
Kp has only regular points
Kp has equation F(x,y,p)=0
x = f(p) and y = g(p) are solutions of the system (S)
With each Kp there is an associated point Qp = Qp( f(p), g(p))
C is the curve with parametric equations x = f(p) en y = g(p)

We have to prove that:
C is the envelope of the set Kp.
Therefore we have to prove that two conditions are fulfilled.

  1. The associated point Qp is on Kp for all p
  2. C is tangent to Kp in point Qp for all p

Proof:

We start from the system (S)

 
   /  F(x,y,p) = 0           (5)
   |
   \  Fp'(x,y,p) = 0       (6)

 x = f(p) and y = g(p) form  a solution of (5). So,

  F(f(p),g(p),p) = 0   for all p     (5')

(5') means that the  point Qp(f(p),g(p)) is on the curve Kp for all p. 
The first condition is fulfilled. We now look at the tangents at point Qp The tangent line in Qp to the curve Kp has direction (Fy'(f(p),g(p),p) , - Fx'(f(p),g(p),p)) The tangent line in Qp to the curve C has direction (f'(p),g'(p)) The two tangent directions are equal <=> Fy'(f(p),g(p),p) g'(p) + Fx'(f(p),g(p),p) f'(p) = 0 We calculate the derivative of (5') with respect to p. Fx'(f(p),g(p),p). f'(p) + Fy'(f(p),g(p),p).g'(p) + Fp'(f(p),g(p),p)=0 Relying on this, we can write the preceding equivalence in another way. The two tangent directions are equal <=> Fp'(f(p),g(p),p)=0 and this is satisfied because x = f(p) and y = g(p) are also solutions of (6) So, C is tangent to Kp in point Qp for all p The second condition is fulfilled.
The curve C is an envelope of the set Kp !

Agreement

If all curves Kp contain a regular fixed point P, then P belongs to the envelope of Kp.
In this special case, we can not speak of a common tangent line.

Envelope of a set of circles

Take a circle C with equation (x - a)2 + (y - b)2 - r2 = 0.
Now suppose that a, b and r are functions of a parameter p. Then, for each value of p, we have a circle Cp.

We'll denote the derivatives, relative to p, of the functions a, b and r for short as a', b' and r'.

The envelope of the set circles is the solution of the system (S).

 
    (x - a)2 + (y - b)2 - r2 = 0

    2(x-a).(-a') + 2(y-b).(-b') - 2 r r'= 0
The second equation is the equation of a line. This line l (blue) is orthogonal to the direction (a',b').
 
             

The locus of the center-point M of the circles Cp have the parametric equations x = a(p) and y = b(p). The tangent line (brown) at that locus has direction (a',b').

Thus, the second equation of the system (S) is a line l that is orthogonal to the tangent line to the locus of the center point of the circles.

The intersection points of the circle and the line l are points of the envelope (green).

Example 1

We start with a set of circles x2 + (y-p)2 = 2p
The envelope of this set is the solution of the system (S)

 
    x2 + (y - p)2 = 2p

    -2(y - p) = 2
If we solve (S) for x and y, we find the parametric equations of the envelope. But, if we eliminate the parameter p, then we find the cartesian equation of the envelope. We find : y =(x2-1)/2. This parabola is tangent to all the circles.
 
               
The chord of the points of contact is perpendicular to the path of the center of the circle.

Example 2

We start with a set of circles x2 + y2 - 2 x cos(p) - 2 y sin(p) = 0
The system (S) is

 
 /  x2 + y2 - 2 x cos(p) - 2 y sin(p) = 0
 |
 \  2 x sin(p) - 2 y cos(p) = 0

<=>

 /   2 x cos(p) + 2 y sin(p) = x2 + y2
 |
 \    x sin(p) - y cos(p) = 0

<=>  ...


<=>     /  sin(p) = -y/2
        \  cos(p) = x/2

We eliminate p

    (x/2)2 + (-y/2)2 = 1

<=> x2 + y2 = 4
This circle is the envelope of the set of circles.
Since the given circles contain the fixed point O(0,0), this point is also an envelope of the circles.

Draw a sketch with some variable circles and the envelope.

Envelope of a slipping segment

Imagine a segment [AB] with fixed length d, slipping on x-axis and y-axis. We'll calculate the envelope of this variable segment.
 
           
The x-coordinate of A is in [0,d], we take A(d.cos(t),0). As the distance from A to B is d, the coordinates of B are (0,d.sin(t)). The parameter is t (see figure)

The equation of the line AB is

 
    x/cos(t) + y/sin(t) = d
To obtain the parametric equations of the envelope if we calculate x and y from the system.
 
    x/cos(t) + y/sin(t) = d

     x sin(t)      y cos(t)
    ---------  -  ----------  = 0
     cos2(t)       sin2(t)
The solution is
 
    x = d cos3(t)
    y = d sin3(t)
This are the parametric equations of an astroid. The contact point P of the segment [AB] and the envelope can be constructed as shown in red on previous figure.
 
           

Envelope of co-axial ellipses

We'll calculate the envelope of co-axial ellipses whose sum of major and minor axes is a constant = d.

Start with the equation of an ellipse

 
    x2/a2 + y2/b2 = 1
with parameters a and b and a + b = d = constant.

At first glance, you may think that there are two parameters, but since they are bounded with a + b = d, there is essentially one parameter. Think a and b as functions of parameter t. Then,

 
   a + b = d  =>  da/dt + db/dt = 0 => db/dt = - da/dt
To obtain the parametric equations of the envelope, we calculate x and y from the system with first equation
 
    x2/a2 + y2/b2 = 1        (1)
We become the second equation if we differentiate the first one relative to t.
 
    -2 x2             -2 y2
    ------.(da/dt) + --------.(db/dt) = 0
     a3                 b3

<=>

     x2/a3  - y2/b3 = 0       (2)
If we calculate x2 and y2 from (1) and (2) we find:
 
     x2 = a3/d  and y2 = b3/d

<=>
     a = (d x2)1/3  and  b = (d y2)1/3
We eliminate a and b by adding these equations. The sum must be d.
 
    d = d1/3.x2/3 + d1/3.y2/3
<=>
    x2/3 + y2/3 = d2/3
and this is the cartesian equation of an astroid.
 
           

envelope of a rolling diameter

A circle with radius r rolls without slipping on a fixed line. A fixed point P of the circle decribes a cycloid with parametric equations
 
  x = r(t - sin(t))
  y = r(1 - cos(t))
You can find here an illustration of the rolling circle.

The center A of this circle has coordinates (rt, r).
Now, consider the variable line PA. We'll find the envelope of this rolling line.

 
   The equations of  PA is

   x cos(t) - y sin(t) + r (sin(t) - t cos(t)) = 0

   The system (S) is

   / x cos(t) - y sin(t) + r (sin(t) - t cos(t)) = 0
   |
   \ x sin(t) + y cos(t) + r t sin(t) = 0

   If we solve this system for x and y, then we have the parametric
   equations of the envelope.

    x = r (t - sin(t) cos(t))
    y = r sin2(t)

   To know the nature of this curve, we transform the equations as follows

    x = (r/2) (2t - 2 sin(t) cos(t))
    y = (r/2) (2 sin2(t))
<=>
    x = (r/2) (2t - sin(2t))
    y = (r/2) (1 + cos(2t))

    Let t' = 2t = t'

    x = (r/2) (t' - sin(t'))
    y = (r/2) (1  + cos(t'))
We find another cycloid described by a fixed point of a rolling circle with radius r/2.

The evolute as an envelope

The evolute of a curve is the envelope of the set of normals to the curve

Here is an illustration of the concept.

Evolute of a parabola

We take the parabola y = a x2, with a positive and constant. A variable point of this parabola is P(t, a t2). The tangent line in P has slope 2at. The normal has slope -1/(2at). The set of all normals can be written as
 
    y - at2 = ( -1/2at) (x-t)
<=>
    2at y - 2 a2 t3  + x +t = 0
<=>
    (2ay + 1) t  - 2 a2 t3  + x = 0
To obtain the parametric equations of the evolute, we calculate x and y from the system
 
    (2ay + 1) t  - 2 a2 t3  + x = 0

    (2ay + 1) - 6 a2 t2 = 0
We find
 
    x = -4 a2 t3

    y = 3 a t2 + 1/2a
To obtain the cartesian equation, we eliminate the parameter t and we find a semi-cubic parabola
 
    16a (y - 1/(2a))3 = 27 x2


           

Exercise -- evolute of an ellipse

Take a variable point P(a cos(t), b sin(t)) of an ellipse x2/a2 + y2/b2 = 1 and calculate the parametric equations of the evolute of the ellipse.

Evolute of an astroid

We start with an astroid with parametric equations
 
    x = cos3(t)
    y = sin3(t)
P(cos3(t), sin3(t)) is a variable point of the astroid. The equation of the normal line in P is
 
    x cos(t) - y sin(t) + sin2(t) - cos2(t) = 0
Now, we calculate the envelope of this set of lines.
The system (S) is
 
  /  x cos(t) - y sin(t) + sin2(t) - cos2(t) = 0
  |
  \  - x sin(t) - y cos(t) - 2 sin(t) cos(t)  + 2 cos(t) sin(t)  = 0
We solve this system for x and y. You will find
 
     x = 3 cos(t) - 2 cos3(t)
     y = 3 sin(t) - 2 sin3(t)
These are the parametric equations of the evolute. When you plot the curve then you can suspect that this evolute is an astroid. It is more difficult to show this. Relative to the standard form, the asteroid is rotated by an angle of 45 degrees. We omit the proof.

Order of contact

Two curves are tangent to each other if and only if both curves share a common tangent line at a common point. We say that there is contact between the two curves at that common point P(xo,yo).

The functions f(x) and g(x) have simple contact in P if and only if

 
    f(xo) = g(xo)  and

    f'(xo) = g'(xo)  and

    f"(xo) is not equal to g"(xo)
This is a contact of order 1.

The functions f(x) and g(x) have contact of order 2 in P if and only if

 
    f(xo) = g(xo)  and

    f'(xo) = g'(xo) and

    f"(xo) = g"(xo) and

    f"'(xo) is not equal to g"'(xo)
....

Osculating circle

K is a curve with equation y = f(x) and P(xo,yo) is a point of K.

All circles have an equation (x - a)2 + (y - b)2 - r2 = 0. This equation contains three parameters a,b and r.

We'll try to find the circle such that the order of contact with K in P is maximum.

Since we have three parameters, normally we can impose three conditions. Then we can have contact of at least order 2.

To calculate y' and y" about the circle, we use implicit differentiation.

 
   (x - a)2 + (y - b)2 - r2 = 0

   (x - a) + (y - b) y' = 0

   1 + y'2 + (y-b) y" = 0
We denote yo' and yo" the values of y' and y" derived from the circle equation taken in P. The three conditions are
 
   yo  = f(xo)

   yo' = f'(xo)

   yo" = f"(xo)
These three conditions are equivalent with
 
   (xo - a)2 + (f(xo) - b)2 - r2 = 0           (1)

   (xo - a) + (f(xo) - b) f'(xo)  = 0            (2)

   1 + f'(xo)2 + (f(xo)  - b) f"(xo) = 0         (3)
This is a system of three equations with three unknowns a,b and r.

we find

 

                       1 + f'(xo)2
    a = xo - f'(xo) -----------------             (4)
                        f"(xo)

                   1 + f'(xo)2
    b = f(xo) +  -----------------                 (5)
                    f"(xo)


           (1 + f'(xo)2)3/2
    r = | ---------------------- |                  (6)
                f"(xo)
We say that this circle is the osculating circle of the curve K in point P.

M(a,b) is the center if the osculating circle and r is the radius.

Osculating circles and envelope

Another approach to the osculating circle

We'll show that
The osculating circle in a point P of a curve K, is the limit position of a circle through point P and two adjacent points P1 an P2 when P1 and P2 tend to P.

Proof:

K has equation y = f(x).

Let P(xo,yo), P1(x1,y1), P(x2,y2) three points of curve K.

A circle C has an equation of the form (x-a)2 + (y-b)2 - r2 = 0.

 
        C contains P, P1, P2
<=>
        (xo - a)2 + (f(xo) - b)2  -  r2 = 0.
        (x1 - a)2 + (f(x1) - b)2  -  r2 = 0.
        (x2 - a)2 + (f(x2) - b)2  -  r2 = 0.
The unknowns are a, b and r. P1 and P2 tend to P, but if we simply change x1, x2 with xo, we can't calculate a,b and r.

Therefore we'll use the temporary function

 
   g(t) = (t - a)2 + (f(t) - b)2  - r2
This function has roots xo, x1, x2.

Using Rolle's theorem, g'(t) has roots x3 and x4 such that xo < x3 < x1 < x4 < x2.
And g"(t) has a root x5 such that x3 < x5 < x4.

We have

 
      g(xo) = 0 ; g'(x3) = 0 ; g"(x5) = 0
<=>
      (xo - a)2 + (f(xo) - b)2  -  r2 = 0

      (x3 - a) + (f(x3) - b).f'(x3) = 0

      1 + f'2(x5) + (f(x5) - b).f"(x5) = 0
Now we take the limit for P1 and P2 tending to P.
Then x3 and x5 --> xo. We get
 
      (xo - a)2 + (f(xo) - b)2  -  r2 = 0

      (xo - a) + (f(xo) - b).f'(xo) = 0

      1 + f'2(xo) + (f(xo) - b).f"(xo) = 0
This is exactly the same system as (1) (2) (3), and therefore C is the osculating circle.

Curvature and the radius of curvature

Consider the tangent lines t and t1 in two neighboring points P and P1 of a curve C. Let (delta s) be the length of the curve from P to P1 and (delta t) is the angle (in radians) enclosed by t and t1.

The average curvature of the curve C from P to P1 is the absolute value of (delta t)/(delta s).

If P1 tends to P, then the average curvature tends to the curvature in point P. So,

 
The curvature of a curve in a point P
                      (delta t)
        =    lim    | ---------- |
          P1 -> P      (delta s)

             dt
        =  |----|
             ds

The inverse of this curvature is called the radius of curvature.

We know from the theory about length
that the length of an elementary part of a curve y = f(x), can be written as

 
        _________
       |
 ds = \| 1 + y'2  dx


       
From this figure we see that (delta t) = (delta h). So dt = dh.
 
 tan(h) = y' => h = arctan(y')

         y" dx
=> dh = ----------
         1 + y'2


  The curvature of a curve in a point P

       dt        dh           y"       dx
   = |----| =  |----| = | ----------. ----- |
       ds        ds        1 + y'2      ds

             y"
   = | ----------------- |
        ( 1 + y'2 )3/2


The radius of curvature =
 

         ( 1 + y'2 )3/2
       | ----------------- |
                y"

This is the same formula as for the radius of the osculating circle.

Catenary

The catenary is the shape of a perfectly flexible chain suspended by its ends and acted on by gravity. In statics it is proved that the equation of the catenary, in a suitable coordinate system, is
 
   y = a cosh(x/a)
If we take the function y = cosh(x), and we transform this curve with the (homothetic) transformation y = y'/a and x = x'/a, then we have a catenary. All catenaries are homothetic transformations of the graph of the cosh function.

Say P(xo,yo) is a variable point of the catenary y = a cosh(x/a).
The slope of the tangent line in P is sinh(xo/a).
The equation of the tangent line is y - yo = sinh(xo/a) .(x - xo).
This tangent line is connected with the tractrix (see below).

Tractrix as a locus

Point P(t,a cosh(t/a)) is a variable point of a catenary. The projection of P on the x-axis is S(t,0). Now we'll calculate the parametric equations of the locus of the intersection points of the tangent line in P and the line through S, perpendicular to that tangent line.

 
The tangent line in P is  y - a cosh(t/a) = sinh(t/a) .(x - t)

The perpendicular through S on the tangent line is  y = - (x - t) / sinh(t/a)
If we solve the system of these two equations for x and y, it will give the parametric equations of the locus.

With a little algebra you'll find

 
   x = t - a tanh(t/a)

         a
   y = ----------
       cosh(t/a)
This locus is called the tractrix associated to the catenary. Point T(t - a tanh(t/a), a/ cosh(t/a) ) is a variable point of the tractrix.

The tangent line in T to the tractrix has slope

 
    dy    dy/dt
    -- =  -----
    dx    dx/dt

with

             -a
    dy/dt = ----------- .sinh(t/a) . (1/a)
            cosh2(t/a)

                     1
    dx/dt = 1 - a ---------- . (1/a)
                  cosh2(t/a)

                   1
          = 1 - ----------
                cosh2(t/a)

              sinh2 (t/a)
          = --------------
              cosh2 (t/a)

and from this

    dy        1
    -- = - ----------
    dx     sinh(t/a)

Since this is the same slope of ST, we see that the tangent line in T to the tractrix is TS. From this, we see that the tangent line in T to the tractrix is orthogonal to the tangent line in P to the catenary . The normal in T to the tractrix is tangent to the catenary. In other words the catenary is the evolute of the tractrix.

See here for a beautiful illustration of this fact.

Moreover, when you calculate the length of segment [TS], you'll find:

 
    |TS| = a = constant
See here for a beautiful illustration of this fact.
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