Themes > Science > Mathematics > Algebra > Foci of a conic section > Topics and Problems > Curves in a plane

..Direction of a tangent line of a curve.
 ..Curve associated with a set curves
 ..Envelope curve
 ..Contact conditions
 ..A set of lines and its envelope
 ..Singular points and the system (S).
..Example 1.
 ..Example 2.
..Converse theorem
 ..Special case
 ..Envelope of a set of circles
 ..Envelope of a slipping segment
 ..Envelope of co-axial ellipses
 ..The evolute as an envelope
..Evolute of a parabola
..Order of contact
 ..Osculating circle
 ..Osculating circles and envelope
 ..Another approach to the osculating circle
 ..Curvature and the radius of curvature
 ..Catenary
 ..Tractrix as a locus


Direction of a tangent line of a curve.

Say P(x,y) is a variable point of the curve.
  • If the curve is given by an equation y = f(x).

    The slope of the tangent line in P is y' = f'(x). A vector with this direction has coordinates (1,f'(x)) or a multiple of that couple.

    Since y' = dy/dx, the direction of the tangent line in P is given by (1, dy/dx) or (dx, dy).

  • If the curve is given by an equation F(x,y)=0.

    y' is implicitly given by : 

     
        Fx'(x,y) + Fy'(x,y) . y' = 0

<=>     Fx'(x,y) + Fy'(x,y) . (dy/dx) = 0

<=>     Fx'(x,y) dx +  Fy'(x,y) dy = 0
    The direction of the tangent line in a variable point P is given by (dx, dy) and that couple is implicitly defined by the last equation.

    So, (Fy'(x,y) , - Fx'(x,y)) gives the direction of the tangent line.

    If we have that Fx'(x,y) = Fy'(x,y) = 0 in a particular point Q, then we say that Q is a singular point of the curve. The other points are regular points.

  • If the curve is given by parameter equations x = f(t) ; y = g(t).

    Now we have 

     
                 dy/dt       g'(t)
        dy/dx = -------- = --------
                 dx/dt       f'(t)
    The direction of the tangent line in P is (f'(t), g'(t)).

Curve associated with a set curves

Think of a curve 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.

With the set of curves, corresponds a curve C. It is the locus of the point Q.

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 the associated point Qp is on Kp, and 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)).

Since Qp is on Kp we have 

 
     F(f(p),g(p),p) = 0   for each p.         (1)
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.

Since (1) hold for each p, we get a new identity if we calculate the derivative, with respect 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.

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
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

A set of lines and its envelope

As an example we take a variable line t. 
 
    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 
 
    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 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.

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, with respect 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.

Exercise: draw some curves Kp and see that y=x is not an envelope curve.

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.

Exercise: draw some curves Kp and see that in this case, y=0 is an envelope curve.

Converse theorem

Here we only consider regular points.
We start with a solution x = f(p) and y = g(p) of the system (S).

We'll show that x = f(p) and y = g(p) are parametric equations of an envelope curve.

We start from the system (S) 

 
   /  F(x,y,p) = 0           (5)
   |
   \  Fp'(x,y,p) = 0       (6)
The equations x = f(p) and y = g(p) define a curve C and 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. Moreover, we have 
 
Fx'(f(p),g(p),p). f'(p) + Fy'(f(p),g(p),p).g'(p) + Fp'(f(p),g(p),p)=0 (7)
From (6) we have 
 
    Fp'(f(p),g(p),p) = 0        and therefore (7) becomes

    Fx'(f(p),g(p),p). f'(p) + Fy'(f(p),g(p),p).g'(p)  = 0  (8)
Since the tangent line in Qp at the curve Kp has direction 
 
    (Fy'(f(p),g(p),p) ,  - Fx'(f(p),g(p),p))
and the tangent line at curve C in point Qp has direction (f'(p),g'(p)), the expression (8) is just the condition for contact.

The curve C is then an envelope of the set Kp.

Special case

If all curves Kp contain a regular fixed point P, then P belongs to the envelope of Kp.

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, with respect 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).

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 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 equations 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 with respect 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. 
 
    d = d1/3.x2/3 + d1/3.y2/3
<=>
    x2/3 + y2/3 = d2/3
and this is the cartesian equation of an astroid.

On the net there is a page that illustrates this theory about the envelope.

See netpage about astroid and envelope and the netpage about envelope.

The evolute as an envelope

The evolute of a curve is the envelope of its set of normals.

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: Take a = 1/4; draw the parabola and a few normals. Then draw the evolute and see how the normals are the tangent lines of the evolute.

Exercise: Take a variable point of an ellipse and calculate the parametric equations of the evolute of the ellipse.

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

  • The equation of the normal in point P(xo,yo) is 
     
  y - f(xo) = ( - 1/ f'(xo)) . (x - xo)

<=>

  (x - xo) + f'(xo).(y - f(xo)) = 0             (7)
    If we replace x and y by a and b, then we find exactly (2). This means that
    The center of the osculating circle is on the normal in P.

  • The equation (7) of the normal is a line, depending on the parameter xo. If we consider xo as a variable parameter, we have a set of lines. The evolute of the curve K with equation y = f(x) is the envelope of this set of normals.

    The points of the evolute are the solutions of the system 

     
  (x - xo) + f'(xo).(y - f(xo)) = 0             (7)

  -1 - f'(xo)2  + (y - f(xo))f"(xo) = 0         (8)
    In view of (2) and (3), we see that (a,b) is a solution of (7) and (8). This means that

    The evolute of a curve K is the locus of the center of the osculating circles and the envelope of all osculating circles is K itself.

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 is  y - a cosh(t/a) = sinh(t/a) .(x - t)

The perpendicular 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(x/a)

         a
   y = ----------
       cosh(x/a)
This locus is called the tractrix associated to the catenary. Point T(t - a tanh(x/a), a/ cosh(x/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(x/a) . (1/a)
            cosh2(x/a)

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

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

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

and from this

    dy        1
    -- = - ----------
    dx     sinh(x/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.

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

 
    |TS| = a = constant


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