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 thereof.
Since y' = dy/dx, the direction of the tangent line in P is given by (1, dy/dx) or (dx, dy).
Example:
We consider the curve y=x^{2}+4x. The direction of the tangent line in P(x,y) is
(dx,dy) or also (dx, (2x+4)dx) or also (1, 2x+4).
In point P(1,5) is the direction of the tangent line (1,6) or also (-1,-6).
y' is implicitly given by :
F_{x}^{'}(x,y) + F_{y}^{'}(x,y) . y' = 0 <=> F_{x}^{'}(x,y) + F_{y}^{'}(x,y) . (dy/dx) = 0 <=> F_{x}^{'}(x,y) dx + F_{y}^{'}(x,y) dy = 0The 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, (F_{y}^{'}(x,y) , - F_{x}^{'}(x,y)) gives the direction of the tangent line.
Example:
We consider the curve x^{2} + 2 y^{2} - 3 xy -3 = 0. The direction of the tangent line in P(x,y) is (4y - 3x, 3y - 2x). In point P(1,2) is the direction of the tangent line (5,4).
If we have that F_{x}^{'}(x,y) = F_{y}^{'}(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.
Example:
We consider the curve x^{2} + 9y^{2} -y^{3} -6x -27y +36 = 0.
The direction of the tangent line in P(x,y) is ( -3y^{2} + 18y -27, -2x + 6).
In point P(3,3) is F_{x}^{'}(3,3) = F_{y}^{'}(3,3) = 0.
Point P(3,3) is a singular point of the curve.
Now we have
dy dy/dt g'(t) ----- = -------- = -------- dx dx/dt f'(t)The direction of the tangent line in P is (f'(t), g'(t)).
Example:
An astroid has parameter equations x = a cos^{3}(t) ; y = a sin^{3}(t).
The direction of the tangent line in P(x,y) is ( - cos(t), sin(t) ).
In point P(a/8, 3a.sqrt(3)/8) is the direction of the tangent line (-1, sqrt(3)).
In point P(a,0) is the direction of the tangent line (1,0).
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
With each K_{p}, we can associate just one point Q_{p}. The position of that associated point Q_{p} depends only on the value of the parameter p.
If p changes smoothly, we have a set of curves K_{p} and the point Q_{p} describes a curve C. It is the locus of the point Q_{p}. C is a curve associated to the set of curves K_{p}.
Examples:
The envelope curve of a set of curves K_{p} is an associated curve C such that
Suppose that x = f(p) and y = g(p) are unknown parametric equations of the envelope curve C of the set. The associated point Q_{p} has coordinates (f(p),g(p)).
First condition : The associated point Q_{p} is on K_{p}
F(f(p),g(p),p) = 0 for each p. (1)Second condition : C is tangent to K_{p} in point Q_{p}
The tangent line at curve C in point Q_{p} has the direction (f'(p),g'(p))
The tangent line at curve K_{p} has the direction
(F_{y}^{'}(x,y,p) , - F_{x}^{'}(x,y,p)).
In point Q_{p} is that direction (F_{y}^{'}(f(p),g(p),p), -F_{x}^{'}(f(p),g(p),p)).
These directions must be the same for each p. Thus,
- f'(p).F_{x}^{'}(f(p),g(p),p) = g'(p).F_{y}^{'}(f(p),g(p),p) for each p. <=> F_{x}^{'}(f(p),g(p),p) . f'(p) + F_{y}^{'}(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.
F_{x}^{'}(f(p),g(p),p). f'(p) + F_{y}^{'}(f(p),g(p),p).g'(p) + F_{p}^{'}(f(p),g(p),p)=0 (3)By virtue of this result (2) becomes
F_{p}^{'}(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 K_{p}.
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 F_{p}^{'}(f(p),g(p),p) = 0This 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 F_{p}^{'}(x,y,p) = 0Conclusion:
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) \ F_{p}^{'}(x,y,p) = 0Note: 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. |
x cos(p) + 2 y sin(p) - 4 = 0The 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) = 0If 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 <=> x^{2}/16 + y^{2}/4 = 1The ellipse is a curve C such that C is tangent to every member of the set of lines.
The circle has equation (x-1)^{2} + y^{2} = 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} + y^{2}) (x - 1)(4-x) cos(t) = ------------------- 2 ((x-1)^{2} + y^{2}) sin^{2}(t) + cos^{2}(t) = 1 <=> y^{2}(4-x)^{2} + (x - 1)^{2}(4-x)^{2} = 4 ((x-1)^{2} + y^{2})^{2} <=> (4-x)^{2} ( y^{2} + (x-1)^{2} ) - 4 ((x-1)^{2} + y^{2})^{2} = 0 <=> ( y^{2} + (x-1)^{2} ) . ((4-x)^{2} - 4 ((x-1)^{2} + y^{2})) = 0 The first factor gives us no curve. The equation of the envelope is (4-x)^{2} - 4 ((x-1)^{2} + y^{2}) = 0 <=> ... <=> x^{2}/4 + y^{2}/3 = 1
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. |
Since the singular point is on the curve K_{p} 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.
F_{x}^{'}(f(p),g(p),p). f'(p) + F_{y}^{'}(f(p),g(p),p).g'(p) + F_{p}^{'}(f(p),g(p),p)=0 (3)Since F_{x}^{'}(f(p),g(p),p) = F_{y}^{'}(f(p),g(p),p) = 0, last equation becomes
F_{p}^{'}(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 K_{p}, is a solution of the system (S). In general, this locus is not an envelope curve of the set curves K_{p}. |
(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 F_{x}^{'}(x,y,p) = - 2(x-p) F_{y}^{'}(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) = 0We 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.
(y-p)^{3} = x^{2} <=> (y-p)^{3} - x^{2} = 0 Here F(x,y,p) = (y-p)^{3} - x^{2} and F_{x}^{'}(x,y,p) = - 2 x F_{y}^{'}(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} - x^{2} = 0 3(y-p)^{2} .(-1) = 0We see that x=0 ; y=p is a solution of that system. It is the y-axis, the locus of all the singular points.
Given:
K_{p} has only regular points
K_{p} has equation F(x,y,p)=0
x = f(p) and y = g(p) are solutions of the system (S)
With each K_{p} there is an associated point Q_{p} = Q_{p}( 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 K_{p}.
Therefore we have to prove that two conditions are fulfilled.
Proof:
We start from the system (S)
/ F(x,y,p) = 0 (5) | \ F_{p}^{'}(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 Q_{p}(f(p),g(p)) is on the curve K_{p} for all p.The curve C is an envelope of the set K_{p} !
The first condition is fulfilled. We now look at the tangents at point Q_{p} The tangent line in Q_{p} to the curve K_{p} has direction (F_{y}^{'}(f(p),g(p),p) , - F_{x}^{'}(f(p),g(p),p)) The tangent line in Q_{p} to the curve C has direction (f'(p),g'(p)) The two tangent directions are equal <=> F_{y}^{'}(f(p),g(p),p) g'(p) + F_{x}^{'}(f(p),g(p),p) f'(p) = 0 We calculate the derivative of (5') with respect to p. F_{x}^{'}(f(p),g(p),p). f'(p) + F_{y}^{'}(f(p),g(p),p).g'(p) + F_{p}^{'}(f(p),g(p),p)=0 Relying on this, we can write the preceding equivalence in another way. The two tangent directions are equal <=> F_{p}^{'}(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 K_{p} in point Q_{p} for all p The second condition is fulfilled.
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} - r^{2} = 0 2(x-a).(-a') + 2(y-b).(-b') - 2 r r'= 0The 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 C_{p} 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).
We start with a set of circles x^{2} + (y-p)^{2} = 2p
The envelope of this set is the solution of the system (S)
x^{2} + (y - p)^{2} = 2p -2(y - p) = 2If 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 =(x^{2}-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.
We start with a set of circles x^{2} + y^{2} - 2 x cos(p) - 2 y sin(p) = 0
The system (S) is
/ x^{2} + y^{2} - 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) = x^{2} + y^{2} | \ 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 <=> x^{2} + y^{2} = 4This circle is the envelope of the set of circles.
Draw a sketch with some variable circles and the envelope.
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) = dTo 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 cos^{2}(t) sin^{2}(t)The solution is
x = d cos^{3}(t) y = d sin^{3}(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.
Start with the equation of an ellipse
x^{2}/a^{2} + y^{2}/b^{2} = 1with 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/dtTo obtain the parametric equations of the envelope, we calculate x and y from the system with first equation
x^{2}/a^{2} + y^{2}/b^{2} = 1 (1)We become the second equation if we differentiate the first one relative to t.
-2 x^{2} -2 y^{2} ------.(da/dt) + --------.(db/dt) = 0 a^{3} b^{3} <=> x^{2}/a^{3} - y^{2}/b^{3} = 0 (2)If we calculate x^{2} and y^{2} from (1) and (2) we find:
x^{2} = a^{3}/d and y^{2} = b^{3}/d <=> a = (d x^{2})^{1/3} and b = (d y^{2})^{1/3}We eliminate a and b by adding these equations. The sum must be d.
d = d^{1/3}.x^{2/3} + d^{1/3}.y^{2/3} <=> x^{2/3} + y^{2/3} = d^{2/3}and this is the cartesian equation of an astroid.
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 sin^{2}(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 sin^{2}(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 of a curve is the envelope of the set of normals to the curve |
y - at^{2} = ( -1/2at) (x-t) <=> 2at y - 2 a^{2} t^{3} + x +t = 0 <=> (2ay + 1) t - 2 a^{2} t^{3} + x = 0To obtain the parametric equations of the evolute, we calculate x and y from the system
(2ay + 1) t - 2 a^{2} t^{3} + x = 0 (2ay + 1) - 6 a^{2} t^{2} = 0We find
x = -4 a^{2} t^{3} y = 3 a t^{2} + 1/2aTo obtain the cartesian equation, we eliminate the parameter t and we find a semi-cubic parabola
16a (y - 1/(2a))^{3} = 27 x^{2}
Take a variable point P(a cos(t), b sin(t)) of an ellipse x^{2}/a^{2} + y^{2}/b^{2} = 1 and calculate the parametric equations of the evolute of the ellipse. |
x = cos^{3}(t) y = sin^{3}(t)P(cos^{3}(t), sin^{3}(t)) is a variable point of the astroid. The equation of the normal line in P is
x cos(t) - y sin(t) + sin^{2}(t) - cos^{2}(t) = 0Now, we calculate the envelope of this set of lines.
/ x cos(t) - y sin(t) + sin^{2}(t) - cos^{2}(t) = 0 | \ - x sin(t) - y cos(t) - 2 sin(t) cos(t) + 2 cos(t) sin(t) = 0We solve this system for x and y. You will find
x = 3 cos(t) - 2 cos^{3}(t) y = 3 sin(t) - 2 sin^{3}(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.
The functions f(x) and g(x) have simple contact in P if and only if
f(x_{o}) = g(x_{o}) and f'(x_{o}) = g'(x_{o}) and f"(x_{o}) is not equal to g"(x_{o})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(x_{o}) = g(x_{o}) and f'(x_{o}) = g'(x_{o}) and f"(x_{o}) = g"(x_{o}) and f"'(x_{o}) is not equal to g"'(x_{o})....
All circles have an equation (x - a)^{2} + (y - b)^{2} - r^{2} = 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} - r^{2} = 0 (x - a) + (y - b) y' = 0 1 + y'^{2} + (y-b) y" = 0We denote y_{o}' and y_{o}" the values of y' and y" derived from the circle equation taken in P. The three conditions are
y_{o} = f(x_{o}) y_{o}' = f'(x_{o}) y_{o}" = f"(x_{o})These three conditions are equivalent with
(x_{o} - a)^{2} + (f(x_{o}) - b)^{2} - r^{2} = 0 (1) (x_{o} - a) + (f(x_{o}) - b) f'(x_{o}) = 0 (2) 1 + f'(x_{o})^{2} + (f(x_{o}) - b) f"(x_{o}) = 0 (3)This is a system of three equations with three unknowns a,b and r.
we find
1 + f'(x_{o})^{2} a = x_{o} - f'(x_{o}) ----------------- (4) f"(x_{o}) 1 + f'(x_{o})^{2} b = f(x_{o}) + ----------------- (5) f"(x_{o}) (1 + f'(x_{o})^{2})^{3/2} r = | ---------------------- | (6) f"(x_{o})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.
y - f(x_{o}) = ( - 1/ f'(x_{o})) . (x - x_{o}) <=> (x - x_{o}) + f'(x_{o}).(y - f(x_{o})) = 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 points of the evolute are the solutions of the system
(x - x_{o}) + f'(x_{o}).(y - f(x_{o})) = 0 (7) -1 - f'(x_{o})^{2} + (y - f(x_{o}))f"(x_{o}) = 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. |
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 P_{1} an P_{2} when P_{1} and P_{2} tend to P. |
Proof:
K has equation y = f(x).
Let P(x_{o},y_{o}), P_{1}(x_{1},y_{1}), P(x_{2},y_{2}) three points of curve K.
A circle C has an equation of the form (x-a)^{2} + (y-b)^{2} - r^{2} = 0.
C contains P, P_{1}, P_{2} <=> (x_{o} - a)^{2} + (f(x_{o}) - b)^{2} - r^{2} = 0. (x_{1} - a)^{2} + (f(x_{1}) - b)^{2} - r^{2} = 0. (x_{2} - a)^{2} + (f(x_{2}) - b)^{2} - r^{2} = 0.The unknowns are a, b and r. P_{1} and P_{2} tend to P, but if we simply change x_{1}, x_{2} with x_{o}, we can't calculate a,b and r.
Therefore we'll use the temporary function
g(t) = (t - a)^{2} + (f(t) - b)^{2} - r^{2}This function has roots x_{o}, x_{1}, x_{2}.
Using Rolle's theorem, g'(t) has
roots x_{3} and x_{4} such that x_{o} < x_{3} < x_{1} < x_{4} < x_{2}.
And g"(t) has
a root x_{5} such that x_{3} < x_{5} < x_{4}.
We have
g(x_{o}) = 0 ; g'(x_{3}) = 0 ; g"(x_{5}) = 0 <=> (x_{o} - a)^{2} + (f(x_{o}) - b)^{2} - r^{2} = 0 (x_{3} - a) + (f(x_{3}) - b).f'(x_{3}) = 0 1 + f'^{2}(x_{5}) + (f(x_{5}) - b).f"(x_{5}) = 0Now we take the limit for P_{1} and P_{2} tending to P.
(x_{o} - a)^{2} + (f(x_{o}) - b)^{2} - r^{2} = 0 (x_{o} - a) + (f(x_{o}) - b).f'(x_{o}) = 0 1 + f'^{2}(x_{o}) + (f(x_{o}) - b).f"(x_{o}) = 0This is exactly the same system as (1) (2) (3), and therefore C is the osculating circle.
The average curvature of the curve C from P to P_{1} is the absolute value of (delta t)/(delta s).
If P_{1} 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 | ---------- | P_{1} -> P (delta s) dt = |----| dsThe inverse of this curvature is called the radius of curvature. |
_________ | ds = \| 1 + y'^{2} dxFrom 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" |
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(x_{o},y_{o}) is a variable point of the catenary y = a cosh(x/a).
The slope of the tangent line in P is sinh(x_{o}/a).
The equation of the tangent line is y - y_{o} = sinh(x_{o}/a) .(x - x_{o}).
This tangent line is connected with the tractrix (see below).
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) cosh^{2}(t/a) 1 dx/dt = 1 - a ---------- . (1/a) cosh^{2}(t/a) 1 = 1 - ---------- cosh^{2}(t/a) sinh^{2} (t/a) = -------------- cosh^{2} (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 = constantSee here for a beautiful illustration of this fact.