This document describes the Witch of Agnesi curve. It begins by introducing the curve and how it is constructed geometrically based on a circle. Vector-valued functions are derived to describe the curve parametrically, and the rectangular equation is obtained by eliminating the parameter. Specifically: 1) Vector-valued functions rA(θ) and rB(θ) are derived to describe points A and B on the curve parametrically. 2) These are combined to obtain the vector-valued function r(θ) for the overall Witch of Agnesi curve. 3) The rectangular equation y=(8a^3)/(x^2+4a^2) is then derived by eliminating

1. WITCH OF AGNESI
Juan Alejandro Álvarez Agudelo - 1115191793
August 27, 2016
Abstract
In this article it exposed a curve which is called curve of Agnesi or witch of Agnesi, this is a
general form from which are derived special cases of functions, what is intended in this work is
explain how it is generated curve, also the deduction of a vector-valued function that describes the
curve of generally form, then the deduction of its rectangular equation, in each case it illustrated
a very simple type of particularly Agnesi curve.
Keywords
Witch of Agnesi, Curve of Agnesi, Vector-valued function.
I. Introduction
In mathematics, the Witch of Agnesi, sometimes called the "Curve of Agnesi" is the
curve dened as follows:
Starting with a xed circle, a point O on the circle is chosen. For any other point B
on the circle, the secant line OB is drawn. The point 2a is diametrically opposite to O.
The line OB intersects the tangent of 2a at the point A. The line parallel to O2a through
A, and the line perpendicular to O2a through B intersect at P. As the point B is varied,
the path of P is the Witch of Agnesi. Also, the curve is asymptotic to the line tangent
to the xed circle through the point O.
Equations
Graph 1: (The Witch of Agnesi with labeled points)
1

2. Suppose the point O is the origin, and that 2a is on the positive y-axis. Suppose the
radius of the circle is a. Then the curve has Cartesian equation y = 8x3
x2+4a2 . Note that if
a = 1/2, then this equation becomes rather simple: y = 1
x2+1 This is the derivative of the
arctangent function. Parametrically, if θ, is the angle between O2a and OB, measured
clockwise, then the curve is dened by the equations x = 2a tan θ, y = 2a cos2
θ. Another
parameterization, with θ, being the angle between OB and the x − axis, increasing anti-
clockwise is x = 2a cot θ, y = 2a sin θ.
Application
The curve has applications to real-life phenomena, applications which have only been
discovered fairly recently during the late twentieth and early twenty-rst centuries. The
Cartesian equation (above) has appeared in the mathematical modeling of some physical
phenomena: the equation approximates the spectral line distribution of optical lines and
x-rays, as well as the amount of power dissipated in resonant circuits. Formally, the
curve is equivalent to the probability density function of the Cauchy distribution. The
cross-section of a smooth hill also has a similar shape. It has been used as the generic
topographic obstacle in a ow in mathematical modeling.
II. Development
Consider a circle of radius a centered on the y − axis at (0, a). Let A be a point on the
horizontal line y = 2a let O be the origin, and let B be the point where the segment OA
intersects the circle. A point P is on the Witch of Agnesi if P lies on the horizontal line
through B and on the vertical line through A.
Graph 2: (Graph of the Witch of Agnesi with geometric details necessary to carry out
the proposed development)
(a)
Show that the point is traced out by the vector-valued function rA (θ) = (2a cot θ)i+(2a)j,
0 θ π, where θ is the angle that OA makes with the positive x-axis.
The tools needed to nd the vector-valued function are purely geometrics:
2

3. Let rA (θ) = xi + yAj, we will calculate x and yA.
tan θ =
2a
x
x =
2a
tan θ
x = 2a cot θ
yA = 2a
So,
rA (θ) = (2a cot θ)i + (2a)j, 0 θ π
(b)
Show that the point B is traced out by the vector-valued function rB (θ) = (a sin 2θ)i +
a(1 − cos 2θ)j, 0 θ π.
The tools needed to nd the vector-valued function are purely geometrics and trigono-
metricals:
Let rB (θ) = Ci + bj, we will calculate C and b.
a2
= C2
+ (a − b)2
[(a − b) + b]
2
= C2
+ (a − b)2
$$$$
(a − b)2
+ 2(a − b)b + b2
= C2
+$$$$
(a − b)2
b [2(a − b) + b] = C2
b [2a − b] = C2
With the equation obtained, along with an equation established by similarity, we can
obtain a new equation that will lead to the solution of the problem, also applying some
trigonometric identities and algebraic processes.
tan θ = b
c = 2a
x
C2
= b [2a − b]
=⇒ tan θ =
b
b (2a − b)
tan θ =
b
b (2a − b)
sin2
θ
cos2 θ
=
b2
b (2a − b)
cos2
θ
sin2
θ
=
2a − b
b
cos2
θ
sin2
θ
=
2a
b
− 1
3

4. cos2
θ
sin2
θ
+
sin2
θ
sin2
θ
=
2a
b
1
sin2
θ
=
2a
b
b = 2a sin2
θ
b = 2a 1 − cos2
θ
b = 2a 1 −
cos 2θ + 1
2
b = 2a
1
2
−
cos 2θ
2
b = a (1 − cos 2θ)
Now,
b
c
=
2a
x
C =
bx
2a
C = ¡a (1 − cos 2θ) x
2¡a
C =
1 − 2 cos2
θ − 1 x
2
C =
¡2 1 − cos2
θ x
¡2
C = sin2
θ 2a cot θ
C = 2a sin θ cos θ
C = a sin 2θ
So,
rB (θ) = (a sin 2θ)i + a(1 − cos 2θ)j, 0 θ π.
(c)
Combine the results of parts (a) and (b) to nd the vector-valued function r (θ) for
the Witch of Agnesi. Use a graphing utility to graph this curve for a = 1.
Since the straight line connecting point A to the point P is perpendicular to the x
axis, then the x component of the vector-valued function r (A) is also the x component
of the vector-valued function of P. Also, since the straight line connecting point B to the
point P is parallel to the x axis, then the y component of the vector-valued function r
(B) is also the y component of the vector-valued function of P. This is:
r (θ) = 2a cot θi + a(1 − cos 2θ)j, 0 θ π
4

5. Then for a = 1:
r (θ) = 2 cot θi + (1 − cos 2θ)j, 0 θ π
Graph 3: (Graph of r (θ) when a = 1)
(d)
Describe the limits limθ→0+ r (θ) and limθ→π− r (θ).
lim
θ→o+
r (θ) = lim
θ→o+
[a2 cot θi + a(1 − cos 2θ)j]
= lim
θ→o+
[a2 cot θi] + lim
θ→o+
[a(1 − cos 2θ)j]
= ∞i
lim
θ→π−
r (θ) = lim
θ→π−
[a2 cot θi + a(1 − cos 2θ)j]
= lim
θ→π−
[a2 cot θi] + lim
θ→π−
[a(1 − cos 2θ)j]
= −∞i
(e)
Eliminate the parameter θ and determine the rectangular equation of the Witch of
Agnesi. Use a graphing utility to graph this function for a = 1 and compare your graph
with that obtained in part (c).
Given that,
5

6. x = 2a cot θ
and
y = 2a sin2
θ = a (1 − cos 2θ)
then we look how to relate x and y, as follows:
cot θ =
cos θ
sin θ
cot θ =
1 − sin2
θ
sin θ
cot2
θ =
1
sin2
θ
− 1
sin2
θ =
1
cot2
θ + 1
after
x = 2a cot θ
y = 2a
cot2 θ+1
=⇒



cot θ = x
2a
y = 2a
x2
4a2 +1
y =
2a
x2
4a2 + 1
y =
2a
x2+4a2
4a2
Finally, the general rectangular equation for Witch Agnesi is:
y =
8a3
x2 + 4a2
and, for a = 1:
y =
8
x2 + 4
6

7. Graph 4: (Graph of y (x) when a = 1)
III. Conclusions
In the present work it is accomplished verify that rA (θ) = (2a cot θ)i + (2a)j, 0 θ π
and rB (θ) = (a sin 2θ)i + a(1 − cos 2θ)j, 0 θ π, so r (θ) = 2a cot θi + a(1 − cos 2θ)j, 0 θ π,
the latter being the general equation of the Witch of Agnesi.
It follows, also the general equation in rectangular coordinates, which corresponds to
the Cartesian equation recorded in the literature, this equation is given by y (x) = 8a3
x2+4a2 ,
where ais a parameter corresponding to the radius of the circle generator of the Witch
of Agne. Although Figure 4 and Figure 3 aren't on the same scale, congruence of both
graphs is evident, ergo, the correspondence between the vector-valued function r (θ) and
the real value function y (x).
While it is trying to prove that the proposed equations in the development describe
the Witch of Agnesi,these functions are not the only which may represent it, besides the
mathematical procedure for reaching the deduction of the equations is not the only.
References
[1] Free internet encyclopedia: https://en.wikipedia.org/wiki/Witch_of_Agnesi
[2] Vector-ValuedFunctionsChapter12-RonLarsonBruceH.Edwards9edition, editorialMcGraw-
Hill
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