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 thecurve de�ned as follows:

Starting with a �xed circle, a point O on the circle is chosen. For any other point Bon 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 throughA, 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 tangentto the �xed circle through the point O.

Equations

Graph 1: (The Witch of Agnesi with labeled points)

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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 = 1x2+1 This is the derivative of the

arctangent function. Parametrically, if θ, is the angle between O2a and OB, measuredclockwise, then the curve is de�ned by the equations x = 2a tan θ, y = 2a cos2 θ. Anotherparameterization, 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 beendiscovered fairly recently during the late twentieth and early twenty-�rst centuries. TheCartesian equation (above) has appeared in the mathematical modeling of some physicalphenomena: the equation approximates the spectral line distribution of optical lines andx-rays, as well as the amount of power dissipated in resonant circuits. Formally, thecurve is equivalent to the probability density function of the Cauchy distribution. Thecross-section of a smooth hill also has a similar shape. It has been used as the generictopographic 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 thehorizontal line y = 2a let O be the origin, and let B be the point where the segment OAintersects the circle. A point P is on the Witch of Agnesi if P lies on the horizontal linethrough B and on the vertical line through A.

Graph 2: (Graph of the Witch of Agnesi with geometric details necessary to carry outthe 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:

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Let rA (θ) = xi + yAj, we will calculate x and yA.

tan θ =2a

x

x =2a

tan θx = 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 canobtain a new equation that will lead to the solution of the problem, also applying sometrigonometric identities and algebraic processes.

{tan θ = b

c =2ax

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− bb

cos2 θ

sin2 θ=

2a

b− 1

3

cos2 θ

sin2 θ+

sin2 θ

sin2 θ=

2a

b1

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) forthe 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 xaxis, then the x component of the vector-valued function r (A) is also the x componentof the vector-valued function of P. Also, since the straight line connecting point B to thepoint 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

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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).

limθ→o+

r (θ) = limθ→o+

[a2 cot θi + a(1− cos 2θ)j]

= limθ→o+

[a2 cot θi] + limθ→o+

[a(1− cos 2θ)j]

= ∞i

limθ→π−

r (θ) = limθ→π−

[a2 cot θi + a(1− cos 2θ)j]

= limθ→π−

[a2 cot θi] + limθ→π−

[a(1− cos 2θ)j]

= −∞i

(e)

Eliminate the parameter θ and determine the rectangular equation of the Witch ofAgnesi. Use a graphing utility to graph this function for a = 1 and compare your graphwith that obtained in part (c).

Given that,

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x = 2a cot θ

andy = 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 θy = 2a

cot2 θ+1

=⇒

cot θ = x

2a

y = 2ax2

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

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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 < πand 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 Witchof Agne. Although Figure 4 and Figure 3 aren't on the same scale, congruence of bothgraphs is evident, ergo, the correspondence between the vector-valued function r (θ) andthe real value function y (x).

While it is trying to prove that the proposed equations in the development describethe Witch of Agnesi,these functions are not the only which may represent it, besides themathematical 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-Valued Functions Chapter 12 - Ron Larson & Bruce H. Edwards 9 edition, editorial McGraw-Hill

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