Witch of Agnesi

Witch of Agnesi
Field: Algebra
Image Created By: John H. Lienhard
Website: Engines of our Ingenuity

Witch of Agnesi

The Witch of Agnesi appears in almost all high school and undergraduate math books. It is a curve that is symmetric about the y-axis and approaches an on the x-axis. The Witch of Agnesi was originally called La versiera di Agnesi, or "The Curve of Agnesi." "Versiera" is very similar to the word "avversiera" in Italian, which means "woman contrary to God". This was interpreted as witch. The name of the curve was mistranslated.

Basic Description

The Witch of Agnesi extends to positive infinity and negative infinity in the horizontal direction.

The following steps show how to construct the Witch geometrically.

 Step 1: Draw a circle with diameter $a$ formed from point A and B. Step 2: Draw tangents at A and B. A tangent is a line that touches the curve of the circle just at one point such that the point is the place where the curve and plane will only meet. Label a point C on tangent B. Label a point D anywhere on the circle. Step 3: Draw a line segment AD, which goes across the inside of the circle and is called a chord. Make a parallel line to the two tangent lines at D. Where it intersects AB, call the point G. Step 4: Extend AD to intersect l 1 at point C. Extend GD to point E. Step 5: Drop a perpendicular line from point C to intersect l 2. The perpendicular dropped from C intersects the point E. Step 6: To create the Witch of Agnesi, drag point D around the circle. As D changes position, the line segment CE will move horizontally. Point E will trace out the Witch.

Trace the Witch of Agnesi below.

The equation for the Witch of Agnesi is the following:

$y= \frac{a^3} {a^2 + x^2}$

A More Mathematical Explanation

Let us derive the Witch equation.

We again create the curve by having a circle with diameter '"`U [...]

Let us derive the Witch equation.

We again create the curve by having a circle with diameter $a$. The diameter is perpendicular to the x-axis. Points A and B are the endpoints of diameter $a$. Point A is on the x-axis.

Let there be a point C such that BC is parallel to the x-axis. This is shown in Image 1. Let D be the point where AC intersects the circle. Refer to Image 2.

A right triangle CED is also formed with leg DE being parallel to the x-axis. Leg CE is perpendicular to the x-axis. Now to generate the curve we let D move around the circle. The Witch is the curved formed by the traces E makes while D moves around the circle. Refer to Image 3.

 Image 1 Image 2 Image 3

Derivation

The triangles formed in the Image 4 are important in the derivation of the Witch. At this point we can use standard rules of geometry to show any relationship within Image 4. Since the angle ADB (we denote this as $\angle ADB$) is it is a right angle.

Image 4

Right triangles CBA and CBD (we denote these triangles as $\triangle CBA$ and $\triangle CBD$) have a common acute angle $\angle ACB$. The angles' measurement in any triangle sum up to 180 o. With subtraction the third angle for both triangles will be the same. Thus the two triangles are similar because all three pair of corresponding angles are the same.

We apply this same logic for right triangles CBA and DEC. Because they are alternate interior angles $\angle ACB$ is congruent to $\angle CDE$ and $\angle DCE$ is congruent to $\angle CAB$. Since the triangles are similar, then their sides are proportional. We can also write this statement like in equation 1.

Eq. 1        $\frac{CB} {DE}=\frac{BA} {EC}$

Let point A be the origin if we place the entire configuration on the Cartesian coordinate system. We can now specify each point. Usually the coordinate axes (the x-axis and the y-axis) are scaled by certain units. In our case, we will not set numerical value but instead use variables. We will now label the coordinate points of D as $\left( \mathrm{i},\mathrm{j} \right)$ and E as $\left( \mathrm{x}, \mathrm{y} \right)$. This means the point D goes $i$ units in the x direction and $j$ units in the y direction.Notice that point D and E are located on the same y coordinate so instead of saying that D is at $\left( \mathrm{i},\mathrm{j} \right)$ we can say it's at $\left( \mathrm{i},\mathrm{y} \right)$. It is easier to have everything in terms of the same variables.

In order to derive the Witch we will use the coordinates of point D and E. We plan to have everything in terms of $x$, $y$ and $a$ (the diameter of the circle or the highest y-value). It is beyond important essential to remember that $a$ is the diameter. You can see in image 4 that $a$ is the length of line AB. We use $x$, $y$ and $a$ to tell where exactly a point is located and the length of a line segment.

Because, we want to have everything in terms of $x$, $y$ and $a$ we must first rewrite $i$ using the necessary variables.

Rewrite Equation 1 using the coordinates of points D and E. Refer to Image 5.

Image 5

Eq. 2        $\frac{x} {x-i}=\frac{a} {a-y}$.

Cross multiply the expression above.

$\frac{x} {x-i}=\frac{a} {a-y}$
$x \left ( a-y \right )= a \left ( x-i \right )$

Distribute the variable $x$ on the left side and the variable $a$ on the right side.

$x a - x y= a x - a i$

Get any terms with $i$ on one side.

$x a - x y - ax = -a i$

Simplify.

$-x y= -ai$

Solve for $i$.

$\frac{xy} {a} = i$

We know the x-coordinate of D (which was once $i$) in terms of the other variables ($x$,$y$,$a$). This will come in handy later.

Refer to Image 6 below.

Image 6

By , we will create three equations looking at $\triangle ADF$ $\triangle BDG$ and $\triangle BAD$.

1. $\overline{AD}^2 = \overline{AF}^2 + \overline{DF}^2$
2. $\overline{DB}^2 = \overline{BG}^2 + \overline{GD}^2$
3. $\overline{AB}^2 = \overline{AD}^2 + \overline{DB}^2$

Use ($i$, $y$, $a$) to rewrite the three equations above.

1. $\overline{AD}^2 = i^2 + y^2$
2. $\overline{DB}^2 = \left (a-y \right )^2 + i^2$
3. $a^2 = \overline{AD}^2 + \overline{DB}^2$

Substituting the first two expression into the third, we get one expression.

$a^2 = i^2 + y^2 + \left (a-y \right )^2 + i^2$

Simplify the expression.

$a^2 = i^2 + y^2 + a^2 - 2ay + y^2 + i^2$

Combine like terms.

$a^2 = 2{i^2} + 2{y^2} + a^2 - 2ay$

Set the expression equal to zero.

$0 = -a^2 + 2{i^2} + 2{y^2} + a^2 - 2ay$

Simplify and get:

$0 = y^2 + i^2 - ay$

Remember, we solved for $i$ in term of $x$, $y$ and $a$.

So, plug $i = \frac{xy} {a}$ into the following expression:

$y^2 + i^2 - ay = 0$
Image 7. Graph of Witch when a=2

Substitute $i$ and get:

$y^2 +\left (\frac{xy} {a} \right )^2 - ay =0$
$y^2 +\frac{\left ( xy \right )^2} {a^2} - ay =0$

Multiply each term by $a^2$ so that every term has the same denominator, which is 1.

$a^2 y^2 + x^2 y^2 - a^3 y =0$

Simplify.

$a^2 y + x^2 y - a^3 =0$

Get all terms with $y$ on the left side.

$a^2 y + x^2 y =a^3$

Factor out $y$ from each term on the left side.

$y ({a^2 + x^2}) =a^3$

Solve for $y$ and you get the Witch in form.

$y= \frac{a^3} {a^2 + x^2}$

Parametric Derivation

We derive the Witch again with a little trigonometry and we will have the form of the Witch.

It's important to know trigonometric identities. In this case, we will use cotangent which is the ratio of the adjacent side and opposite side of a right triangle.

Image 8

Looking at $\triangle \mathrm{A}\mathrm{F} \mathrm{C}$ in the Image 8.

$\cot \theta = \frac{x} {a}$

So,

$x = \mathrm{a} \cot \theta$

And, we rewrite the Cartesian form of the Witch using our new expression for x.

$y= \frac{a^3} {a^2 + \left ( \mathrm{a} \cot \theta \right )^2}$

Do the operation within the parentheses.

$y= \frac{a^3} {a^2 + \left ( \mathrm{a}^2 \cot^2 \theta \right )}$

Factor out the $a^2$ from both terms in the denominator.

$y= \frac{a^3} {a^2 \left ( \cot^2 \theta + 1 \right )}$

Use the Pythagorean Identity: $\cot^2 \theta + 1 = \csc^2 \theta$

$y= \frac{a^3} {a^2 \csc^2 \theta}$

Use the Reciprocal Identity: $\frac{1} {\csc \theta} = \sin \theta$

$y= \mathrm{a} \sin^2 \theta$

The Witch of Agnesi is sometimes presented in another form in math textbooks because the derivation of the Witch is based on different assumptions. In our case, we had a circle with diameter a. Other texts have a circle with radius a, and diameter 2a. The part of the equation that includes variable x does not change. When the diameter is 2a the Cartesian equation of the Witch is:

$y= \frac{8 a^3} {4 a^2 + x^2}$

The Witch has a parametric equation of:

$y= \mathrm{2 a} \sin^2 \theta$

The two representations of the Witch curve are equivalent. We will show this by substituting $b$ for $a$ in the equation we derived and $2a$ in the other form of the Witch.

The following is the result when $a = b$:

\begin{alignat}{2} y & = \frac{a^3} {a^2 + x^2} \\ & = \frac{b^3} {b^2 +x^2} \\ \end{alignat}

The following is the result when $2 a = b$:

\begin{alignat}{2} y & = \frac{8 a ^3} {4 a^2 + x^2} \\ & = \frac{b^3} {b^2 +x^2} \\ \end{alignat}

The two expressions are equivalent.

$\frac{b^3} {b^2 +x^2} = \frac{b^3} {b^2 +x^2}$

Why It's Interesting

History

Maria Gaetana Agnesi (1718-1799) was the most famous female mathematician in the 18th century. She was best known for her contributions to differential calculus, in particular, for her study of the cubic curve known as the Witch of Agnesi. Despite prejudice, she persevered in mathematics. In 1750, she was appointed to the chair of the mathematics department at the University of Bologna. She rejected the offer and instead became the caretaker of her 20 siblings.

At a young age Agnesi was very involved in academics. At the age of nine she could speak Italian, Greek, Latin, Hebrew and French. She also published a paper on higher education for women. By age 20, Agnesi was publishing mathematics papers and deriving mathematical formulas. Agnesi's work, Instituzioni Analytiche ad Uso Della Gioventu Italiana (Analytic Insitutions for the use of Young Italians) was published in 1748 in two volumes. It was written in Italian so that it could be accessible to the public.

Agnesi is known for her peculiar methods of solving problems. After a long day of thinking and trying to solve the problem, she would leave her notebooks on her desk and go to sleep, and in the morning, her problems would often be solved. During the night, she would sleepwalk to her desk and solve the problem. Agnesi could not recall waking up and writing the solution, so she was always surprised that in the morning the answer was right in front of her. She just needed to relax a little bit. This unusual way of solving problems is how she discovered the formula for the Witch.

Agnesi is referred to as the Witch, not because of her personality but because she is associated with the curve. The curve is called the Witch because of a mistranslation from Italian to English. Agnesi originally used the word versorio, which means turning. When her major work was published, John Colson translated her work into English and confused versorio with versiera (witch). This mishap rendered her work as the Witch of Agnesi. It is actually ironic that the curve is associated with Agensi because she was not the first to study the curve. It was first investigated in 1703 by Guido Grandi.

References

Johnson, Art. Famous Problems and Their Mathematicians. Englewood, CO: Teacher Ideas, 1999. Print. p.134

• Stewart, J. (2009). Calculus Early Transcendentals. Mason, Ohio: Cengage Learning.