Tuesday, 16 August 2016

Abstract paintings with velocity fields

The following images show a graphical representation of the flow of velocity fields. In them you can observe the behavior of particles moving with respect to the velocity field. I made these images with the program GeoGebra and I used filters from Snapseed.

$\mathbf v=(x-y,x+y)$

$\mathbf v=\left(-\dfrac{y}{x^2+y^2},\dfrac{x}{x^2+y^2}\right)$

$\mathbf v=(x,-y)$

Guess who is $\mathbf v$

$\mathbf v=(-x+xy-x^2,-xy+y)$

$\mathbf v=\left(\dfrac32\cos y,\dfrac32\,\text{sen } x\right)$

$\mathbf v=(x^2-y^2,2xy)$

$\mathbf v=\left(\dfrac32\cos y,\dfrac32\text{sen} x-y\right)$

$\mathbf v=\left(-1-\dfrac{x}{(x^2+y^2)^{3/4}},-1-\dfrac{y}{(x^2+y^2)^{3/4}}\right)$

If you have time to observe the behavior of the flow defined by the velocity field and want to make abstract paintings, then click the image below or on the link.

Saturday, 23 July 2016

Vector fields: Examples

Vector fields arise very naturally in physics and engineering applications from physical forces: gravitational, electrostatic, centrifugal, etc. For example, the vector field defined by the function
$\mathbf{F}(x,y,z)=-w_0\left(\frac{x}{\left(x^2+y^2+z^2\right)^{3/2}},\frac{z}{\left(x^2+y^2+z^2\right)^{3/2}},\frac{z}{\left(x^2+y^2+z^2\right)^{3/2}}\right),$
where $w_0$ is a real number, is associated with gravity and electrostatic attraction. The gravitational field around a planet and the electric field around a single point charge are similar to this field. The field points towards the origin (when $w_0>0$) and is inversely proportional to the square of the distance from the origin.

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Simulation

Gravitational/Electrostatic field: Click on the image (or link below) to run the simulation.

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Another important example is the velocity vector field $\mathbf{v}$ of a steady-state fluid flow. The vector $\mathbf{v}(x, y)$ measures the instantaneous velocity of the fluid particles (molecules or atoms) as they pass through the point $(x, y)$. Steady-state means that the velocity at a point $(x, y)$ does not vary in time -even though the individual fluid particles are in motion. If a fluid particle moves along the curve $\mathbf{x}(t) = (x(t), y(t))$, then its velocity at time $t$ is the derivative
$\mathbf{v}= \frac{d\mathbf{x}}{dt}$
of its position with respect to $t$. Thus, for a time-independent velocity vector field
$\mathbf{v}(x, y) = ( v_1(x, y), v_2(x, y) )$
the fluid particles will move in accordance with an autonomous, first order system of ordinary differential equations
$\frac{dx}{dt}= v_1(x, y),\qquad \frac{dy}{dt}= v_2(x, y)$

According to the basic theory of systems of ordinary differential equations, an individual particle's motion $\mathbf{x}(t)$ will be uniquely determined solely by its initial position $\mathbf{x}(0) = \mathbf{x}_0$. In fluid mechanics, the trajectories of particles are known as the streamlines of the flow. The velocity vector $\mathbf{v}$ is everywhere tangent to the streamlines. When the flow is steady, the streamlines do not change in time. Individual fluid particles experience the same motion as they successively pass through a given point in the domain occupied by the fluid.

Examples of velocity vector fields of steady-state fluids flow are the following:

1. Rigid body rotation: $$\mathbf{v}(x,y)=(-wy,wx),\quad w\in \mathbb R.$$
2. Stagnation point: $$\mathbf{v}(x,y)=(kx,-ky), \quad k\in \mathbb R.$$
3. Vortex: $$\mathbf{v}(x,y)=\left(-\dfrac{y}{x^2+y^2},\dfrac{x}{x^2+y^2}\right).$$
4. Source and Sink: $$\mathbf{v}(x,y)=\dfrac{q}{2\pi}\left(\dfrac{x}{x^2+y^2},\dfrac{y}{x^2+y^2}\right),\quad q\in \mathbb R.$$

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Simulation

The  following simulations show the steady-state fluid flows defined by the above velocity fields. Click on the image (or link below) to access the simulations.

 Rigid body rotation Link: Here Stagnation point Link: Here Vortex Link: Here Source and Sink Link: Here

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Wednesday, 20 July 2016

Flux in 2D (Part 2)

3. Outward flux across a closed plane curve

 Recall that: If $C$ is a piecewise-smooth, simple closed curve, the net outward flux of a vector field $\mathbf v =v_1(x,y)\,\mathbf i +v_2(x,y)\,\mathbf j$ across $C$ is given by $$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n\, dS$$ where $\mathbf n$ is a unit vector normal to $C$, directed outward from the region bounded by $C$.

Example 1. Suppose that $C$ is a circle of radius $r$, centre the origin. Let $\mathbf v=x\,\mathbf i+y\,\mathbf j$. In this case, the unit normal vector to $C$ is in the same direction as $\mathbf v$. Thus
$$\mathbf v \cdot \mathbf n =|\mathbf v|=r.$$
It follows that
$$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n dS=\oint_C r\,dS=r\oint_C dS=2\pi r^2.$$

The following applet shows a dynamic view of Example 1. Change the radius of the circle to check the general solution.

Example 2. Consider the same circle $C$ of radius $r$, centre the origin. But now we have $\mathbf v=-y\,\mathbf i+x\,\mathbf j$. In this case, $\mathbf v$ is perpendicular to the unit normal vector $\mathbf n$. Thus
$$\mathbf v \cdot \mathbf n =0.$$
Therefore
$$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n\, dS=0.$$

The following applet shows a dynamic view of Example 2. Change the radius of the circle to check the general solution.

Observation: The previous examples make sense physically. In the Example 1, the fluid is spewing out the origin, and thus flowing through $C$. On the other hand, in Example 2, the fluid is spinning around the origin. None of it crosses $C$.

 Flux form of Green's theorem If $C$ is a positively oriented simple closed curve enclosing a region $D$ and $\mathbf v =v_1(x,y)\,\mathbf i +v_2(x,y)\,\mathbf j$ then $$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n \,dS=\iint_D\text{div}(\mathbf v )\,dA$$ where $\text{div} (\mathbf v)=\dfrac{\partial v_1}{\partial x}+\dfrac{\partial v_2}{\partial y}$.

Example 3. Consider again the example of a circle $C$ of radius $r$, centre at the origin, and the velocity vector $\mathbf v =x\, \mathbf i +y\, \mathbf j$. Since
$$\text{div}(\mathbf v)=1+1=2,$$
then
$$\text{Net outward flux}=\oint_C \mathbf v\cdot \mathbf n \,dS=\iint_D\text{div}(\mathbf v )\,dA=\iint_D 2\,dA=2\pi r^2.$$

If we move the circle away from the origin, the flux is still $2\pi r^2$. However, computing $\oint_C \mathbf v \cdot \mathbf n \,dS$ becomes quite hard.

The following applet shows a dynamic representation of Example 3. Move the circle around. Observe that the flux remains constant. You can also observe the flux for a semicircle by dragging the slider for the partial path.

Key Concept

In the context of fluids the divergence measures how much fluid is being added (or taken away); these are known as sources (or sinks). For $\mathbf v =x\, \mathbf i +y\, \mathbf j$, fluid is being added everywhere (imagine rain falling on the ground and then flowing from the origin). For $\mathbf v =-y\, \mathbf i +x\, \mathbf j$ the divergence is zero. No fluid is being added or removed, there are no sources or sinks.

Flux in 2D

1. What is flux in 2D?

 Recall that: The flux of any two dimensional vector field $\mathbf v =v_1(x,y)\,\mathbf i +v_2(x,y)\,\mathbf j$ across a plane curve $C$ is defined by $$\text{Flux}=\int_C\mathbf v \cdot \mathbf n\, dS$$ where $\mathbf n$ is a unit normal vector to $C$.

A particular context that helps to understand the definition of flux is considering the velocity field of a fluid. Let $C$ be a plane curve and let $\mathbf v$ be a velocity vector  in the plane. Now imagine that $C$ is a membrane across which the fluid flows, but does not impede the flow of the fluid. In other words, $C$ is an idealised membrane invisible to the fluid. In this context, the flux of $\mathbf v$ across $C$ is the quantity of fluid flowing through $C$ per unit time, or the rate of flow.

2. Flux across line segments

The following applet shows a representation of a fluid flowing through a line segment. Activate the boxes to show the field and flow. Observe what happens to the flux when you change the angle, length or position of the segment.

Example 1. Consider a fluid flowing with a velocity vector $\mathbf v =-y\, \mathbf i + x\, \mathbf j$. Calculate the flux of $\mathbf v$ across the line $C=\{(x,y)|\,x=2, 2\leq y\leq 6\}$.

First, we parametrise the line as
$$\mathbf r (t)=2\,\mathbf i +(4t+2)\,\mathbf j \qquad (0\leq t\leq 1).$$
Then we have that $\mathbf r ' (t)=4\,\mathbf j.$ A unit tangent vector to $C$ is then given by
$$\mathbf T=\frac{\mathbf r ' (t)}{|\mathbf r ' (t)|}=\frac{4\,\mathbf j}{4}=\mathbf j$$
The unit vector normal is then
$$\mathbf n =\mathbf T \times \mathbf k = \mathbf j \times \mathbf k = \mathbf i.$$
With this parametrisation,
$$\mathbf v = -(4t+2)\,\mathbf i+ 2 \,\mathbf j \qquad \implies \qquad \mathbf v \cdot \mathbf n= -4t-2.$$
Note also that within the integral the infinitesimal element of arc can be expressed in terms of the parametrisation as $dS=|\mathbf r ' (t)|dt=4\,dt$ in this case. The flux across $C$ is then
$$\int_0^1(-16t-8)dt=-8(t^2+t)\Big|_0^1=-16.$$

 Recall that: If $\mathbf r (t)=x(t)\,\mathbf i +y(t)\,\mathbf j$ is a parametrisation of $C$ for $a\leq t\leq b$, then $$\text{Flux}=\int_a^b\left(v_1(t)\dot{y}-v_2(t)\dot{x}\right)dt.$$

You can check the previous result in the following applet, which shows a representation of the fluid flowing with a velocity vector $\mathbf v =-y\, \mathbf i + x\, \mathbf j$. Play with the applet changing the position of the endpoints of the line segment. Observe what happens to the flux.

Activity 1: Use the applet to answer the following questions.

1. What is the flux across the line segment oriented from $(0,0)$ to $(2,3)$?

2. What is the flux if the line segment is oriented from $(2,3)$ to $(0,0)$?

3. What is the flux if the line segment is oriented from $(-2,-2)$ to $(2,2)$ or from $(-3,3)$ to $(3,-3)$?

The general case:

Calculate the flux across the curve defined as the line segment oriented from $(a,b)$ to $(c,d)$, with $a,b,c,d$ real constants. Check the values obtained in the previous activity with the general solution.

Example 2. Consider a fluid flowing with a velocity vector $\mathbf v =x\, \mathbf i -y \, \mathbf j$. Calculate the flux of $\mathbf v$ across the line segment from $(-1,1)$ to $(-1,3)$ followed by another line segment from $(-1,3)$ to $(-2,5)$.

To calculate the flux, we shall calculate the flux across each of the two curves which make the complete curve. That is, we shall calculate
$$\text{Flux}=\int_{C_1}\mathbf F \cdot \mathbf n_1 dS+\int_{C_2}\mathbf F \cdot \mathbf n_2 dS.$$

For $C_1$:

First, we parametrise the line as
$$\mathbf r (t)=-\mathbf i +(2t+1)\,\mathbf j \qquad (0\leq t\leq 1).$$
Then we have that $\mathbf r ' (t)=2\,\mathbf j.$ A unit tangent vector to $C_1$ is then given by
$$\mathbf T=\frac{\mathbf r ' (t)}{|\mathbf r ' (t)|}=\frac{2\,\mathbf j}{2}=\mathbf j$$
The unit vector normal is then
$$\mathbf n_1 =\mathbf T \times \mathbf k = \mathbf j \times \mathbf k = \mathbf i.$$
With this parametrisation,
$$\mathbf v = -\mathbf i-(2t+1) \,\mathbf j \qquad \implies \qquad \mathbf v \cdot \mathbf n_1= -2.$$
Note also that $dS=|\mathbf r ' (t)|dt=2\,dt$ in this case. The flux across $C$ is then
$$\int_0^1-2dt=-2t\Big|_0^1=-2.$$

For $C_2$:

We parametrise the line as
$$\mathbf r (t)=-(t+1)\,\mathbf i +(2t+3)\,\mathbf j \qquad (0\leq t\leq 1).$$
Then we have that $\mathbf r ' (t)=-\mathbf i+2\,\mathbf j.$ A unit tangent vector to $C_2$ is then given by
$$\mathbf T=\frac{\mathbf r ' (t)}{|\mathbf r ' (t)|}=\frac{-\mathbf i+2\,\mathbf j}{\sqrt{5}}.$$
The unit vector normal is then
$$\mathbf n_2 =\mathbf T \times \mathbf k = \frac{2\mathbf i+\mathbf j}{\sqrt{5}}.$$
With this parametrisation,
$$\mathbf v = -(t+1)\,\mathbf i- (2t+3) \,\mathbf j \qquad \implies \qquad \mathbf v \cdot \mathbf n_2= \frac{-4t-5}{\sqrt{5}}.$$
Note also that $dS=|\mathbf r ' (t)|dt=\sqrt{5}\,dt$ in this case. The flux across $C$ is then
$$\int_0^1(-4t-5)dt=\left(-2t^2-5\right)\Big|_0^1=-7.$$
Therefore the flux is $-2-7=-9$.

You can check the previous result in the following applet, which shows a representation of the fluid flowing with a velocity vector $\mathbf v =x\, \mathbf i - y\, \mathbf j$. Play with the applet changing the position of the endpoints of the line segmen. Observe what happens to the flux.

Activity 2:  Within the applet.

1. Move around the three points that define the line segments. Observe what happens to the flux.

2. If we fix the endpoints of the path and we just move the point in the middle, you can observe that the flux remains constant. Why?

3. Finally, drag the three points to create the path from $(3,0)$ to $(3,6)$ and then to $(0,6)$. Observe that the flux is zero. Why? What happens if you move the endpoints of the path along the axis.

Next>

Key Concepts

A key concept expressed in terms of line integrals is flux. Flux measures the rate that a field crosses a given line. The idea of flux is especially important for Green’s theorem, and in higher dimensions for Stokes’ theorem and the divergence theorem.

Key Equations

Calculating a scalar line integral:

$$\int_C f(x,y)dS=\int_a^bf\left(x(t),y(t)\right)\sqrt{(x'(t))^2+(y'(t))^2}.$$

Calculating a vector line integral:

$$\int_C\mathbf F (\mathbf r)\cdot \mathbf d \mathbf r =\int_C\left(F_1dx+F_2dy\right)=\int_a^b \mathbf F (\mathbf r (t) ) \cdot \mathbf {r'}(t) dt.$$

Calculating flux:

$$\int_C \mathbf F \cdot \mathbf n dS=\int_C \left(F_1dy-F_2dx\right).$$

Flux across a line segment

Consider a two-dimensional flow of a fluid with the  velocity field  $$\mathbf{v}=-y\, \mathbf{i} +x\,\mathbf{j}.$$

The aim of this activity is to investigate the physical meaning of the flux and circulation of $\mathbf{v}$ across a line segment.

(a) Calculate the flux of $\mathbf{v}$ across the following line segments:

1. $C_1$: from $A=(0,-1)$ to $B=(0,1)$.
2. $C_2$: from $A=(0,3)$ to $B=(-4,0)$.
3. $C_3$: from $A=(0,-1)$ to $B=(0,2)$.
4. $C_4$: from $A=(-2,0)$ to $B=(0,2)$.

Confirm your answers using the following applet:

(b) In part (a) you should have found that in some cases the flux was equal to  zero. Use the same applet (above) to investigate where you need to put the endpoints of the line segment in order to obtain a flux equal to zero. For example, use the applet to define the line segment from $A=(-2,-3)$ to $B=(2,3)$ or from $A=(-5,2)$ to $B=(2,5)$, and observe what happens to the flux in each case.

1. Describe a general condition required for the flux across the line segment to be zero.

2. Explain what happens physically.

Vector fields

What is a vector field?

In general, a vector field is a function that assigns vectors to points in space.

A vector field in the $xy$ plane is a vector function of 2 variables:
$\mathbf{F}(x,y)=\left(F_1(x,y),F_2(x,y)\right)=F_1(x,y)\mathbf{i}+F_2(x,y)\mathbf{j}$

The best way to picture a vector field is to draw the arrow representing the vector $\mathbf{F}(x,y)$ starting at the point $(x,y)$. Of course, it's impossible to do this for all points $(x,y)$, but we can gain a reasonable impression of $\mathbf{F}$ by doing it for a few representative points in $\mathbb R^2$.

Similarly a vector field in 3-D is a vector function of 3 variables:
\begin{eqnarray*}
\mathbf{F}(x,y,z)&=&\left(F_1(x,y,z),F_2(x,y,z),F_3(x,y,z)\right)\\&=&F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}
\end{eqnarray*}

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Simulation

Some computer algebra systems are capable of plotting vector fields in two or three dimensions. They give a better impression of the vector field than is possible by hand because the computer can plot a large number of representative vectors. The simulations below are tools for plotting vector fields in two and three dimensions.

Click on the image (or link) to run the simulation

2D Vector Field
http://ggbm.at/cXgNb58T

3D Vector Field
http://ggbm.at/KKB2Ndez

Classification of conics

1. Introduction

Consider the  equation of the form
$ax^2+by^2+cxy+dx+ey+f=0$
where $a,b,\ldots ,f$ are real numbers, and at least one of the numbers $a,b,c$ is not zero. An equation of this type is called a quadratic equation in $x$ and $y$, and
$ax^2+by^2+cxy$
is called the associated quadratic form.

Graphs of  quadratics equations are known as conics, or conic sections. The most important conics are ellipses, circles, hyperbolas and parabolas; these are called non-degenerate conics. The remaining conics are called degenerated and include single points and pairs of lines.

A conic is said to be in standard position relative to the coordinate axes if its equation can be expressed in one of the following forms:
\begin{align}
& \bullet \;\;\dfrac{x^2}{k^2}+\dfrac{y^2}{l^2}=1; \;\;k,l>0,  \nonumber \\ % no number is shown
& \bullet \;\;\dfrac{x^2}{k^2}-\dfrac{y^2}{l^2}=1\;\; \text{or}\;\; \dfrac{x^2}{k^2}-\dfrac{y^2}{l^2}=1; \;\;k,l>0, \label{conics}\\ % there is a number
& \bullet\;\; x^2=ky\;\; \text{or}\;\; y^2=kx;\;\; k\neq 0. \nonumber % no number
\end{align}

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Simulation 1. Click on the image (or link) to explore non-degenerate conics in standard position.

Conics in standard position, click Here

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Observe that no conic in standard position has an $xy$-term (that is, a cross-product term) in its equation; the presence of an $xy$-term in the equation of a non-degenerate conic indicates that the conic is rotated out of standard position (Figure 1).

 Figure 1. Rotated

Also no conic in standard position has both an $x^2$ and a $x$ term or both $y^2$ and a $y$ term. If there is no cross-product term, the occurrence of either of these pairs in the equation of a non-degenerate conic indicates that the conic is translated out of standard position (Figure 2).

 Figure 2. Translated.
The occurrence of either of these pairs and a cross-product term usually indicates that the conic is both rotated and translated out of standard position (Figure 3).

 Figure 3. Rotated and translated.

2. Identifying conics

One technique for identifying the graph of a non-degenerate conic that is not in standard position consists of rotating and translating the $xy$-coordinate axes to obtain an $uv$-coordinate system relative to which the conic is in standard position. Once this is done, the equation of the conic in the $uv$-system will have one of the forms given in (\ref{conics}) and can then easily be identified.

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Simulation 2. Click on the image (or link below) to explore the rotation and translation of conics.

Rotation and translation of conics, click Here

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Example: Completing the square and translating.

Since the quadratic equation
$2x^2+y^2-12x-4y+18=0$
contains $x^2$-, $x$-, $y^2$-, and $y$-terms but no cross-product term, its graph is a conic that is translated out of standard position but not rotated. This conic can be brought into standard position by suitably translating coordinate axes. To do this, first collect $x$-terms and $y$-terms. This yields
$(2x^2-12x)+(y^2-4y)+18=0 \quad \text{or}\quad 2(x^2-6x)+(y^2-4y)=-18.$
By completing the squares on the two expressions in parentheses, we obtain
\begin{eqnarray}\label{uno}
\end{eqnarray}
If we translate the coordinate axes by means of the translation equations
$u=x-3,\quad v=y-2$
then (\ref{uno}) becomes
$2u^2+v^2=4\quad \text{or}\quad \frac{u^2}{2}+\frac{v^2}{4}=1$
which is the equation of an ellipse in standard position in the $uv$-system.

2.1 Eliminating the cross-product term

We shall now show how to identify conics that are rotated out of standard position. Now consider a conic $C$ whose equation in $xy$-coordinates is $ax^2+by^2+cxy+dx+ye+f=0$. This equation  can always be written in matrix form:
$\big(x\;\;y\big) \Bigg( \begin{matrix} a & c/2 \\ c/2 & b \end{matrix}\Bigg) \Bigg( \begin{matrix} x \\ y \end{matrix}\Bigg) + \big(d\;\;e\big) \Bigg( \begin{matrix} x \\ y \end{matrix}\Bigg) +f=0$
or
$\mathbf{x}^{T}A\mathbf{x}+K\mathbf{x}+f=0$
where
$A=\Bigg( \begin{matrix} a & c/2 \\ c/2 & b \end{matrix}\Bigg),\quad \mathbf{x}= \Bigg( \begin{matrix} x \\ y \end{matrix}\Bigg) \quad \text{and}\quad K=\big(d\;\;e\big).$

We would like to rotate the $xy$-coordinate axes so that the equation of the conic in the new $uv$-coordinate system has no cross-product term. This can be done as follows.

Step 1. Find a matrix
$P=\Bigg( \begin{matrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{matrix}\Bigg)$
that orthogonally diagonalises the matrix $A$.

Step 2. Interchange the columns of $P$, if necessary, to make $\text{det}(P) = 1$. This ensures that the orthogonal coordinate transformation
\begin{eqnarray}\label{dos}
\begin{matrix}
x  \\
y
\end{matrix}\Bigg)=P \Bigg(
\begin{matrix}
u  \\
v
\end{matrix}\Bigg)
\end{eqnarray}
is a rotation.

Step 3. To obtain the equation for $C$ in the $uv$-system, substitute (\ref{dos}) into the matrix form of the equation. This yields
$\left(P\mathbf{v} \right)^{T}A\left(P\mathbf{v} \right)+K\left(P\mathbf{v} \right)+f=0$
or
\begin{eqnarray}\label{tres}
\mathbf{v} ^{T}\left(P^TAP\right)\mathbf{v} +\left(KP \right)\mathbf{v}+f=0.
\end{eqnarray}

Since $P$ orthogonally diagonalises $A$,
$P^TAP=\Bigg( \begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix}\Bigg)$
where $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$. Thus (\ref{tres}) can be rewritten as
$\big(u\;\;v\big) \Bigg( \begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix}\Bigg) \Bigg( \begin{matrix} u \\ v \end{matrix}\Bigg) + \big(d\;\;e\big)\Bigg( \begin{matrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{matrix}\Bigg) \Bigg( \begin{matrix} u \\ v \end{matrix}\Bigg) +f=0$
or
$\lambda_1u^2+\lambda_2u^2+d^{\prime}u+e^{\prime}v+f=0$
(where $d^{\prime}=dp_{11}+ep_{21}$ and $e^{\prime}=dp_{12}+ep_{22}$). This equation has no cross-product term.

Example: Eliminating the Cross-Product Term.

Consider  the conic $C$ whose equation is $5x^2-4xy+8y^2-36=0$.

To describe this conic, first we write the quadratic equation in its matrix form. That is
\begin{eqnarray}\label{cuatro}
\mathbf{x}^{T}A\mathbf{x}-36=0
\end{eqnarray}
where
$$A=\Bigg( \begin{matrix} 5 & -2 \\ -2 & 8 \end{matrix}\Bigg)$$

The characteristic equation of $A$ is
$$\text{det}(A-\lambda I)=\left| \begin{matrix} 5-\lambda & -2 \\ -2 & 8-\lambda \end{matrix}\right|=(\lambda-9)(\lambda-4)=0$$
so the eigenvalues of $A$ are $\lambda=4$ and $\lambda=9$. We leave it for the reader to show that orthonormal bases for the eigenvalues are
$\lambda=4:\quad \mathbf{v}_1= \Bigg( \begin{matrix} 2/\sqrt{5} \\ 1/\sqrt{5} \end{matrix}\Bigg),\qquad \lambda=9:\quad \mathbf{v}_2= \Bigg( \begin{matrix} -1/\sqrt{5} \\ 2/\sqrt{5} \end{matrix}\Bigg)$
Thus
$$P=\Bigg( \begin{matrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{matrix}\Bigg)$$
orthogonally diagonalises $A$. Moreover, $\text{det}(P)=1$, and thus the orthogonal coordinate transformation
\begin{eqnarray}\label{cinco}
\mathbf{x} =P\mathbf{v}
\end{eqnarray}
is a rotation. Substituting (\ref{cinco}) into (\ref{cuatro}) yields
$(P\mathbf{v})^{T}A(P\mathbf{v})-36=0\quad\text{or}\quad \mathbf{v}^T(P^TAP)\mathbf{v}-36=0$

Since
$P^TAP=\Bigg( \begin{matrix} 4 & 0 \\ 0 & 9 \end{matrix}\Bigg)$
this equation can be written as
$\big(u\;\;v\big) \Bigg( \begin{matrix} 4 & 0 \\ 0 & 9 \end{matrix}\Bigg) \Bigg( \begin{matrix} u \\ v \end{matrix}\Bigg) -36=0$
or
$4u^2+9v^2-36=0\quad\text{or}\quad \frac{u^2}{9}+\frac{v^2}{4}=1$
which is the equation of the ellipse in standard form.

3. Exercise

Translate and rotate the coordinate axes, if necessary, to put the conic in standard position. Name the conic and give its equation in the final coordinate system.

1. $9x^2-4xy+6y^2-10x-20y-5=0$
2. $3x^2-8xy-12y^2-30x-64y=0$
3. $2x^2-4xy-y^2-4x-8y+14=0$
4. $21x^2+6xy+13y^2-114x+34y+73=0$

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Simulation 3. You can check your results in the following simulation. Click on the image (or link below) to access the simulation.

Classification of conics, click Here

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