## Posts

### Abstract paintings with velocity fields

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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.

https://www.geogebra.org/m/JPUBhFgs

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

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

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

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.

https://www.geogebra.org/m/JPUBhFgs

### Vector fields: Examples

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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.

Link: Here

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\[

\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.

Link: Here

----------------------------------------------------------------…

### Flux in 2D (Part 2)

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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\,\mathb…

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\,\mathb…

### Flux in 2D

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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 a…

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 a…

### Flux across a line segment

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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 se…

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 se…

### Vector fields

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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 …

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 …