# Mathema

Random notes about mathematics with dynamic and interactive activities

Random notes about mathematics with dynamic and interactive activities

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

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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.

**Simulation**

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

Link: Here

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Another important example is the velocity vector field $\mathbf{…

\[

\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.

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

Link: Here

----------------------------------------------------------------------------------------------------------------------

Another important example is the velocity vector field $\mathbf{…

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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 $C$.

$$\mathbf v \cdot \mathbf n =|\mathbf v|=r.$$

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.

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The

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…

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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.

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:

1. Describe a general condition required for the flux across the line se…

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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*}

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…

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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

ax^2+by^2+cxy

\]

is called the

Graphs of quadratics equations are known as

A conic is said to be in

& \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 n…

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