- Both functions $f$ and $g$ are defined for all real numbers.
- Both functions $f$ and $g$ are periodic with period $2p$ where $p\gt 0$.
- The function $f$ is even and the function $g$ odd.
- The function $f$ is a shift of $g$ to the left by $p/2$, that is $f(x) = g(x+p/2)$ for all real numbers $x.$
- For all real numbers $x$ we have $-1 \leq f(x) \leq 1$ and $-1 \leq g(x) \leq 1.$
- The functions $f$ and $g$ satisfy $f(0) =1$ and $g(0) =0.$
- If the function $f$ is differentiable at $0$, then it must satisfy $f'(0) =0.$ If the function $g$ is differentiable at $0,$ then it must satisfy $g'(0) =1.$
- Today we continued Mathematica explorations of probabilities. We went through Probabilities.nb notebook. Towards the end of Problem 1 in that notebook I used the recursive definition to calculate the exact values for the expected values. I explained more details how recursive definition in Mathematica in 20220308_Recursions_Sphears.nb.
- The next project consists of exploring probabilities using Mathematica. I suggest several problems in Explore_Probabilities.nb. Two problems in this notebook, random quadratic equations and random matrices involve geometric probabilities: random quadratic equations in $\mathbb{R}^3$ and random matrices in $\mathbb{R}^4$. To illustrate how geometric probabilities work, in 20220308_Recursions_Sphears.nb I illustrate probabilistic approach to unit spheres in spaces $\mathbb{R}^2$, $\mathbb{R}^3$, $\mathbb{R}^4$, $\mathbb{R}^5$ and $\mathbb{R}^6.$ We will talk further about this on Thursday.
- Today we started exploring probabilities with Mathematica. I started by designing a model for the Monty Hall Problem. It is in 20220303_MH_problem.nb. I will finish by making a function which will model a single Monty Hall Problem simulation, so that it is easy to perform thousands, or even millions, Monty Hall problem simulations.
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Below I present a proof of Morley's Miracle Theorem.
I start with the background knowledge from trigonometry and triangle geometry that I use in the proof.
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The following trigonometric identity holds for all $x, y \in \mathbb{R}$ \[ \sin(x+y) \sin(x-y) = (\sin x )^2 - (\sin y )^2. \] It is almost like a difference of squares formula.
Here is a proof: (at each step, for easier simplification, I order terms in alphabetic order, both in products and in sums) \begin{align*} \sin(x+y) \sin(x-y) & = \bigl( (\cos x) (\sin y) + (\cos y) (\sin x) \bigr) \bigl( - (\cos x) (\sin y) + (\cos y) (\sin x) \bigr) \\ & = - ( \cos x )^2 ( \sin y )^2 + ( \cos y )^2 ( \sin x )^2 \\ & = - \bigl( 1 - (\sin x)^2\bigr) \bigl( \sin y \bigr)^2 + \bigl( 1- (\sin y)^2 \bigr) \bigl( \sin x \bigr)^2 \\ & = - ( \sin y )^2 + (\sin x )^2 ( \sin y )^2 + ( \sin x )^2 - ( \sin x )^2 ( \sin y )^2 \\ & = ( \sin x )^2 - (\sin y )^2. \end{align*}
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For all $\theta$ we have \[ \sin(3\theta) = 4 (\sin \theta) \sin\left(\theta + \tfrac{\pi}{3}\right) \sin\left(\theta + \tfrac{2\pi}{3}\right) \]
Proof. Let $\theta$ be arbitrary real number. Then \begin{align*} 4 (\sin \theta) \sin\left(\theta + \tfrac{\pi}{3}\right) \sin\left(\theta + \tfrac{2\pi}{3}\right) & = 4 (\sin \theta) \sin\left(\theta + \tfrac{\pi}{3}\right) \sin\left(\tfrac{\pi}{3} - \theta \right) \\ & = 4 (\sin \theta) \left( \tfrac{3}{4} - (\sin\theta)^2 \right) \\ & = 3 (\sin \theta) - 4 (\sin\theta)^3 \\ & = \sin(3\theta). \end{align*}
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Law of Cosines. Let $\alpha, \beta, \gamma$ and $a, b, c,$ be positive real numbers. Assume that $\alpha+\beta+\gamma = \pi.$ The following statements are equivalent:
- There exist a triangle whose angles are $\alpha, \beta, \gamma,$ and whose sides opposite to these angles, respectively, are $a, b, c.$
- $\displaystyle a^2 = b^2 + c^2 - 2bc \cos\alpha.$
- $\displaystyle b^2 = c^2 + a^2 - 2ca \cos\beta.$
- $\displaystyle c^2 = a^2 + b^2 - 2ab \cos\gamma.$
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Law of Sines. Let $\alpha, \beta, \gamma$ and $a, b, c,$ and $R$ be positive real numbers. Assume that $\alpha+\beta+\gamma = \pi.$ The following statements are equivalent:
- There exist a triangle whose angles are $\alpha, \beta, \gamma,$ whose sides opposite to these angles, respectively, are $a, b, c,$ and whose circumradius is $R.$
- $\displaystyle \frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.$
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Proposition. Let $\alpha, \beta, \gamma$ be positive real numbers such that $\alpha+\beta+\gamma = \pi.$ Let $\triangle ABC$ be a triangle with the angle at $A$ equal to $\alpha$ and the sides $\overline{AC} = \sin \beta,$ $\overline{AB} = \sin \gamma.$ Then $\overline{BC} = \sin \alpha,$ the angle at $B$ is equal to $\beta$ and the angle at $C$ is equal to $\gamma.$
Proof. Denote by $a$ the length of the line segment $BC$ and calculate it using the Law of Cosines: \begin{alignat*}{2} a^2 & = (\sin \beta)^2 + (\sin \gamma)^2 -2 (\sin \beta)(\sin \gamma) (\cos \alpha) & & \quad\text{factor $\sin \gamma$} \\ & = (\sin \beta)^2 + (\sin \gamma) \bigl( (\sin \gamma) - 2 (\sin \beta) (\cos \alpha) \bigr) & & \quad \text{$\gamma = \pi - (\alpha+\beta)$} \\ & = (\sin \beta)^2 + \sin( \alpha + \beta ) \bigl( \sin( \alpha + \beta ) - 2 (\sin \beta) (\cos \alpha) \bigr) & & \quad \text{$\sin \gamma = \sin (\alpha+\beta)$ } \\ & = (\sin \beta)^2 + \sin( \alpha + \beta ) \sin( \alpha - \beta ) & & \quad \text{angle addition} \\ & = (\sin \alpha)^2 & & \quad \text{the first proven identity} \end{alignat*} Hence, $a = \sin\alpha.$
Denote by $\beta'$ and by $\gamma'$ the angles at $B$ and $C$, respectively. By the Law of Sines: \[ 1 = \frac{\sin\alpha}{\sin\alpha} = \frac{\sin\beta}{\sin(\beta')} = \frac{\sin\gamma}{\sin(\gamma')}. \] Consequently, \[ \sin\beta = \sin(\beta') \quad \text{and} \quad \sin\gamma = \sin(\gamma'). \] Since $\beta, \beta', \gamma, \gamma' \in (0,\pi)$, the preceding condition yields \[ \beta = \beta' \ \ \text{or} \ \ \beta = \pi - \beta' \quad \text{and} \quad \gamma = \gamma' \ \ \text{or} \ \ \gamma = \pi - \gamma'. \] Since we have $\alpha+\beta+\gamma = \pi$ and $\alpha+\beta'+\gamma' = \pi$, that is, $\beta+\gamma = \beta'+\gamma',$ it is not possible that $\beta = \pi - \beta'$ and $\gamma = \pi - \gamma'$. Hence, either $\beta = \beta'$ or $\gamma = \gamma'.$ But either of these two conditions, together with $\beta+\gamma = \beta'+\gamma',$ implies the other one. Thus, the angle at $B$ equals to $\beta$ and the angle at $C$ equals to $\gamma.$
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The picture below shows a triangle $ABC$ with the side lengths $a, b, c$ and the angle at $A$ equal to $3\alpha$, the angle at $B$ equal to $3\beta$, the angle at $C$ equal to $3\gamma.$
Below we present a proof that the orange triangle $A_eB_eC_e$ is equilateral.
- Since the $3\alpha, 3 \beta, 3\gamma$ are the angles of $\triangle ABC$ we have \[ 3\alpha + 3 \beta + 3\gamma = \pi, \qquad \text{that is} \qquad \alpha + \beta + \gamma = \frac{\pi}{3} \]
- By the Law of Sines applied to $\triangle ABC$ we have \[ \frac{a}{\sin(3\alpha)} = \frac{b}{\sin(3\beta)} = \frac{c}{\sin(3\gamma)} = 2R \] where $R$ is the circumradius of $\triangle ABC$. Consequently, \[ a = 2 R \sin(3\alpha), \quad b = 2 R \sin(3\beta), \quad c = 2 R \sin(3\gamma). \]
- In this item we consider $\triangle AB_eC.$ The angles of this triangle are \[ \measuredangle CAB_e = \alpha, \quad \measuredangle ACB_e = \gamma, \quad \measuredangle AB_eC = \pi - (\alpha+\gamma) = \pi - \left(\frac{\pi}{3} - \beta\right) = \frac{2\pi}{3} + \beta. \] By the Law of Sines applied to $\triangle AB_eC$ we have \[ \frac{b}{\sin\left(\tfrac{2\pi}{3} + \beta\right)} = \frac{\overline{CB_e}}{\sin \alpha}. \] Consequently, since $b = 2 R \sin(3\beta)$, we have \[ \overline{CB_e} = \frac{2 R \sin(3\beta) \sin\alpha}{\sin\left(\beta+\tfrac{2\pi}{3}\right)} = \frac{8 R (\sin \beta) \sin\left(\beta+\tfrac{\pi}{3}\right) \sin\left(\beta+\tfrac{2\pi}{3}\right) \sin\alpha}{\sin\left(\beta+\tfrac{2\pi}{3}\right)} = 8 R (\sin \alpha) (\sin\beta) \sin\left(\beta+\tfrac{\pi}{3}\right). \]
- Similarly, from $\triangle BA_eC,$ we obtain \[ \overline{CA_e} = 8 R (\sin \alpha) (\sin\beta) \sin\left(\alpha+\tfrac{\pi}{3}\right). \]
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Now consider $\triangle A_eCB_e.$ For its sides $\overline{CB_e}$ and $\overline{CA_e}$ we calculated \[ \overline{CB_e} = \bigl(8 R (\sin \alpha) (\sin\beta) \bigr) \sin\left(\beta+\tfrac{\pi}{3}\right) \quad \text{and} \quad \overline{CA_e} = \bigl( 8 R (\sin \alpha) (\sin\beta) \bigr) \sin\left(\alpha+\tfrac{\pi}{3}\right), \] and the angle enclosed by these two sides equals to $\gamma.$
By the Side-Angle-Side rule for similarity, the triangle $\triangle A_eCB_e$ is similar to $\triangle A'C'B'$ with the angle at $C'$ equal to $\gamma$ and the sides $\overline{C'B'} = \sin\left(\beta+\tfrac{\pi}{3}\right),$ $\overline{C'A'} = \sin\left(\alpha+\tfrac{\pi}{3}\right).$ Since \[ \alpha+\tfrac{\pi}{3} + \beta+\tfrac{\pi}{3} + \gamma = \pi, \] we can apply Proposition and deduce that the angle at $A'$ is $\beta+\tfrac{\pi}{3},$ the angle at $B'$ is $\alpha+\tfrac{\pi}{3}$ and $\overline{A'B'} = \sin\gamma.$
Since $\triangle A_eCB_e$ is similar to $\triangle A'C'B'$ we deduce that its angle at $A_e$ is $\beta+\tfrac{\pi}{3},$ the angle at $B_e$ is $\alpha+\tfrac{\pi}{3}$ and \[ \overline{A_eB_e} = \bigl(8 R (\sin \alpha) (\sin\beta) \bigr) (\sin\gamma) = 8 R (\sin \alpha) (\sin\beta) (\sin\gamma). \]
- Analogous considerations for $\triangle C_eAB_e$ would yield that the angle at $C_e$ is $\beta+\tfrac{\pi}{3},$ the angle at $B_e$ is $\gamma+\tfrac{\pi}{3}$ and \[ \overline{C_eB_e} = 8 R (\sin \alpha) (\sin\beta) (\sin\gamma). \] Analogous considerations for $\triangle A_eBC_e$ would yield that the angle at $A_e$ is $\gamma+\tfrac{\pi}{3},$ the angle at $C_e$ is $\alpha+\tfrac{\pi}{3}$ and \[ \overline{A_eC_e} = 8 R (\sin \alpha) (\sin\beta) (\sin\gamma). \]
- Since we proved \[ \overline{A_eB_e} = \overline{A_eC_e} = \overline{C_eB_e} = 8 R (\sin \alpha) (\sin\beta) (\sin\gamma), \] the orange triangle $\triangle A_eB_eC_e$ is equilateral.
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Today we illustrated
Morley's Miracle Theorem.
I continued the notebook from Tuesday 20220222_Triangle.nb.
The picture above is the was created late on Friday. I will keep improving the file. Ideally, one would add sufficient amount of information to the picture so that a proof of the theorem "emerges" from the picture, or at least a glimpse of a proof.
As before, the construction that we did is implemented in the Mathematica command Manipulate[].
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Today we continued with geometric explorations of a triangle. I continued the notebook from Tuesday 20220222_Triangle.nb. I this notebook I presented a construction of the circumscribed circle of a triangle. The orange point below is called circumcenter of a triangle.
It is nice to implement this construction in the Mathematica command Manipulate[].
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For the assignment, based on what we already did, you should be able to construct the
centroid point of a triangle. The orange point below is called the centroid of a triangle. It is the intersection of the three medians, the line segment joining a vertex of a triangle to the midpoint of the opposite side.
It is nice to implement this construction in the Mathematica command Manipulate[].
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For the assignment, based on what we will do Friday, you should be able to construct the
incircle of a triangle. The orange point below is called incenter of a triangle.
It is nice to implement this construction in the Mathematica command Manipulate[].
- In conclusion, your task on the assignment is to produce four manipulations: three pictures presented here and the fourth manipulation that I will presented on Friday.
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An optional manipulation that you can try to construct is the
orthocenter point of a triangle. The orange point below is called the orthocenter of a triangle. It is the intersection of the three altitudes.
It is nice to implement this construction in the Mathematica command Manipulate[].
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Yesterday I presented some explorations of the following function:
\[
(x,y) \mapsto x\sqrt{1-y^2} + y \sqrt{1-x^2}, \qquad (x,y) \in [-1,1] \times[-1,1].
\]
of two variables. A clearly related function of one variable is
\[
x \mapsto x\sqrt{1-x^2}, \qquad x \in [-1,1].
\]
The graph of this function is below:
This function reminds me of one period of the sine function. It would be even nicer if the maximum and minimum were $1$ and $-1$ as follows:
But we lost that the derivative is $1$ at $x=0.$ We can correct this by the change of the independent variable $x\mapsto x/2$:
This is the graph of the function
\[
x \mapsto x\sqrt{1-\frac{x^2}{4}}, \qquad x \in [-2,2].
\]
The maximum of this function is $1$ at $x=\sqrt{2}$ and the minimum is $-1$ at $x=-\sqrt{2}.$ In fact, the restriction of the preceding function on the interval between its minimum and its maximum looks very much like the corresponding part of the sine function:
Now introduce the periodic extension of this restriction:
and change the sign of every second part to make the function continuous:
And finally, we shift the sine to get the cosine:
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The corresponding funny unit cirdle is:
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The corresponding funny unit sphere is:
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The corresponding funny torus is:
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It might be interesting to rotate the funny circle in its plane as it travels around the $z$-axes while making the preceding funny torus. In the picture below the rotating funny circle rotates in its plane by $1/4$ of the full rotation, while making a full rotation around $z$-axes. This is a very subtle change, as you can see below:
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It might be interesting to rotate the funny circle in its plane as it travels around the $z$-axes. In the picture below the rotating funny circle rotates in its plane by $1/2$ of the full rotation, while making a full rotation around $z$-axes. This is a subtle change, as you can see below:
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Further we rotate the funny circle in its plane as it travels around the $z$-axes. In the picture below the rotating funny circle rotates in its plane by $3/4$ of the full rotation, while making a full rotation around $z$-axes. This is a more visable change, as you can see below:
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In the next picture we rotate the funny circle in its plane by $1$ full rotation, while making a full rotation around $z$-axes:
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In the next picture we rotate the funny circle in its plane by $2$ full rotations, while making one full rotation around $z$-axes:
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In the next picture we rotate the funny circle in its plane by $3$ full rotations, while making one full rotation around $z$-axes:
- All these explorations are in the Mathematica notebook Function2var.nb. You can find this notebook in our shared Dropbox folder \307_Files\2022\.
- Trying to solve a geometry problem, I run across a function of two variables that attracted my attention: \[ (x,y) \mapsto x\sqrt{1-y^2} + y \sqrt{1-x^2}, \qquad (x,y) \in [-1,1] \times[-1,1]. \] I wrote a Mathematica notebook with my explorations of this function Function2var.nb. You can find this notebook in our shared Dropbox folder \307_Files\2022\. Below I post some graphs and animations from this notebook.
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This is the graph of this function. The white circles are the unit circles in the planes $x = -1,$ $x = 1,$ $y = -1,$ $y = 1,$ $z = -1,$ $z = 1,$ and centered at the axes piercing the corresponding plane.
This graph, together with its reflection across the $xy$-plane make an interesting picture:
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Below is an animation of the preceding graph. The animating function is
\[
t \mapsto 2 t \sqrt{1-t^2} \qquad t \in \left[-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right].
\]
Place the cursor over the image to start the animation.
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Here is Mathematica ContourPlot[] of the function
\[
(x,y) \mapsto x\sqrt{1-y^2} + y \sqrt{1-x^2}, \qquad (x,y) \in [-1,1] \times[-1,1].
\]
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Another interesting aspect of this function is that we can think of y as time and look at the transformation of the graph of
\[
x \mapsto x \sqrt{1-y^2}+y \sqrt{1-x^2} \quad x \in [-1,1] \quad \text{with a fixed $y$,}
\]
where $y$ changes from time $y=-1$ to time $y=1.$ Notice that at $y=-1$ the graph is the lower half of the unit circle and at $y=1$ the graph is the upper half of the unit circle.
Place the cursor over the image to start the animation.
Place the cursor over the image to start the animation.
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Yesterday in class I wrote the notebook 20220215_Explore_Funny_Trig.nb. In this notebook I explored the funny trigonometric functions that we introduced earlier in the context of "solids of revolution" studied in calculus. After the class yesterday I added more explorations. It occurred to me that the functions NgonCos[t,n] and NgonSin[t,n] have a drawback that they are not piecewise linear. Therefore, I introduced another parametrization of the regular $n$-gon using the functions
- nNgCos[t,n] and nNgSin[t,n] (new-ngon-cosine and new-ngon-sine) These are functions that parameterize regular polygon with $n$ sides and each is piecewise linear.
- Enjoy the beauty of funny trigonometry.
- Read the notebook TheBeautyOfTrigonometry_12.nb. Also read the notebook 20220211_FunnyTrig.nb which I wrote on Friday and also today. Both of these files are in our shared Dropbox folder \307_Files\2022\.
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In 20220211_FunnyTrig.nb I introduced several variations on the theme of trigonometric functions Cos[t] and Sin[t]. Sometimes I call these functions "funny trigonometric" functions. The main features of the "funny trigonometric" functions are as follows:
Functions $f$ and $g$ are called a funny cosine and a funny sine, respectively, if they satisfy the following properties:
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In 20220211_FunnyTrig.nb I introduced the following funny cosine and funny sine:
- LiCos[t] and LiSin[t] (line-cosine and line-sine)
- TrCos[t] and TrSin[t] (trapezoidal-cosine and trapezoidal-sine)
- NgonCos[t,n] and NgonSin[t,n] (ngon-cosine and ngon-sine) These are functions that parameterize regular polygon with $n$ sides. It is true that not all of these functions are funny cosine and funny sine. However, one can play with them as if they were.
- The objective of your first project is to read TheBeautyOfTrigonometry_12.nb and reconstruct all, or almost all of the content there with some funny trigonometric funtions. You can use ones listed here, or create some on your own.
- To produce quality documents of any kind, one should look how one's tool handles COLORS. The most popular digital encoding of color is RGB color model. That is what I use in Mathematica. In fact I use Mathematica command RGBColor[r,g,b] where r, g, b are real numbers between 0 and 1. Since we use three numbers here, one can identify a color with a vector. Then all colors live in the unit cube. In this case that cube is called the Color Cube. So, colors become an application of vectors, I love this application so much that I wrote a webpage to celebrate it: Color Cube.
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Inspired by the parametrization of the unit circle by the functions $\cos$ and $\sin$ presented yesterday, today I derived a parametrization of a regular $n$-gon inscribed in the unit circle. Here $n$ is a positive integer greater than $2.$ In class I derived the functions analogous to Cos[t] and Sin[t].
Here is a sketch of that derivation which I did in class. We derived the parametrization functions to be
- The final product of this work are the graphs of the first nine regular $n$-gons inscribed in the unit circel: the equilateral triangle, the square, and the pentagon, the hexagon, the heptagon, the octagon, the nonagon, the decagon, the hendecagon, the dodecagon, see the Wikipedia page.
- In the notebook TheBeautyOfTrigonometry_12.nb I explore trigonometric functions Cos[t] and Sin[t] and how they are used in different settings in Mathematics. The most important feature of the functions Cos[t] and Sin[t] is that they provide a parametrization of the unit circle. The animation below illustrates how this parametrization is obtained.
-
The main components of the animation are as follows
- The black circle is the unit circle.
- The values of t $\in [0,2\pi]$ are represented by orange lengths. In each scene there are three orange lengths: the horizontal length, the vertical length and the arc length on the unit circle.
- The values of Cos[t] are green horizontal line segments.
- The values of Sin[t] are blue vertical line segments.
- The principle presented above is the general principle of parametric equations of curves in the plane. Given two functions, one green and one blue, taken as the coordinates of a moving point, they do trace a curve in the plane. It is of interest to consider the converse question: given a curve in the plane, for example an octagon inscribed in the unit circle, find a green and a blue function that do trace that octagon.
Place the cursor over the image to start the animation.
- The Mathematica part of the class will start on Tuesday, February 8, 2022 in BH 215 at 2:00 pm.
- I already sent you an email with the instructions how to set-up a Dropbox folder for your homework assignments and how to gain access to a shared Dropbox folder with the class materials. I hope that I will soon have the instructions on how to gain remote access to BH 215 lab computers.
- The information sheet
- We will use
- To get started with Mathematica see my Mathematica page. Please watch the videos that are on my Mathematica page before the first class. Watching the movies is essential for being able to organize your homework notebooks well! I will not discuss the basics of notebook structuring in class.