Chapter 5.1 The Elementary Notion of Area

Chapter 5 is all about the principle of finite approximation.

Finite Approximation

The principle of finite approximation is to understand infinite things and things with infinitely many parts, study their approximation by things that are finite and have finitely many parts.

Our introdution to this principle will be with understanding area and motion.

Motion and Area

Since ancient times, mathematicians have studied the properties of area and motion. These ideas are deeply rooted in the human experience of the world. However, natural language is not precise enough to explore these ideas.

In fact, only at the very end of the 19th century did mathematics finally become sufficiently developed to provide a framework for reasoning to support the investigation of the ideas that Euclid studied about 2,500 years earlier.

This part of the course focuses on developing a precise language to support reasoning that uses the principle of finite approximation. The language that this reasoning requires is the language of sequences and sequential convergence.

The principle of finite approximation is critical to the study of motion, change, and area. It is a powerful way to use and to extend human intuition about finite things and things that have finitely many parts to infinite things and to things with infinitely many parts.

This is a subset, \(R\), of the plane that is bounded by a curve \(\partial R\):

The curve \(\partial R\) is the boundary of \(R\) and it decomposes the plane into three parts, a part that is inside the curve, a part that is outside the curve, and the curve itself.

If we think of the curve to be the base of a tank with vertical walls that are one unit of length tall and you filled the tank with water to the top, you could use the measurement of the volume of water to determine the area of the base

This gives the intuition that the region \(R\) has area that is bounded by \(\partial R\).

However, what do the above statements mean? What is a curve? What is area? What does it mean for a curve to bound a region?

Although natural language is not precise enough for studying mathematics, the intuition it supports is essential. These words in our natural language facilitate the communication of ideas about human interaction with the environment.

Development of mathematical language is an incremental process that requires patience, care, and great attention to detail.

Path, Curve, Simple Curve, Simple Closed Curve, Parameterization

A path is a function from an interval to \(\mathbb R^d\).

A curve in \(\mathbb R^d\) is the range (we will often prefer to say image) of a path. For now, take \(d\) to be 2.

A simple curve, \(\Gamma\), is the image of a path, \(c\), where

• \(c\) is one-to-one, that is, \(c(x) = c(y)\) implies that \(x = y\).

A simple closed curve, \(\Gamma\), is the image of a path, \(c\), where:

• \(c\) is defined on an interval \([a,b]\), where \([a,b)\) is not empty;

• \(c\) is one-to-one on \([a,b)\);

• \(c(a) = c(b)\).

In both cases, the path \(c\) is a parameterization of \(\Gamma\), it parameterizes \(\Gamma\).

Use the images below to understand the definitions:

In Example 1, we will see our first example of a parameterization of a curve. In this case, the curve is the graph of the function \(f(x)=x^2\) restricted to a closed interval. Pay attention to how we can create more than one parameterization to a curve by changing the time domain using the principle of transformation.

Example 1

Take \(f\) to be the function from \([-1,2]\) to \(\mathbb R\) that is given by \[f(x)=3x^2.\] Find parameterizations for \(f\) (viewed as a subset of \(\mathbb R^2\)) that have domains equal to the following intervals:

1. \([-1,2]\);

2. \([0,1]\).

Is \(f\) a simple curve? Is \(f\) a simple closed curve?

The function \(f\) looks like this.

1. One parameterization for \(f\) is \(c \colon [-1,2]\to \mathbb R^2\), where \(c(t)=(t,3t^2)\).

2. Use transformation to find a parameterization. First, transform the interval \([0,1]\) to the interval \([-1,2]\). For example, \[ [0,1]\to\text{(scale by 3)}\to [0,3]\to\text{(translate by }-1)\to[-1,2].\] Mathematically, \(t\) transforms by \(T_{-1}\circ S_3(t)=3t-1\). So \(c(t)=(t,3t^2)\) becomes \(c(t)=(3t-1,3(3t-1)^2)\). A parameterization \(C \colon [0,1]\to \mathbb R^2\) is \[C(t)=(3t-1,3(3t-1)^2).\]

Our ultimate goal is to use curves and parameterizations to identify a way to describe the idea of inside and outside. This is very difficult to do in general.

However, the idea that a point is inside or outside a disk can be precisely understood as a statement about the point being in a certain feasible set.

Inside or Outside Disk

For any disk \(D\) of radius \(r\) with center \(p\), a point \(q\) is, respectively, inside \(D\) or outside \(D\) if \[ ||p-q|| < r \quad \text{or if}\quad ||p-q|| > r.\]

The disk \(D\) of radius \(r\) with center \(p\) is the set of all \(q\) in \(\mathbb R^2\) such that \[||p-q||\leq r.\]

The set \(\partial D\) (read: “boundary of \(D\)”) is the circle of radius \(r\) with center \(p\), and is a simple closed curve.

Use the pictures below to visually understand the idea of inside, outside and on the boundary of a disk.

In the next example, construct a parameterization of a circle. Use the principle of transformation to simplify the construction.

Example 2

Take \(D\) to be the disk of radius \(2\) with center at \((1, 3)\). Find a path \(c\) that parameterizes the circle \(\partial D\). Be sure that \(c\) moves clockwise around \(\partial D\).

Recall that \(c\colon [0,2\pi]\to \mathbb{R}^2\) given by \(c(t)=(\cos(t),\sin(t))\) parameterizes the unit circle in a counter-clockwise pattern. Transform it to get a parameterization for \(\partial D\): \[\ (\cos(t),\sin(t))\to\text{ scale by 2 for radius}\to (2\cos(t),2\sin(t))\to\text{ translate by }V=\langle 1,3\rangle\text{ to get center}\to (2\cos(t)+1,2\sin(t)+3) \]

This can be expressed as \(\langle 1,3\rangle +2c(t).\)

The new path parameterizes \(\partial D\), but does so counter-clockwise. Reverse the direction by changing the angle to negative angles. Therefore the final answer is \(C\colon [0,2\pi]\to\mathbb{R}^2\) \[C(t)=(2\cos(-t)+1,2\sin(-t)+3).\]

Because an ellipse is an asymmetrically scaled circle, we can use what we know about circles to define inside and outside for an ellipse.

Inside and Outside of Ellipse

A solid ellipse is an asymmetrically scaled disk and an ellipse is an asymmetrically scaled circle and the boundary of a solid ellipse.

If \(E\) is a solid ellipse and \(D\) is a disk so that \[S(D) = E,\] for some asymmetric scaling \(S\), then a point \(p\) is:

• inside \(E\) if and only if \(p\) is equal to \(S(q)\) for some \(q\) that is inside \(D\);

• outside \(E\) if and only if \(p\) is equal to \(S(q)\) for some \(q\) that is outside \(D\);

• in \(\partial E\) if and only if \(p\) is equal to \(S(q)\) for some \(q\) that is in \(\partial D\).

Use the picture below to understand the definition.

In the next problem, recognize the scaling hidden in the ellipse to simplify the problem of finding a point inside an ellipse to a problem of scaling the appropriate circle and point in the disk. That will also help with finding a path that parameterizes the ellipse.

Example 3

Take \(E\) to be the solid ellipse that is given by the equation \[\frac{(x-2)^2}{4}+(y-1)^2 \leq 4.\]

1. Find a point that is inside \(E\).

2. Find a path \(c\) that parameterizes the ellipse \(\partial E\).

An ellipse is a transformed version of a circle. Take \(D\) to be the disk centered at \((0,0)\) with radius \(r=2\), which is the set of all points \((x,y)\) for which \(x^2+y^2\leq 4\).

Take \(S=X_2\), the asymmetric scaling in the \(x\)-axis by \(2\).

Then \(S(D)=\{(2x,y)\colon (x,y)\in D\}=\{(2x,y)\colon x^2+y^2\leq 4\}.\) Set \(X=2x\). Solve for \(x\) to get \(\frac{X}{2}=x\). Rewrite so \(S(D)=\{(2x,y)\colon x^2+y^2\leq 4\}\) becomes \(S(D)=\left\{(X,y)\colon \left(\frac{x}{2}\right)^2+y^2\leq 4\right\}\) or \(S(D)=\left\{(X,y)\colon \frac{x^2}{4}+y^2\leq 4\right\}.\)

Thus \(\frac{x^2}{4}+y^2\leq 4\) is a solid ellipse centered at \((0,0)\).

To get the desired center, translate by \(V=\langle 2,1\rangle\) to get

\[ \begin{align*} V+S(D)&=\left\{\langle 2,1\rangle +(x,y)\colon \frac{x^2}{4}+y^2\leq 4\right\}\\ &=\left\{(x+2,y+1)\colon \frac{x^2}{4}+y^2\leq 4\right\}\\ &=\left\{(X,Y)\colon \frac{(X-2)^2}{4}+(Y-1)^2\leq 4\right\}&&\text{set }X=x+2\text{ and }Y=y+1 \end{align*} \]

the solid ellipse desired.

In summary, compute \(\langle 2,1\rangle+X_2\) to get the desired solid ellipse. Use this to find points inside \(E\).

1. Pick any point in \(D\) the disk centered at \((0,0)\) with radius \(2\). For example \((1,1)\). Now transform it: \(\langle 2,1\rangle +X_2(1,1)=(4,2)\) will be a point in \(E\). Confirm it works by doing the following calculation: \[ \begin{align*} \frac{(4-2)^2}{4}+(2-1)^2&=\frac{2^2}{4}+1^2\\ &=\frac{4}{4}+1\\ &=2\\ &\leq 4 \end{align*} \]

2. To find a path \(c\) that parameterizes the ellipse \(\partial E\), take the parameterization of the circle centered at \((0,0)\) with radius \(2\) and transform it:

\[\begin{align*} (2\cos(t),2\sin(t))&\to\text{ assymetric scale $x$ by 2}\\&\to (4\cos(t),2\sin(t))\\&\to\text{ translate by }V=\langle 2,1\rangle\text{ to get center}\\&\to (4\cos(t)+2,2\sin(t)+1) \end{align*}\]

Therefore the final answer is \(C\colon [0,2\pi]\to\mathbb{R}^2\) \[C(t)=(4\cos(t)+2,2\sin(t)+1).\]

General paths and their images are too poorly behaved to conform to physical intuition about what paths should be. To say anything meaningful requires that we initially restrict to a very basic collection of paths, which we will expand as we develop our mathematical tools.

First, we define an epoch.

Epoch

An epoch in an interval \([a,b]\), where \(a<b\), is an \(n\)-tuple \((t_1, \dots, t_n)\) so that:

• For each \(k\) in \(\{1,\dots,n-1\}\), \(t_k<t_{k+1}\)

• \(t_1=a\)

• \(t_n=b\)

Use the picture to understand what an epoch is.

The paths that we specialize to are ones that created by connecting linear pieces.

Continuous Piecewise Linear Path

A continuous, piecewise linear path \(\ell\) is a path with domain \([a,b]\), where \(a<b\), with the property that there is an epoch \((t_1, \dots, t_n)\) and linear functions \(\ell_1,\ell_2,\dots,\ell_{n-1}\) so that

• for each \(k\) in \(\{1, \dots, n-1\}\), \(\ell=\ell_k\) each interval \([t_k, t_{k+1}]\)

• for each \(k\) in \(\{1, \dots, n-2\}\), \(\ell_k(t_{k+1})=\ell_{k+1}(t_{k+1})\)

The ordered vertex set of \(\ell\) is the \(n\)-tuple \((\ell(t_1), \dots, \ell(t_n))\).

The vertex set of \(\ell\) is the set of distinct entries of the ordered vertex set of \(\ell\)

Use the picture below to visualize the definition.

In this example, we will construct a piecewise linear path with a given vertex set and domain. However, we do not specify the epoch in the problem. This means that there can be more than one piecewise linear path.

Example 4

Construct a continuous, piecewise linear path with domain equal to \([1, 4]\) and ordered vertex set equal \(\big((2, 0), (1, 1), (2, 4), (5, 2)\big)\).

How would your answer change if you wanted the domain to be \([0,5]\)?

Construct linear paths that go from one vertex to the another vertex by finding vectors that translate from one vertex to the other.

• Path from \((2,0)\) to \((1,1)\): \(t\langle -1,1\rangle+(2,0)\)

• Path from \((1,1)\) to \((2,4)\): \(t\langle 1,3\rangle+(1,1)\)

• Path from \((2,4)\) to \((5,2)\): \(t\langle 3,-2\rangle+(2,4)\).

Now stitch them together so that the domain of our continuous, piecewise linear path is \([1,4]\):

\[ c(t)=\begin{cases} (t-1)\langle -1,1\rangle+(2,0)&\text{ if }1\leq t<2\\ (t-2)\langle 1,3\rangle+(1,1)&\text{ if }2\leq t<3\\ (t-3)\langle 3,-2\rangle+(2,4)&\text{ if }3\leq t\leq 4\\ \end{cases} \]

Change the time intervals so get an answer that is defined on \([0,5]\). Here is one example:

\[ c(t)=\begin{cases} t\langle -1,1\rangle+(2,0)&\text{ if }0\leq t<1\\ \frac{(t-1)}{2}\langle 1,3\rangle+(1,1)&\text{ if }1\leq t<3\\ \frac{(t-3)}{2}\langle 3,-2\rangle+(2,4)&\text{ if }3\leq t\leq 5\\ \end{cases} \]

We do not yet have a way of defining what it means for a point to be inside a simple closed curve \(\Gamma\).

Restrictions must be made on such curves for a definition of inside and outside to be meaningful.

However, even without a precise definition, our intuition can help us to identify what inside and outside should be given a graphical representation of a curve.

First, understand whether or not the following curves form a bounded region by graphing the curves.

Example 5

Take \(\Gamma\) to be the curve in the plane that is the image of the path \(c\). Determine graphically whether or not \(\Gamma\) encloses a bounded region in the plane for these choices of paths:

1. \(c\colon [0,3\pi]\to\mathbb R^2\) by \(c(t)=(2t,\cos(t))\)

2. \(c\colon [0,\pi]\to\mathbb R^2\) by \(c(t)=(3\cos(2t),\sin(2t))\)

1. Not a bounded region

2. A bounded region: Ellipse

A simple example of a continuous piecewise linear path that we can easily study is a rectangle. Here is how to construct a rectangle:

• Take \(p\) to be a point in \(\mathbb R^2\) and take \(V\) to be a unit vector.
• Take \(a\) and \(b\) to be positive real numbers.
• The set \(S\) that is given by \[S = \{p, aV+p, bV_\perp+aV+p, bV_\perp +p\}\] is the vertex set of a rectangle.

Use the picture below to visualize the construction:

Using the definition of a piecewise linear path, we can define what a rectangle is.

Rectangle and Vertex Set

Take \(c\) to be the piecewise linear path that is given by \[c(t) = \begin{cases}taV + p&\text{ } 0\leq t <1\\(t-1)bV_\perp + aV+p&\text{ } 1\leq t <2\\-(t-2)aV + bV_\perp+aV+p&\text{ } 2\leq t <3\\-(t-3)bV_\perp + bV_\perp+p&\text{ } 3\leq t \leq 4.\end{cases}\hspace{3in}\] where \(a\) and \(b\) are real numbers and \(V\) is a unit vector.

Define \(\partial R\), the boundary of \(R\), to be the set \(c([0, 4])\)—The path \(c\) parameterizes \(\partial R\).

The union of all line segments with endpoints in \(\partial R\) is rectangle \(R\) with vertex set \(S\).

Note: Every rectangle is the feasible set for a system of linear inequalities.

A point \(p\) is inside \(R\) if it is in \(R\) but not in \(\partial R\) and is outside \(R\) if it is not in \(R\).

Use the next example to understand the definition of a rectangle via its parameterization, its description as a feasible set, and by finding a point inside the rectangle. It will be helpful to remember that all the angles inside of a rectangle are 90 degrees. So use vectors to help complete the example.

Example 6

Take \(R\) to be the rectangle with vertices \((2,5)\), \((5,6)\), \((3,12)\), and \((0,11)\).

1. Find a piecewise linear function \(c\) that parameterizes \(\partial R\) and simulate the motion of a particle whose position at time \(t\) is \(c(t)\).

2. Describe \(R\) as the feasible set for a system of inequalities.

3. Find a point that is inside \(R\).

1. Calculate \(V\) by computing \((5,6)-(2,5)\) and creating a unit vector: \(V=\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle\), \(p=(2,5)\), and \(V_\perp=\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle\). Calculate \(a\) and \(b\), which are the side lengths of the rectangle: \(a=\|(5,6)-(2,5)\|=\sqrt{10}\) and \(b=\|(3,12)-(5,6)\|=\|(-2,6)\|=\sqrt{40}\). The piecewise linear function is

\[ c(t) = \begin{cases}t\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle + (2,5)&\text{if } 0\leq t <1\\(t-1)\sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle + \sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle+(2,5)&\text{if } 1\leq t <2\\-\sqrt{10}(t-2)\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle + \sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle+\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle+(2,5)&\text{if } 2\leq t <3\\-\sqrt{40}(t-3)\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle + \sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle+(2,5)&\text{if } 3\leq t \leq 4.\end{cases} \]

2. The rectangle is all \((x,y)\) that satisfy a system of inequalities. Here are two ways to do it:

   1. Create linear equations for the line that passes through \((2,5)\) and \((5,6)\), the line that passes through \((5,6)\) and \((3,12)\), the line that passes through \((3,12)\) and \((0,11)\), and the line that passes through \((0,11)\) and \((2,5)\). Then write the inequalities

   • the lines that passes through \((2,5)\) and \((5,6)\) is \(y=\frac{1}{3}(x-2)+5\)

   • the lines that passes through \((5,6)\) and \((3,12)\) is \(y=-3(x-5)+6\)

   • the lines that passes through \((3,12)\) and \((0,11)\) is \(y=\frac{1}{3}(x-3)+12\)

   • the lines that passes through \((0,11)\) and \((2,5)\) is \(y=-3(x-0)+11\)

     So the system of inequalities: \[ \begin{cases} y\geq \frac{1}{3}(x-2)+5\\ y\leq-3(x-5)+6\\ y\leq\frac{1}{3}(x-3)+12\\ y\geq-3(x-0)+11 \end{cases} \]

   2. Use the vector paths, rewrite them as linear equations, and then write the inequalities.

   • \(t\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle + (2,5)=t\langle 3,1 \rangle+(2,5)\) can be written as \(y=\frac{1}{3}(x-2)+5\)

   • \[\begin{align*}&(t-1)\sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle + \sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle+(2,5)\\&=t\left\langle -\frac{\sqrt{40}}{\sqrt{10}},\frac{3\sqrt{40}}{\sqrt{10}}\right\rangle +\left(\frac{\sqrt{40}}{\sqrt{10}}+5,-\frac{3\sqrt{40}}{\sqrt{10}}+6\right)\end{align*}\] can be written as \(y=\frac{\frac{3\sqrt{40}}{\sqrt{10}}}{-\frac{\sqrt{40}}{\sqrt{10}}}\left(x-\frac{\sqrt{40}}{\sqrt{10}}-5\right)-3\frac{\sqrt{40}}{\sqrt{10}}+6\)

   • \[\begin{align*}&-\sqrt{10}(t-2)\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle + \sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle+\sqrt{10}\left\langle \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right\rangle+(2,5)\\&=\langle-3t,-t \rangle+\left(-\frac{\sqrt{40}}{\sqrt{10}}+11,3\frac{\sqrt{40}}{\sqrt{10}}+8\right)\end{align*}\] becomes \(y=\frac{1}{3}\left(x+\frac{\sqrt{40}}{\sqrt{10}}-11\right)+3\frac{\sqrt{40}}{\sqrt{10}}+8\)

   • \[\begin{align*}&-\sqrt{40}(t-3)\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle + \sqrt{40}\left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}}\right\rangle+(2,5)\\&=t\left\langle\frac{\sqrt{40}}{\sqrt{10}},-\frac{3\sqrt{40}}{\sqrt{10}} \right\rangle+\left(-\frac{3\sqrt{40}}{\sqrt{10}}-\frac{\sqrt{40}}{\sqrt{10}}+2,\frac{9\sqrt{40}}{\sqrt{10}}+\frac{3\sqrt{40}}{\sqrt{10}}+5\right)\end{align*}\] can be written as \(y=\frac{\frac{-3\sqrt{40}}{\sqrt{10}}}{\frac{\sqrt{40}}{\sqrt{10}}}\left(x+\frac{3\sqrt{40}}{\sqrt{10}}+\frac{\sqrt{40}}{\sqrt{10}}-2\right)+\frac{9\sqrt{40}}{\sqrt{10}}+\frac{3\sqrt{40}}{\sqrt{10}}+5\)

   So the system of inequalities: \[ \begin{cases} y\geq \frac{1}{3}(x-2)+5\\ y\leq\frac{\frac{3\sqrt{40}}{\sqrt{10}}}{-\frac{\sqrt{40}}{\sqrt{10}}}\left(x-\frac{\sqrt{40}}{\sqrt{10}}-5\right)-3\frac{\sqrt{40}}{\sqrt{10}}+6\\ y\leq\frac{1}{3}\left(x+\frac{\sqrt{40}}{\sqrt{10}}-11\right)+3\frac{\sqrt{40}}{\sqrt{10}}+8\\ y\geq\frac{\frac{-3\sqrt{40}}{\sqrt{10}}}{\frac{\sqrt{40}}{\sqrt{10}}}\left(x+\frac{3\sqrt{40}}{\sqrt{10}}+\frac{\sqrt{40}}{\sqrt{10}}-2\right)+\frac{9\sqrt{40}}{\sqrt{10}}+\frac{3\sqrt{40}}{\sqrt{10}}+5 \end{cases} \]

3. A point inside the rectangle is any \((x,y)\) that satisfies the system of inequalities given in part b. For example \((4,7)\) would work.

Now that we understand the idea of inside and outside for a rectangle, we can understand the idea of area.

Area of Rectangles

Take \(\partial R\) to be the boundary of a rectangle \(R\) with vertex set \(\{p_1, p_2, p_3, p_4\}\).

Take \(c\) to be a continuous piecewise linear parameterization of \(\partial R\) with ordered vertex set \((p_1, p_2, p_3, p_4, p_1)\).

Take \((t_1, t_2, t_3, t_4, t_5)\) to be an epoch for the domain of \(c\) so that for each \(i\) in \(\{1,2,3,4,5\}\), \[c(t_i) = p_i, \quad {\rm where} \quad p_5 = p_1.\]

Since \(\partial R\) is a closed curve, it is not necessary to specify the last entry in its oriented vertex set, so simply write \((p_1, p_2, p_3, p_4)\) and omit the last entry.

Henceforth, adopt this notation for the ordered vertex set of any simple, closed, continuous, piecewise linear path.

A parameterization can “trace” the rectangle in many ways, but we can simplify the classification by defining an orientation. For a rectangle, here is how we can define it. It has to do with if there is a relationship between two vectors.

Positively Oriented and Negatively Oriented Rectangle

Take \(\partial R\) to be the boundary of a rectangle \(R\) with vertex set \(\{p_1, p_2, p_3, p_4\}\) and \(c\) to be a continuous piecewise linear parameterization of \(\partial R\) with ordered vertex set \((p_1, p_2, p_3, p_4)\).

Take \(V\) to be the vector \(p_2-p_1\).

The path \(c\) is positively oriented if \[p_3 - p_2 = ||p_3-p_2||\hat{V}_\perp\] and negatively oriented if \[p_3 - p_2 = -||p_3-p_2||\hat{V}_\perp.\]

The ordered vertex set \((p_1, p_2, p_3, p_4)\) is positively oriented or negatively oriented if \(c\) is, respectively, positively oriented or negatively oriented.

Simplify language and refer to any continuous, piecewise linear path that parameterizes the boundary of a rectangle as a rectangular path, and its image as a rectangular boundary.

Use the picture below to understand orientation.

Solidify your understanding with the next example.

Example 7

Take \(R\) to be the rectangle with ordered vertex set \(((5,1), (8, 2), (6, 8), (3, 7))\). Is \(R\) positively oriented?

Take \(p_1=(5,1)\), \(p_2=(8,2)\), and \(p_3=(6,8)\).

Take \(V=p_2-p_1=(8,2)-(5,1)=\langle 3, 1 \rangle .\) Then \(V_\perp=\langle-1,3\rangle\) and \(\hat{V}_\perp=\left\langle-\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}} \right\rangle\)

To see if \(R\) is positively oriented, check if \(p_3-p_2=||p_3-p_2||\hat{V}_\perp\).

Notice that \[p_3-p_2=(6,8)-(8,2)=\langle -2,6\rangle,\] \[\|p_3-p_2\|=\sqrt{6^2+2^2}=\sqrt{40},\] and \[|p_3-p_2||\hat{V}_\perp=\sqrt{40}\left\langle-\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}} \right\rangle=\left\langle -\frac{\sqrt{40}}{\sqrt{10}},\frac{3\sqrt{40}}{\sqrt{10}}\right\rangle=-\left\langle\frac{-2\sqrt{10}}{\sqrt{10}},6\frac{\sqrt{10}}{\sqrt{10}} \right\rangle=\langle -2, 6\rangle.\]

Hence \[p_3 - p_2 = ||p_3-p_2||\hat{V}_\perp\] so \(R\) is positively oriented.

Another way to specify orientation is by encoding the information in the boundary. This definition tells us how to do it.

Oriented Line Segement and Oriented Boundary

Denote by \(\overrightarrow{pq}\) the oriented line segment with endpoints \(p\) and \(q\).

Any path that parameterizes the line segment \(\overline{pq}\) that is determined by \(\overrightarrow{pq}\) must start at \(p\) and end at \(q\).

A path \(c\) that parameterizes the boundary of a rectangle \(R\) specifies an order on each line segment of the boundary by the order in which it reaches each of the endpoints and it specifies an ordered vertex set \((p_1, p_2, p_3, p_4)\) for \(\partial R\).

Denote by \(\partial_oR\) the oriented boundary of \(R\), where \[\partial_oR = \left\{\overrightarrow{p_1p_2}, \overrightarrow{p_2p_3}, \overrightarrow{p_3p_4}, \overrightarrow{p_4p_1}\right\}.\]

The oriented boundary \(\partial_oR\) is a positively (negatively) oriented boundary if the path \(c\) is positively (negatively) oriented.

A point \(p\) is inside \(R\) if and only if \(p\) lies to the left of any oriented line segment of \(\partial_o R\).

Understand the definition by completing the next example. One thing to note is that finding whether a point is inside or outside a rectangle can be done in at least two ways. One is a transformation approach and the other is using the feasible set definition. For this example, we use the transformation approach.

Example 8

The set \(\{(4,1), (7,3), (5,6), (2,4)\}\) is a rectangular vertex set. Construct the positively oriented boundary of this rectangle and the negatively oriented boundary of this rectangle. Determine whether or not the point \(p\) is inside or outside \(R\), where \(p\) is given by:

1. \(p=(6,4)\)

2. \(p=(3,5)\)

A positively oriented boundary is \[\left\{\overrightarrow{(4,1)(7,3)}, \overrightarrow{(7,3)(5,6)}, \overrightarrow{(5,6)(2,4)}, \overrightarrow{(2,4)(4,1)}\right\}.\]

A negatively oriented boundary is \[\left\{\overrightarrow{(4,1)(2,4)}, \overrightarrow{(2,4)(5,6)}, \overrightarrow{(5,6)(7,3)}, \overrightarrow{(7,3)(4,1)}\right\}.\]

To determine if a point \(p\) is inside the rectangle, check if it lies to the left any oriented line segment of the oriented boundary. In some cases, this may be easier to see by translating the rectangle so that the lower left vertex is \((0,0)\) and one of its sides is parallel to the \(x\)-axis.

1. The lower left vertex is \((4,1)\). Translate all vertices by \(\langle -4,-1\rangle\). The new vertices are \((0,0)\), \((3,2)\), \((1,5)\), and \((-2,3).\)

2. Now rotate \((3,2)\) to the \(x\)-axis. Do that by first finding the appropriate angle: \[p=\left(\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}\right).\] The angle is going to be \(p^{-1}=\left(\frac{3}{\sqrt{13}},-\frac{2}{\sqrt{13}}\right).\) Thus \((3,2)\star p^{-1}\) will rotate \((3,2)\) to the \(x\)-axis. Rotate the rest of the vertices by \(p^{-1}\):

• \((3,2)\star p^{-1}=\left(3\cdot \frac{3}{\sqrt{13}}-2\cdot\frac{-2}{\sqrt{13}},3\cdot\frac{-2}{\sqrt{13}}+2\cdot\frac{3}{\sqrt{13}}\right)=\left(\frac{13}{\sqrt{13}},0\right)\).

• \((1,5)\star p^{-1}=\left(1\cdot \frac{3}{\sqrt{13}}-5\cdot\frac{-2}{\sqrt{13}},1\cdot \frac{-2}{\sqrt{13}}+5\cdot\frac{3}{\sqrt{13}}\right)=\left(\frac{13}{\sqrt{13}},\frac{13}{\sqrt{13}}\right)\)

• \((-2,3)\star p^{-1}=\left(-2\cdot\frac{3}{\sqrt{13}}-3\cdot\left(\frac{-2}{\sqrt{13}}\right),-2\cdot\left(\frac{-2}{\sqrt{13}}\right)+3\cdot \frac{3}{\sqrt{13}}\right)=\left(0,\frac{13}{\sqrt{13}}\right)\)

1. To determine if \((6,4)\) is in \(R\), perform the translation and rotation from before to get \((V+(6,4))\star p^{-1}=\left(\frac{12}{\sqrt{13}},\frac{5}{\sqrt{13}}\right)\). Compare the coordinates of this point with the vertices \((0,0)\), \(\left(\frac{13}{\sqrt{13}},0\right)\), \(\left(\frac{13}{\sqrt{13}},\frac{13}{\sqrt{13}}\right)\),\(\left(0,\frac{13}{\sqrt{13}}\right)\). This is going to be inside of the rectangle. Hence the original point \((6,4)\) is inside of the original rectangle.

2. To determine if \((3,5)\) is in \(R\), perform the translation and rotation from before to get \((V+(3,5))\star p^{-1}=\left(\frac{5}{\sqrt{13}},\frac{14}{\sqrt{13}}\right)\). Compare the coordinates of this point with the vertices \((0,0)\), \(\left(\frac{13}{\sqrt{13}},0\right)\), \(\left(\frac{13}{\sqrt{13}},\frac{13}{\sqrt{13}}\right)\),\(\left(0,\frac{13}{\sqrt{13}}\right)\). The \(y\)-coordinate is bigger than the \(y\)-coordinate of the upper vertices, so it will not be inside of \(R\). Hence the original point \((3,5)\) is outside of the original rectangle.

Because we have concretely defined what it means to be inside and outside of a rectangle, we can define the area of a rectangle.

Area of Rectangle

For any rectangle \(R\) with side lengths equal to \(a\) and \(b\), define the area of \(R\), \(\mathcal A(R)\), to be the product \[\mathcal A(R) = ab.\]

This notion of area in the plane agrees with the way in which we measure area. It is not changed by rigid motions.

In the next example, continue to solidify your understanding of rectangles by calculating the area and also determining a point that is inside of the rectangle.

Example 9

Take \(R\) to be the rectangle with vertices \((2,1), (7,6), (-8,21), (-13,16)\)

1. Construct both a positively and negatively oriented rectangular boundary for \(R\).

2. Find a point \(p\) that is inside \(R\) and show that \(p\) is inside of \(R\).

3. Calculate the area that is bounded by \(\partial R\).

1. A positively oriented boundary is \[\left\{\overrightarrow{(2,1)(7,6)}, \overrightarrow{(7,6)(-8,21)}, \overrightarrow{(-8,21)(-13,16)}, \overrightarrow{(-13,16)(2,1)}\right\}.\] A negatively oriented boundary is \[\left\{\overrightarrow{(2,1)(-13,16)}, \overrightarrow{(-13,16)(-8,21)}, \overrightarrow{(-8,21)(7,6)}, \overrightarrow{(7,6)(2,1)}\right\}.\]

2. To determine if a point \(p\) is inside the rectangle, check if it lies to the left any oriented line segment of the oriented boundary. In some cases, this may be easier to see by translating the rectangle so that the lower left vertex is \((0,0)\) and one of its sides is parallel to the \(x\)-axis.

   1. The lower left vertex is \((2,1)\). Translate all vertices by \(\langle -2,-1\rangle\). The new vertices are \((0,0)\), \((5,5)\), \((-10,20)\), and \((-15,15).\)
   2. Now rotate \((5,5)\) to the \(x\)-axis. Find the appropriate angle: \[p=\left ( \frac{5}{\sqrt{50}},\frac{5}{\sqrt{50}}\right) \]. The angle is going to be \(p^{-1}=\left ( \frac{5}{\sqrt{50}},-\frac{5}{\sqrt{50}}\right).\) Rotate all vertices by \(p^{-1}\).
   • \((5,5)\star p^{-1}=\left(\frac{50}{\sqrt{50}},0\right)\)
   • \((-10,20)\star p^{-1}=\left(\frac{50}{\sqrt{50}},\frac{150}{\sqrt{50}}\right)\)
   • \((-15,15)\star p^{-1}=\left(0,\frac{150}{\sqrt{50}}\right)\)

Based on the transformed vertices, a point \(q=(q_x,q_y)\) will be inside the rotated rectangle so long as \(0< q_x<\frac{50}{\sqrt{50}}\) and \(0<q_y<\frac{150}{\sqrt{50}}\). For example \(q=\left(\frac{1}{\sqrt{50}},\frac{1}{\sqrt{50}}\right)\) will be in the rotated rectangle. To get the corresponding point in the original rectangle, “reverse” the transformation: \(\langle 2,1\rangle+ (q\star p)=\left(2,\frac{10}{\sqrt{50}}+1\right)\). Thus \(q=\left(2,\frac{10}{\sqrt{50}}+1\right)\) is inside of the rectangle with vertices \((2,1), (7,6), (-8,21), (-13,16)\).

3. To find \(\mathcal{A}(R)\), find the side lengths of \(R\). Calculate the distance between the appropriate vertices of either the original rectangle or rotated rectangle. One side length will be \[a=\|(7,6)-(2,1)\|=\|\langle 5,5\rangle\|=\sqrt{50}\] while the other will be \[b=\|(-8,21)-(7,6)\|=\|\langle-15,15\rangle\|=\sqrt{450}.\] So the area is \[\mathcal{A}(R)=\sqrt{50}\cdot\sqrt{450}=150.\]

Triangles

The final geometric shape we study is a triangle. In some ways, a triangle is simpler than a rectangle. In other ways, it is more challenging. First, understand how we define a triangle as well as understanding how we define inside and outside.

Inside and Outside of a Triangle

Denote by \(\partial R\) the union of all line segments with endpoints in \(\{p_1, p_2, p_3\}\), where \(p_1\), \(p_2\), and \(p_3\) are distinct points in the plane.

The triangle with vertex set \(\{p_1, p_2, p_3\}\) is the set \(R\) that is given by \[R = \big\{t(q_2 -q_1) + q_1\colon t\in [0,1] \;{\rm and}\; q_1, q_2 \in \partial R\big\}.\]

The set \(\partial R\) is the boundary of \(R\).

A point is inside \(R\) if it is in \(R\) and not in \(\partial R\).

A point is outside \(R\) if it is not in \(R\).

Just as with rectangles, any triangle is the feasible set for a system of linear inequalities.

Use the picture below to understand the information given above.

Just like with a rectangle, we can define an orientation for the triangle.

Orientated Triangular Boundary

Given a triangle with vertex set \(\{p_1, p_2, p_3\}\), an oriented triangular boundary \(\partial_oR\) is the set \[\partial_oR = \left\{\overrightarrow{p_1p_2}, \overrightarrow{p_2p_3}, \overrightarrow{p_3p_1}\right\}.\]

First practice constructing a continuous piecewise linear parameterization for a given triangle.

Example 10

Take \(R\) to be the triangle with ordered vertex set \(\big((1,5), (3,-1), (7,2)\big)\). Find a continuous, piecewise linear parameterization \(\ell\) for the boundary of \(R\), where the domain of \(\ell\) is equal to \([0,5]\). Is \(\partial R\) a simple closed curve?

Create linear paths that go to each vertex:

• The linear path from \((1,5)\) to \((3,-1)\) is \(t\langle 2,-6 \rangle+(1,5)\).

• The linear path from \((3,-1)\) to \((7,2)\) is \(t\langle 4,3 \rangle+(3,-1)\).

• The linear path from \((7,2)\) to \((1,5)\) is \(t\langle -6,3 \rangle+(7,2)\).

Now create a piecewise linear parameterization by modifying the \(t\) variable. Here is one example

\[ \ell(t)=\begin{cases} \frac{t}{2}\langle 2,-6\rangle +(1,5) &\text{if }0\leq t<2\\ \frac{(t-2)}{2}\langle 4,3 \rangle+(3,-1) &\text{if } 2\leq t<4\\ (t-4)\langle -6,3 \rangle+(7,2) &\text{if }4\leq t <5 \end{cases} \]

Based on this, conclude that \(\partial R\) is a simple closed curve.

Finally, determine whether a point is inside or outside of a triangle.

Example 11

Take \(R\) to be the triangle with vertex set \(\big\{(0,0), (2,3), (5,0)\big\}\). Determine whether or not the following points are inside \(R\), outside \(R\), or in \(\partial R\):

1. \((1,1)\);

2. \((4,1)\).

First find a way to describe the points on \(\partial R\)

• \((2t,3t)\) where \(t\in [0,1]\) describes points on the line segment from \((0,0)\) to \((2,3)\)
• \((3s+2,-3s+3)\) where \(s\in[0,1]\) describes points on the line segment from \((2,3)\) to \((5,0)\)
• \((5u,0)\) where \(u\in[0,1]\) describes points on the line segment from \((0,0)\) to \((5,0)\).

1. The point \((1,1)\) is \(R\) provided there are points on the boundary for which the line segment that connects the two points crosses \((1,1).\) There are many ways to go about it, but here is one way.

• The point \((1,0)\) is on the line segment from \((0,0)\) to \((5,0)\): take \(u=\frac{1}{5}\), then \((5u,0)\) becomes \((1,0)\)
• The point \(\left(1,\frac{3}{2}\right)\) is on the line segment from \((0,0)\) to \((2,3)\): take \(t=\frac{1}{2}\), then \((2t,3t)\) becomes \(\left(1,\frac{3}{2}\right)\).
• Obtain the point \((1,1)\) by creating the line segment that connects \(q_1=(1,0)\) and \(q_2=\left(1,\frac{3}{2}\right)\) together \(t(q_2 -q_1) + q_1=t\left\langle 0,\frac{1}{2}\right\rangle+(1,0)=\left(1,\frac{3}{2}t\right)\). Find \(t\) so that \((1,1)=\left(1,\frac{3}{2}t\right)\). The time value is \(t=\frac{2}{3}\), which means that \((1,1)\) is inside \(R\).

2. The point \((4,1)\) is in \(\partial R\) because it is on \((3s+2,-3s+3)\), where \(s\in[0,1]\), which describes points on the line segment from \((2,3)\) to \((5,0)\). Find \(s\) so that \((3s+2,-3s+3)=(4,1)\). Solve to get \(s=\frac{2}{3}\).

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