$$ \newcommand{\RR}{\mathbb{R}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\EE}{\mathbb{E}} \newcommand{\HH}{\mathbb{H}} \renewcommand{\SS}{\mathbb{S}} \newcommand{\DD}{\mathbb{D}} \newcommand{\pp}{^{\prime\prime}} \newcommand{\p}{^\prime} \newcommand{\proj}{\operatorname{proj}} \newcommand{\area}{\operatorname{area}} \newcommand{\len}{\operatorname{length}} \newcommand{\acc}{\operatorname{acc}} \newcommand{\ang}{\sphericalangle} \newcommand{\map}{\mathrm{map}} \newcommand{\SO}{\operatorname{SO}} \newcommand{\dist}{\operatorname{dist}} \newcommand{\length}{\operatorname{length}} \newcommand{\uppersum}[1]{{\textstyle\sum^+_{#1}}} \newcommand{\lowersum}[1]{{\textstyle\sum^-_{#1}}} \newcommand{\upperint}[1]{{\textstyle\smallint^+_{#1}}} \newcommand{\lowerint}[1]{{\textstyle\smallint^-_{#1}}} \newcommand{\rsum}[1]{{\textstyle\sum_{#1}}} \newcommand{\partitions}[1]{\mathcal{P}_{#1}} \newcommand{\erf}{\operatorname{erf}} \newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}} \newcommand{\smat}[1]{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)} $$

Assignment 6

Problems

Right Angles

In this set of two problems, we make use of the fact that we can finally measure angles rigorously in our new geometry, to reprove an important fact we already know, and to prove the one remaining postulate of Euclid: the 4th.

Exercise 1 Prove that rectangles exist, using all of our new tools! (Ie write down what you know to be a rectangle, explain why each side is a line segment, parameterize it to find the tangent vectors at the vertices, and use the dot product to confirm that they are all right angles).

To prove Euclid’s 4th postulate, we need to first phrase it more precisely than Euclids original all right angles are equal.

Proposition 1 (Euclids’ Postulate 4) Given the following two configurations:

• A point \(p\), and two orthogonal unit vectors \(u_p,v_p\) based at \(p\)
• A point \(q\), and two orthogonal unit vectors \(a_q,b_q\) based at \(q\)

There is an isometry \(\phi\) of \(\EE^2\) which takes \(p\) to \(q\), takes \(u_p\) to \(a_q\), and \(v_p\) to \(b_q\).

Exercise 2 Prove Euclids’ forth postulate holds in the geometry we have built founded on calculus.

Hint: there’s a couple natural approaches here.

• You could directly use Exercise Moving from p to q. from a previous homework to move one point to the other and line up one of the tangent vectors. Then deal with the second one: can you explain why its either already lined up, or will be after one reflection?
• Alternatively, you could show that every right angle can be moved to the “standard right angle” formed by \(\langle 1,0\rangle, \langle 0,1\rangle\) at \(O\). Then use this to move every angle to every other, transiting through \(O\)

Measurement of the Circle

The half angle identities played a crucial role in Archimedes’ ability to compute the preimeter of \(n\) gons in his paper The Measurement of the Circle. Indeed, to calculate the circumference of an inscribed \(n\)-gon, its enough to be able to find \(\sin\tau/(2n)\):

The side-length of an inscribed \(n\)-gon is \(2\sin\frac{\tau}{2n}\), found via bisecting the side to form a right triangle. The perimeter of the \(n\)-gon is just \(n\) times this.

By repeatedly bisecting the sides, we can start with something we can directly compute - like a triangle, and repeatedly bisect to compute larger and larger \(n\) gons.

Archimedes’ method: repeatedly doubling the number of sides of the \(n\)-gon to get polygons approaching the circle.

In the book, I use the half-angle identities to compute the exact value of \(\sin\tau/12\). Follow that example further, to retrace the steps of Archimedes.

Exercise 3 Continue to bisections until you can compute \(\sin(\tau/(2\cdot 96))\). What is the perimeter of the regular \(96\)-gon (use a computer to get a decimal approximation, after your exact answer).

Explain how we know that this is provably an underestimate of the true length, using the definition of line segments.

Optional: Be brave - and go beyond Archimedes! Compute the circumference of the 192-gon.

Quadrature of the Parabola

This is also a two-problem seriess, where we complete Archimedes famous Quadrature of the Parabaola using modern tools from calculus. Archimedes problem was about a parabolic segment: that is, the region enclosed by a parabola and a line segment connecting two points on the parabola. Instead of working in complete generality like Archimedes, we will be content to just study a special case in this problem. We will look at the parabola \(y=x^2\), and the parabolic segment cut out by this and the line connecting \((-1,1)\) to \((2,4)\).

The Parabolic Segment in this Problem

Recall Archimedes main result: the area of this parabolic segment is \(4/3\)rds the area of the largest inscribed triangle, which is the triangle whose base is the line segment, and third vertex lies on the parabola at the point where the tangent line is parallel to the base.

The overall segment’s are is 4/3rds that of this triangle.

Exercise 4  

• Write down a formula for the area of the triangle whose third vertex lies at \((x,x^2)\)

Hint: instead of finding the height of the triangle to use \(\tfrac{1}{2}bh\), can you use the fact that the determinant of a matrix calculates the area of a parallelogram whose sides are the column vectors, and that the area of a the triangle you want is half a parallelogram?

• Use Calculus I to find the point \(x\) where the inscribed triangle has maximal area. Then show that Archimedes was right: the slope of the tangent line to the parabola at this point is exactly the same as the slope of the line segment forming the triangle’s base!

This gives us the starting point: the area for which archimedes wishes to compare the parabolic segment. Next - we need to find the parabolic segment’s area. We could of course follow Archimedes’ original method (and if you choose to, this can be your class project!) But here, we will use our modern tools and confirm the answer with calculus:

Exercise 5 Compute the area of the parabolic segment (via integration, as the area between two curves). Show that its exactly \(4/3\)rds the area of the triangle!

The Area and Circumference Constants

A circle has a circumference constant: the ratio of its radius to its circumference, which we’ve named \(\tau\). But it also has an *area constant: the ratio of its area to the square of its radius, which we’ve named \(\pi\).

It was Archimedes who first showed that these two constants were intimately related, by finding that \(\tau = 2\pi\). Here we will again use our modern tools to reprove Archimede’s result.

Tau is the circumference of the unit circle \(x^2+y^2=1\). We can parameterize the top half of this circle via \[\gamma(t)=(t,\sqrt{1-t^2})\]

And then compute its arclength via the integral \[\frac{\tau}{2}=\int_{-1}^1 \|\gamma^\prime(t)\|dt=\int_{-1}^1\frac{1}{\sqrt{1-t^2}}dt\]

But we can also write down the area of the circle as an integral: the top half of the circle is \(y=\sqrt{1-x^2}\) and the bottom half is \(y=-\sqrt{1-x^2}\) so the area is

\[\pi = \int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dydx=\int_{-1}^1 2\sqrt{1-x^2}dx\]

Your goal in this problem is to show these two integrals are equal to one aother!

Exercise 6 Prove that \[\int_{-1}^1\frac{1}{\sqrt{1-t^2}}dt=\int_{-1}^1 2\sqrt{1-x^2}dx\] Thus showing thtat \(\frac{\tau}{2}=\pi\).

Hint: Do \(u\)-substitutions to the integrals to make them into the same integral. The goal isn’t to evaluate them and get a number! This is just a Calc II problem - but a tricky one, so here’s one outline you could follow:

1. Rewrite the area integrand \(\sqrt{1-x^2}\) as \(\frac{1-x^2}{\sqrt{1-x^2}}\). Use properties of integrals to break this into two integrals, and see \[\pi = \tau -\int_{-1}^1\frac{2x^2}{\sqrt{1-x^2}}dx\]

2. Now we just have to evaluate this new integral: Do the \(u\)-substitution \(u=\sqrt{1-x^2}\) to this, to show that \[\int_{-1}^1 \frac{2x^2}{\sqrt{1-x^2}}dx=\int_{-1}^1 2\sqrt{1-u^2}du=\pi\] (This \(u\)-sub requires some work: you’ll need at some point to solve for \(x\) in terms of \(u\)!)

3. Now just assemble the pieces! You never completed a single integral, but you still managed to prove that \(\tau =2\pi\).

Solutions

Right Angles

Rectangles: Freya

1

Quadrature of the Parabola:

Problem 4: Katie

We are asked to find the area of the maximal triangle between a parabola and line. We don’t know yet what third point will maximize the area of our triangle so we will let \(p:(x,x^2)\) represent the point we want. Next we want to find the area of this triangle, but how can we do this? One way, is to find two vectors that create a parallelogram with twice the area of our triangle. Let’s use the vectors: \(V:<3,3>\) and \(U:<x+1,x^2-1>\), be those vectors. Thus visually we have:

The desired triangle inscribed in the parabola.

These vectors create the following parallelogram:

A parallelogram, with twice the area of the desired triangle.

Thus, the area of the triangle is \(\frac{1}{2} det\begin{pmatrix} x+1 & 3 \\ x^2 -1 & 3 \end{pmatrix} = \frac{1}{2}(3(x+1)-(3(x^2-1)))=\frac{3}{2}(-x^2+x+2).\) So we found out that the area of our triangle is \(\frac{3}{2}(-x^2+x+2)\). But we want this to be the maximum triangle, and to maximize it we must take the derivative and sit it equal to zero: \[(\frac{3}{2}(-x^2+x+2))'=(-\frac{3}{2}x^2+\frac{3}{2}x+3)'= -3x+\frac{3}{2}=0 \] Thus, \[x = \frac{1}{2}\]

Therefore, we found that \((\frac{1}{2}, \frac{1}{4})\) is the point where our triangle has maximum area!

Next, we want to show Archimedes was right and show that the tangent vector to \((\frac{1}{2}, \frac{1}{4})\) has the same slope as \(V = <3,3>\). We can do this by finding a tangent vector to our curve: \[(x,x^2)'=<1,2x>\] This vector at \(x = \frac{1}{2}\) is \(<1,1>\). Notice that \(<1,1>\) has the same slope as \(V\)! Therefore we have shown what we need to!

Problem 5: Daniel