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Assignment 4

Problems

Parameterization Invariance

Below are four different curves which all trace out the same set of points in the plane: the segment of the $x$ axis between $0$ and $4$.

\[\alpha(t)=(t,0)\hspace{1cm}t\in[0,4]\] \[\beta(t)=(2t,0)\hspace{1cm}t\in[0,2]\] \[\gamma(t)=(t^2,0)\hspace{1cm}t\in[0,2]\]

Because these all describe the same set of points, we of course want them to have the same length! But our definition of the length function involves integrating infinitesimal arclengths (derivatives), and these curves don’t all have the same derivative! Thus, to really make sure our definition makes sense, we need to check that it doesn’t matter which parameterization we use, we will always get the same length.

Exercise 1 Check these three parameterizations of the segment of the $x$-axis from $0$ to $4$ all have the same length.

After doing this exercise, read the proof of Theorem Length & Parameterization Invariance (which follows this exercises’ original location in the text): you don’t have to write anything here, but it’s good to see how do to this in general with the chain rule!

Non-Isometries

Exercise 2 Write down a linear map that sends both $\langle 1,0\rangle$ and $\langle 0,1\rangle$ to unit vectors, but is not an isometry.

This shows there’s not a shortcut to checking something is an isometry by just seeing what happens to the basis vectors!

Composition and Inversion of Isometries

Exercise 3 If $\phi$ and $\psi$ are two isometries of $\EE^2$, prove that both the composition $\phi\circ\psi$ is an isometry, and the inverse $\phi^{-1}$ is an isometry.

Remember, you will need to explain why at every point $p\in \EE^2$, these maps do not change the lengths of tangent vectors. This will probably involve the multivariable chain rule, whether you do it in words, or in equations!

Homogenity and Isotropy

In class we built a couple different sorts of isometries from the basic ones we construted by hand (translations and rotations about 0). In this exercise, you are to prove the existence of another very useful isometry. We will use this homework problem all the time

Exercise 4 (Moving from $p$ to $q$.) Given any two pairs $p,v_p$ and $q,w_q$ of points $p,q$ in Euclidean space and unit tangent vectors $v_p\in T_p\EE^2$, $w_q\in T_q \EE^2$ based at them, prove that there exists an isometry taking $v_p$ to $w_q$.

Hint: try to combine pieces we know about, and prove the result does what you need by applying it both to the point $p$, and applying its derivative to the vector $v$.

Lines of Symmetry

In this exercise we will investigate the third potential definition of line, which involves isometries.

Definition 1 (Line of Symmetry) A fixed point of an isometry $\phi\colon \EE^2\to\EE^2$ is a point $p$ with $\phi(p)=p$.

A curve $\gamma$ is called a line of symmetry of $\EE^2$ if there exists an isometry which fixes $\gamma(t)$ for all $t$.

In this exercise, you show that the curves which are lines of symmetry are exactly the same as the curves which are lines under Archimedes’ definition!

Exercise 5 (Reflections in Any Line)

Show that map $\phi(x,y)=(x,-y)$ is an isometry of $\EE^2$. Explain why this shows that the $x$-axis is a line of symmetry of the plane.
Show that every curve which is distance-minimizing in the plane is also a line of symmetry. Hint: given an isometry that reflects in the $x$ axis, can you build an isometry that reflects in any other line? Consider moving the line to the $x$ axis, reflecting, and then moving back.

Equilateral Triangles Revisited

In this question we will revisit two problems from Greek geometry. That is we will be re-proving things we knew before, so we know they are still true in our new foundations!

This problem requires the distance function on the Euclidean plane, which we did not get to in class on Thursday, but will cover on Tuesday. However - you all areadly know the distance function so you can absolutely complete the homework now if you like!

Definition 2 If $p=(x,y)$ and $q=(a,b)$ are two points in the Euclidean plane, the distance from $p$ to $q$ is the length of the shortest curve connecting them. Working this out, we find the familiar pythagorean theorem:

\[\dist(p,q)=\sqrt{(x-a)^2+(y-b)^2}\]

First, we re-prove the very first proposition of Euclid, the existence of an equilateral triangle. Then we redo your earlier homework problem on finding a smaller equilateral triangle inside of it, of half the side lengths (but this is much easier with our new tools!).

Exercise 6

Beginning with the segment $[0,\ell]$ along the $x$-axis, construct an equilateral triangle by finding the coordinates of a point $p=(x,y)\in\EE^2$ which is equidistant from both endpoints of the segment.
Re-prove that inside of this equilateral triangle, you can inscribe a smaller one with exactly half the side length. Hint: just find where the vertices should be, and then measure the distances between them!

OPTIONAL Problems

These problems are optional and you will not to turn them in. But - they are excellent reviews of calculus, and applications of the material we are learning now!

Length of a Parabola

Arclength integrals give a good opportunity to practice a lot of Calculus II integration techinques. Even for relatively simple curves like the parabola, the answers can be quite nontrivial!

Exercise 7 (The Length of a Parabola) Find the length of the parabola $y=x^2$ between from $x=0$ to $x=a$, following the steps below.

Paramterize the curve as $c(t)=(t,t^2)$, show the arclength integral is $L(a)=\int_{[0,a]}\sqrt{1+4t^2}$
Perform the trigonometric substitution $x=\frac{1}{2}\tan\theta$ to convert this to some multiple of the integral of $\sec^3(\theta)$.
Let $I=\int\sec^3(\theta)d\theta$ and do integration by parts with $u=\sec\theta$ and $dv=\sec^2\theta$.
After parts, use the trigonometric identity $\tan^2\theta=\sec^2\theta-1$ in the resulting integral to get another copy of $I=\int\sec^3\theta d\theta$ to appear.
Get both copies of $I$ to the same side of the equation and solve for it! To check your work at this stage, you should have found that \[\int\sec^3\theta d\theta = \frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|\]
Relate this back to your original integral, and undo the substitution $x=\frac{1}{2}\tan\theta$: can you use somet trigonometry to figure out what $\sec\theta$ is?
Finally, you have the antiderivative in terms of $x$! Now evaluate from $0$ to $a$.

Minimizing a Function by Minimizing its Square

Here’s a problem that’s straight up single variable calculus, but turns out to be a quite useful “trick” in geometry! Oftentimes we want to minimize a function in geometry (like arclength, or distance) but this turns out to be technically hard because of the square root. One might wonder - what happens if I square the function, and try to minimize that instead? That will have an easier formula (no roots), but will I get the right answer?

This exercise shows, yes you will!

Exercise 8 (Minimizing the Square: A Very Useful Trick!) Let $f(x)$ be a differentiable positive function of one variable, and let $s(x)=f(x)^2$ be its square. Show that the minima of $s(x)$ and $f(x)$ occur at the same points, by following the steps below:

First, assume $x=a$ is the location of a minimum of $f$. What does the first and second derivative test tell you about the values $f^\prime(a)$ and $f^{\prime\prime}(a)$? Use this, together with the fact that $f(a)>0$ to show that $x=a$ is also the location of a minimum of $s$ (using the second derivative test).
Conversely, assume $x=a$ is the location of a minimum of $s(x)$. Now, you know information about the derivatives $s^\prime(a)$ and $s^{\prime\prime}(a)$. Use this to conclude information about $f^\prime(a)$ and $f^{\prime\prime}(a)$ to show that $a$ is a minimum for $f$ as well.