Introduction
Even if the title is in Japanese, this post isn’t primarily in Japanese. I might add more content if I have time.
Problem 1: https://blog.yexca.net/en/archives/183
Problem 2: This post
Problem 3: https://blog.yexca.net/en/archives/188
Well, motivation is often crushed by reality, but I’m recovering faster this time. Hopefully, things will get better.
Regarding this problem, it seems to belong to a specific course on vector calculus. I didn’t get it at first until I remembered 3b1b’s Essence of Linear Algebra and the general representation of vectors in math.
In fact, the problem defines almost everything you need. If these weren’t covered in your curriculum, you’d basically have to understand the definitions on the spot and solve it. Being able to solve this is impressive.
Also, since these are all proofs, I can’t guarantee the process is 100% correct.
Problem
Source: https://www.i.u-tokyo.ac.jp/edu/entra/examarchive.shtml
The copyright of the problem belongs to the University of Tokyo. It is quoted here for educational purposes only, non-commercially.
Consider a smooth curve $\boldsymbol{p}=(p(t),q(t))(t \in [a,b])$ in the $xy$-plane. The length $l_{a^{'},b^{'}}$ of $\boldsymbol{p}$ from time $t=a^{'} $ to $b^{'}$ is defined as:
$$ l_{a^{'},b^{'}} = \int_{a^{'}}^{b^{'}} \sqrt{(\frac{\mathrm{d} p}{\mathrm{d} t})^2+(\frac{\mathrm{d} q}{\mathrm{d} t})^2} \mathrm{d} t $$The total length $l_{a,b}$ of $\boldsymbol{p}$ is denoted by $L$. Assume $\frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} t} \neq (0,0)$. If we represent the length $l_{a,t}$ of $\boldsymbol{p}$ from time $a$ to $t$ as the variable $s=s(t)$, we can view $\boldsymbol{p}$ as a curve with parameter $s \in [0,L]$. We also call $s$ “time.” Answer the following questions.
(1) Show the following equality:
$$ \sqrt{(\frac{\mathrm{d} p}{\mathrm{d} s})^2+(\frac{\mathrm{d} q}{\mathrm{d} s})^2} = 1 $$(2) Let $\theta = \theta(s)$ be the angle between the tangent vector $\frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} s} = (\frac{\mathrm{d} p}{\mathrm{d} s},\frac{\mathrm{d} q}{\mathrm{d} s})$ of $\boldsymbol{p}$ at time $s$ and the $x$-axis. Show the following equality:
$$ \frac{\mathrm{d} p}{\mathrm{d} s} \frac{\mathrm{d}^2 q}{\mathrm{d} s^2}-\frac{\mathrm{d} q}{\mathrm{d} s}\frac{\mathrm{d}^2 p}{\mathrm{d} s^2} = \frac{\mathrm{d} \theta}{\mathrm{d} s} $$In the following, assume the curve $\boldsymbol{p}$ is a smooth closed curve that forms the boundary of a convex set $\boldsymbol{K}$, and $\boldsymbol{p}$ moves around $\boldsymbol{K}$ counter-clockwise.
(3) Explain why $\frac{\mathrm{d} \theta}{\mathrm{d} s} \ge 0$ holds at any time $s$.
(4) A point $\boldsymbol{x}=(x,y)$ not included in $\boldsymbol{K}$ can be uniquely represented by time $s \in [0,L]$ and the distance $r$ between $\boldsymbol{x}$ and $\boldsymbol{K}$ as:
$$ \boldsymbol{x} = \boldsymbol{p} (s) + r \boldsymbol{u} (s) $$Here, $\boldsymbol{u} (s)$ is the unit normal vector of $\boldsymbol{p}$ at time $s$, pointing outside $\boldsymbol{K}$. For such $\boldsymbol{x} = (x,y)$, show the following equality:
$$ \left | det \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial r} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial r} \end{pmatrix} \right | =1 +r\frac{\mathrm{d} \theta}{\mathrm{d} s} $$(5) For a non-negative real number $D$, let $K_D$ be the set of points within distance $D$ from $K$. Show that the area $A_D = \iint_{K_D} \mathrm{d}x\mathrm{d}y$ of $K_D$ can be expressed using the area $A$ of $K$ and the total length $L$ of $\boldsymbol{p}$ as:
$$ A_D = A + LD + \pi D^2 $$Solution
This description is essentially arc length parametrization, which uses the curve’s arc length as the parameter. Specifically, it uses the distance from the starting point to a point on the curve as the new parameter, making the parametrization clean and reflective of geometric properties.
Question 1
Since the curve time $t \in [a,b]$, the total length $L$ is:
$$ l_{a,b} = \int_{a}^{b} \sqrt{(\frac{\mathrm{d} p}{\mathrm{d} t})^2+(\frac{\mathrm{d} q}{\mathrm{d} t})^2} \mathrm{d} t $$And $s$ is the arc length from $a$ to $t$:
$$ s = \int_{a}^{t} \sqrt{(\frac{\mathrm{d} p}{\mathrm{d} t})^2+(\frac{\mathrm{d} q}{\mathrm{d} t})^2} \mathrm{d} t $$Thus, $s$ represents the arc length from the starting point $a$ to the current point, so:
$$ \frac{\mathrm{d} s}{\mathrm{d} t} = \sqrt{(\frac{\mathrm{d} p}{\mathrm{d} t})^2+(\frac{\mathrm{d} q}{\mathrm{d} t})^2} $$Since $\boldsymbol{p}$ is a curve with parameter $s$, and $s$ is a function of $t$:
$$ \frac{\mathrm{d} p}{\mathrm{d} t} = \frac{\mathrm{d} p}{\mathrm{d} s} \frac{\mathrm{d} s}{\mathrm{d} t} \\ \frac{\mathrm{d} q}{\mathrm{d} t} = \frac{\mathrm{d} q}{\mathrm{d} s} \frac{\mathrm{d} s}{\mathrm{d} t} $$Substituting these:
$$ \begin{align} \frac{\mathrm{d} s}{\mathrm{d} t} &=\sqrt{(\frac{\mathrm{d} p}{\mathrm{d} s} \frac{\mathrm{d} s}{\mathrm{d} t})^2+(\frac{\mathrm{d} q}{\mathrm{d} s} \frac{\mathrm{d} s}{\mathrm{d} t})^2} \\ &=\sqrt{(\frac{\mathrm{d} p}{\mathrm{d} s})^2+(\frac{\mathrm{d} q}{\mathrm{d} s})^2} \frac{\mathrm{d} s}{\mathrm{d} t} \end{align} $$Since $\frac{\mathrm{d} p}{\mathrm{d} t} \ne 0$, it follows that $\frac{\mathrm{d} s}{\mathrm{d} t} \ne 0$:
$$ \sqrt{(\frac{\mathrm{d} p}{\mathrm{d} s})^2+(\frac{\mathrm{d} q}{\mathrm{d} s})^2} = 1 $$This proof shows that under arc length parametrization, the velocity vector $\frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} s}$ has a constant length of $1$.
Question 2
Generally, for a unit vector with an angle $\theta$ relative to the $x$-axis, the components are:
$$ v_x = \cos(\theta) \\ v_y = \sin(\theta) $$From Q1, we know $\begin{vmatrix} \frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} s} \end{vmatrix} = 1$, so:
$$ \begin{align} \frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} s} &= (\frac{\mathrm{d} p}{\mathrm{d} s},\frac{\mathrm{d} q}{\mathrm{d} s}) \\ &= (\cos(\theta(s)), \sin(\theta(s))) \end{align} $$Differentiating the components with respect to $s$:
$$ \begin{align} \frac{\mathrm{d}^2 p}{\mathrm{d} s^2} &= -\sin(\theta(s)) \cdot \frac{\mathrm{d} \theta}{\mathrm{d} s} \\ \frac{\mathrm{d}^2 q}{\mathrm{d} s^2} &= \cos(\theta(s)) \cdot \frac{\mathrm{d} \theta}{\mathrm{d} s} \end{align} $$Therefore:
$$ \begin{align} \frac{\mathrm{d} p}{\mathrm{d} s} \frac{\mathrm{d}^2 q}{\mathrm{d} s^2}-\frac{\mathrm{d} q}{\mathrm{d} s}\frac{\mathrm{d}^2 p}{\mathrm{d} s^2} &= \cos(\theta(s)) \cdot \cos(\theta(s)) \cdot \frac{\mathrm{d} \theta}{\mathrm{d} s} - \sin(\theta(s)) \cdot (-\sin(\theta(s)) \cdot \frac{\mathrm{d} \theta}{\mathrm{d} s}) \\ &= (\cos^2\theta(s) + \sin^2\theta(s)) \cdot \frac{\mathrm{d} \theta}{\mathrm{d} s} \\ &= \frac{\mathrm{d} \theta}{\mathrm{d} s} \end{align} $$Question 3
Since $\boldsymbol{p}$ is a smooth convex curve, the change in the tangent angle is continuous and does not reverse (it’s monotonic). Because it moves counter-clockwise, the angle $\theta$ with the $x$-axis is constantly increasing. Therefore, its derivative is always non-negative: $\frac{\mathrm{d} \theta}{\mathrm{d} s} \ge 0$.
Question 4
Since $\frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} s}$ is the unit tangent vector and $\boldsymbol{u} (s)$ is the outward unit normal vector:
$$ \boldsymbol{u} (s) = (\frac{\mathrm{d} q}{\mathrm{d} s},-\frac{\mathrm{d} p}{\mathrm{d} s}) $$Take the partial derivatives of $\boldsymbol{x}$ with respect to $s$ and $r$:
$$ \begin{align} \frac{\partial \boldsymbol{x}}{\partial s} &= \frac{\mathrm{d} \boldsymbol{p}}{\mathrm{d} s} + r\frac{\mathrm{d} \boldsymbol{u}}{\mathrm{d} s} \\ \frac{\partial \boldsymbol{x}}{\partial r} &= \boldsymbol{u} (s) \end{align} $$Components for $x$:
$$ \begin{align} \frac{\partial x}{\partial s} &= \frac{\mathrm{d} p}{\mathrm{d} s} + r\frac{\mathrm{d}^2 q}{\mathrm{d} s^2} \\ \frac{\partial x}{\partial r} &= \frac{\mathrm{d} q}{\mathrm{d} s} \end{align} $$Components for $y$:
$$ \begin{align} \frac{\partial y}{\partial s} &= \frac{\mathrm{d} q}{\mathrm{d} s} - r\frac{\mathrm{d}^2 p}{\mathrm{d} s^2} \\ \frac{\partial y}{\partial r} &= -\frac{\mathrm{d} p}{\mathrm{d} s} \end{align} $$The Jacobian matrix becomes:
$$ \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial r} \ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial r} \end{pmatrix}
\begin{pmatrix} \frac{\mathrm{d} p}{\mathrm{d} s} + r\frac{\mathrm{d}^2 q}{\mathrm{d} s^2} & \frac{\mathrm{d} q}{\mathrm{d} s} \ \frac{\mathrm{d} q}{\mathrm{d} s} - r\frac{\mathrm{d}^2 p}{\mathrm{d} s^2} & -\frac{\mathrm{d} p}{\mathrm{d} s} \end{pmatrix} $$
Calculating the determinant:
$$ \begin{align} \det\begin{pmatrix} \frac{\mathrm{d} p}{\mathrm{d} s} + r\frac{\mathrm{d}^2 q}{\mathrm{d} s^2} & \frac{\mathrm{d} q}{\mathrm{d} s} \\ \frac{\mathrm{d} q}{\mathrm{d} s} - r\frac{\mathrm{d}^2 p}{\mathrm{d} s^2} & -\frac{\mathrm{d} p}{\mathrm{d} s} \end{pmatrix} &= (\frac{\mathrm{d} p}{\mathrm{d} s} + r\frac{\mathrm{d}^2 q}{\mathrm{d} s^2})(-\frac{\mathrm{d} p}{\mathrm{d} s})-(\frac{\mathrm{d} q}{\mathrm{d} s} - r\frac{\mathrm{d}^2 p}{\mathrm{d} s^2})(\frac{\mathrm{d} q}{\mathrm{d} s}) \\ &= -(\frac{\mathrm{d} p}{\mathrm{d} s})^2 -r\frac{\mathrm{d}^2 q}{\mathrm{d} s^2}\frac{\mathrm{d} p}{\mathrm{d} s}-(\frac{\mathrm{d} q}{\mathrm{d} s})^2+r\frac{\mathrm{d}^2 p}{\mathrm{d} s^2}\frac{\mathrm{d} q}{\mathrm{d} s} \\ &= -[(\frac{\mathrm{d} p}{\mathrm{d} s})^2+(\frac{\mathrm{d} q}{\mathrm{d} s})^2]-r(\frac{\mathrm{d}^2 q}{\mathrm{d} s^2}\frac{\mathrm{d} p}{\mathrm{d} s}-\frac{\mathrm{d}^2 p}{\mathrm{d} s^2}\frac{\mathrm{d} q}{\mathrm{d} s}) \end{align} $$Using the results from Q1 and Q2:
$$ \begin{align} \left | \det\begin{pmatrix} \frac{\mathrm{d} p}{\mathrm{d} s} + r\frac{\mathrm{d}^2 q}{\mathrm{d} s^2} & \frac{\mathrm{d} q}{\mathrm{d} s} \\ \frac{\mathrm{d} q}{\mathrm{d} s} - r\frac{\mathrm{d}^2 p}{\mathrm{d} s^2} & -\frac{\mathrm{d} p}{\mathrm{d} s} \end{pmatrix} \right | &= \left | -1-r\frac{\mathrm{d} \theta}{\mathrm{d} s} \right | \\ &= 1+r\frac{\mathrm{d} \theta}{\mathrm{d} s} \end{align} $$Note: The matrix involved here is the Jacobian matrix .
Question 5
This proof is easiest to understand with a diagram.
- The blue area is $A$, the area of $K$.
- The green areas are “rectangles” extending from the edges of $K$. Their total area is $L \times D$.
- The pink areas are the corners. Because the curve is convex, smooth, and closed, these circular sectors sum up to exactly one circle with radius $D$. The area of this circle is $\pi D^2$.
So, $A_D = A + LD + \pi D^2$. This is actually a specific case of Steiner’s Formula .
Wrote with ChatGPT