曲線・曲面 - ryamadaの遺伝学・遺伝統計学メモ

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作者: ryamada
発売日: 2017/01/27
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---
title: "Calculus5 曲線・曲面 Curved line and surface"
author: "ryamada"
date: "2017年1月22日"
output: 
  html_document:
    toc: true
    toc_depth: 6
    number_section: true
---

```{r setup, include=FALSE}
library(knitr)
opts_chunk$set(echo = TRUE)
library(rgl)
knit_hooks$set(rgl = hook_webgl)
```

# 曲線の長さ Length of curve

$$
y = f(x) = x + \frac{1}{2}x^2
$$

```{r}
x <- seq(from=0,to=3,length=100)
y <- x+ 1/2*x^2
plot(x,y,type="l")
```



$$
\frac{d}{dx}f(x) = 1 + x
$$
微分すると、単位x増分当たりのyの増加量がわかる。それらは、微小直角三角形。

Differentiation gives increase of y along with unit increase in x. They are two edges of a right triangle.

三角形 (x,y) ,(x+dx,y),(x+dx,y+dy)は直角三角形の3頂点。
曲線の長さはこの三角形の斜辺の積分

(x,y) ,(x+dx,y),(x+dx,y+dy) are three vertices of a right triangle.
The length of curve is the integral of the obliques.

$$
\int_{0}^3 \sqrt{dx^2+dy^2} = \int_0^3 \sqrt{(\frac{dx}{dx})^2+(\frac{dy}{dx})^2}dx = \sqrt{1+(\frac{dy}{dx})^2}dx \\
$$

積分できるのか？
Can we integrate this?

$$
\int_0^3 \sqrt{2+2x+x^2}dx\\
=[\sqrt{2+2x+x^2}(x+1) + \sinh^{-1}(x+1)]^3_0
$$

Mathematicaを使えば…。

Check Mathematica http://www.wolframalpha.com/examples/Math.html for "Calculus AND ANALYSIS"
and http://d.hatena.ne.jp/ryamada22/20170128 

積分できるかできないかわからなくても近似計算はできる。

Even when the integral is not easy shape, computational calculation is possible.

```{r}
integrate(function(x){(2+2*x+x^2)^(1/2)},0,3)
```

# Exercise 1

## Exercise 1-1

$ y = f(x) = e^{x^2}$ の $x \in [-1,1]$の長さを計算せよ。

Calculate the length of $ y = f(x) = e^{x^2}$ for $x \in [-1,1]$.

# 曲線のパラメタ表示 Parameteric expression of curve.

$$
x = \cos{t}\\
y = \sin{t}\\
t \in [0,2\pi)
$$

```{r}
t <- seq(from=0,to=1,length=100)*2*pi
t <- t[-length(t)]
x <- cos(t)
y <- sin(t)
plot(x,y,type="l")
matplot(cbind(x,y),type="l")
```

$$
L(\tau) = \int_0^\tau \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt
$$

$$
\frac{dx}{dt} = -\sin{t}\\
\frac{dy}{dt} = \cos{t}\\
L(\tau) = \int_0^\tau \sqrt{(\sin{t})^2 + (\cos{t})^2} dt\\
=\int_0^\tau 1 dt\\
=[t]^\tau_0 = \tau
$$

## 同じ軌道、スピードが違う Same trajectory but different speed

$$
x = \cos{t^2}\\
y = \sin{t^2}\\
t \in [0,\sqrt{2\pi})
$$

```{r}
t <- seq(from=0,to=1,length=100)*sqrt(2*pi)
t <- t[-length(t)]
x <- cos(t^2)
y <- sin(t^2)
plot(x,y,type="l")
matplot(cbind(x,y),type="l")
```

$$
L(\tau) = \int_0^\tau \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt\\
= \int_0^\tau \sqrt{(-2t\sin(t^2))^2+(2t\cos(t^2))^2}dt\\
= \int_0^\tau 2t \sqrt{\sin^2{(t^2)}+\cos^2{(t^2)}} dt\\
= \int_0^\tau 2t dt\\
= [t^2]_0^\tau\\
= \tau^2
$$

# Exercise 2

## Exercise 2-1

曲線 
$$
x = \cos{t}\\
y = \sin{t}\\
t \in [0,2\pi)
$$
を描き、その曲線の上に適当に多数の点を取り、その点での進行方向ベクトル$(\frac{dx}{dt},\frac{dy}{dt})$を描け。

Draw the curve and put its speed vectors $(\frac{dx}{dt},\frac{dy}{dt})$ on many points on the curve.

## Exercise 2-2

$$
x = \cos{t^2}\\
y = \sin{t^2}\\
t \in [0,\sqrt{2\pi})
$$
について同様のことをせよ。

Do the same for this parameterization.

## Exercise 2-3

3次元空間の曲線
$$
x = r\cos{t}\\
y = r\sin{t}\\
z = r^2\\
r = 0.1t+0.1\\
t \in [0,100]
$$
```{r,rgl=TRUE}
t <- seq(from=0,to=1,length=1000)*100
r <- 0.1*t+0.1
x <- r*cos(t)
y <- r*sin(t)
z <- r^2
plot3d(x,y,z,type="l")
```

$\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}$を計算し、曲線の長さとしてその積分をRのintegrate()関数を使って計算せよ。

This is a curve in 3D space. Calculate $\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}$ and calculate the length of curve with R's integrate() function.

また、曲線の長さを折れ線の長さの集まりとして
```{}
t <- seq(from=0,to=1,length=n)*100
```
$n=10,100,1000,10000,100000$について計算せよ。

For multiple n values, calculate the length of broken lines that approximate the curve.

# 単位円の面積 Area of a unit circle.

単位円の面積は、The following is the 1/4 of unit circle's area.

$$
y = \sqrt{(1-x^2)}\\
y = 0 \\
x \in [0,1]
$$
の4倍として求めることができる。

$$
4 \int_0^1 \sqrt{(1-x^2)}dx
$$

この積分を解くのは難しい。
It's no easy to solve this.

```{r}
integrate(function(x){4*sqrt(1-x^2)},0,1)
```

微小底辺$dl$を円周上に持ち、高さが半径1の三角形の和として計算することもできる。

中心からの半径方向を$\theta \in [0,2\pi)$とすれば

The area of circle is the sum of small triangles whose base lines are small arcs $dl$ of the circle and whose hight are the radius.
The directions of the triangles are parameterized as $\theta \in [0,2\pi)$.

$$
\int_0^{2\pi} \frac{1}{2}\times dl \\
dl = \sqrt{(\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2}d\theta = d\theta 
$$
これは、円周の長さ$2\pi$を底辺の長さとする三角形の面積。

It corresponds to the area of triangle whose base is a whole circle.

別の方法は円を半径が0から1に変化する円のあつまりとみなすものである。

Another way is to consider the area of circle as the sum of cicles with radius ranging from 0 to 1.

半径$r$での微小面積は$2\pi r \times dr$.

The small area attached to the circle with radius r is $2\pi r \times dr$.

$$
\int_0^1 2\pi r dr = 2\pi \int_0^1 r dr = 2\pi [\frac{1}{2}r^2]^1_0 = \pi
$$
# 曲面 Curved surface

以下のような曲面の面積はどうなるか？

How do we approach to the area of the curved surface as below?

$$
z = \sin{x}+\cos{y}\\
x, y \in [-5,5]
$$
```{r,rgl=TRUE}
x <- y <- seq(from=-1,to=1,length=100) * 5
xy <- as.matrix(expand.grid(x,y))
z <- sin(xy[,1])+cos(xy[,2])
xlim <- max(xy[,1])-min(xy[,1])
ylim <- max(xy[,2])-min(xy[,2])
zlim <- max(z)-min(z)
plot3d(xy[,1],xy[,2],z,aspect=c(xlim,ylim,zlim))
```

(x,y),(x+dx,y),(x,y+dy),(x+dx,x+dy)が作る「斜面」の面積$dS$を足し合わせる。

Small areas that are slopes over (x,y),(x+dx,y),(x,y+dy),(x+dx,x+dy), $dS$, should be summed up.

これを斜めになった平行四辺形の面積とみなす。
They are considered small parallelograms.

平行四辺形の第1の辺は$(x,y,z)$と$(x+dx,y,z+\frac{\partial x}{\partial z}dx)$とが作るベクトル。
第2の辺は$(x,y,z)$と$(x,y+dy+z+\frac{\partial y}{\partial z}dy)$。

The parallelogram's two edges are; one connecting $(x,y,z)$と$(x+dx,y,z+\frac{\partial x}{\partial z}dx)$ and 
and the other connecting $(x,y,z)$ and $(x,y+dy+z+\frac{\partial y}{\partial z}dy)$.

2つのベクトルは They are:
$$
v_1 = (dx,0,\frac{\partial z}{\partial x}dx)\\
v_2 = (0,dy,\frac{\partial z}{\partial y}dy)
$$

2つのベクトル$v_1,v_2$が作る平行四辺形の面積は
The area of parallelogram is;

$$
dS = ||v_1||v_2|\sin{\theta}|
$$
$$
\cos{\theta} = \frac{(v_1,v_2)}{|v_1||v_2|}
$$
ただし、$(v_1,v_2)$は内積。
where $(v_1,v_2)$ is inner product.

したがって Therefore,

$$
dS = ||v_1||v_2|\sqrt{1-(\frac{(v_1,v_2)}{|v_1||v_2|})^2}|\\
= \sqrt{|v_1|^2|v_2|^2 - (v_1,v_2)^2}
$$

ここで Using 
$$
|v_1|^2 = (1+(\frac{\partial z}{\partial x})^2dx^2\\
|v_2|^2 = (1+(\frac{\partial z}{\partial y})^2dy^2\\
(v_1,v_2) =(\frac{\partial z}{\partial x}\frac{\partial z}{\partial y})^2dxdy
$$

以上より、Then,
$$
dS = \sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2} dxdy
$$

## Exercise 3

### Exercise 3-1 

2つの式を比較せよ。

Compare two formulas.

$$
\int_{0}^3 \sqrt{dx^2+dy^2} = \int_0^3 \sqrt{(\frac{dx}{dx})^2+(\frac{dy}{dx})^2}dx = \sqrt{1+(\frac{dy}{dx})^2}dx \\
$$

$$
dS = \sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2} dxdy
$$


## 曲面の前に、平面 Before using this to curved surface, apply it to the flat one.

$$
z = ax + by + c
$$

$$
\frac{\partial z}{\partial x}=a\\
\frac{\partial z}{\partial y}=b\\
dS = \sqrt{1+a^2+b^2}dxdy \\
\int_{x_0}^{x_1} \int_{y_0}^{y_1} dS dxdy = \sqrt{1+a^2+b^2} \times (x_1-x_0)(y_1-y_0)
$$
単なる、平行四辺形の面積。

Simple parallelogram's area.

## 単位球面積 Surface area of unit sphere.

$$
x^2+y^2+z^2=1\\
z = \pm \sqrt{1-(x^2+y^2)}
$$
$ z= \sqrt{1-(x^2+y^2)}$ , $0 \le x^2+y^2 \le 1;x,y \ge 0$を扱うことにする。

Calculate a part of the sphere of $ z= \sqrt{1-(x^2+y^2)}$ , $0 \le x^2+y^2 \le 1;x,y \ge 0$.

$$
dS = \sqrt{1+(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2} dxdy\\
\frac{\partial z}{\partial x}= \frac{1}{2}\frac{-2x}{\sqrt{1-(x^2+y^2)}}\\
\frac{\partial z}{\partial y} = \frac{1}{2}\frac{-2y}{\sqrt{1-(x^2+y^2)}}\\
$$

$$
dS = \sqrt{1+\frac{x^2}{1-(x^2+y^2)}+\frac{y^2}{1-(x^2+y^2)}}dxdy\\
= \sqrt{\frac{1}{1-(x^2+y^2)}}dxdy
$$

$$
S = \int_{x^2+y^2 \le 1} dS = \int_{x^2+y^2 \le 1} \sqrt{\frac{1}{1-(x^2+y^2)}}dxdy
$$

これは解けないが、うまくパラメタ表示を変えることで解けるようにできる。

This is not easy but converting to different parameterization, then it can  be.

$$
x = r \cos{\theta}\\
y = r \sin{\theta}\\
0 \le r \le 1\\
0 \le \theta \le 2\pi\\
$$

$(r\cos{\theta},r\sin{\theta}),((r+dr)\cos{\theta},(r+dr)\sin{\theta}),(r\cos{(\theta+d\theta)},r\sin{(\theta+d\theta)}),((r+dr)\cos{(\theta+d\theta)},(r+dr)\sin{(\theta+d\theta)})が作る微小面積は

Small area should be changed accordingly.

$$
dr \times (r \times d\theta)
$$
したがって$dxdy$に相当するのは$rdrd\theta$

Means $dxdy$ is converted to $rdrd\theta$.

ゆえに
$$
S = \int_{x^2+y^2 \le 1} \sqrt{\frac{1}{1-(x^2+y^2)}}dxdy\\
= \int_{0 \le r \le 1, 0\le \theta \le 2\pi} \sqrt{\frac{1}{1-r^2(\cos^2\theta + \sin^2{\theta})}}rdrd\theta\\
=\int_0^{2\pi} 1 d\theta \int_0^1 r\sqrt{\frac{1}{1-r^2}}dr
$$

$$
\frac{d}{dr} \sqrt{1-r^2} = \frac{1}{2}(-2r) \sqrt{\frac{1}{1-r^2}}\\
= -r\sqrt{\frac{1}{1-r^2}}
$$

したがって
$$
S = 2\pi
$$
これは球の上半分であるから、球面全体の表面積は、$4\pi$.

This is the upper half's area, therefore the are of unit sphere is $4\pi$.

## 曲面 Curverd surface


$$
z = \sin{x}+\cos{y}\\
x, y \in [-5,5]
$$
```{r,rgl=TRUE}
x <- y <- seq(from=-1,to=1,length=100) * 5
xy <- as.matrix(expand.grid(x,y))
z <- sin(xy[,1])+cos(xy[,2])
xlim <- max(xy[,1])-min(xy[,1])
ylim <- max(xy[,2])-min(xy[,2])
zlim <- max(z)-min(z)
plot3d(xy[,1],xy[,2],z,aspect=c(xlim,ylim,zlim))
```

$$
\frac{\partial z}{\partial x} = -\sin{x}\\
\frac{\partial z}{\partial y} = \cos{y}
$$


2つのベクトルは　The vecors are;
$$
(dx,0,-\sin{x}dx)\\
(0,dy,\cos{y}dy)
$$


したがって Then

$$
dS = ||v_1||v_2|\sqrt{1-(\frac{(v_1,v_2)}{|v_1||v_2|})^2}|\\
= \sqrt{|v_1|^2|v_2|^2 - (v_1,v_2)^2}
$$

ここで Using
$$
|v_1|^2 = (1+\sin^2{x})dx^2\\
|v_2|^2 = (1+\cos^2{y})dy^2\\
(v_1,v_2) = -\sin{x}\cos{y}dxdy
$$

以上より、Therefere
$$
dS = \sqrt{(1+\sin^2{x})(1+\cos^2{y})-\sin^2{x}\cos^2{y}} dxdy
= \sqrt{1+(\sin^2{x}+\cos^2{y})}dxdy
$$

解けないけれど、計算機的には計算できる。
Computationally calculable.


## Exercise 4

## Exercise 4-1

Mathematicaのサイトを使ってこの面積を計算せよ。

Calculate the area using Mathematica's site.