MIT 18.02 Multivariable Calculus, Fall 2007

Up: OCW,OpenCourseWare,MOOC,MassiveOpenOnlineCourse, 미적분,calculus > 다변수미적분,multivariable_calculus

1. Lec 1
2. Lec 2
3. Lec 3
4. Lec 4
5. Lec 5
6. Lec 6
7. Lec 7
8. Lec 8
9. Lec 9
10. Lec 10
11. Lec 11

1. Lec 1 ¶

https://www.youtube.com/watch?v=PxCxlsl_YwY

벡터,vector 내적,inner_product과 코사인법칙,cosines_law
$\vec{A},\vec{B}$ 사이의 각이 θ이고
$\vec{C}=\vec{A}-\vec{B}$ 일 때,

$|\vec{C}|^2=|\vec{A}|^2+|\vec{B}|^2-2|\vec{A}| |\vec{B}| \cos\theta$

pf.
$|\vec{C}|^2=\vec{C}\cdot\vec{C}=(\vec{A}-\vec{B})\cdot(\vec{A}-\vec{B})$

$=\vec{A}\cdot\vec{A}-\vec{A}\cdot\vec{B}-\vec{B}\cdot\vec{A}+\vec{B}\cdot\vec{B}$
$=|\vec{A}|^2+|\vec{B}|^2-2\vec{A}\cdot\vec{B}$

Sign of $\vec{A}\cdot\vec{B}$ :

＞ 0 if θ < 90° (예각)
＝ 0 if θ = 90° (직각) (A⊥B) (직교성,orthogonality)
＜ 0 if θ > 90° (둔각)

평면,plane 방정식

$x+2y+3z=0$

은, (계수를 보면)

$\vec{OP}=\langle x,y,z \rangle$ 와 $\vec{A}=\langle 1,2,3 \rangle$

을 내적한

$\vec{OP}\cdot\vec{A}=0$ 즉 $\vec{OP}\bot\vec{A}$

꼴이다.

Remember: $\vec{A}\cdot\vec{B}=0$

$\Leftrightarrow\,\cos\theta=0$
$\Leftrightarrow\,\theta=90\textdegree$
$\Leftrightarrow\,\vec{A}\bot\vec{B}$

[edit]

2. Lec 2 ¶

https://www.youtube.com/watch?v=9FLItlbBUPY

Yesterday: dot product

$\vec{A}\cdot\vec{B}=\sum a_ib_i = |\vec{A}| |\vec{B}| \cos(\theta)$

i.e.

$\cos(\theta)=\frac{\vec{A}\cdot\vec{B}}{|\vec{A}| |\vec{B}|}$

Component of $\vec{A}$ along direction $\hat{u}$ (단위벡터,unit_vector)

$=|\vec{A}| \cos(\theta)$
$=|\vec{A}| |\hat{u}| \cos(\theta)$
$=\vec{A}\cdot \hat{u}$

x방향으로 $\vec{A}$ 가 있고, 1사분면 적당한 방향으로 $\vec{B}$ 가 있고, y축 방향으로 $\vec{A}{}'$ 가 있는 그림. A, B 사잇각은 θ이고, B와 A' 사잇각은 θ'임.
$\vec{A}$ 를 90°만큼 돌린 것은 $\vec{A}{}'$
$\theta'=\frac{\pi}{2}-\theta$
$\cos(\theta')=\sin(\theta)$

$\vec{A}=\langle a_1,a_2 \rangle$ 이면
$\vec{A}{}'=\langle -a_2,a_1 \rangle$ 이다.

$|\vec{A}|\cdot|\vec{B}|\sin(\theta)$
$=|\vec{A}{}'| |\vec{B}| \cos(\theta')$
$=\vec{A}{}' \cdot \vec{B}$
$=\langle -a_2,a_1 \rangle \cdot \langle b_1, b_2 \rangle$
$=a_1b_2-a_2b_1$
$=\det(\vec{A},\vec{B})$
$=\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix}$
= determinant of $\vec{A}$ and $\vec{B}$
= 같은 점에서 시작하는 $\vec{A}$ 와 $\vec{B}$ 가 이루는 평행사변형,parallelogram의 면적

사인법칙 얘기
±(A와 B사이의 평행사변형의 넓이) =

$|\vec{A}| |\vec{B}|\sin\theta=\det(\vec{A},\vec{B})$

±(삼각형의 넓이) =

$\frac12|\vec{A}| |\vec{B}|\sin\theta=\frac12\det(\vec{A},\vec{B})$

따라서,

면적 = 행렬식,determinant의 절대값
area = absolute value of determinant

Determinant in space: 세 벡터.
$\det(\vec{A},\vec{B},\vec{C})=\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}$

$=a_1\begin{vmatrix}b_2&b_3\\c_2&c_3\end{vmatrix}-a_2\begin{vmatrix}b_1&b_3\\c_1&c_3\end{vmatrix}+a_3\begin{vmatrix}b_1&b_2\\c_1&c_2\end{vmatrix}$

Geometrically,

$\det(\vec{A},\vec{B},\vec{C})$
= ±(volume of 평행육면체,parallelepiped)

Cross-product of 2 vectors in 3-space
Def.

$\vec{A}\times\vec{B}$ (is a vector)
$=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}$
$=\begin{vmatrix}a_2&a_3\\b_2&b_3\end{vmatrix}\hat{i}-\begin{vmatrix}a_1&a_3\\b_1&b_3\end{vmatrix}\hat{j}+\begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix}\hat{k}$

Thm.

$\bullet\;|\vec{A}\times\vec{B}|=$ area of parallelogram
$\bullet\;\operatorname{dir}(\vec{A}\times\vec{B})=$ ⊥(parallel) to the plane of the parallelogram, with right-hand rule

Another look at volume
$\det(\vec{A},\vec{B},\vec{C})=\vec{A}\cdot(\vec{B}\times\vec{C})$ 이다. 이유는?

[edit]

3. Lec 3 ¶

https://www.youtube.com/watch?v=bHdzkFrgRcA

Matrices (행렬,matrix)
Often: linear relations(linear_relation - 선형관계,linear_relation) between variables
Ex: change of coordinate systems

Entries in matrix product AX: dot products between rows of A and columns of X
A: 3×3 matrix
X: column vector ↔ 3×1 matrix

What AB represents: do transformation B, then transformation A.
(AB)X = A(BX)
(matrix product is associative)
Note: AB ≠ BA

항등행렬,identity_matrix(curr. 단위행렬,unit_matrix) I
IX=X
$I_{3\times 3}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatirx}$
일반적으로
$I_{n\times n}=$ (diagonal elements는 모두 1, 나머지는 모두 0인 행렬)

Inverse matrix (역행렬,inverse_matrix)
Inverse of A: matrix M such that

AM = I
MA = I

Need: A square matrix n×n

M = A⁻¹

Solution to

AX = B

X = A⁻¹B

AX = B

⇓

A⁻¹(AX) = A⁻¹B

⇓

X = A⁻¹B

$A^{-1}=\frac1{\det(A)}\operatorname{adj}(A)$
adj는 adjoint matrix. (참조:

Adjugate_matrix)

TODO: adjoint adjugate adjunct 차이?

Steps: (on a 3×3 example)
$A=\begin{bmatrix}2&3&3\\2&4&5\\1&1&2\end{bmatrix}$

① Minors:
$\begin{bmatrix}3&-1&-2\\3&1&-1\\3&4&2\end{bmatrix}$
(예를 들어 위 행렬의 (1,1)의 3은 $\begin{vmatrix}4&5\\1&2\end{vmatrix}$ 로 계산한 것.)

② Cofactors: (여인수,cofactor)
flip signs in checkerboard
＋－＋
－＋－
＋－＋ (＋ : leave alone, － : flip sign)
$\begin{bmatrix}3&1&-2\\-3&1&1\\3&-4&2\end{bmatrix}$

③ Transpose: switch rows and columns (전치,transposition or 전치,transpose... curr 전치행렬,transpose_matrix)
$\begin{bmatrix}3&-3&3\\1&1&-4\\-2&1&2\end{bmatrix} \;\;\leftarrow\text{adj}(A)$

④ Divide by determinant of A
$|A|=\begin{vmatrix}2&3&3\\2&4&5\\1&1&2\end{vmatrix}=3$ 이므로
$A^{-1}=\frac13\begin{bmatrix}3&-3&3\\1&1&-4\\-2&1&2\end{bmatrix}$

이것이 선형계,linear_system를 푸는 단계.

[edit]

4. Lec 4 ¶

https://www.youtube.com/watch?v=YBajUR3EFSM

평면의 방정식

평면,plane

$ax+by+cz=d$

의 법선벡터,normal_vector는

$\vec{N}=\langle a,b,c \rangle$

1) 원점을 지나고 법선벡터 $\vec{N}=\langle 1,5,10 \rangle$ 인 평면은?
평면 위 점 $P=(x,y,z)$ 라 하면 (P is in plane)

$\Leftrightarrow\; \vec{OP}\cdot\vec{N}=0$
$\Leftrightarrow\; x+5y+10z=0$

2) Plane through $P_0(2,1,-1)$ and $\bot\vec{N}=\langle 1,5,10\rangle$
P is in plane

$\Leftrightarrow\; \vec{P_0P}\cdot\vec{N}=0$
$\Leftrightarrow\; \langle x-2,y-1,z+1\rangle \cdot \langle 1,5,10 \rangle=0$
$\Leftrightarrow\; (x-2)+5(y-1)+10(z+1)=0$
$\Leftrightarrow\; x+5y+10z=-3$

[edit]

5. Lec 5 ¶

https://www.youtube.com/watch?v=57jzPlxf4fk

사이클로이드,cycloid

[edit]

6. Lec 6 ¶

https://www.youtube.com/watch?v=0D4BbCa4gHo

Cycloid (wheel radius 1, at unit speed):

$\vec{r}(t)=\langle t-\sin t,1-\cos t\rangle$

Velocity vector:

$\vec{v}=\frac{d\vec{r}}{dt}=\langle\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\rangle$

$=\langle 1-\cos t,\sin t\rangle$

Speed(scalar):

$|\vec{v}|=\sqrt{(1-\cos t)^2+\sin^2 t}$

$=\sqrt{1-2\cos t+\cos^2t+\sin^2 t}$
$=\sqrt{2-2\cos t}$

Acceleration(vector):

$\vec{a}=\frac{d\vec{v}}{dt}$

e.g. cycloid:

$\vec{a}=\langle\sin t,\cos t\rangle$

단위접벡터,unit_tangent_vector (See also 곡률,curvature)

$\vec{v}=\frac{d\vec{r}}{dt}=\frac{d\vec{r}}{ds}\cdot\frac{ds}{dt}=\hat{T}\frac{ds}{dt}=\hat{T}|\vec{v}|$
Velocity has

direction: tangent to trajectory $\hat{T}$
length: speed ds/dt

[edit]

7. Lec 7 ¶

https://www.youtube.com/watch?v=U1EcnfTKXJ0

기초 linear algebra

[edit]

8. Lec 8 ¶

https://www.youtube.com/watch?v=dK3NEf13nPc

Function of 1 variable:

f(x)
ex. sin x

Function of 2 variables:

f(x, y)
ex. x² + y²
ex2. f(x,y)=1-x²-y²

contour_plot 얘기

Approximation formula - 근사,approximation

$f(x)\approx f(x_0)+f'(x_0)(x-x_0)$

선형근사,linear_approximation

편미분,partial_derivative

$\frac{\partial f}{\partial x}(x_0,y_0)=\lim_{\small\Delta x\to 0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}$

$\frac{\partial f}{\partial y}(x_0,y_0)=\lim_{\small\Delta y\to 0}\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}$

[edit]

9. Lec 9 ¶

https://www.youtube.com/watch?v=UYe98CcxPbs

Approximation formula:
If we change

$x\to x+\Delta x$
$y\to y+\Delta y$
$z=f(x,y)$ then
$\Delta z\approx f_x\Delta x+f_y \Delta y$

이 공식을 정당화하기 위해: 접평면,tangent_plane to $z=f(x,y)$ 를 생각
Know: $f_x,f_y$ are 기울기,slopes of 2 접선,tangent_lines
$\text{If }\frac{\partial f}{\partial x}(x_0,y_0)=a\;\Rightarrow\;L_1=\begin{cases}z=z_0+a(x-x_0)\\y=y_0\end{cases}$
$\text{If }\frac{\partial f}{\partial y}(x_0,y_0)=b\;\Rightarrow\;L_2=\begin{cases}z=z_0+b(y-y_0)\\x=x_0\end{cases}$
$L_1,L_2$ are both tangent to the graph $z=f(x,y).$
Together they determine a 평면,plane:

$z=z_0+a(x-x_0)+b(y-y_0)$

Approximation formula says: graph of $f$ is close to its 접평면,tangent_plane.

Application of Partial Derivatives
Optimization Problems
⤳ find min/max(최대최소) of a function $f(x,y)$

At a local min. or max(극소, 극대. see 극값,extremum),

$f_x=0\text{ and }f_y=0$
⇔ 접평면,tangent_plane to graph $z=f(x,y)$ is horizontal!

Def. $(x_0,y_0)$ is 임계점,critical_point of $f$ if $f_x(x_0,y_0)=0\text{ and }f_y(x_0,y_0)=0$

Ex. $f(x,y)=x^2-2xy+3y^2+2x-2y$
$f_x=2x-2y+2=0$
$f_y=-2x+6y-2=0$
더하면 $4y=0,y=0,x=-1$ 이렇게 1 critical point $(x,y)=(-1,0)$

Least-square interpolation //나중에 제곱,square>최소제곱,least_square 내삽,interpolation과 연결
Given experimental 자료,data:

(x₁, y₁), (x₂, y₂) … (x_n, y_n)

Find best fit line - best $a$ and $b$
minimizing total square deviation

$y = ax + b$

Deviation for each data point: (편차,deviation)

$y_i-(ax_i+b)$

Minimize

$D(a,b)=\sum_{i=1}^{n}\left[y_i-(ax_i+b)\right]^2$

편미분하면,

$\frac{\partial D}{\partial a}=\sum_{i=1}^{n}2(y_i-(ax_i+b))\cdot(-x_i)=0$
$\frac{\partial D}{\partial b}=\sum_{i=1}^{n}2(y_i-(ax_i+b))\cdot(-1)=0$

$\begin{cases}\sum_{i=1}^{n}(x_i^2a+x_ib-x_iy_i)=0\\ \sum_{i=1}^{n}(x_ia+b-y_i)=0\end{cases}$

$\begin{cases}\left(\sum_{i=1}^{n}x_i^2\right)a+\left(\sum_{i=1}^{n}x_i\right)b=\sum_{i=1}^{n}x_iy_i\\\left(\sum_{i=1}^{n}x_i\right)a+nb=\sum_{i=1}^{n}y_i\end{cases}$
2×2 linear system, solve for (a, b)

[edit]

10. Lec 10 ¶

https://www.youtube.com/watch?v=3_goGnJm5sA

Recall: 임계점,critical_points of $f(x,y):$ where $f_x=0\text{ and }f_y=0$

Question: how do we decide between

local min (극소)
local max (극대)
saddle?

Question: how do we find global min/max?
↳ there occur either at a critical point, or on boundary / at infinity (??? 뜻 정확히)

Second derivative test // derivative_test, curr goto 판정법,test#s-2.3
First consider $w=ax^2+bxy+cy^2$
Example: $w=x^2+2xy+3y^2=(x+y)^2+2y^2$

In general, if $a\ne0$

$w=a\left( x^2 + \frac{b}{a}xy \right) + cy^2$
$=a\left(x + \frac{b}{2a}y \right)^2 + \left( c-\frac{b^2}{4a} \right) y^2$
$=\frac{1}{4a}\left[ 4a^2\left( x+\frac{b}{2a}y \right)^2 + (4ac-b^2)y^2 \right]$ - sum of two squares

3 cases: