'''MIT 18.02 Multivariable Calculus, Fall 2007'''

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

[[TableOfContents]]

= Lec 1 =
https://www.youtube.com/watch?v=PxCxlsl_YwY

[[벡터,vector]] [[내적,inner_product]]과 [[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: $\vec{A}\cdot\vec{B}=0$
 $\Leftrightarrow\,\cos\theta=0$
 $\Leftrightarrow\,\theta=90\textdegree$
 $\Leftrightarrow\,\vec{A}\bot\vec{B}$

= Lec 2 =
https://www.youtube.com/watch?v=9FLItlbBUPY

Yesterday: [[스칼라곱,scalar_product,dot_product|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 [[방향,direction|direction]] $\hat{u}$ ([[단위벡터,unit_vector]])
 $=|\vec{A}| \cos(\theta)$
 $=|\vec{A}| |\hat{u}| \cos(\theta)$
 $=\vec{A}\cdot \hat{u}$

https://i.imgur.com/QwUuKBA.png
----
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 $\vec{A}{}'$
$\theta'=\frac{\pi}{2}-\theta$
$\cos(\theta')=\sin(\theta)$

https://i.imgur.com/WJUaq3ym.png

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

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

[[벡터곱,vector_product,cross_product|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?
https://i.imgur.com/qLHagcjl.png

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

https://i.imgur.com/OI5yE5al.png
https://i.imgur.com/oRStDaMl.png

Matrices ([[행렬,matrix]])
Often: linear relations([[linear_relation]] - [[선형관계,linear_relation]]) between variables
Ex: change of coordinate systems
https://i.imgur.com/PxvL5gYl.png
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

https://i.imgur.com/BGZ1yQFl.png
What AB represents: do [[변환,transformation|transformation]] B, __then__ transformation A.
(AB)X = A(BX)
(matrix product is [[결합법칙,associativity|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)

https://i.imgur.com/T1pQ96ml.png

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
is
 X = A^^−1^^B

AX = B
 ⇓
A^^−1^^(AX) = A^^−1^^B
 ⇓
X = A^^−1^^B

$A^{-1}=\frac1{\det(A)}\operatorname{adj}(A)$
adj는 adjoint matrix. (참조: [[WpEn: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 [[행렬식,determinant|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_system]]를 푸는 단계.
= Lec 4 =
https://www.youtube.com/watch?v=YBajUR3EFSM

평면의 방정식

[[평면,plane]]
 $ax+by+cz=d$
의 [[법선벡터,normal_vector]]는
 $\vec{N}=\langle a,b,c \rangle$

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

= Lec 5 =
https://www.youtube.com/watch?v=57jzPlxf4fk

[[사이클로이드,cycloid]]

= 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 

= Lec 7 =
https://www.youtube.com/watch?v=U1EcnfTKXJ0

기초 linear algebra

= 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-x^^2^^-y^^2^^

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]]

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

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

= 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,slope]]s of 2 [[접선,tangent_line]]s
$\text{If }\frac{\partial f}{\partial x}(x_0,y_0)=a\;\Rightarrow\;L_1=\begin{cases}z=z_0+a(x-x_0)\\y=y_0\end{cases}$
$\text{If }\frac{\partial f}{\partial y}(x_0,y_0)=b\;\Rightarrow\;L_2=\begin{cases}z=z_0+b(y-y_0)\\x=x_0\end{cases}$
$L_1,L_2$ 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 [[편미분,partial_derivative|Partial Derivatives]]
[[최적화,optimization|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,interpolation]]과 연결
Given experimental [[자료,data]]:
 (x,,1,,, y,,1,,), (x,,2,,, y,,2,,) … (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)

= Lec 10 =
https://www.youtube.com/watch?v=3_goGnJm5sA

Recall: [[임계점,critical_point]]s 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:
 1) $4ac-b^2<0$ ⇒ [[안장점,saddle_point]]
  one term ≥ 0, the other ≤ 0
 2) $4ac-b^2=0$
  Ex. $w=x^2$ (doesn't depend on $y$ at all)
 3) $4ac-b^2>0$
20m

= Lec 11 =