4. 편미분

4.1 서론

미분이란

(1)
\begin{align} y=f(x) \qquad {{d} \over {dx}} y = {d \over {dx}} f(x) = \mathrm{기울기} \end{align}

그런데 변수가 여러개일 때의 미분은?

(2)
\begin{equation} z = f(x, y) \end{equation}

e.g.

(3)
\begin{equation} z = f(x, y) = x^2 y - e^{xy} \end{equation}
(4)
\begin{align} f_x = \left( {{\partial f} \over {\partial x}} \right)_y = {{\partial z} \over {\partial x}} = 2xy - ye^{xy} \end{align}
(5)
\begin{align} f = x^2 - y^2 = r^2 \cos^2 \theta - r^2 \sin^2 \theta = r^2 - 2y^2 = 2x^2 - r^2 \end{align}
(6)
\begin{align} {{\partial^2 f} \over {\partial y \partial x}} = {\partial \over {\partial y}} \left( {{\partial f} \over {\partial x}} \right)_y = \left[ {\partial \over {\partial y}} (2 xy - y e^{xy} ) \right]_x = 2x - e^{xy} - xye^{xy} \end{align}
(7)
\begin{align} {{\partial^2 f} \over {\partial x \partial y}} = {{\partial^2 f} \over {\partial y \partial x}} \end{align}

4.2 이원 멱급수

(8)
\begin{align} f(x, y) & = (x-a)^n (y-b)^n \\ & \qquad \qquad a_{ab} \equiv f(a, b) \\ & = a_{00} + a_{10} (x-a) + a_{01} (y-b) + a_{00} (x-a)^2 + a_{11} (x-a)(x-b) + a_{02} (y-b)^2 + \cdots \\ {{\partial f} \over {\partial x}} & = a_{10} + 2 a_{20} (x-a) + a_{11} (x-b) + \cdots \\ & \qquad \qquad a_{10} = {\partial \over {\partial x}} f(a , b), \qquad a_{01} = {\partial \over {\partial y}} f(a, b) \\ {{\partial^2 f} \over {\partial x^2}} & = 2 a_{20} + \cdots + (x-a)^n (y-b)^n \\ & \qquad \qquad a_{20} = {1 \over 2} {{\partial^2 } \over {\partial x^2}} f(a, b) \end{align}
(9)
\begin{align} f(x, y) & = f(a, b) + \left[ f_x (a, b) (x-a) + f_y (a, b) (y-b) \right] + \left[ {1 \over 2} f_{xx} (a, b) (x-a)^2 + f_{xy} (a, b) (x-a)(y-b) + {1 \over 2} f_{yy} (a, b) (y-b)^2 \right] + \cdots \\ & = \left[ (x-a) {\partial \over {\partial x}} + (y-b) {\partial \over {\partial y}} \right] f(a, b) \qquad \qquad = \left[ (x-a) {{\partial } \over {\partial x}} + (y-b) {{\partial} \over {\partial y}} \right]^2 f(a, b) \\ & = \sum_{n=0}^\infty {1 \over {n!}} \left[ (x-a){\partial \over {\partial x}} + (y-b) {\partial \over {\partial y}} \right]^n f(a, b) \end{align}

4.3 전미분

(10)
\begin{align} y=f(x), \qquad {{dy} \over {dx}} = \lim_{\Delta x \rightarrow 0} {{\Delta y } \over {\Delta x}} \end{align}
wjsalqns1.png

접선의 기울기 = $dy / dx$

(11)
\begin{align} {{\Delta y} \over {\Delta x}} = {{dy } \over {dx}} + \epsilon \end{align}
(12)
\begin{align} z = f(x, y), \qquad dz = \left( {{\partial z} \over {\partial x}} \right) dx + \left( {{\partial z} \over {\partial y}} \right) dy \end{align}
wjsalqns2.png
(13)
\begin{align} f (x, y) & = 0 \\ E & = (x+dx, y+dy) \\ RE & = f(x+dx , y+dy) - f(x, y) = \Delta z \\ BE & = \left( {{dz } \over {dx} } \right)_y dx + \left( {{dz} \over {dy}} \right)_x dy = dz \\ \Delta z & = f(x+dx, y+dy) - f(x + dx, y) \implies & y\ 방향\ 변화\ \quad \left[ {\partial \over {\partial y}} f(x+dx, y) \right] dy \\ & \quad + f(x + dx, y) - f(x, y) \qquad \implies & x\ 방향\ 변화\ \quad \left[ {\partial \over {\partial x}} f(x, y) \right] dx \\ & \approx \left[ {{\partial} \over {\partial y}} f(x, y) \right] dy + \left[ {{\partial} \over {\partial x}} f(x, y) \right] dx + \epsilon \end{align}
(14)
\begin{align} u & = f(x, y, z, \cdots ) \\ du & = \left( {{\partial f} \over {\partial x}} \right) dx + \left( {{\partial f} \over {\partial y}} \right) dy + \left( {{\partial f} \over {\partial z}} \right) z + \cdots \end{align}

Chain rule

(15)
\begin{align} y & = \ln \sin 2x \\ {{dy} \over {dx}} & = {1 \over {\sin x}} {d \over {dx} } \sin x \\ & = {1 \over { \sin x}} \cos 2x {d \over {dx}} (ex) \\ & = 2 \end{align}
(16)
\begin{align} y = \ln u, \quad u = \sin v, \quad v = 2x \\ {{dy} \over {dx}} = {{dy} \over {du}} {{du} \over {dv}} {{dv} \over {dx}} \qquad \end{align}

e.g.

(17)
\begin{align} z & = xy = 2t^2 \sin t \\ & \qquad x = 2t^2, y = \sin t \\ z & = f(x, y) = f(x(t), y(t)) \\ dx & = y dx + x dy = \left( {{\partial z} \over {\partial x}} \right)_y dx + \left( {{\partial z} \over {\partial y}} \right)_x dy \end{align}

e.g.

(18)
\begin{align} x + e^x = t = x(t) \qquad \qquad \\ {{dx} \over {dt}} + e^xx {{dx} \over {dt}} = 1 \qquad \quad {{dx} \over {dt}} & = {1 \over {1 + e^x}} \\ {{d^2 x } \over {dt^2 }} + e^x \left( {{dx} \over {dt}} \right)^2 + e^x {{\partial^2 x } \over {d t^2}} = 0, \qquad {{d^2 x} \over {dt^2}} & = {{ - e^x} \over {1 + e^x}} \left( {{dx} \over {dt}} \right)^2 \end{align}

e.g.
$x^3 - 3y^3 + xy + 21 = 0$$(x, y) = (1, 2)$ 에서의 접선

(19)
\begin{align} 3 x^2 + 9y^2 {{dy} \over {dx}} + x {{dy } \over {dx}} + y & = 0 \\ 3x^2 +y + (x - 9y^2 ) {{dy} \over {dx}} & = 0 \end{align}

=z f(x, y) = f(x(s, t) , y(s,t))의 경우

(20)
\begin{align} z & = xy, \quad \begin{cases} x & = \sin (s + t) \longrightarrow dx & = \cos (s + t) (ds + dt) \\ y & = s - t \longrightarrow dz & = dx-dt \end{cases} \\ dz & = x dy + ydx \\ & = x (ds - dt) + y \left[ \cos (s+t) \right] (ds + dt) \\ & = \left[ x + y \cos (s+t) \right] ds + \left[ -x + y (s+t) \right] dt \\ & = \left( {{dz} \over {ds}} \right)_t ds + \left( {{dz} \over {dt}} \right)_s dt \end{align}
(21)
\begin{align} u & = u(x(s, t), y(s, t), z(s, t) ) \\ {{\partial u} \over {\partial t}} & = {{\partial u} \over {\partial x}} {{\partial x} \over {\partial t}} + {{\partial u} \over {\partial y}} {{\partial y } \over {\partial t}} + {{\partial u} \over {\partial z}} {{\partial z} \over {\partial t}} \end{align}

e.g.$\begin{cases} z & = x^2 + xy \\ x^2 + y^3 & = 5t +5 \\ x^3 - y^2 & = s^2 + t^2 \end{cases}$

(22)
\begin{align} dz = 2x dx + y dx + x dy & = (2x+y) dy + x\ dy \\ 2xdx + 3y^2 dy & = s\ dt + t\ ds \\ 3x^2 dx - 2y dy & = 2 s\ ds + 2t\ dt \\ \begin{pmatrix} 2x & 3y^2 \\ 3x^2 & -2y \end{pmatrix} \begin{pmatrix} dx \\ dy \end{pmatrix} & = \begin{pmatrix} s\ dt + t\ ds \\ 2s\ ds + 2t\ dt \end{pmatrix} \\ dx & = {{ \begin{vmatrix} s\ dt + t\ dx & 3y^2 \\ 2t\ ds + 2s\ dt & -2y \end{vmatrix} } \over { \begin{vmatrix} 2x & 3y^2 \\ 3x^2 & -2y \end{vmatrix} }} \end{align}
(23)
\begin{align} \left( {{\partial s} \over {\partial z}} \right)_x = ? \qquad \qquad \\ s = s(z, x), \quad y = y(z, x), \quad t = t(z, x) \\ z = x^2 + xy \qquad dz = 2x\ dx + y\ dz + x\ dy \\ x\ dy = - (2x + y) dx) + dz \\ dy = - {{ (2x + y} \over x} dx + {1 \over x} dz = \left( {{\partial y} \over {\partial x}} \right)_z dx + \left( {{\partial dy} \over {\partial dz}} \right)_x dz \end{align}

4.4 미소량을 이용한 어림

(24)
\begin{align} df \cong \Delta f = f(x + \Delta x) - f(x) \cong f' (x) \Delta x = f' (x) dx. \end{align}

예제 1: 다음의 어림값 구하기.

(25)
\begin{align} {1 \over \sqrt{0.25 - 10^{-20}}} - {1 \over \sqrt{0.25}} \end{align}
(26)
\begin{align} f(x) & = {1 \over \sqrt{x}} \\ \Delta f & = f(0.25 - 10^{-20} ) - f(0.25) \\ & \quad \begin{cases} x & = 0.25 \\ dx & = -10^{-20} \end{cases} \longrightarrow df = d {1 \over \sqrt{x}} \end{align}
(27)
\begin{align} d {1 \over \sqrt{x}} = -{1 \over 2} x^{-{3 \over 2}} dx = \left( - {1 \over 2} \right) (0.25)^{-{3 \over 2}} ( -10^{-20} ) = 4 \times 10^{-20} \end{align}

예제 2: $n$이 매우 클 때 다음을 증명하기

(28)
\begin{align} {1 \over n^2} - {1 \over {(n+1)^2}} \cong {2 \over n^3} \end{align}
(29)
\begin{align} f(x) & = {1 \over x^2} \\ \Delta f & = f(n) - f(n+1), \\ & \quad \begin{cases} x & = n \\ dx & = -1 \end{cases} \longrightarrow df = d {1 \over x^2} \end{align}
(30)
\begin{align} d \left( {1 \over x^2} \right) = - {2 \over x^3} dx = - {2 \over n^3} (-1) = {2 \over n^3} \end{align}

4.5 연쇄법칙과 함수의 함수 미분

예제 1: $y= \ln \sin 2x$

(31)
\begin{align} {{dy} \over {dx}} = {1 \over {\sin 2x}} \cdot {d \over {dx}} ( \sin 2x ) = {1 \over {\sin 2x}} \cdot \cos 2x \cdot {d \over {dx}} (2x) = 2 \cot 2x \end{align}
(32)
\begin{align} y = \ln u, \qquad u = \sin v, \qquad v = 2x \\ {{dy} \over {dx}} = {{dy} \over {du}} {{du} \over {dv}} {{dv} \over {dx}} \qquad \end{align}

예제 2: $z = 2t^2 \sin t$

(33)
\begin{align} z = xy, \qquad x = 2t^2, \qquad y= \sin t \end{align}
(34)
\begin{align} {{dz} \over {dt}} & = y {{dx} \over {dt}} + x{{dy} \over {dt}} \\ & = {{\partial z} \over {\partial x}} {{dx} \over {dt}} + {{\partial z} \over {\partial y}} {{dy} \over {dt}} \\ \Delta z & = {{\partial z} \over {\partial x}} \Delta x + {{\partial z} \over {\partial y}} \Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y \end{align}

여기서 $\Delta x, \Delta y$와 더불어 $\epsilon_1, \epsilon_2 \rightarrow 0$.

이 식을 $\Delta t$로 나누고 $\Delta t \rightarrow 0$ 접근하게 하면 식 (34)를 얻는다.

4.6 음함수 미분

예제 1: $x + e^x = t$

(35)
\begin{align} {{dx} \over {dt}} & + e^x {{dx} \over {dt}} = 1 \\ {{dx} \over {dt}} & = {1 \over {1 + e^x}} \end{align}

이때 미소량 방법을 사용하면 $dx + e^x dx = dt$, 양변을 $dt$로 나누기.

음함수 미분법을 사용하면 더 높은 차수의 미분도 구할 수 있다.

(36)
\begin{align} {{d^2 x} \over {dt^2}} & +e^x {{d^2 x} \over {dt^2}} + e^x \left( {{dx} \over {dt}} \right)^2 = 0 \\ {{d^2 x} \over {dt^2}} & = {{-e^x \left( {{dx} \over {dt}} \right)^2 } \over {1 + e^x}} = {{- e^x} \over { (1+e^x)^3 }} \end{align}

음함수 미분은 복잡한 식으로 표현되는 곡선의 기울기를 구할 때 좋다.

예제 2: $x^3 - 3 y^3 + xy + 21 = 0$의 점 $(1, 2)$에서의 기울기

(37)
\begin{align} 3x^2 - 9y^2 {{dy} \over {dx}} + x {{dy} \over {dx}} + y = 0, & \qquad \longleftarrow x = 1, y=2\ 대입 \\ 3 - 36 {{dy} \over {dx}} + {{dy} \over {dx}} +2 = 0, & \qquad {{dy} \over {dx}} = {5 \over 35} = {1 \over 7} \end{align}

접선에 대한 방정식은

(38)
\begin{align} {{y-2} \over {x-1}} = {1 \over 7}, \implies x - 7y + 13 = 0 \end{align}

4.7 더 많은 연쇄법칙

$z = f( x(s), y(t))$일 경우 $\partial z / \partial s, \partial z / \partial t$ 구하기

예제 1:

(39)
\begin{align} z = xy, \qquad x = \sin (s + t), \qquad y=s-t \end{align}
(40)
\begin{align} dz =y dx + x dy, \qquad dx = \cos (s + t) ( ds + dt), \qquad dy = ds - dt \end{align}
(41)
\begin{align} dz & = y \cos (s+t) (ds + dt) + x(ds - dt) \\ & = [ y \cos (s + t) + x ] ds + [y \cos (s + t) - x ] dt \end{align}
(42)
\begin{align} {{\partial z} \over {\partial t}} & = y \cos (s + t) - x \\ {{\partial z} \over {\partial s}} & = y \cos (s + t) - x \\ \therefore\ dz & = {{\partial z} \over {\partial s}} ds + {{\partial z} \over {\partial t}} dt \end{align}

예제 2:

(43)
\begin{align} u = x^2 + 2xy - y \ln z, \qquad x = s + t^2, \qquad y = s - t^2, \qquad z = 2t \end{align}
(44)
\begin{align} du & = 2x dx + 2x dy + 2y dx - {y \over z} dz - \ln z dy \\ & = (2x + 2y)(ds + 2t dt) + (2x - \ln z) (ds - 2t dt) - {y \over 2} (2 dt) \\ & = (4x + 2y - \ln z) ds + \left( 4yt + 2t \ln z - {{2y} \over z} \right) dt \end{align}
(45)
\begin{align} {{\partial u} \over {\partial s}} & = 4x + 2y - \ln z \\ {{\partial u } \over {\partial t}} & = 4yt + 2t \ln z - {{2y} \over z} \end{align}

한 개의 미분만을 원한다면 다른 하나를 0으로 두어 간단히 할 수 있다. 예컨대 $\partial u / \partial t$ 만을 원한다면 $ds = 0$,

(46)
\begin{align} du_s & = (2x + 2y) (2t dt) + (2x - \ln z) (-2t dt) - {y \over z} (2 dt) \\ & = \left( 4yt + 2t \ln z - {{2y} \over z} \right) dt \end{align}

첨자 $s$$s$를 상수로 고정시켰다는 의미.

미소량 대신 미분 연쇄법칙을 사용하면

(47)
\begin{align} {{\partial u} \over {\partial t}} & = {{\partial u} \over {\partial x}} {{\partial x} \over {\partial t}} + {{\partial u} \over {\partial y}} {{\partial y} \over {\partial t}} + {{\partial u} \over {\partial z}} {{\partial z} \over {\partial t}} \\ & = (2x + 2y) (2t) + (2x - \ln z) (-2t) + \left( - {y \over z} \right) 2 \\ & = 4yt +2t \ln z - {{2y } \over z} \end{align}

4.8 최대, 최소값 문제에서 편미분의 응용

예제 1: Dup tent

부피가 $V$로 주어진 천막을 재료를 최소로 사용해 제작하기

(48)
\begin{align} V & = \omega \times \omega \tan \theta \times l = \omega^2 \tan \theta = \mathrm{const.} \\ S & = {1 \over 2} \times 2 \omega \times \omega \tan \theta \times 2 + 2 \times l \times { \omega \over {\cos \theta }} \\ S & = S(u, \theta ) \\ & {{\partial S } \over {\partial \omega }} = {{\partial S } \over {\partial \theta }} = 0 \end{align}

4.9 제한조건이 있는 최대 최소문제; 라그랑주 곱수

$y = 1 - x^2$을 참조하는 실 위의 점과 원점의 거리 $d$의 최소값 찾기

(49)
\begin{align} d & = \sqrt{ x^2 + y^2 } \\ f (x, y) & = x^2 + y^2 \end{align}

방법 1:

(50)
\begin{align} f(x, y) & = x^2 + (1 - x^2) ^2 = x^4 - x^2 + 1 \\ {{\partial f} \over {\partial x}} & = 4 x^3 - 2 x = 4x \left( x^2 - {1 \over 2} \right) \implies x = 0, x = \pm {1 \over \sqrt{2}} \\ {{\partial^2 f} \over {\partial x^2}} & = 12 x^2 - 2 \implies \begin{cases} -2, & \quad x = 0 \quad \mathrm{(max)} \\ +4 & \quad x^2 = {1 \over 2} \quad \mathrm{(min)} \end{cases} \end{align}

라그랑주 곱수

$\phi (x , y) = \mathrm{const.}$인 조건이 있을 때 $f(x, y)$의 최대 최소값

(51)
\begin{align} f(x) \qquad {{df} \over {dx}} = 0 \longrightarrow & \begin{cases} df & = 0 = {{\partial f} \over {\partial x}} dx + {{\partial f} \over {\partial dy}} dy \\ d \phi & = 0 = {{ \partial \phi } \over {\partial x}} dx + {{\partial \phi } \over {\partial y}} dy \end{cases} \\ & d(f - \lambda \phi ) = \left[ {{\partial f} \over {\partial x}} + \lambda {{\partial \phi } \over {\partial x}} \right] dx + \left[ {{\partial f} \over {\partial y}} + \lambda {{\partial \phi } \over {\partial y}} \right] dy = 0 \end{align}
(52)
\begin{align} \begin{cases} {{\partial F} \over {\partial x }} = {{\partial f} \over {\partial x}} + \lambda {{ \partial \phi } \over {\partial x}} & = 0 \\ {{\partial F} \over {\partial y}} = {{\partial f} \over {\partial y}} +\lambda {{\partial \phi } \over {\partial y}} & = 0 \\ \qquad \qquad \phi (x, y) & = \mathrm{const.} \end{cases} \qquad (x, y): \mathrm{독립\ 변수}, \qquad F \equiv f + \lambda \phi \end{align}

방법 2: 라그랑주 곱수 적용

(53)
\begin{align} f(x, y) & = x^2 + y^2 \qquad \phi (x, y) = y+ x^2 = 1 \\ F & = f + \lambda \phi = x^2 + y^2 + \lambda (y + x^2 ) \\ {{\partial F} \over {\partial x}} & = 2x + 2 \lambda x = 2 \lambda (1 + \lambda ) = 0 \implies x = 0, \lambda = -1 \quad (\mathrm{i.e.} y = 1) \\ {{\partial F} \over {\partial y}} & = 2y + \lambda = 0 \implies y = - {1 \over 2} , x^2 = {1 \over 2} \end{align}

변수가 세 개일 때의 라그랑주 곱수
$f (x, y, z) \quad \mathrm{under} \quad \phi (x, y, z) = \mathrm{const.}$

(54)
\begin{align} df = 0 & = {{\partial f} \over {\partial x}} dx + {{\partial f} \over {\partial y}} dy + {{\partial f} \over {\partial z}} dz \\ d \phi = 0 & = {{\partial \phi } \over {\partial x}} dx + {{\partial \phi} \over {\partial y}} dy + {{\partial \phi} \over {\partial z}} dz \\ F & \equiv f + \lambda \phi \\ dF & = \left[ {{\partial f} \over {\partial x}} + \lambda {{\partial \phi} \over {\partial x}} \right] dx + \left[ {{\partial f} \over {\partial y}} + \lambda {{\partial \phi} \over {\partial y}} \right] dy + \left[ {{\partial f} \over {\partial z}} + \lambda {{\partial \phi} \over {\partial z}} \right] dz = 0 \end{align}

e.g.
타원체 ${x^2 \over a^2 } + {y^2 \over b^2 } + {z^2 \over c^2} = 1$ 안에 직육면체의 모서리가 $x, y, z$ 축과 평행하게 넣을 때, 직육면체의 최대 부피는? (직육면체와 타원체의 접점이 $( x,y, z)$)

(55)
\begin{align} V & = 8 xyz \\ F & \equiv 8xyz + \lambda \left( {x^2 \over a^2 } + {y^2 \over b^2 } + {z^2 \over c^2} \right) \\ {{\partial F}\over {\partial x}} & = 8yz + {{2x \lambda } \over a^2 } = 0 \\ {{\partial F}\over {\partial y}} & = 8xz + {{2y \lambda } \over b^2 } = 0 \\ {{\partial F}\over {\partial z}} & = 8xy + {{2z \lambda } \over c^2 } = 0 \\ & \longrightarrow 24 xyz + 2 \lambda \left( {x^2 \over a^2 } + {y^2 \over b^2 } + {z^2 \over c^2} \right) = 0 \\ & xyz = - {1 \over 12} \lambda, \\ & \lambda = -12 xyz \\ & 8yz + {{2x} \over a^2 } ( -12 xyz) = xyz \left( 1 - {{3x^2} \over a^2} \right) = 0 \\ & x^2 = {1 \over 3} a^2 \end{align}
(56)
\begin{align} \phi (x, y, z) & = \mathrm{const.} \\ d \phi = 0 & = {{\partial \phi } \over {\partial x}} dx + {{\partial \phi } \over {\partial y}} dy = {{\partial \phi } \over {\partial z}} dz = - ( \vec{\nabla} \phi ) d \vec{r} = 0 \\ & = \left( {{\partial \phi} \over {\partial x}}, {{\partial \phi } \over {\partial y}} , {{\partial \phi } \over {\partial z}} \right) \cdot (dx, dy, dz) = 0 \end{align}
(57)
\begin{align} d \vec{r} & = (dx, dy, dz) = \mathrm{in\ 접하는\ 평면} \\ \partial f & = ( \vec{\nabla} f) \cdot \vec{r} = 0 \\ & \vec{\nabla} f \perp d \vec{r} \implies \vec{\nabla} \phi \parallel \vec{\nabla} f \end{align}

조건이 여러 개일 때: 라그랑주 곱수를 여러 개 만들면 됨

(58)
\begin{align} f(x, y, z) & \\ \phi_1 (x, y, z) & = \mathrm{const.} \\ \phi_2 (x, y, z) & = \mathrm{const.} \\ F & = f + \lambda_1 phi_1 +\lambda_2 \phi_2 \\ dF & = \left( {{\partial f} \over {\partial z}} +\lambda_1 {{\partial \phi_1 } \over {\partial z}} + \lambda_2 {{\partial \phi_2} \over {\partial z}} \right) dz + \left( {{\partial f} \over {\partial x}} + \lambda_1 {{\partial \phi_1 } \over {\partial x}} \lambda_2 {{\partial \phi_2} \over {\partial x}} \right) dx + \\ & \qquad + \left( {{\partial f} \over {\partial y}} + \lambda_1 {{\partial \phi_1 } \over {\partial y}} + \lambda_2 {{\partial \phi_2} \over {\partial y}} \right) dy = 0 \end{align}

e.g. $xy = 6$$7x + 24z = 0$의 교점과 원점의 거리 최소값 찾기

(59)
\begin{align} f & = x^2 + y^2 + z^2 \\ F & = f + \lambda_1 2y + \lambda_2 ( 7x +24 z) \\ & {{\partial F} \over {\partial x}} = {{\partial F } \over {\partial y}} = {{\partial F} \over {\partial z}} = 0 \end{align}

4.10 경계점 문제

  1. $x$가 어떤 범위 안에서 제한되는 경우
  2. $f(x)$ 또는 $f'(x)$에 특이점이 존재할 경우
10-1.png 10-2.png
10-3.png 10-4.png

예제 3: 평면 $x^2 -4yz = 8$ 위의 점 중 원점에 가장 가까운 점 찾기

(60)
\begin{align} f(x, y, z) & = x^2 + y^2 + z^2, \qquad x^2 = 8 + 4yz \ge 0 \\ f(y,z) & = 8 + 4xy + y^2 + z^2 \end{align}
(61)
\begin{align} {{\partial f} \over {\partial y}} & = 4z + 2y = 0, \\ {{\partial f} \over {\partial z}} & = 4y + 2z = 0, \\ & y = z = 0, \quad x = \pm 2 \sqrt{2} \\ & (x, y, z) = (\pm 2 \sqrt{2} , 0, 0) \\ & zy = -2, \quad x= 0 \end{align}
(62)
\begin{align} f(y, z) & = y^2 + z^2 = y^2 + \left {{-2} \over y} \right)^2 = y^2 + {4 \over y^2} \\ {{\partial f} \over {\partial y}} & = 0 = 2y - i {1 \over y^3} = 0 \\ & y^4 = 4, \quad y = \pm \sqrt{2}, \quad z= \mp \sqrt{2} \\ & (x, y, z) = (0 , \pm \sqrt{2}, \mp \sqrt{2}) \end{align}
(63)
\begin{align} F & = x^2 + y^2 + z^2 + \lambda ( x^2 - 4yz) \\ {{\partial F} \over {\partial x}} & = 2x + 2 \lambda x = 2x (1 + \lambda ) = 0 \implies x = 0 \quad \mathrm{or} \quad \lambda = -1 \\ {{\partial F} \over {\partial y}} & = 2y - 4 \lambda z = 0 \implies y = 2 \lambda z \\ {{\partial F} \over {\partial z}} & = 2z - 4 \lambda y = 0 \implies y = {1 \over {2 \lambda }} z \end{align}

(i) $y=z=0, \quad x = \pm 2 \sqrt{2}$
(ii) $\lambda = \pm 1 /2 , \quad x = 0, \quad y = \pm \sqrt{2}, \quad z = \mp \sqrt{2}$

예제 4: $x^2 + y^2 = 1$일 때 $x, y > 0$에 대해 $y-x$의 최소값 구하기

(64)
\begin{align} y-x & = k, \\ y & = x +k \\ F & = y - x + \lambda ( x^2 + y^2) \\ {2 \over {4 \lambda^2 }} & = 1, \qquad \lambda^2 = {1 \over 2} {{\partial F} \over {\partial x}} & = -1 + 2 \lambda x = 0, \qquad x = {1 \over {2 \lambda}} = \pm {1 \over \sqrt{2}} \\ {{\partial F} \over {\partial y}} & = 1 + 2 \lambda y = 0, \qquad y = - {1 \over {2 \lambda }} = \mp {1 \over \sqrt{2}} \\ \end{align}

4.11 변수변환

라플라스 방정식을 극좌표로 쓰기:

(65)
\begin{align} {{\partial ^2 F} \over {\partial x^2 }} + {{\partial^2 F } \over {\partial y^2}} & = 0 \\ {1 \over r} {\partial \over {\partial r}} \left( r {{\partial F } \over {\partial r} } \right) +{1 \over r^2 } {{\partial^2 F} \over {\partial \theta^2}} & = 0 \end{align}
(66)
\begin{align} x & = r \cos \theta, \qquad y = r \sin \theta \\ & \quad F (x, y) = F(r, \theta) \\ r & = r(x, y), \qquad \theta = \theta (x, y) \end{align}
(67)
\begin{align} {{\partial} \over {\partial x}} F (r, \theta) & = {{\partial F } \over {\partial x}} + {{ \partial F} \over {\partial \theta}} {{\partial \theta } \over {\partial x}} \\ & = {{\partial F }\over {\partial r}} \cos \theta + {{\partial F \over {\partial \theta}} \left( - {{ \sin \theta } \over r} \right) \\ \left( {{\partial } \over {\partial x}} \right)^2 & = \left( \cos \theta {\partial \over {dr} } - {{ \sin \theta} \over r} {\partial \over {\partial \theta} \right) \left( \right) F \end{align}

르장드르 변환

(68)
\begin{align} f = f(x, y), \qquad g = f - qy \\ \end{align}
(69)
\begin{align} f(x, y) & = g(x, z) \\ & p = \left( {{\partial f} \over {\partial x}} \right)_y \qquad q = \left( {{\partial f} \over {\partial y}} \right)_x \end{align}
(70)
\begin{align} df & = {{\partial f} \over {\partial x}} dx +{{\partial f} \over {\partial y}} dy \\ & = p dx + q dy \\ dy & = df - ydq - qdy \\ & = p dx - ydq \end{align}
(71)
\begin{align} g & = g(x, q) \\ & p = \left( {{\partial f} \over {\partial x}} \right)_y \qquad -y = \left( {{\partial g} \over {\partial q}} \right)_x \end{align}
(72)
\begin{align} \left( {{\partial p} \over {\partial q}} \right)_x & = {{\partial^2 g} \over {\partial x \partial q}} \\ \left( {{ \partial y} \over {\partial x}} \right)_q & = - {{\partial^2 g} \over {\partial q \partial x}} \end{align}

4.12 적분의 미분; 라이프니츠 규칙

(73)
\begin{align} {d \over {dx}} \int_z^x f(t) dt & = {d \over {dx}} \left[ F(x) - F(a) \right] \\ & = {{dF} \over {dX}} = f(x) \\ \int_x^a f(t) dt & = - \int_a^x f(t) dt = - f(x) \end{align}
(74)
\begin{align} {d \over {dx}} \int_{u(x)}^{v(x)} f(t) dt = f(v) {{dv} \over {dx}} - f(u) {{du} \over {dx}} \end{align}

예제 3:

(75)
\begin{align} {d \over {dx}} \int_{x^2}^{ \sin^{-1} x } {{\sin t } \over t} dt & = {{ \sin (\sin^{-1} x ) } \over {\sin^{-1} x}} {d \over {dx}} \sin^{-1} x - {{ \sin (x^2) } \over x^2} {d \over {dx}} x^2 \\ & = {x \over { \sin^{-1} x}} {1 \over \sqrt{1 - x^2}} - {2 \over x} \sin (x^2) \\ & \left( \sin^{-1} x = y, \qquad x = \sin y \right) \end{align}

라이프니츠 규칙

(76)
\begin{align} {d \over {dx}} \int_{u(x)}^{v(x)} f(x,t) dt & = \int_u^v {{\partial f(x,t) } \over {\partial x}} dt + f(x, v) {{dv} \over {dx}} - f(x, u) {{du} \over {dx}} \end{align}
(77)
\begin{align} {d \over {dx}} \int_u^v f(x, t) dt = \int_u^v {{\partial f(x, t)} \over {\partial x}} dt \end{align}

예제 4:

(78)
\begin{align} \int_0^\infty dt t^{2n + 1} e^{- k t^2} & \\ \int_0^\infty dt t e^{- kt^2 } & = \int_0^\infty {1 \over 2} dx e^{-kx} = \left. {1 \over {2 (-k)}} e^{- kx} \right\rvert_0^\infty = {1 \over {2k}} \\ \int_0^\infty dt t^3 e^{-kt^2} & = - {{\partial } \over {\partial k}} \int_0^\infty dt t e^{-k t^2} = {1 \over {2 k^2}} \\ \therefore\ \int_0^\infty dt t^{2n + 1} e^{- k t^2} & = {{n!} \over {2 k^{n+1} }}, \quad n \in \mathbb{Z} \end{align}

예제 5:

(79)
\begin{align} {d \over {da}} \int_0^1 {{t^n -1 } \over {\ln t}} dt, & \qquad a > -1 \\ \quad t^a = e^{a \ln t}, & \qquad {d \over {da}} t^a = \ln t e^{-a \ln t} = t^a \ln t \\ {d \over {da}} \int_0^1 {{t^n -1 } \over {\ln t}} dt & = \int_0^1 dt {{ {\partial \over {\partial a}} (t^a) } \over { \ln t}} = \int_0^1 dt\ t^a = {1 \over a+1} \end{align}