Transcript Chapter 4 Partial differentiation

Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 4 Partial differentiation (편미분)
Lecture 12 Introduction of partial differentiation
1. Introduction
- Differentiation
y  f  x   y 
dy
dx
: Slope of the curve y  f x  or rate of y with respect to x
ex.
- Time rates such as velocity, acceleration, and rate of cooling of a hot body.
- Rate of change of volume of a gas with applied pressure (P, V)
- Rate of decrease of fuel in your car tank with distance travelled (l, Q)
- Differential equation
- Finding Max. or Min.
- Partial differentiation
z  f x, y 
When we want to find the slope of z with respect to y, keeping x constant,
we can use the partial differentiation.
z
y x const .
(or
z
with respect tox with constanty)
x y const .
z
z
y x  const .
y
x
z
x
y  const .
- High order partial derivatives
 z  2 z

,
x x x 2
 z  2 z

,
x y xy
cf. Otherexpressions,
 2 z
3 z

,
x xy x 2y
z x , f x , f1
z  f x, y   x3 y  e xy
Example
f z

 f x  z x  f1  3x 2 y  ye xy ,
x x
f z

 f y  z y  f 2  x3  xexy ,
y y
2 f
2z

 f yx  z yx  f 21  3 x 2  e xy  xyexy ,
xy xy

etc.
(Caution)
x  r cos ,
y  r sin 
z  x2  y2 ,
z  r 2 cos2   r 2 sin 2  ,
 z   z   z 
  ,   ,   ?
 r   r  x  r  y
 z 
2
2
   2r cos   sin  ,
 r 
z  2x2  x2  y 2  2x2  r 2 ,
z  x2  y2  2 y 2  r 2  2 y2 ,


 z 
   2r
 r  x
 z 
   2r
 r  y
The symbol z / r x is usually read “the partial of z with respect to r, with x
held constant”. However, the important point to understand is that the notation
means that z has been written as a function of the variables r and x only, and then
differentiated with respect to r.
2. Power series in two variables (두 변에 대한 멱급수)
Example 1.



x3
y2
x3 xy2
f x, y   sin x cos y   x   1 
   x  
 .
3!
2!
3! 2!



Example 2
2
3

x  y  x  y 
ln1  x  y   x  y  


2
3
x2
y 2 x3
y3
2
2
 x  y   xy 
  x y  xy 

2
2
3
3
- General expression
First, express the function with the power series and then determine the coefficients.
f ( x, y)  a00  a10 ( x  a)  a01 ( y  b)  a20 ( x  a)2  a11 ( x  a)( y  b)
 a02 ( y  b)2  a30 ( x  a)3  a21 ( x  a)2 ( y  b)
 a12 ( x  a)( y  b)2  a03 ( y  b)3  
Finding partial derivatives,
f x  a10  2a20 ( x  a)  a11 ( y  b)  
f y  a01  a11 ( x  a)  2a02 ( y  b)  
f xx  2a20  termscontaining( x  a) and/or ( y  b)
f xy  a11  termscontaining( x  a) and/or ( y  b)
f (a, b)  a00 ,
f xx (a, b)  2a20 ,
f x (a, b)  a10 ,
f xy (a, b)  a11 ,
f y (a, b)  a01 ,
etc.
Then,
f ( x, y)  f (a, b)  f x (a, b)(x  a)  f y (a, b)( y  b)

1
[ f xx (a, b)(x  a)2  2 f xy (a, b)(x  a)( y  b)  f yy (a, b)( y  b)2 ]
2!
Using a simpler form, x – a = h and y – b = k.
1
2
2
[
f
(
a
,
b
)
h

2
f
(
a
,
b
)
hk

f
(
a
,
b
)
k
]
xy
yy
second-order terms, 2! xx
2
1 
 
 h  k  f (a, b)
2!  x
y 
similarly, third-order terms
3
1 
 
1
 h  k  f (a, b)  [h 3 f xxx (a, b)  3h 2 k f xxy (a, b)  ]
3!  x
y 
3!
Finally,
n
1 
 

f ( x, y )    h  k  f (a, b)
y 
n  0 n!  x

3. Total differentials (전 미분)
- Single variables
dy
y
 lim
dx x  0 x
dy d
y' 

f ( x)  dy  ydx
dx dx
- Two variables and more
z  f x, y   dz 
z
z
dx  dy
x
y
u  f x, y, z,  du 
f
f
f
dx  dy  dz  
x
y
z
4. Approximations using differentials (미소량을 이용한 어림)
Example 1.
f x  
1
0.25  10 20
1
x

1
0.25


f  f 0.25  10 20  f 0.25  f 0.25  x  f 0.25  10 20

 1
 1
f x      x  3 / 2  f 0.25    0.25 3 / 2
 2
 2

1
0.25  10 20

1
 4  10 20.
0.25
- Example 2.
f x  
1
1
2


n2 n  12 n3
1
2













f

f
n

f
n

1

f
n

x

f
n

1

x2
n3
2

 f x    3 
n 

- Example 3. reduced mass
 1  m11  m21
If m_1 is increased by 1%, what fractional change in m_2 leaves  unchanged?
dm1  0.01m1
1


1
1

    2 d  m1 2 dm1  m2 2 dm2
m1 m2
0  m1 2 dm1  m2 2 dm2   unchanged
dm2
dm1
0.01m1




m22
m12
m12
or
ex. m1  m2  m2  0.01m2 ,
dm2
 0.01m2 / m1.
m2
3m1  m2  m2  0.03m2 .
cf . yx   ydx
Example 4.
R
kl
r2
Relative error rate: 5 % in the length measurement
and 10 % for the radius measurement
dl / l  0.05,
dr / r  0.1
ln R  ln k  ln l  2 ln r
dR dl
dr
dR dl
dr
 2
 largest

2
 0.05  20.1  0.25.
R
l
r
R
l
r
cf . yx   ydx
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 4 Partial differentiation
Lecture 13 Chain rule
5. Chain rule or differentiating a function of a function
(연쇄법칙과 함수의 함수 미분하기)
Example 1.
y  lnsin 2 x  
dy
?
dx
dy
1
d
1
d

 sin 2 x  
 cos 2 x  2 x   2 cot 2 x.
dx sin 2 x dx
sin 2 x
dx
y  ln u,
where u  sin v &
dy dy du dv

dx du dv dx
v  2 x.
‘chain rule’
Example 2.
z  2t 2 sin t 
dz
?
dt
dz
 4t sin t  2t 2 cos t
dt
z  xy,
where x  2t 2 & y  sin t ,
dz
dx
dy
y x
dt
dt
dt
dz z dx z dy 
z
z 
 y  , x  



dt x dt y dt 
x
y 
dz 
z
z
dx  dy
x
y
Example 3.
z  x y , where
y  tan 1 t , x  sin t ,
dz
?
dt
dz z dx z dy
1


 yx y 1  cost  x y ln x 
.
dt x dt y dt
1 t2
6. Implicit differentiation (음함수 미분)
Example 1.
dx d 2 x
xe t 
,
?
dt dt2
x
We realized that x is a function and just differentiate each term of the equation
with respect to t (implicit differentiation).
dx
dx
dx
1
 ex
1 

dt
dt
dt 1  e x
cf . dx  e x dx  dt 
dx
 ex  1
dt
2
2
dx x dx
d 2x
differentiating again
x d x
x  dx 
e
 1    

e

e
  0
dt
dt
dt2
dt2
 dt 

d 2 x  e x  dx
 ex
dt


.
x 3
dt2
1  ex
1 e
2


This problem is even easier if we want only the numerical values of the
derivatives at a point.
dx
dx
1st )
 1
 1 or
dt
dt
dx 1
 ,
dt 2
d 2x
d 2x
1
2nd)

1


1

   0.
2
2
dt
dt
2
7. More chain rule (더 많은 연쇄법칙)
Example 1.
z  xy , x  sin s  t , y  s  t ,
z z
,
?
s t
dz  ydx  xdy, dx  coss  t ds  dt, dy  ds  dt.
dz  y cos(s  t )(ds  dt)  x(ds  dt)
 [ y cos(s  t )  x]ds  [ y cos(s  t )  x]dt
For ds  0 ( s  const.),
For dt  0 (t  const.),
z
 y coss  t   x
t
z
 y coss  t   x.
s
Example 2.
u  x 2  2 xy  y ln z , x  s  t 2 , y  s  t 2 , z  2t ,
u u
,
?
s t
- Using the differentials,
du  2 xdx  2 xdy  2 ydx 
y
dz  ln zdy
z
 2 x  2 y ds  2tdt  2 x  ln z ds  2tdt 
y
2dt
z
2y 

 4 x  2 y  ln z ds   4 yt  2t ln z 
dt.
z


u
 4 x  2 y  ln z ,
s
u
2y
 4 yt  2t ln z  .
t
z
cf. dus ds  0
 2 x  2 y 2tdt  2 x  ln z  2tdt 
y
2dt   4 yt  2t ln z  2 y dt.
z
z 

- Using the derivatives,
u u x u y u z



t x t y t z t
 

 
 

  x 2  2 xy  y ln z  s  t 2    x 2  2 xy  y ln z  s  t 2 
 x
 t
  y

 t

  
  x 2  2 xy  y ln z  2t 
 z
 t 
2y
 y
 2 x  2 y 2t   2 x  ln z  2t     2  4 yt  2t ln z  .
z
 z
cf. Using the matrix form,
 u

 s
u   u
  
t   x
u
y
u 

z 
 x

 s
 y

 s
 z

 s
x 

t 
y 

t 
z 

t 
Example 3.
z  x  y, x2  y 2  t 2 , x sin t  yey ,
dz / dt  ?
dz  dx  dy
2 xdx  2 ydy  2tdt,

sin tdx  x costdt  ye y  e y dy


 xdx  ydy  tdt,
 x

 

y
sin tdx   y  1e dy   x costdt
 sin t

 dx   t



dt


y 


  y  1e  dy    x cost 
y
tdt
y
 x costdt   y  1e y  t  y  1e y  xy cost
dx 

dt,
y
x
y
 x y  1e  y sin t
sin t   y  1e y
similarly for dy.
‘A computer may save us some time with the algebra.’
Example 4.
z  x2  xy, x2  y3  st  5, x3  y 2  s 2  t 2 ,
z / s, z / t  ?
dz  2 xdx  xdy  ydx
2 xdx  3 y 2 dy  sdt  tds
3x 2 dx  2 ydy  2sds  2tdt.
dx 
sdt  tds
3y2
2sds  2tdt  2 y
2x
3x 2
2
3y
 2y
 2 ys  6ty dt   2 yt  6sy ds ,

In thisway, dzds, dt 
2
 4 xy  9 x 2 y 2
z z
,
s t
2
similarly for dy.
Let’s skip Example 5.
Example 6. Rectangular vs. polar coordinates. (reciprocal)
x  r cos
r  x2  y2
y  r sin 
  t an1
   1
i)
  tan
x x 
ii)
y
x
constant y
y
 y / x2
y




x  1  ( y 2 / x2 )
r2
x


r cos   r sin    y
 
i) and ii)-1 are different!!
constant r

y
y
r2
 x 
2
 y cot   y( csc  )  2  2 2  
  
sin  y / r
y
   y 
This is a general rule: partial derivatives u / v & v / u are not usually
reciprocals; they are reciprocals if the other independent variables (besides u or v)
are the same in both cases.
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 4 Partial differentiation
Lecture 14 Max. & Min.
26/15
8. Application of partial differentiation to maximum and minimum problems
(최대, 최소값 문제에서 편미분의 응용)
- dy/dx=0 is a sufficient condition for max. or min. of f(x).
min. (concave)
max. (convex)
(d2y/dx2 > 0)
(d2y/dx2 < 0)
inflection
(d2y/dx2 = 0)
- To minimize z = f(x,y),
z
z
0 &
 0.
x
y
cf. saddle point
27/15
Example. A pup tent of given volume, V, with ends but no floor, is to be made
using the least possible material. find the proportions.
V
1
 2w  l  w tan  w2l tan  fixed
2

 only twovariablesof ( w, l , ) are independent.
Material area A  2w2 tan 
 2w2 tan 
2w
l
2lw
cos
2w  V
2V
V



2
csc .  l  2
 2
  2w tan 

cos  w tan 
w
w
tan



To minimize A,
A
2V csc 
 4w tan  
 0,
2
w
w
V csc V csc cot
w 

2 tan
sec2 
3
or
A
2V
 2w2 sec 2  
csc  cot   0.

w
cos
cos cos2 

.
2 sin 2 
sin 2 
sin   0, cos  0  cos2   1 / 2,   45  V  w2l , 2 w  l / 2 .
28/15
9. Maximum and minimum problems with constants; Lagrange multipliers
(제한조건이 있는 최대 최소값 문제 ; Lagrange 곱수)
Example 1. shortest distance
y  1  x2 ,
min of d  x 2  y 2 ?
Let's use d 2  f  x 2  y 2 .
- Methods: (a) elimination,
(b) implicit differentiation,
(c) Lagrange multipliers
(a) Elimination (제거방법)

f  x2  y 2  x2  1  x2

2
 x2  1  2x2  x4  x4  x2  1
df
 4 x 3  2 x  0  x  0, x   1 / 2
dx
 2 at x  0
d2 f
2
 12x  2  
dx2
4 at x   1 / 2
(relative max.)
(min. y  1 / 2).
29/15
(b) Implicit differentiation (음함수 미분)
f  x2  y2
dy
df
 2x  2 y .
dx
dx
df
 2 x  4 xy  y  1  x 2  dy  2 xdx
df  2 x  4 xy dx, or
dx
df  2 xdx  2 ydy, or
df
 2 x  4 xy  0. (conditionfor minimization)
dx
T o minimize f ,


2 x  4 x 1  x 2  0  x  0, x   1/ 2
2
d2 f
d2y
 dy 
 2  2   2 y 2 .
2
dx
dx
 dx 

dy d 2 y 
 We easily know the values of
, 2 . 
dx dx 

30/15
Example 2. Shortest distance from the origin to the plane x  2 y  2 z  3.
f  x 2  y 2  z 2  3  2 y  2 z   y 2  z 2
2
f f

 0.
y z
For minimization,
f
 23  2 y  2 z   2  2 y  0,
y
 yz
2
f min
elimination 
2
f
 23  2 y  2 z   2  2 z  0
z
2
1
 x
3
3
2
1  2  2
            1.
3  3  3
cf. Equation of plane, ax+by+cz=d
If (a,b,c) is a unit vector, abs(d) is a distance from the origin.
31/15
(c) Lagrange Multipliers
f ( x, y ),  ( x, y )  const.
‘two functions’
df 
f
f
dx  dy
x
y
d 


dx 
dy constraint  
x
y

conditionfor max.or min.
 f
 
 
 f

dy  0


dx






x 
y 
 x
 y
f

f


 0,

0
x
x
y
y
‘single function’
 condition for max.or min.of F  x, y   f  x, y     x, y 
cf. valid for more than variables, ex. (x,y,z)
32/15
- Using the Lagrange multipliers,
f  x, y   x 2  y 2 ,
  x, y   y  x 2  1

F x, y   f    x 2  y 2   y  x 2

F
 0  2 x    2 x  2 x1   .  x  0,   1.
x
F
 2 y    0.
y
x0 
y  1  x2
y  1,   2
1
2
  1  y  , x  
1
.
2
33/15
Example 3. Find the volume of the largest rectangular parallelepiped (that is box)
2
2
2
with edges parallel to the axes, inscribed in the ellipsoid, x  y  z  1
2
2
2
a
b
c
2
2
2
x
y
z
  x, y , z   2  2  2  1
a
b
c
 x2 y 2 z 2 
F x, y, z   f    8 xyz    2  2  2 
b
c 
a
F
2x
F
2y
F
2z
 8 yz    2  0,
 8 xz    2  0,
 8 xy    2  0.
x
a
y
b
z
c
Multiplying each equation with the other variable, and then, adding all three,
 x2 x2 x2 
3  8 xyz  2  2  2  2   3  8 xyz  2  0    12xyz.
b
c 
a
F
2x
2x
1
 8 yz    2  8 yz  12xyz  2  0  x 2  a 2 .
x
a
a
3
1
1
8abc
Similarly, y 2  b 2 , z 2  c 2  Maximum volume 8 xyz 
3
3
3 3
34/15
- More constraints
f ( x, y, z, w), 1 x, y, z, w  const., 2 x, y, z, w  const.
df 
f
f
f
f
dx  dy  dz 
dw  0,
x
y
x
w
d1 
1



dx  1 dy  1 dz  1 dw  0,
x
y
z
w
d2 
2



dx  2 dy  2 dz  2 dw  0
x
y
z
w
35/15
F  f  11  2 2
dF  df  1 d1  2 d2
 f

 

 
 f
   1 1  2 2 dx    1 1  2 2 dy
x
x 
y
y 
 x
 y

 

 
 f
 f
   1 1  2 2 dz  
 1 1  2 2 dw  0
z
z 
w
w 
 z
 w
f


f


 1 1  2 2  0
 1 1  2 2  0
x
x
x
y
y
y
f


f


 1 1  2 2  0,
 1 1  2 2  0
z
z
z
w
w
w
To find the maximum or minimum of f subject to the conditions Φ1=const.
and Φ2=const., define F=f + λ1 Φ1+ λ2 Φ2 and set each of the partial derivatives
of F equal to zero. Solve these equation and the Φ equation for the variables and
the λ’s.
36/15
Example 4. Minimized distance from the origin to the intersection of
xy  6, 7 x  24z  0
F  f  11  22
 x 2  y 2  z 2  1 7 x  24z   2 xy
F
F
F
 2 x  71  2 y  0,
 2 y  2 x  0,
 2 z  241  0.
x
y
z
x  12 / 5,
y  5 / 2, z  7 / 10  d  5 / 2 .
37/15
10. Endpoint or boundary point problems (끝점 혹은 경계점 문제)
- Besides the extreme points, we should check the boundary points (or lines).
case I
case II
case III
case IV
38/15
Example 1. A piece of wire 40 cm long is to be used to form the perimeters of a
square and a circle in such a way as to make the total area (of a square and circle) a
maximum.
1 

A  r 2  10  r 
2 

2
r
dA
1  1 

 
 2r  210  r      2r 1    10  0.
dr
2  2 
4


 
r 1    5, r  2.8,
4

(40-2r)/4
A  56.
d2A
 

2

1    0 : Min.
2
dr
4

Considering the values at the boundary points,
At r  0, A  100.
only square
At 2r  40, r  20 /  ,
A  400/   127.
only circle**
39/15
Example 2. The temperature in a rectangular plate bounded by the lines,
x  0, y  0, x  3, y  5
T  xy2  x2 y  100
(3,5)
- Differentiating,
y=5
T
T
 y 2  2 xy  0,
 2 xy  x 2  0  x  y  0, T  100.
x
y
x=3
- Boundary check
1) x  3
(0,0)
T  3 y 2  9 y  100
dT
3
1
 6 y  9  0  y  , T  93 .
dy
2
4
min.
2) y  5
T  25x  5 x 2  100
max.
dT
5
1
 25  10x  0  x  , T  131 .
dx
2
4
and check corners. At (3,5), T= 130.
40/15
H. W. (Due 5/21)
Chapter 4
4-15, 6-4, 9-11, 10-9
41/15
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 4 Partial differentiation
Lecture 15 Change of variables
11. Change of variables (변수 변환)
Sometimes, we can make the differential equation simpler by changing variables.
Example 1. Make the change of variables.
2F 1 2F
r  x  vt, s  x  vt,
 2 2 0
2
x
v t
F F r F s F F






 (  )F
x
r x s x r s
r s
F F r F s
F
F




v
v
 v(  ) F
t
r t s t
r
s
r s
Here, we can use the operation notation,





x
r
s



 v(  )
t
r s
Then,
 2 F  F

 F F
2F
2F 2F
 ( )  (  )(

) 2 2

r s r s
rs s 2
x 2 x x
r
2
 2 F  F
 
F F
2F 2F
2  F
 ( )  v (  )v (

) v ( 2 2
 2)
2
t t
r s r s
rs s
t
r
2F 1 2F
2F
 2  2 2 4
0
rs
x
v t
cf. compare with the original eq,
2F
 F
 ( )  0  F  f s   g r 
rs r s
 F  f (r )  g (s)  f ( x  vt)  g ( x  vt)
2F 1 2F
 2 2 0
2
x
v t
cf.
2F
 F
F
 ( )0 
 some function with only s.
rs r s
s
F  f s   const. ( can be a function without s.)
F  f s   g r 
Example 2. Laplace equation
 2
cf. Schrodinger eq.: Hˆ   E , Hˆ  
 U
2m
2F 2F
 2 0
2
x
y
1  F
1 2F
(r
) 2
0
2
r r r
r 
 x  r cos
for 
 y  r sin 
cylindrical
2F 2F 2F 1 2
1
1 

1 2
cf .
 2  2 
rF  2 ( 2
sin 
F 2
F)
2
2
2
x
y
z
r r
r sin  

sin  
 x  r sin  cos

for  y  r sin  sin 
 z  r cos

spherical
(i)
F F x F y
F
F


 cos
 sin 
r x r y r
x
y
F F x F y
F
F


 r sin 
 r cos
 x  y 
x
y
 cos

  r sin 
 F   F 
sin   x   r 




F

F
r cos 
 


 y  






 F 

  cos

x


 F    r sin 
 y 


sin  

r cos 
1
 F 


 r 
 F 


  
(ii)
F F r F


x r x 
F F r F


y r y 
For convenience,

F sin 
 cos

x
r
r

F cos
 sin 

y
r
r
F
 cos
x
F
H
 sin 
y
G
 2 F G
G sin  G

 cos

x 2
x
r
r 
 2 F H
H cos H


sin


2
y
x
r
r 
F

F

F sin  F

r
r 
F cos F

r
r 
2F 2F
G
H 1 
H
G 


cos


sin


cos


sin



2
2
x
y
r
r r 

 
1)
1)
2)
G
 2 F sin   2 F sin  F
 cos 2 
 2
r
r
r r
r 
H
 2 F cos  2 F cos F
 sin  2 
 2
r
r
r r
r 
G
H
2F 2F
2
2
 cos
 sin 
 (cos   sin  ) 2  2
r
r
r
r
2F 2F
G
H 1 
H
G 


cos


sin


cos


sin



2
2
x
y
r
r r 

 
1)
2)
2
2
2) G  cos  F  sin  F  sin   F  cos F

r
r
r  2
r 
H
2F
F cos  2 F sin  F
 sin 
 cos



r
r
r  2
r 
1
H
G  1 F 1  2 F
  cos
 sin 

)
 (
2
r

  r r r 
Finally,
 2 F  2 F 1  F
1 2F
 2  2 
(r
) 2
x
y
r r r
r  2
12. Differentiation of integrals; Leibniz’ rule
(적분의 미분 ; Leibniz 규칙)
dF x 
f x  

dx

x
a
f t dt  F t  a  F x   F a .
x
d vx 
dv
du






f
t
dt

f
v

f
u
.

u

x

dx
dx
dx
v f
d vx 
dv
du






f
x
,
t
dt

f
x
,
v

f
x
,
u

dt .


u

x

u
dx
dx
dx
x
‘Leibniz’ rule’