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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. ex. - Time rates such as velocity, acceleration, and rate of cooling of a hot body. - PowerPoint PPT Presentation
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Chapter 4 Partial differentiation ( 편미분 )
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 12 Introduction of partial differentiation
1. Introduction
xyxfy
dx
dyyxfy
respect to with of rateor curve theof Slope :
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.
- Differentiation
- Partial differentiation
yxfz ,
When we want to find the slope of z with respect to y, keeping x constant, we can use the partial differentiation.
)constant with respect towith or (..
yxx
z
y
z
constyconstx
x
y
z
.constxy
z
.constyx
z
1
2
322
2
2
,, s,expressionOther cf.
etc. , , ,
ffz
yx
z
yx
z
xyx
z
y
z
xx
z
x
z
x
xx
- High order partial derivatives
Example xyeyxyxfz 3,
,3 21
xyxx yeyxfzf
x
z
x
f
,32
xyyy xexfzf
y
z
y
f
,3 221
22xyxy
yxyx xyeexfzfyx
z
yx
f
(Caution)
rr
zyryyxz
rr
zrxyxxz
rr
zrrz
y
x
2,22
2,22
,sincos2,sincos
22222
22222
222222
?,,,
sin,cos
22
yx r
z
r
z
r
zyxz
ryrx
The symbol 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.
xrz /
2. Power series in two variables ( 두 변에 대한 멱급수 )
Example 1.
.!2!3!2
1!3
cossin,2323
xyxx
yxxyxyxf
Example 2
3322
321ln
322
322
32
yxyyx
xyxy
xyx
yxyxyxyx
- General expression
303
212
221
330
202
112
20011000
)())((
)()()()(
))(()()()(),(
byabyaxa
byaxaaxabya
byaxaaxabyaaxaayxf
First, express the function with the power series and then determine the coefficients.
)()(2 112010 byaaxaaf x
)(2)( 021101 byaaxaaf y
)(and/or )( containing terms2 20 byaxaf xx
)(and/or )( containing terms11 byaxaf xy
etc. , ),( , 2),(
, ),( , ),( , ),(
1120
011000
abafabaf
abafabafabaf
xyxx
yx
Finding partial derivatives,
0
),(!
1),(
n
n
bafy
kx
hn
yxf
]))(,())()(,(2))(,([!2
1
))(,())(,(),(),(
22 bybafbyaxbafaxbaf
bybafaxbafbafyxf
yyxyxx
yx
]),(),(2),([!2
1 22 kbafhkbafhbaf yyxyxx
),(!2
12
bafy
kx
h
]),( 3),([!3
1),(
!3
1 23
3
bafkhbafhbafy
kx
h xxyxxx
Using a simpler form, x – a = h and y – b = k.
second-order terms,
similarly, third-order terms
Finally,
Then,
3. Total differentials ( 전 미분 )
dxydyxfdx
d
dx
dyy )('
x
y
dx
dyx
0lim
- Single variables
- Two variables and more
dyy
zdx
x
zdzyxfz
,
dzz
fdy
y
fdx
x
fduzyxfu ,,,
4. Approximations using differentials ( 미소량을 이용한 어림 )
Example 1. 25.0
1
1025.0
120
.10425.0
1
1025.0
1
25.02
125.0
2
1
1025.025.025.01025.0
1
20
20
2/32/3
2020
fxxf
fxffff
xxf
- Example 2. 322
2
1
11
nnn
3
32
2
211
1
nxf
nnfxnfnfnff
xxf
- Example 3. reduced mass1
21
11 mm
If m_1 is increased by 1%, what fractional change in m_2 leaves unchanged?
./01.0or01.0
unchanged 0
111
01.0
122
221
121
122
2
22
212
1
22
212
12
21
11
mmm
dm
m
m
m
dm
m
dm
dmmdmm
dmmdmmdmm
mdm
.03.03,01.0 ex. 22212221 mmmmmmmm
dxyxycf .
Example 4.2r
klR
Relative error rate: 5 % in the length measurement
and 10 % for the radius measurement
1.0/,05.0/ rdrldl
.25.01.0205.02largest 2
ln2lnlnln
r
dr
l
dl
R
dR
r
dr
l
dl
R
dR
rlkR
dxyxycf .
Chapter 4 Partial differentiation
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 13 Chain rule
5. Chain rule or differentiating a function of a function ( 연쇄법칙과 함수의 함수 미분하기 )
Example 1. ?2sinln dx
dyxy
.2cot222cos2sin
12sin
2sin
1xx
dx
dx
xx
dx
d
xdx
dy
dx
dv
dv
du
du
dy
dx
dy
xvvuuy
.2&sinwhere,ln
‘chain rule’
Example 2. ?sin2 2 dt
dzttz
y
zx
x
zy
dt
dy
y
z
dt
dx
x
z
dt
dz
dt
dyx
dt
dxy
dt
dz
tytxxyz
,
,sin&2where, 2
dyy
zdx
x
zdz
ttttdt
dzcos2sin4 2
Example 3. ?,sin,tanwhere, 1
dt
dztxtyxz y
.1
1lncos
21
txxtyx
dt
dy
y
z
dt
dx
x
z
dt
dz yy
6. Implicit differentiation ( 음함수 미분 )
Example 1. ?, 2
2
dt
xd
dt
dxtex x
xx
edt
dx
dt
dxe
dt
dx
1
11
We realized that x is a function and just differentiate each term of the equation with respect to t (implicit differentiation).
1. xx edt
dxdtdxedxcf
.11
01
3
2
2
2
2
2
2
2
2again atingdifferenti
x
x
xdtdxx
xxx
e
e
e
e
dt
xd
dt
dxe
dt
xde
dt
xd
dt
dxe
dt
dx
This problem is even easier if we want only the numerical values of the derivatives at a point.
.02
111)2,
2
111)1
2
2
2
2
dt
xd
dt
xdnd
dt
dxor
dt
dx
dt
dxst
7. More chain rule ( 더 많은 연쇄법칙 )
Example 1. ?,,,sin,
t
z
s
ztsytsxxyz
.,cos, dtdsdydtdstsdxxdyydxdz
.cos.),(0For
cos.),(0For
])cos([])cos([
)())(cos(
xtsys
zconsttdt
xtsyt
zconstsds
dtxtsydsxtsy
dtdsxdtdstsydz
Example 2.
?,,2,,,ln2 222
t
u
s
utztsytsxzyxyxu
.2
ln2422ln2222
0 cf.
.2
ln24,ln24
.2
ln24ln24
22ln2222
ln222
dtz
yztytdt
z
ytdtzxtdtyx
dsdu
z
yztyt
t
uzyx
s
u
dtz
yztytdszyx
dtz
ytdtdszxtdtdsyx
zdydzz
yydxxdyxdxdu
s
- Using the differentials,
- Using the derivatives,
.2
ln2422ln2222
2ln2
ln2ln2
2
2222
z
yztyt
z
ytzxtyx
tt
zyxyxz
tst
zyxyxy
tst
zyxyxx
t
z
z
u
t
y
y
u
t
x
x
u
t
u
cf. Using the matrix form,
t
z
s
z
t
y
s
y
t
x
s
x
z
u
y
u
x
u
t
u
s
u
Example 3. ?/,sin,, 222 dtdzyetxtyxyxz y
dttx
t
dy
dx
eyt
yx
tdtxdyeytdx
tdtydyxdx
dyeyetdtxtdx
tdtydyxdx
dydxdz
yy
yy
cos1sincos1sin
,
cossin
,222
‘A computer may save us some time with the algebra.’
.for similarly ,
sin1
cos1
1sin
1cosdydt
tyeyx
txyeyt
eyt
yxeytdtx
ytdt
dx y
y
y
y
Example 4. ?/,/,,5, 2223322 tzsztsyxstyxxyxz
t
z
s
zdtdsdz
dyyxxy
dssyytdttyys
yx
yx
ytdtsds
ytdssdt
dx
tdtsdsydydxx
tdssdtdyyxdx
ydxxdyxdxdz
,, way,In this
.for similarly ,94
6262
23
32
222
3
.2223
32
2
22
22
2
2
2
2
2
Let’s skip Example 5.
Example 6. Rectangular vs. polar coordinates. (reciprocal)
sin
cos
ry
rx
x
y
yxr
1
22
tan
222
21
)/(1
/tan i)
r
y
xy
xy
x
y
xx
yrrx
sincos ii)
y
r
ry
yyyy
x
y
2
2222
/sin)csc(cot
i) and ii)-1 are different!!
constant y
constant r
This is a general rule: partial derivatives are not usually
reciprocals; they are reciprocals if the other independent variables (besides u or v)
are the same in both cases.
uvvu /&/
26/15
Chapter 4 Partial differentiation
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 14 Max. & Min.
27/15
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)inflection
(d2y/dx2 > 0) (d2y/dx2 < 0)(d2y/dx2 = 0)
cf. saddle point
- To minimize z = f(x,y),
.0&0
y
z
x
z
28/15
t.independen are ),,( of variablesonly two
fixedtantan22
1 2
lw
lwwlwV
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.
w2 l
To minimize A,
.0cotcsc2
sec2,0csc2
tan4 222
w
Vw
A
w
Vw
w
A
tan.csc
2tan2
tancos
2tan2
cos
2tan2 area Material
22
22
2
w
Vl
w
Vw
w
Vww
lwwA
.2/2,45,2/1cos0cos,0sin
.sin
coscos
sin2
cos
sec
cotcsc
tan2
csc
22
2
2
223
lwlwV
orVV
w
29/15
9. Maximum and minimum problems with constants; Lagrange multipliers ( 제한조건이 있는 최대 최소값 문제 ; Lagrange 곱수 )
Example 1. shortest distance
. use sLet'
?ofmin,1
222
222
yxfd
yxdxy
- Methods: (a) elimination,
(b) implicit differentiation,
(c) Lagrange multipliers
(a) Elimination ( 제거방법 )
).2/1min.(2/1at4
max.) relative(0at2212
2/1,0024
1211
22
2
3
2442222222
yx
xx
dx
fd
xxxxdx
df
xxxxxxxyxf
30/15
(b) Implicit differentiation ( 음함수 미분 )
xdxdyxyxyxdx
dfdxxyxdf
dx
dyyx
dx
dfydyxdxdf
yxf
2142or,42
.22or,22
2
22
2/1,00142
on)minimizatifor condition (.042, minimize To
2
xxxxx
xyxdx
dff
. , of values theknoweasily We.222
2
2
2
22
2
2
dx
yd
dx
dy
dx
ydy
dx
dy
dx
fd
31/15
Example 2. Shortest distance from the origin to the plane .322 zyx
.13
2
3
2
3
1
3
1
3
2
0222232,0222232
.0 on,minimizatiFor
neliminatio223
222
min
222222
f
xzy
zzyz
fyzy
y
f
z
f
y
f
zyzyzyxf
cf. Equation of plane, ax+by+cz=d
If (a,b,c) is a unit vector, abs(d) is a distance from the origin.
32/15
(c) Lagrange Multipliers
yxyxfyxF
yy
f
xx
f
dyyy
fdx
xx
f
dyy
dxx
d
dyy
fdx
x
fdf
constyxyxf
,,, of min.or max.for condition
0,0
0
constraint
min.or max.for condition
.),(),,(
‘two functions’
‘single function’
cf. valid for more than variables, ex. (x,y,z)
33/15
.02
.1,0.12220
,
1,,,222
222
yy
F
xxxxx
F
xyyxfyxF
xyyxyxyxf
- Using the Lagrange multipliers,
.2
1,
2
11
2,10
xy
yx
21 xy
34/15
Example 3. Find the volume of the largest rectangular parallelepiped (that is box)
with edges parallel to the axes, inscribed in the ellipsoid, 12
2
2
2
2
2
c
z
b
y
a
x
.02
8,02
8,02
8
8,,
1,,
222
2
2
2
2
2
2
2
2
2
2
2
2
c
zxy
z
F
b
yxz
y
F
a
xyz
x
F
c
z
b
y
a
xxyzfzyxF
c
z
b
y
a
xzyx
.120283283 2
2
2
2
2
2
xyzxyzc
x
b
x
a
xxyz
Multiplying each equation with the other variable, and then, adding all three,
33
88 volumeMaximum
3
1,
3
1Similarly,
.3
10
2128
28
2222
2222
abcxyzczby
axa
xxyzyz
a
xyz
x
F
35/15
- More constraints
0
,0
,0
.,,,.,,,,),,,,(
22222
11111
21
dww
dzz
dyy
dxx
d
dww
dzz
dyy
dxx
d
dww
fdz
x
fdy
y
fdx
x
fdf
constwzyxconstwzyxwzyxf
36/15
0
22
11
22
11
22
11
22
11
2211
2211
dwwww
fdz
zzz
f
dyyyy
fdx
xxx
f
dddfdF
fF
0,0 22
11
22
11
www
f
zzz
f
0 22
11
xxx
f 0 22
11
yyy
f
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.
37/15
Example 4. Minimized distance from the origin to the intersection of 0247,6 zxxy
.2/510/7,2/5,5/12
.0242,02,072
247
1221
21222
2211
dzyx
zz
Fxy
y
Fyx
x
F
xyzxzyx
fF
38/15
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
39/15
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.
.56,8.2,54
1
.0104
122
1
2
11022
2
110
22
Arr
rrrdr
dA
rrA
.:04
122
2
Mindr
Ad
**circleonly .127/400,/20,402At
squareonly .100,0At
Arr
Ar
Considering the values at the boundary points,
r
(40-2r)/4
40/15
Example 2. The temperature in a rectangular plate bounded by the lines,5,3,0,0 yxyx10022 yxxyT
.100,002,02 22
Tyxxxyy
Txyy
x
T
(0,0)
(3,5)
x=3
y=5
.4
1131,
2
501025
100525
5)2
.4
193,
2
3096
10093
3)1
2
2
Txxdx
dT
xxT
y
Tyydy
dT
yyT
x
and check corners. At (3,5), T= 130.
- Differentiating,
- Boundary check
max.
min.
41/15
H. W. (Due 5/21)
Chapter 4
4-15, 6-4, 9-11, 10-9
Chapter 4 Partial differentiation
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
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.
01
, , 2
2
22
2
t
F
vx
Fvtxsvtxr
Fsrs
F
r
F
x
s
s
F
x
r
r
F
x
F)(
Fsr
vs
Fv
r
Fv
t
s
s
F
t
r
r
F
t
F)(
Here, we can use the operation notation,
srx
)(sr
vt
2
22
2
2
2
2
2))(()(s
F
sr
F
r
F
s
F
r
F
srx
F
xx
F
)2()()()(2
22
2
22
2
2
s
F
sr
F
r
Fv
s
F
r
Fv
srv
t
F
tt
F
041 2
2
2
22
2
sr
F
t
F
vx
F
01
2
2
22
2
t
F
vx
Fcf. compare with the original eq,
Then,
rgsfFs
F
rsr
F
0)(2
)()()()( vtxgvtxfsgrfF
cf.
rgsfF
sconstsfF
ss
F
s
F
rsr
F
.)without function a becan (.
.only ith function w some0)(2
Example 2. Laplace equation
02
2
2
2
y
F
x
F0
1)(
12
2
2
F
rr
Fr
rr
sin
cos
ry
rxfor
cos
sinsin
cossin
rz
ry
rx
for
)sin
1sin
sin
1(
11 .
2
2
2222
2
2
2
2
2
2
2
FFr
rFrrz
F
y
F
x
Fcf
cylindrical
spherical
cf. Schrodinger eq.: Um
HEH 2
2ˆ ,ˆ
y
Fr
x
Fr
y
y
Fx
x
FF
y
F
x
F
r
y
y
F
r
x
x
F
r
F
cossin
sincos
Fr
F
rry
Fx
F
Fr
F
y
Fx
F
rr
1
cossin
sincos
cossin
sincos
(i)
(ii)
F
rr
F
y
F
y
r
r
F
y
F
F
rr
F
x
F
x
r
r
F
x
F
cossin
sincos
For convenience,
F
rr
F
y
FH
F
rr
F
x
FG
cossin
sincos
H
rr
H
x
H
y
F
G
rr
G
x
G
x
F
cossin
sincos
2
2
2
2
GH
rr
H
r
G
y
F
x
Fsincos
1sincos
2
2
2
2
F
rr
F
rr
F
r
H
F
rr
F
rr
F
r
G
2
2
2
2
2
2
2
2
coscossin
sinsincos
2
2
2
222 )sin(cossincos
r
F
r
F
r
H
r
G
1) 2)
1)
F
r
F
rr
F
r
FH
F
r
F
rr
F
r
FG
sincoscossin
cossinsincos
2
22
2
22
)1
(1
sincos1
2
2
F
rr
F
r
GH
r
2
2
22
2
2
2 1)(
1
F
rr
Fr
rry
F
x
FFinally,
2)
GH
rr
H
r
G
y
F
x
Fsincos
1sincos
2
2
2
2
1) 2)
12. Differentiation of integrals; Leibniz’ rule ( 적분의 미분 ; Leibniz 규칙 )
.
.
dx
duuf
dx
dvvfdttf
dx
d
aFxFtFdttfdx
xdFxf
xv
xu
x
a
x
a
‘Leibniz’ rule’
.,,,
v
u
xv
xudt
x
f
dx
duuxf
dx
dvvxfdttxf
dx
d
