engineering maths

1. Partial differentiation

2. 1. Introduction
 
  xyxfy
dx
dy
yxfy
respect towithofrateorcurvetheofSlope: 

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

3. - 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.
)constantwithrespect towithor(
..
yx
x
z
y
z
constyconstx 




x
y
z
.constx
y
z



.constyx
z



4. 1
2
322
2
2
,,s,expressionOthercf.
etc.,,,
ffz
yx
z
yx
z
xyx
z
y
z
xx
z
x
z
x
xx





















- High order partial derivatives
Example   xy
eyxyxfz  3
,
,3 2
1
xy
xx yeyxfzf
x
z
x
f






,3
2
xy
yy xexfzf
y
z
y
f







,3 2
21
22
xyxy
yxyx xyeexfzf
yx
z
yx
f







5. (Caution)
 
r
r
z
yryyxz
r
r
z
rxyxxz
r
r
z
rrz
y
x
2,22
2,22
,sincos2,sincos
22222
22222
222222


























 

?,,,
sin,cos
22


























yx r
z
r
z
r
z
yxz
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  /

6. 2. Power series in two variables
Example 1.
  .
!2!3!2
1
!3
cossin,
2323
 












xyx
x
yx
xyxyxf
Example 2
       








3322
32
1ln
3
22
322
32
y
xyyx
xy
xy
x
yx
yxyx
yxyx

7. - General expression



3
03
2
12
2
21
3
30
2
02
11
2
20011000
)())((
)()()()(
))(()()()(),(
byabyaxa
byaxaaxabya
byaxaaxabyaaxaayxf
First, express the function with the power series and then determine the coefficients.
 )()(2 112010 byaaxaafx
 )(2)( 021101 byaaxaafy
)(and/or)(containingterms2 20 byaxafxx 
)(and/or)(containingterms11 byaxafxy 
etc.,),(,2),(
,),(,),(,),(
1120
011000
abafabaf
abafabafabaf
xyxx
yx


Finding partial derivatives,

8. 














0
),(
!
1
),(
n
n
baf
y
k
x
h
n
yxf
]))(,())()(,(2))(,([
!2
1
))(,())(,(),(),(
22
bybafbyaxbafaxbaf
bybafaxbafbafyxf
yyxyxx
yx


]),(),(2),([
!2
1 22
kbafhkbafhbaf yyxyxx 
),(
!2
1
2
baf
y
k
x
h 










]),(3),([
!3
1
),(
!3
1 23
3











bafkhbafhbaf
y
k
x
h xxyxxx
Using a simpler form, x – a = h and y – b = k.
second-order terms,
similarly, third-order terms
Finally,
Then,

9. 3. Total differentials
dxydyxf
dx
d
dx
dy
y  )('
x
y
dx
dy
x 


 0
lim
- Single variables
- Two variables and more
  dy
y
z
dx
x
z
dzyxfz





 ,
   








 dz
z
f
dy
y
f
dx
x
f
duzyxfu ,,,

10. 4. Approximations using differentials
Example 1. 25.0
1
1025.0
1
20

 
 
       
   
.104
25.0
1
1025.0
1
25.0
2
1
25.0
2
1
1025.025.025.01025.0
1
20
20
2/32/3
2020






















fxxf
fxffff
x
xf

11. - Example 2.   322
2
1
11
nnn



          
  







3
32
2
2
11
1
n
xf
n
nfxnfnfnff
x
xf


12. - Example 3. reduced mass
1
2
1
1
1 
 mm
If m_1 is increased by 1%, what fractional change in m_2 leaves  unchanged?
 
./01.0or
01.0
unchanged0
111
01.0
12
2
2
2
1
1
2
1
1
2
2
2
2
2
21
2
1
2
2
21
2
1
2
21
11
mm
m
dm
m
m
m
dm
m
dm
dmmdmm
dmmdmmd
mm
mdm










.03.03,01.0ex. 22212221 mmmmmmmm 
  dxyxycf .

13. Example 4. 2
r
kl
R 
Relative error rate: 5 % in the length measurement
and 10 % for the radius measurement
1.0/,05.0/  rdrldl
  .25.01.0205.02largest2
ln2lnlnln


r
dr
l
dl
R
dR
r
dr
l
dl
R
dR
rlkR
  dxyxycf .

14. Partial differentiation

15. 5. Chain rule or differentiating a function of a function
Example 1.   ?2sinln 
dx
dy
xy
    .2cot222cos
2sin
1
2sin
2sin
1
xx
dx
d
x
x
x
dx
d
xdx
dy

dx
dv
dv
du
du
dy
dx
dy
xvvuuy

 .2&sinwhere,ln
‘chain rule’

16. Example 2. ?sin2 2

dt
dz
ttz




















y
z
x
x
z
y
dt
dy
y
z
dt
dx
x
z
dt
dz
dt
dy
x
dt
dx
y
dt
dz
tytxxyz
,
,sin&2where, 2

dy
y
z
dx
x
z
dz






tttt
dt
dz
cos2sin4 2


17. Example 3. ?,sin,tanwhere, 1
 
dt
dz
txtyxz y
.
1
1
lncos 2
1
t
xxtyx
dt
dy
y
z
dt
dx
x
z
dt
dz yy







 

18. 6. Implicit differentiation
Example 1. ?, 2
2

dt
xd
dt
dx
tex x
x
x
edt
dx
dt
dx
e
dt
dx


1
1
1
We realized that x is a function and just differentiate each term of the equation
with respect to t (implicit differentiation).
1.  xx
e
dt
dx
dtdxedxcf
 
 
.
11
01
3
2
2
2
2
2
2
2
2
againatingdifferenti
x
x
x
dt
dxx
xxx
e
e
e
e
dt
xd
dt
dx
e
dt
xd
e
dt
xd
dt
dx
e
dt
dx












 
This problem is even easier if we want only the numerical values of the
derivatives at a point.
.0
2
1
11)2,
2
1
11)1 2
2
2
2







dt
xd
dt
xd
nd
dt
dx
or
dt
dx
dt
dx
st

19. 7. More chain rule
Example 1.   ?,,,sin, 





t
z
s
z
tsytsxxyz
   .,cos, dtdsdydtdstsdxxdyydxdz 
 
  .cos.),(0For
cos.),(0For
])cos([])cos([
)())(cos(
xtsy
s
z
consttdt
xtsy
t
z
constsds
dtxtsydsxtsy
dtdsxdtdstsydz











20. Example 2.
?,,2,,,ln2 222






t
u
s
u
tztsytsxzyxyxu
       
 
 
        .
2
ln2422ln2222
0cf.
.
2
ln24,ln24
.
2
ln24ln24
22ln2222
ln222
dt
z
y
ztytdt
z
y
tdtzxtdtyx
dsdu
z
y
ztyt
t
u
zyx
s
u
dt
z
y
ztytdszyx
dt
z
y
tdtdszxtdtdsyx
zdydz
z
y
ydxxdyxdxdu
s























- Using the differentials,

21. - Using the derivatives,
        .
2
ln2422ln2222
2ln2
ln2ln2
2
2222
z
y
ztyt
z
y
tzxtyx
t
t
zyxyx
z
ts
t
zyxyx
y
ts
t
zyxyx
x
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

22. Example 3. ?/,sin,, 222
 dtdzyetxtyxyxz y
 
   
dt
tx
t
dy
dx
eyt
yx
tdtxdyeytdx
tdtydyxdx
dyeyetdtxtdx
tdtydyxdx
dydxdz
yy
yy







































cos1sincos1sin
,
cossin
,222
 
 
 
 
.forsimilarly,
sin1
cos1
1sin
1cos
dydt
tyeyx
txyeyt
eyt
yx
eytdtx
ytdt
dx y
y
y
y







23. Example 4. ?/,/,,5, 2223322
 tzsztsyxstyxxyxz
   
 
t
z
s
z
dtdsdz
dy
yxxy
dssyytdttyys
yx
yx
ytdtsds
ytdssdt
dx
tdtsdsydydxx
tdssdtdyyxdx
ydxxdyxdxdz















,,way,In this
.forsimilarly,
94
6262
23
32
222
3
.2223
32
2
22
22
2
2
2
2
2

24. Let’s skip Example 5.
Example 6. Rectangular vs. polar coordinates. (reciprocal)


sin
cos
ry
rx


x
y
yxr
1
22
tan



222
2
1
)/(1
/
tani)
r
y
xy
xy
x
y
xx













 
yrr
x








sincosii)
 
y
r
ry
yy
yy
x
y
2
222
2
/sin
)csc(cot 

















i) and ii)-1 are different!!
constant y
constant r

25. 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. 26/15
Partial differentiation

27. 8. Application of partial differentiation to maximum and minimum problems
- 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. 28/15
t.independenare),,(ofvariablesonly two
fixedtantan2
2
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,
.0cotcsc
2
sec2,0
csc2
tan4 22
2










w
V
w
A
w
V
w
w
A




















tan
.csc
2
tan2
tancos
2
tan2
cos
2
tan2areaMaterial
2
2
2
2
2
w
V
l
w
V
w
w
Vw
w
lw
wA

.2/2,45,2/1cos0cos,0sin
.
sin
coscos
sin2
cos
sec
cotcsc
tan2
csc
22
2
2
22
3
lwlwV
or
VV
w












29. 29/15
9. Maximum and minimum problems with constants; Lagrange multipliers
Example 1. shortest distance
.usesLet'
?ofmin,1
222
222
yxfd
yxdxy


- Methods: (a) elimination,
(b) implicit differentiation,
(c) Lagrange multipliers
(a) Elimination
 









).2/1min.(2/1at4
max.)relative(0at2
212
2/1,0024
1211
2
2
2
3
2442222222
yx
x
x
dx
fd
xxxx
dx
df
xxxxxxxyxf

30. 30/15
(b) Implicit differentiation (음함수 미분)
   xdxdyxyxyx
dx
df
dxxyxdf
dx
dy
yx
dx
df
ydyxdxdf
yxf
2142or,42
.22or,22
2
22




  2/1,00142
on)minimizatiforcondition(.042,minimizeTo
2


xxxxx
xyx
dx
df
f












 .,ofvaluestheknoweasilyWe.222 2
2
2
22
2
2
dx
yd
dx
dy
dx
yd
y
dx
dy
dx
fd

31. 31/15
Example 2. Shortest distance from the origin to the plane .322  zyx
   
   
.1
3
2
3
2
3
1
3
1
3
2
0222232,0222232
.0on,minimizatiFor
neliminatio223
222
min
222222

































f
xzy
zzy
z
f
yzy
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. 32/15
(c) Lagrange Multipliers
 
 
     yxyxfyxF
yy
f
xx
f
dy
yy
f
dx
xx
f
dy
y
dx
x
d
dy
y
f
dx
x
f
df
constyxyxf
,,,ofmin.ormax.forcondition
0,0
0
constraint
min.ormax.forcondition
.),(),,(






























































 ‘two functions’
‘single function’
cf. valid for more than variables, ex. (x,y,z)

33. 33/15
   
   
 
.02
.1,0.12220
,
1,,,
222
222












y
y
F
xxxx
x
F
xyyxfyxF
xyyxyxyxf
- Using the Lagrange multipliers,
.
2
1
,
2
1
1
2,10


xy
yx


2
1 xy 

34. 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
 
 
.0
2
8,0
2
8,0
2
8
8,,
1,,
222
2
2
2
2
2
2
2
2
2
2
2
2

















c
z
xy
z
F
b
y
xz
y
F
a
x
yz
x
F
c
z
b
y
a
x
xyzfzyxF
c
z
b
y
a
x
zyx



.120283283 2
2
2
2
2
2
xyzxyz
c
x
b
x
a
x
xyz 





 
Multiplying each equation with the other variable, and then, adding all three,
33
8
8volumeMaximum
3
1
,
3
1
Similarly,
.
3
1
0
2
128
2
8
2222
22
22
abc
xyzczby
ax
a
x
xyzyz
a
x
yz
x
F






35. 35/15
- More constraints
   
0
,0
,0
.,,,.,,,,),,,,(
2222
2
1111
1
21








































dw
w
dz
z
dy
y
dx
x
d
dw
w
dz
z
dy
y
dx
x
d
dw
w
f
dz
x
f
dy
y
f
dx
x
f
df
constwzyxconstwzyxwzyxf






36. 36/15
02
2
1
1
2
2
1
1
2
2
1
1
2
2
1
1
2211
2211




























































dw
www
f
dz
zzz
f
dy
yyy
f
dx
xxx
f
dddfdF
fF


















0,0 2
2
1
1
2
2
1
1 

















www
f
zzz
f 







02
2
1
1 








xxx
f 


 02
2
1
1 








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. 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
21
222
2211












dzyx
z
z
F
xy
y
F
yx
x
F
xyzxzyx
fF





38. 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. 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.
.56,8.2,5
4
1
.010
4
12
2
1
2
1
1022
2
1
10
2
2

































Arr
rrr
dr
dA
rrA





.:0
4
122
2
Min
dr
Ad









 
  **
circleonly.127/400,/20,402At
squareonly.100,0At


 Arr
Ar
Considering the values at the boundary points,
r
(40-2r)/4

40. 40/15
Example 2. The temperature in a rectangular plate bounded by the lines,
5,3,0,0  yxyx
10022
 yxxyT
.100,002,02 22






Tyxxxy
y
T
xyy
x
T
(0,0)
(3,5)
x=3
y=5
.
4
1
131,
2
5
01025
100525
5)2
.
4
1
93,
2
3
096
10093
3)1
2
2






Txx
dx
dT
xxT
y
Tyy
dy
dT
yyT
x
and check corners. At (3,5), T= 130.
- Differentiating,
- Boundary check
max.
min.

41. 41/15

42. Partial differentiation
Change of variables

43. 11. Change of variables
Sometimes, we can make the differential equation simpler by changing variables.
Example 1. Make the change of variables.
0
1
,, 2
2
22
2







t
F
vx
F
vtxsvtxr
F
srs
F
r
F
x
s
s
F
x
r
r
F
x
F
)(
























F
sr
v
s
F
v
r
F
v
t
s
s
F
t
r
r
F
t
F
)(
























Here, we can use the operation notation,
srx 







)(
sr
v
t 








44. 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
2
2
2
2
s
F
sr
F
r
F
v
s
F
r
F
v
sr
v
t
F
tt
F



























04
1 2
2
2
22
2










sr
F
t
F
vx
F
0
1
2
2
22
2






t
F
vx
F
cf. compare with the original eq,
Then,
   rgsfF
s
F
rsr
F








0)(
2
)()()()( vtxgvtxfsgrfF 

45. cf.
 
   rgsfF
sconstsfF
s
s
F
s
F
rsr
F













.)withoutfunctionabecan(.
.onlyithfunction wsome0)(
2

46. Example 2. Laplace equation
02
2
2
2






y
F
x
F
0
1
)(
1
2
2
2









F
rr
F
r
rr 






sin
cos
ry
rx
for











cos
sinsin
cossin
rz
ry
rx
for
)
sin
1
sin
sin
1
(
11
. 2
2
2222
2
2
2
2
2
2
2
FF
r
rF
rrz
F
y
F
x
F
cf


 


















cylindrical
spherical
cf. Schrodinger eq.: U
m
HEH  2
2
ˆ,ˆ 

47. y
F
r
x
F
r
y
y
Fx
x
FF
y
F
x
F
r
y
y
F
r
x
x
F
r
F







































cossin
sincos























































































F
r
F
rr
y
F
x
F
F
r
F
y
F
x
F
rr
1
cossin
sincos
cossin
sincos
(i)

48. (ii)














































F
rr
F
y
F
y
r
r
F
y
F
F
rr
F
x
F
x
r
r
F
x
F
cos
sin
sin
cos
For convenience,
























F
rr
F
y
F
H
F
rr
F
x
F
G
cos
sin
sin
cos




























H
rr
H
x
H
y
F
G
rr
G
x
G
x
F
cos
sin
sin
cos
2
2
2
2

49. 


























GH
rr
H
r
G
y
F
x
F
sincos
1
sincos2
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
coscos
sin
sinsin
cos
2
2
2
2
22
)sin(cossincos
r
F
r
F
r
H
r
G











 
1) 2)
1)

50. 











































F
r
F
rr
F
r
FH
F
r
F
rr
F
r
FG
sincos
cossin
cossin
sincos
2
22
2
22
)
1
(
1
sincos
1
2
2





















F
rr
F
r
GH
r
2
2
22
2
2
2
1
)(
1














F
rr
F
r
rry
F
x
F
Finally,
2)



























GH
rr
H
r
G
y
F
x
F
sincos
1
sincos2
2
2
2
1) 2)

51. 12. Differentiation of integrals; Leibniz’ rule
           
  
 
    .
.
dx
du
uf
dx
dv
vfdttf
dx
d
aFxFtFdttf
dx
xdF
xf
xv
xu
x
a
x
a




‘Leibniz’rule’  
 
    .,,,  


v
u
xv
xu
dt
x
f
dx
du
uxf
dx
dv
vxfdttxf
dx
d