CHAPTER 5 PARTIAL DERIVATIVES INTRODUCTION SMALL INCREMENTS & RATES OF CHANGE IMPLICIT FUNCTIONS...
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Transcript of CHAPTER 5 PARTIAL DERIVATIVES INTRODUCTION SMALL INCREMENTS & RATES OF CHANGE IMPLICIT FUNCTIONS...
CHAPTER 5PARTIAL DERIVATIVES
INTRODUCTIONSMALL INCREMENTS & RATES OF CHANGE
IMPLICIT FUNCTIONSCHAIN RULE
JACOBIAN FUNCTIONHESSIAN FUNCTIONSTATIONARY POINT
INTRODUCTION
• Consider the following functions where
are independent variables.
• If we differentiate f with respect variable , then we assume thati. as a single variableii. as constants
nxxxf ,...,, 21
nxxx ,...,, 21
ix
ix
nii xxxxx ,...,,,..., 1121
NotationIf • First order partial derivatives:
• Second order partial derivatives:
,, yxff
y
f
x
f
and
x
f
yxy
fy
f
xyx
fy
f
yy
fx
f
xx
f
2
2
2
2
2
2
Example 1Write down all partial derivatives of the following function
xyyxyxyxf ln2cos2, 233
Example 1Write down all partial derivatives of the following function
SolutionFirst order PD
xyyxyxyxf ln2cos2, 233
xyyxyxy
fx
yyyx
x
f
ln22sin43
2cos23
23
232
Second order PD
xyxyx
xyyxyxy
y
f
yy
fx
yxy
x
yyyx
x
x
f
xx
f
ln22cos86
ln22sin43
6
2cos23
3
23
2
2
2
23
232
2
2
Second order PD (mixed partial)
x
yyyx
x
yyyx
y
x
f
yxy
fx
yyyx
xyyxyxx
y
f
xyx
f
22sin49
2cos23
22sin49
ln22sin43
22
232
2
22
23
2
In example 1, we observed that
This properties hold for all functions provided that certain smoothness properties are satisfies.
The mixed partial derivative must be equal whenever f is continuous.
xy
f
yx
f
22
Example 2Write down all partial derivatives of the following functions:
yxxyxxeyxfii
xxyexyxfixy
y
2
342
sin,235,
Solution 2
340340
801210
3206310
235,
42
42
422
24
2
2
4224
342
yy
yy
yy
y
xexy
fxe
yx
f
exy
fxe
x
f
xexy
fxyxe
x
f
xxyexyxfi
Solution 2
xxyxxyyxxeexyxy
f
xxyxxyyxxeexyyx
f
xyxexy
f
yxyyxyxyyeexyx
f
xxyxexy
f
xyxyxyexyex
f
yxxyxxeyxfii
xyxy
xyxy
yx
xyxy
xy
xyxy
xy
2cos2sin2
2cos2sin2
sin
2cos2sin2
cos
2cos
sin,
222
222
332
2
222
2
222
2
SMALL INCREMENTS & RATES OF CHANGE
Notation for small increment is Let then
i. A small increment in z, is given by
Where are small increments of the stated variablesii. Rate of change z wrt time, t is given by
,,...,, 21 nxxxfz
.
nxxx ,...,, 21
z
nn
xx
zx
x
zx
x
zz
...2
21
1
t
x
x
z
t
x
x
z
t
x
x
z
t
z n
n
...2
2
1
1
Example 3
The measurements of closed rectangular box are length, x = 5m, width y = 3m, and height, z = 3.5m, with a possible error of in each measurement.
What is the maximum possible error in the calculated value of the volume, V and the surface, S area of the box?
cm10
1
Solution 3
Volume of rectangular box:Possible error of the volume:
xyzV
x
y
z
V
zz
Vy
y
Vx
x
VV
Solution 3
Possible error of the volume:
xyz
Vxz
y
Vyz
x
VxyzV
43000
1.0351.05.351.05.33
zxyyxzxyz
zz
Vy
y
Vx
x
VV
Solution 3
Surface Area of rectangular box:Possible error of the volume:
yzxzxyA 222
x
y
z
A
zz
Ay
y
Ax
x
AA
Solution 3
Possible error of the surface area:
yxz
Azx
y
Azy
x
AyzxzxyA
222222
222
4601.032521.05.32521.05.3232
222222
zyxyzxxzy
zz
Ay
y
Ax
x
AA
Example 4
The radius r of a cylinder is increasing at the rate of 0.2cms-1 while the height, h is increasing at 0.5cms-1. Determine the rate of change for its volume when r=8cm and h =12cm.
Solution 4Volume of cylinder:
Rate of change:
hrV 2
3
1
t
r
r
V
t
h
h
V
t
V
h
r
125.0
82.0
?,
hdt
dh
rdt
drt
V
(1)
(2)
Solution 4From (1)Differentiate partially wrt t:
Substitute in (2):
2
3
1
3
2r
h
Vrh
r
V
4.70
2.01283
25.08
3
1
2.03
25.0
3
1
2
2
rhrt
V
IMPLICIT FUNCTIONS
DefinitionLet f be a function of two independent variables x and y, given by
constant. To determine the derivative of this implicit function:LetHence,
ccyxf ,,
yz
xz
dx
dy
dy
dy
y
z
dx
xd
x
z
x
z0
dx
dy
., cyxfz
Example 5Assume that y is a differentiable of x that satisfies the given function. Find using implicit differentiation.
2sin
323
2
342
yxxyxxeii
xxyexi
xy
y
dx
dy
Solution 5(i) Let
Then,
Therefore ,
323 342 xxyexz y
xexy
z
xyxex
z
y
y
34
632
42
24
xex
xyxe
dx
dyy
y
34
63242
24
THE CHAIN RULE
DefinitionLet z be a function of two independent variables x and y, while x and y are functions of two independent variables u and v.
The derivatives of z with respect to u and v as follows: Hence,
dv
dy
y
z
dv
xd
x
z
v
z
du
dy
y
z
du
xd
x
z
u
z
Example 6Let , where and
Find and
yexyz x22 vvux 32 2 uuvy 32
u
z
v
z
Solution 6
yexyz x22 x
x
exy
z
yeyx
z
2
2
2
22
vvux 32 2
32
4
2
uv
x
uvu
x
uuvy 322
3
6
12
uvv
y
vu
y
Solution 6Therefore
3222232223
2222
322233223
322
22
22
322613222623222
32212128122422
uuv
xx
uuv
xx
euuveuuuvuvexuyey
v
y
y
z
v
x
x
z
v
z
euveuuvuvvexuvyey
u
y
y
z
u
x
x
z
u
z
JACOBIAN FUNCTION
DefinitionLet be n number of functions of n variables
nfff ...,,, 21 nxxx ...,,, 21
nnn
n
n
xxxff
xxxffxxxff
,...,,
,...,,,...,,
21
2122
2111
JACOBIAN FUNCTION
Jacobian for this system of equations is given by:
OR
n
n
nn
n
n
x
f
x
f
x
f
x
f
x
f
x
fx
f
x
f
x
f
J
21
22
2
2
1
11
2
1
1
n
nnn
n
n
x
f
x
f
x
f
x
f
x
f
x
fx
f
x
f
x
f
J
11
2
2
2
1
2
1
2
1
1
1
Example 7Given and , determine the Jacobian for the system of equation.
yeu x 2sin2 yev x 2cos2
Solution 7Given and , determine the Jacobian for the system of equation.
yeu x 2sin2 yev x 2cos2
y
yyyy
yeye
yeyey
v
y
ux
v
x
u
J
xx
xx
4cos42cos2sin42cos42sin4
2sin22cos2
2cos22sin2
22
22
22
22
INVERSE FUNCTIONS FOR PARTIAL DERIVATIVES
DefinitionLet u and v be two functions of two independent variables x and y. . . Partial derivatives and are given by:
yxfvyxfu ,,, ,,,v
x
u
y
u
x
v
y
Jxu
v
y
J
yu
v
x
Jxv
u
y
J
yv
u
x
Example 8Given and , Find and
yeu x 2sin2 yev x 2cos2
,,,v
x
u
y
u
x
v
y
Solution 8Given and , Find and
yeu x 2sin2 yev x 2cos2
,,,v
x
u
y
u
x
v
y
y
ye
v
y
y
ye
v
x
y
ye
u
y
y
ye
u
x
xx
xx
4sin2
2sin
4sin2
2cos
4sin2
2cos
4sin2
2sin
22
22
Example 9Let and Find and
2223 43,3 vuxyxyxz
u
z
v
z
.52 vuy
HESSIAN FUNCTION
DefinitionLet f be a function of n number of variables . Hessian of f is given by the following determinant:
2
2
2
22
1
12
2
2
22
22
12
12
1
2
21
22
21
12
n
n
nn
n
n
n
n
x
f
xx
f
xx
f
xx
f
x
f
xx
fxx
f
xx
f
x
f
H
nxxx ...,,, 21
HESSIAN FUNCTION
Hessian of a function of 2 variables:Let f be a function of 2 independent variables x and y. Then the Hessian of f is given by:
22
2
2
2
2
2
22
2
2
2
yx
f
y
f
x
f
y
f
xy
fyx
f
x
f
H
HESSIAN FUNCTION
Stationary PointDefinition Given a function . The stationary point of occurs when and
Properties of Stationary Point
22
2
2
2
2
2
22
2
2
2
yx
f
y
f
x
f
y
f
xy
fyx
f
x
f
H
yxff , yxff ,0
x
f0
y
f
HESSIAN FUNCTION
Properties of Stationary Pointa) If H<0, then stationary point is a SADDLE
POINT
b) If H>0a) MAXIMUM POINT if
b) MINIMUM POINT if
c) If H=0, then TEST FAILS or NO CONCLUSION
0,02
2
2
2
y
f
x
f
0,02
2
2
2
y
f
x
f
Example 10Find and classify the stationary points of
223 23, yxyxxyxf
Solution 10Find stationary point(s):
1022
22
10263
2632
2
yxyx
yxy
f
yxx
yxxx
f
Substitute (2) in (1)
Stationary points:
3
4,0
3
4,0
0430430263
2
2
xx
yy
yyyyyyy
3
4,
3
4,0,0
Find the Hessian function:
4662
2
2
66
2
2
2
2
2
xH
yx
fy
f
xx
f
Determine the properties of SP:
Point Hessian: Conclusion
SP is a maximum point
SP is a saddle point
0,0
3
4,
3
4
4662 xH
02
06
08
2
2
2
2
y
fx
fH
08 H