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Objectives :
• Identify domain and range of function of two and three variables.
• Sketch graphs and level of curves of functions of two and three variables
• Compute first and second partial derivatives.
Definition :
A function of two variables is a rule f thatassigns to each ordered pair (x,y)
in a set D a unique number z = f (x,y).
The set D is called the domain of the
function, and the corresponding values
of z = f (x,y) constitute the range of f .
Find the domains and range of the following functions and evaluate f at the given points.
1a) , ;
1
Eva1uate 3, 2
6: , 1 0, 1 , 3, 2
2
, ,
b) , ;2
Eva1uate 2,3 , 2,1
x yf x y
x
f
Answer D x y x y x f
z f x y range is z z
xyf x y
x y
f f
EXAMPLE 1
Find the domains and range of the following functions and evaluate f at the given points.
2 2
2 2
2
c) , 25 ;
Eva1uate 2,3 , 7, 4
: , 25 ,
, , 0 5
) , 3ln
Eva1uate 3, 2
f x y x y
f f
Answer D x y x y
z f x y range is z z
d f x y y x
f
Continue…
A set of points where f is a constant is
called a level curve. A set of level
curves is called contour map.
Figure 13.1.4 (p. 909)
Figure 13.1.8 (p. 911)
Figure 13.1.10 (p. 912)
Table 13.1.2a (p. 913)
Table 13.1.2b (p. 913)
Table 13.1.2c (p. 913)
Table 13.1.2d (p. 913)
Table 13.1.2e (p. 913)
Table 13.1.2f (p. 913)
2 2
2 2 2 2
, 9
or
3
f x y x y
z x y
2 2
, sin2
x yf x y
Sketch the level curves of the function
2 2, 9
0,1,2,3
f x y x y
for k
EXAMPLE 2
2 2, 9f x y x y
Limits Along Curves
For a function one variable there two one-sided limits at a point
namely
reflecting the fact that there are only two directions from which x
can approach
0x
)(lim and )(lim00
xfxfxxxx
0x
Function of
2 variables
Function of
3 variables
))(),((lim),(lim0
)C along(00 ),(),(
tytxfyxfttyxyx
))(),(),((lim),,(lim0
)C along(000 ),,(),,(
tztytxfzyxfttzyxzyx
EXAMPLE 3
Find the limit of along 22
),(yx
xyyxf
axis- the)( xa
axis- the)( yb
xyc line the)(
xyd line the)(
2 parabola the)( xye
The process of differentiating a function of several variables with respect to one of its variables while keeping the other variable(s) fixed is called partial differentiation, and the resulting derivative is a partial derivative of the function.
The derivative of a function of a single variable f is defined to be the limit of difference quotient, namely,
Partial derivatives with respect to x or y are defined similarly.
0limx
f x x f xf x
x
If , then the partial derivatives of f with
respect to x and y are the functions and ,
respectively, defined by
and
provided the limits
exists.
,z f x y
xf yf
0
, ,, limx
x
f x x y f x yf x y
x
0
, ,, limy
y
f x y y f x yf x y
y
We can interpret partial derivatives as rates of change.
If , then represents the rate of
change of z with respect to x when y is fixed. Similarly,
is the rate of change of z with respect to y when x is
fixed.
),( yxfz xz /
yz /
If find and
Solution :
,24),( 22 yxyxf )1,1(xf ).1,1(yf
xyxf x 2),(
2)1,1( xf
yyxf y 4),(
4)1,1( yf
EXAMPLE 4
If calculate and .,sin),(
yx
xyxf
x
f
y
f
Implicit Differentiation (i)
then, variableone offunction a as function
abledifferenti a defines implicitly 0) If : Theorem
xy
F(x,y
),(
),(
yxF
yxF
dx
dy
y
x
EXAMPLE 5
1543
by defined implicitly is offunction a as if , Find
35 x-xy-y
xydx
dy
35
512
35
)512(
),(
),(
Thus
1543),( so
,1543 have We
4
2
4
2
35
35
y
x
y
x
yxF
yxF
dx
dy
x-xy-yyxF
x-xy- y
y
x
Solution :
EXAMPLE 6
Find and if z is defined implicitly as a function of
x and y by the equation
x
z
y
z
054 )(
16 )(
3222
333
yzzxyzxb
xyzzyxa
Implicit Differentiation (ii)
If z = F (x, y, z) then and
z
x
F
F
x
z
z
y
F
F
y
z
EXAMPLE 7
yx
f
y
f
xff yyx
2
)(x
2
2
)(x
f
x
f
xf
xf xxx
xy
f
x
f
yff xxy
2
)(y
2
2
)(y y
f
y
f
yff yyy
yxxyxy
fy
f
xxyxyxx
fx
f
xxyyxxyy
fy
f
yxyyxxyx
fx
f
f
xyxyxyxy
fyxxyyxyxx
f
f
yxyxyxfffff
yyy
yyx
xxy
xxx
yx
yyyxxyxx
2422
32422
3233
2333
42243233432
432
,
6)3()(
46)3()(
46)42()(
122)42()(
are of derivative second The
3)( , 42)(
are of derivativefirst The
),( if ,, Find
EXAMPLE 8
2323 2),( yyxxyxf Find the second partial derivatives of
xy
yyyxxyxx eyxyxfffff )(),(given if Find ,,,
EXAMPLE 9
EXAMPLE 10
thenexist, and , of sderivative
partial with )( and )( ),( If : Theorem
hgf
x,yhvx,yguu,vfz
,
,
y
v
v
z
y
u
u
z
y
z
x
v
v
z
x
u
u
z
x
z
variables twooffunction for diagram threeThe
yvxyu
yxvyux
v
v
z
x
u
u
z
x
z
xy
vyx
x
vxy
y
uy
x
u
yxvxyu
vv
zu
u
z
vuz
yzxz
yxvxyuv
sin43
)sin2(2)(3
obtain weHence
cosy , sin2 ,2 ,
,sin and fromSimilarly
2 and 3
obtain we, From
. and rulechain a Use
sin, and,uz Suppose
22
22
22
22
2
23
2233
Solution :
EXAMPLE 11
yxyxxyxy
yvxxyuy
z
yxyx
yxyxyxy
yvxyux
z
yxvxyu
yvxxyu
yxvxyuy
v
v
z
y
u
u
z
y
z
2sin6
cos)sin(2)(6
cos26
sin43
sin)sin(4)(3
sin43
obtain we,sin and ngSubstituti
cos26
)cos(2)2(3
.continue..
453
2222
22
2362
2222
22
22
22
22
If where and ,
find when t = 0 .
,3 42 xyyxz tx 2sin ty cos
dt
dz
Solution :
The Chain Rule gives
we calculate the derivatives, since we get
dt
dy
y
z
dt
dx
x
z
dt
dz
432 yxyx
z
32 12xyxy
z
,3 42 xyyxz
EXAMPLE 12
Then from and , we get
So,
Therefore
tdt
dx2cos2
tx 2sin ty cos
tdt
dysin
)sin)(12()2cos2)(32( 324 txyxtyxydt
dy
y
z
dt
dx
x
z
dt
dz
)sin)(cos2sin122(sin)2cos2)(cos3cos2sin2( 324 tttttttt
6)0)(00()2)(30(0
tt
z
rules. theseformulate help todconstructe
becan diagrams treeand , variablesofnumber any
of functions composite toapplied becan ruleChain
322 and ,12 ,43 with 3
if find torulechain theUse
tztytxyzxw
dtdw
Solution :
w
x
y
z
t
t
tdt
dz
z
w
dt
dy
y
w
dt
dx
x
w
dt
dw
EXAMPLE 13
16)320(3
)12(9)(6)43(12
9612
)3(3)2(3)6(2
obtain wediagram, tree theFrom
2
232
2
2
t-tt
ttttt
ytztx
tyztx
dt
dz
z
w
dt
dy
y
w
dt
dx
x
w
dt
dw
The pressure P (in kilopascals), volume V (in liters), and
temperature T (in kelvins) of a mole of an ideal gas are related by
the equation Find the rate at which the pressure is
changing when the temperature is 300 K and increasing at a rate
0.1 K/s and the volume is 100 L and increasing at a rate of 0.2 L/s.
Solution :
If t represent the time elapsed in seconds, then at the given
instant we have
.2.0 ,100 ,1.0 ,300 dt
dVV
dt
dTT
.31.8 TPV
EXAMPLE 14
dt
dV
V
T
dt
dT
Vdt
dV
V
P
dt
dT
t
P
dt
dP2
31.831.8
The chain rule gives
The pressure is decreasing at a rate of about 0.042 kPa/s.
04155.0
)2.0(100
)300(31.8)1.0(
100
31.82
Figure 13.6.2 (p. 960)
Theorem 13.6.3 (p. 961)
Figure 13.6.5 (p. 964)
Theorem 13.6.5 (p. 964)
Objective :
• Compare absolute extrema and local extrema.
• Locate critical points and determine its
classification using second partial derivatives test.
Definitions 13.8.1 and 13.8.2 (p. 977)
Figure 13.8.2 (p. 977)
function. a of values
minimumor maximumeither refer to extremum wordThe
point. saddleor point minimum point, maximum-
:iespossibilit three
has )(graph of point) stationary(or point Critical
x,y fz
R.in )(every for )( )(such that D,Rregion
a exists thereif ,function theof minimum local a is )((ii)
R.in )(every for )( )(such that D,Rregion
a exists thereif ,function theof maximum local a is )( (i)
Din b)(a, that and D,domain ain defines
variables twooffunction a is )( Suppose
x,yx,yfa,bf
fa,bf
x,yx,yfa,bf
fa,bf
x,yfz
Definition
Point (a,b,f (a,b)) is a Local Maximum
Point (a,b,f (a,b)) is a Local Minimum
A pair (a,b) such that and is
called a critical point or stationary point. To find out
whether a critical point will give f (x, y) a local maximum or
a local minimum, or will give a saddle point, we use
theorem : Second Derivative Test.
0),( baf x
0),( baf y
0)G( if made becan sconclusion No (iv)
0)G( ifpoint saddle a is )),(,,( (iii)
0),( and 0)G( if minimum local a is ),( (ii)
0),( and 0)G( if maximum local a is ),( (i)
Then
)],([),(),(),( ),(
),( ),()G(
Let R.in point critical a is
b)(a, and Rregion aon sderivative partial second continuous
has that variables twooffunction a is )( that Suppose
2
a,b
a,bbafba
bafa,bbaf
bafa,bbaf
bafbafbafbafbaf
bafbafa,b
x,yf
xx
xx
xyyyxx
yyxy
xyxx
Test Derivative Second : Theorem
xyyxyxf
yxyxyxf
yxy
yxf
111),( )iii(
16),( )ii(
3),( (i)
44
23
Find the critical points of the following functions and
determine whether f (x,y) at that point is a local maximum or a
local minimum, or value at a saddle point.
EXAMPLE 15
Solution :
40)2(2)G(
2 0 2
)G( and derivative second theFind : 3 Step
1y 01
0 02
uslysimultaneo 0 and 0 Solve :2 Step
1 , 2
sderivative partialfirst theFind : 1 Step
3),( )i(
2
2
2
y
23
yyfffx,y
yfff
x,y
y
xx-
ff
y fxf
yxy
yxf
xyyyxx
yyxyxx
yx
x
point maximum a is )3
2(0,-1, , Therefore
3
210
3
)1()1,0(
02)1,0(
04)1(4G(0,-1) (0,-1)At
point saddle a is )3
2(0,1,- Therefore,
3
210
3
1)1,0(
04)1(4G(0,1) (0,1)At
(0,-1) and (0,1)point Test the : 4 Step
3
f
f
-
f
-
xx
.on valueextremum
theisfunction afor extremum Absolute
region.given only
on )( of valueextremum thefind to
useful more isIt ns.applicatio practicalmost for
enough not is extremum local thefindingClearly
region given any
x,yf
Definition
region. in the minimum absolute theis minimum local the
then region, in thepoint saddle no and maximum local
no is thereand minimum local aonly assumes )( If (ii)
region. in the maximum absolute theis maximum local the
then region, in thepoint saddle no and minimum local
no is thereand maximum local aonly assumes )( If (i)
R.region ain continuous is )( Suppose
x,yf
x,yf
x,yf
Theorem
.0,0:),(region in the
6),( of extremum absolute theFind 33
yxyxR
xyyxyxf
Solution :
(2,2)point
critical theonly test weR,region theoutside is (0,0) Since (2,2). and
(0,0) points criticalobtain weusly,simultaneo equations theSolving
063 and 063
have weThus
usly.simultaneo 0),( and 0),(such that )( Find
points. critical Find : 1 Step
22
xyyx
yxfyxfx,y yx
EXAMPLE 16
12 )2,2( thus6 )(
6 )2,2( thus6 )(
12)22( thus6)(
(2,2) and (2,2) (2,2), Compute : 2 Step
xyyy
xyxy
xxxx
yyxyxx
fyx,yf
fx,yf
, fx x,yf
fff
minimum. local a is
8)2)(2(622)2,2(
Therefore
012)2,2( and 0108)6()12(21G(2,2)
point. critical heclassify t and G(2,2) Compute : 3 Step
33
2
f
f-- xx
R.in minimum absolute theis
-8(2,2) minimum local thetherefore
R,in maximum local no is thereSince
f
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