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PARTIAL DERIVATIVES
(8, 9, 10 , 11)
Functions of Several VariablesLesson 8
Objectives:
At the end of the lesson you should be able to:
1. Define a function of two variables.
2. Evaluate a function.
3. Find the domain and range of a function.
4. Sketch a function.
5. Find the limit of a function.
6. Find the point of discontinuity of a function.
UTP/JBJ 1
Functions of Several Variables
A function of two variables is a rule that assigns to each
ordered pair of real number (x,y) in a set D (Domain) a
real unique number denoted by f (x,y).
The set D is the domain of f and its range R is the set of
values that f takes on, that is
UTP/JBJ 2
}),(/),({ Dyxyxf
Definition
UTPJBJ
Function is composed of two parts these are the Range and the Domain.
For examples,
?)
?)
?)1,()
?)1,1()
:
).1ln(),(
Ranged
Domainc
efb
fa
isWhat
yxyxf
3
?
1),( 2
fofrangetheisWhat
yxyxf
).,,()
).,,()
).4,2,2()
)25ln(),,( 222
zyxgofrangetheFindc
zyxgofdomainFindb
gEvaluatea
zyxzyxg
Domain of a FunctionGiven the following functions find the domain D:
xyoryx
yxyxf
,0
,),(
x
y
UTP/JBJ 4
a.
More Illustrative examples!!!
/UTPJBJ 5
22222
22
2404
4
53),(.
yxoryx
yx
yxyxfb
x
y
22
Level curves (contour curves) is a set of points
joined together at a constant elevation. It is a
method of visualizing a given function.
The other methods are arrow diagrams and graphs.
The level curves of a function of two variables are the
curves with equations f ( x, y ) = k , where k is a constant
( in the range of f ).
UTP/JBJ 6
Definition
Level Curves
Sketch some level curves of the function
.4),( 22 yxyxg
0,4/
1
4),(22
22
kk
y
k
x
yxyxgk
Let k = 1 , 2 , 4 y
x
UTP/JBJ 7
Limits
UTP/JBJ 8
Definition:
3
),(),(21
22
11
),(),(
,)
.),(lim,
),(),(),()
),(),(),()..
,),(lim
Lyxalongc
existnotdoesyxfthenLL
CalongbayxasLyxfifband
CalongbayxasLyxfifaei
Lyxf
bayx
bayx
Example 1: Limit
2),(lim,)
1),0(lim)
1)0,(lim)
:
lim
),(),(
),0(),(
)0,(),(
22
2
)0,0(),(
xxfxyLetc
yfb
xfa
Solution
yx
yx
xxyx
yyx
xyx
yx
What is the Limit ?UTP/JBJ 9
Evaluate
Example 2: Limit
18
18lim
)66(cos)3(6lim)2(coslim
:
)2(coslim
)3,6(),(
)3,6(),()3,6(),(
)3,6(),(
yx
yxyx
yx
yxxy
Solution
yxyx
Find the
UTP/JBJ 10
Do Nos. 5, 9, 17, and 18 on page 944
Continuity
UTP/ JBJ 11
Definition:
A function f of two variables is continuous at (a,b) if
f is defined on the Domain D at every point (a,b) on
D that is,
),(),(lim),(),(
bafyxfbayx
Or simply, the surface which is the graph of a continuous function has no hole or break on it.
UTP/JBJ 12
Example: Continuity
Find h(x,y)=g(f(x,y)) and the set on which h is continuous.
22
2
2
2
2
0/),(,
1
1)),((),(
:
.),(,1
1)(
xyoryx
yxyxD
yx
yxyxfgyxh
Solution
yxyxft
ttg
Solution:
UTP/JBJ 13
See the graph!!!
x
y
Example: Continuity
22
),( yxeyxF yx
Determine the set of points where F(x,y) is continuous
0)( 2 yx To be continuous, the values of
UTP/JBJ 14
Solve Nos. 28 and 34 on page 945
Partial Derivatives Lesson 9
Objectives:
At the end of the lesson you should be able to:
1. Define partial derivative.
2. Find the partial derivative of a function.
3. Define Chain Rule
4. Use the Chain Rule
5. Define Implicit Differentiation.
6. Use the Implicit Differentiation
UTP/JBJ 1
Partial Derivatives
UTP/JBJ 2
The function is a surface. The partial
derivatives of f at (a,b,c) are the slopes of the tangents to the two curves
),( yxfz
.21 CandC
x
y
z
1C
2C
1T2T
)0,,( baP
P(a,b,c)
Partial Derivative Rules
UTP /JBJ 3
1. Consider y as a constant and differentiate f(x,y) with
respect to x thus, the result is
2. Consider x as a constant and differentiate f(x,y) with
respect to y thus, the result is
x
zf
x
f
x
yxfx
),(
y
zf
y
f
y
yxfy
),(
Rules:
Higher Partial Derivatives
UTP /JBJ 4
Some notations are the following:
yx
f
yx
ff
xy
f
xy
ff
y
f
yy
ff
x
f
xx
ff
xy
yx
yy
xx
2
2
2
2
2
2
)(
)()(
)(
)()(
)(
)()(
)(
)()(
UTP/JBJ 5
Examples: Partial Derivative
1. Given
.,2),( 432xxxffindyxyxyxf
yxf xxx 48
Your Answer is
UTP/JBJ 6
2. Given .),(2
yxyx ffindeyxf
Example: Partial Derivative
22
22
2
3
2
2
22
)2()()2(
)(
yxyxyx
yxyxyx
yxx
eyxyef
xyeyyef
yef
Solution:
Is ?xyyx ff
UTP/ JBJ 7
3. Given :
.
,),,( 33445
zyxffind
zyzyxxzyxf
Example : Partial Derivative
233
333
3434
48
16
45
zyxf
zyxf
zyxxf
zyx
yx
x
Your solution is
What is the value of ?)2,2,1( zyxf
1536)4)(8(48)2()2)(1(48 233 zyxf
Practice Tasks
Solve the following as indicated:
1.
2.
3.
4.
.)1,2(,)0,1(;43),( 232 yx ffyyxxyxf
.;;),(y
f
x
f
y
xeyxf yx
.;;ln),(2
2
2
232
y
f
x
fxyyxyxf
UTP/JBJ 8
.;;),cos(),( 43yyyxyyx ffyxyxyxf
Partial Derivates of Three Variables:
1.
2.
.;4),,( 23zyxfyxzxyzyxf
zzyyxy fyzx
y
zezyxf ;sin),,(
22
Chain Rule
UTP/JBJ 9
s
y
y
z
s
x
x
z
s
z
t
y
y
z
t
x
x
z
t
z
tshytsgxwhereyxfzIfCase
t
dy
y
z
dt
dx
x
z
dt
dzthenthyand
tgxwhereyxfzIfCasedt
dx
dx
dy
dt
dythen
tgxandxfyIf
),(,),(),(.2
),(
)(,),(.1
.
),(,)(
The Tree Diagram for the Chain Rule
z
x y
t t
z
x y
u v u v
Let’s go to the examples!!!
UTP/JBJ 10
z
x y t
u vu v
u v
Examples: The Chain Rule
1.
2.
.;,, 2222
dt
dzeyexyxz tt
t
z
s
zesyesx
y
xz tt
,;1,,
Show the tree diagram for these
Let us solve the examples!
UTP/JBJ 11
Answers:
1. 22222 yxyeext
z tt
2. 22;
y
xse
y
se
t
z
y
xe
y
e
s
z tttt
UTP/JBJ 12
Practice More !
Solve Nos. 22-24 , page 974
Implicit Differentiation The Rules are:
y
x
F
F
yFxF
x
y
z
y
F
F
zFyF
y
z
z
x
F
F
zFxF
x
z
UTP/JBJ 13
Example: Implicit Differentiation
)(arctan zyzxwherey
zand
x
z
Find
Solution:
a) yyz
yz
yzyF
F
x
z
z
x
1)(
1)(
1)(1
12
2
2
b) yyz
z
yzy
yzz
F
F
y
z
z
y
2
2
2
)(1)(1
1
)(1
UTP/JBJ 14
Example : Implicit Differentiation
Find )(ln zxzywherey
zand
x
z
Your answers are:
a) 1)(
1
zxyF
F
x
z
z
x
b) 1)(
)(
zxy
zxz
F
F
y
z
z
y
UTP/JBJ 15
Do Nos. 41 and 44 on page 957
Worded Problem
The temperature at a point (x,y) on a flat metal plate is given by , where T is in and x,y in meters.
Find the rate of change of temperature with respect to
distance at the point (2,1) in the x-direction and in the y-
direction.
)1(
60),(
22 yxyxT
C0
You are to find )1,2()1,2( yx TandT
UTP/JBJ 16
3
10
3
20
)1(
120,
)1(
120222222
yx
yx
TT
yx
yT
yx
xT
Negative values mean that the temperatures are decreasing!!!
UTP/JBJ 17
Do Nos. 77 and 80 on page 958
Worded Problem
If resistors of ohms are connected in
parallel to make an R-ohm resistor, the values of R can
be found by the equation . Find the value
of .
321 ,, RandRR
321
1111
RRRR
ohmsRandRRwhenR
R9045,30 321
2
To solve must be constants and use implicit differentiation .
31 RandR
9
12
2
R
R
F
F
R
R
UTP/JBJ 18
Example on Tangent LinesFind the point of intersection of the tangent lines to the curve
at the points where t = 0 and t = 0.5. What is the angle of intersection of these two tangent lines?
ttttr cos,sin2,sin)(
Note: The tangent line at t=0, is the line thru a point with position vector r(0) and in the direction of the tangent vector r’(0).
Likewise, the tangent line at t =0.5 is the line thru a point with position vector r(0.5) and in the direction of the tangent vector
r’(0.5).
UTP/JBJ 19
The angle of intersection of the two curves is the angle between the two tangent vectors to the curves at the point of intersection.
Use:
ba
ba cos
UTP/JBJ 20
Tangent Planes and Linear Approximations
Lesson 10Objectives
At the end of the lesson you should be able to:
1. Define tangent planes.
2. Find an equation of the tangent plane to the surface.
3 Find the equation of the normal line to the surface.
4. Define linear approximation.
5. Find the linear approximation.
6. Define differential .
7. Find the differential.UTP/JBJ 1
Tangent Planes
Definition
Tangent plane to the surface S at the point P is defined to be the plane that contains both tangent lines and .1T
2T
z
y
x
1T
2T1C
2C
*))(,())(,( 0000000 yyyxfxxyxfzz yx
Tangent Plane
* Equation of the tangent
plane to the surface
z = f (x,y) at the point
),,( 0000 zyxP
UTP/JBJ 2
P
Example 1: Tangent Plane
Find the equation of the tangent plane to the given surface
at the point (-1,2,4).yyxz 24 22
y
z
x
028
)2(2)1(84
)()(
222
88);,(
000
zyx
yxz
yyfxxfzz
yf
xfyxfz
yx
y
x
UTP/JBJ 3
0P
NL
Example 2: Normal Line
Find the equation of the normal line (NL) on the said
point of the given surface in Example 1.
Recall the vector equation of a line. You need a point and a direction for the line.
Do you have these two?
cbatzyxzyx
vtrr
,,,,,, 000
0
1,2,84,2,1,, tzyx
Your vector equation of the normal line!!
UTP/JBJ 4
Example 3: Tangent Plane
Find the points on the ellipsoid where
the tangent plane is parallel to the plane 3x - y + 3z = 1
132 222 zyx
z
y
x
0Pczzf
byyf
axxf
zyxzyxf
z
y
x
0
0
0
222
66
44
22
132),,(
Recall two parallel vectors!!
a = k v
UTP/JBJ 5
kzky
kzkykx
kzyx
zyxcbazyx
00
000
000
000000
;2
1
33;2;3
3,1,33,2,
3,2,2,,6,4,2
Substitute these values to the ellipsoid and you get the
required point on the surface of the solid. The values are:
)28.,14.,85.(),,(28.0 0000 zyxPthenk
UTP/JBJ 6
Do Nos. 1-3 on page 966
Linear ApproximationDefinition:
Linear approximation L(x,y) of f (x,y) at the point (a,b) is
the function defining the z-values on the tangent plane.
Recall the equation of the tangent plane
Where the point given
is substituted to the above formula, you get
))(,())(,( 0000000 yyyxfxxyxfzz yx
)),(,,(),,( 000 bafbazyx
))(,())(,(),( bybafaxbafbafz yx
UTP/JBJ 7
Example 4: Linearization
Find the linear approximation of the function
and use it to approximate f (6.9, 2.06).
)2,7()3ln(),( atyxyxf
13
)2(3)7(10
))(,())(,( 0000000
yxz
yxz
yyyxfxxyxfzz yx
33
3)2,7(
13
1)2,7(
yxf
yxf
y
x
28.01)06.2(39.6)06.2,9.6( fz
Then approximate
UTP/JBJ 8
Example 5: Linear Approximation
Compute the linear approximation of
at (0,0).
yxexyxf 2
2),(
1)0,0(;222)0,0(22
yxy
yxx efexf
yxz
yxz
yyyxfxxyxfzz yx
21
)0(1)0(21
))(,()(),( 0000000
UTP/JBJ 9
Take note !
Linear approximation is also called tangent plane approximation
Differential
UTP/JBJ 10
For a function y = f (x), the differential d x is an independent
variable and it is a real number.
The differential d y is defined as d y = f’(x)d x.
x
yy = f (x)
dyy
x
yd
y
x increment in xchange in height of the curvechange in height of the tangent line
Total DifferentialFor , a differentiable function, d z is called the
total differential wherein dx and dy are independent
variables. Total differential is defined by
),( yxfz
dyy
zdx
x
zdz
dyyxfdxyxfdz yx
),(),(
* Its value is approximately equal to increment in z
*
)( z
To find the increment in z, use),(),( yxfyyxxfz
UTP/JBJ 11
Example 6: Total Differential
Let and x changes from (3,-1) to (2.96, -0.95).
Find the total differential and the increment in z. Is there a
difference between the two?
22 3yxyxz
a) Increment in z )( z
72.0
1528.14
)1(3)1(33)95.0(3)95.0)(96.2()96.2(
3)(3))(()(2222
2222
z
z
z
yxyxyyyyxxxxz
UTP/JBJ 12
b) Total differential d z
73.0
45.028.0
)05.0)(9()04.0)(7(
)6(2
dz
zd
dz
dyyxdxyxzd
dyy
zdx
x
zzd dx = 2.96-3 =-0.04
dy = -0.95+1 = 0.05
Proceed to application!
UTP/JBJ 13
How much is the difference?
Example 7: Application for Differentials
The length and width of a rectangle are 30 cm. and 24 cm., respectively, with an error in measurement of at most 0.1cm in each. Use differential to estimate the maximum error in the evaluated area of the rectangle.
UTP/JBJ 14
Area = L W
24.5 cmdA
LdWWdLdA
dWW
AdL
L
AdA
WLA
Example 8: Application for Differentials
Use differentials to estimate the amount of tin in a closed can with diameter 8 cm. and height 12 cm. if the tin is 0.04 cm. thick.
UTP/JBJ 15
hrV 2cmdh
cmdr
08.0
04.0
Use:
..08.16
2 2
2
cmcudV
dhrdrrhdhh
Vdr
r
VdV
hrV
Practice More
On page 967 Do Nos. 32, 34 and 35.
UTP/JBJ 16
Directional Derivatives and Gradient Vectors
Lesson 11
Objectives:
At the end of the lesson you should be able to:
1. Define directional derivative.
2. Find directional derivative.
3. Define gradient vector.
4. Find gradient vector.
5. Solve applications for directional derivative and gradient vector.
UTP/JBJ 1
Directional Derivative
Directional derivative is the rate of change of a function of
two or more variables in a given direction.
Let z = f (x,y), then the partial derivatives are
defined as
yx ff ,
h
yxfhyxfyxf
h
yxfyhxfyxf
hy
hx
),(),(lim),(
),(),(lim),(
0000
000
00000
000
and represents the rates of change of z in the x- and y-
directions of the unit vectors i and j.
UTP/JBJ 3
Definition
The directional derivative of f at in the direction of a
unit vector u=<a,b>
is
If this limit exists.
),( 00 yx
h
yxfhbyhaxfyxfD
hu
),(),(lim),( 0000
000
Comparing Definition 2 to the first, if u = i= <1,0>,then
and if u=j=<0,1> , then .
It means that the partial derivatives of f with respect to x
and y are special cases of directional derivatives.
xi ffD
yj ffD
UTP/JBJ 4
Theorem
If f is a differentiable function of x and y, then f has a
directional derivative in the direction of the unit vector u = <a,b>
and
byxfayxfyxfD yxu ),(),(),(
Lets have an example!
UTP/JBJ 5
See the illustration for you to visualize
Example 1: Directional Derivative
Find the directional derivative of at
the given point (1,2) and u is the unit vector given by angle , .
23 43),( yxyxyxf
9.3),(
6sin)43(
6cos)33[(),(
),(),(),(
2
yxfD
yxyxyxfD
byxfayxfyxfD
u
u
yxu
2
1,
2
3
sin,cos
u
uFind u=<a,b>And then use the formula
UTP/JBJ 6
Try Another
6
Directional Derivative in the Space
Definition
The directional derivative of f at in the direction
of the unit vector u= (a,b,c) is
),,( 000 zyx
h
zyxfhczhbyhaxfzyxfD
hu
),,(),,(lim),,( 000000
0000
if this limit exists.
Hence the formula that can be used is,
czyxfbzyxfazyxfzyxfD zyxu ),,(),,(),,(),,(
UTP/JBJ 7
Example 2: Directional Derivative
Find the directional derivative of g(x,y,z) = (x +2y +3z) at the
given point ( 1,1,2) in the direction of v = 2j-k.
2!/
zyxf
zyxf
zyxf
u
zyx322
3;
32
1;
322
1
5
1,2,0
56
1
52
1
53
20),,(
2
1,
3
1,
6
1
5
1,2,0),,(
zyxfD
zyxfD
u
u
Unit vector
UTP/JBJ 8
2. Given are the following:
3
1,
3
1,
3
1
);1,2,1(;),,( 32
u
andPzyxzyxf
a) Find gradient of f at the given point.
b) Find the rate of change of f at P in the direction of the unit vector.
UTP/JBJ 9
The Gradient VectorDefinition:
If f is a function of two variables x and y, then the gradient of f
is the vector function
jy
fi
x
ffff yx
,
If f is a function of three variables x,y, and z, then the
gradient of f is the vector function
k
z
fj
y
fi
x
fffff zyx
,,
UTP/JBJ 10
Example 3: Gradient Vector
Find the gradient of
0,3)3,1(
ln,)3,1(
,)3,1(
:
).3,1(ln),()
f
xx
yf
fff
Solution
Patxyyxfa
yx
UTP/JBJ 11
Lets Go to the Next Problem !!
Example 4: Gradient Vector
Find the gradient of
).2,0,3(),,( 2 Patexzyxf zy
Try and solve this!!
UTP/JBJ 12
Directional Derivative and Gradient Vector
Directional derivative can be found using the gradient
vector.These formulas are the following:
a) For f (x,y),
b) For f (x,y,z),
baffufyxfD yxu ,,),(
cbafffzyxfD
ufzyxfD
zyxu
u
,,,,),,(
),,(
UTP/JBJ 13
Practice Task
Do the following problems. Find the directional derivatives at the given point and and a given vector:
1. The function at (3,4) in the
direction of the vector <-1.2>.
Guided Questions?
a) What is your u?
b) What are the values for ?
c)
)ln(),( 22 yxyxf
)4,3(;)4,3( yx ff
?)4,3( fDu
UTP/JBJ 14
b) The temperature T in the metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,2,2) is 120 degrees Centigrade.
What is the rate of change of T at (1,2,2) in the direction toward the point (2,1,3)?
How to solve???
UTP/JBJ 15
Maximal Direction Property of the Gradient
Suppose a function f is differentiable at the point and
that the gradient of f at the point is not equal to zero then,
a) The largest value of the directional derivative at
is and the unit vector points in the direction of
b) The smallest value of the directional derivative at
is and occurs when the unit vector points in the direction of .
0P
fDu 0P
0f .0f
fDu
0P0f
0f
UTP/JBJ 16
Example 5 : Maximum and Minimum Rates of Change
Find the maximum and minimum rates of change of the
function at the point (1,3).22),( yxyxf
UTP/JBJ 17
Maximum /Minimum rates of change
yxyx ffffff ,,
Example 6: Maximum/Minimum Rate of Change
In what direction as increasing most rapidly
at the point P(2,1)? What is the maximum rate of change?
In what direction is it decreasing most rapidly?
xyxeyxf 2),(
UTP/JBJ 18
Do the following:
On page 987
Nos. 10, 16, 26, 32 and 34
PRACTICE MORE!!
UTP/JBJ 19
![Ch-8, Integrations or Anti-Derivatives [ ch-8 ] + differentiations [ ch-6 ]](https://img.pdfslide.us/doc/110x75/577cdb401a28ab9e78a7bada/ch-8-integrations-or-anti-derivatives-ch-8-differentiations-ch-6-.jpg)
