EE107 SP 02B Directional Derivatives Directional Derivatives
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Transcript of EE107 SP 02B Directional Derivatives Directional Derivatives
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7-1
EE107 Lec2B
Directional Derivatives
7.1 DIRECTIONAL DERIVATIVES
Directional derivatives give the rate of change of a function in a given direction. Consider
the following figure.
i,jand kare unit vector in the direction of positive x, y, zaxis respectively. A vectorrcan
be defined as zyxzyx ,,kjir vector from 0 to zyx ,, .
Therefore, we have
ir hzyhx ,, , jr hzhyx ,, , kr hhzyx ,, .
By using the limit definition, we have
h
fhf
h
zyxfzyhxfr
x
fzyx
x
f
hh
)()(lim
,,,,lim,,
00~
rir
h
fhf
h
zyxfzhyxfr
y
fzyx
y
f
hh
)()(lim
,,,,lim,,
00~
rjr
h
fhf
h
zyxfhzyxfr
z
fzyx
z
f
hh
)()(lim
,,,,lim,,
00~
rkr
x
f
,
y
f
and
z
f
measure the rate of change of f in the direction of i,jand k
respectively.
z
xy
(x,y,z)
(0,0,1)
(1,0,0) (0,1,0)
0ij
r
k
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7-2
DEFINITION 7.1. Let kjiu cba be any given unit vector and point zyxp ,, .
Then, the directional derivative of fin the direction of u at the point p, denoted by
pfDu , is given by
h
zyxfhczhbyhaxfpfDh
u,,,,lim
0
pfDpx
fi
, pfDp
y
fj
, pfDp
z
fk
.
It is difficult to find the directional derivatives by Definition 7.1. We need a TOOL.
DEFINITION 7.2. Iff is a function, of which the partial derivatives exist at the point
zyxp ,, , then the gradient off at the pointpis defined by
kji pz
fp
y
fp
x
fpf
.
Recall that a unit vectoruhaving the same direction as r is given by rr
u1
.
THEOREM 7.1. If
zyxf ,, is a differentiable function at any given point
zyxp ,, , then the directional derivative offat point pin the direction of any given
unit vector kjiu cba is
zyxfczyxfbzyxfa
pfpfD
zyx
u
,,,,,,
u
THEOREM 7.2. The maximum value of pfDu at the point yxp , is pf
and it occurs in the direction of pf .
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7-3
EXAMPLE 7.1. (A) Find the directional derivative of xyezzyxf 2,, at the point
3,2,1 in the direction of the vector kjir 53 . (B) What is the maximum rate of
increase at the point 3,2,1 ?
SOLUTION. (A):
22122 18233,2,1,, eefyezzyxf xxy
x
22122 9133,2,1,, eefxezzyxf yxy
y
221 6323,2,12,, eefzezyxf yxy
z
By Definition 7.2, kjikji 236369183,2,12222
eeeef . A unit
vectoruhaving the direction of kjir 53 is
kjirr
u 5335
11
.
By Theorem 7.1,
35
15
35
30
35
9
35
54
5335
1
69183,2,13,2,1
2222
222
eeee
eeeffDu kjikjiu
(B) By Theorem 7.2, the maximum rate of increase offat 3,2,1 is
2444 2136813243,2,1 eeeef .
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7-4
EXAMPLE 7.2. Suppose a rectangular coordinate system is located in space such that the
temperature Tat the point zyx ,, is given by 222100 zyxT .
(a) Find the rate of change of T at the point 2,3,1 in the direction of the
vector kjir .
(b) In what direction from the point 2,3,1 does T increase most rapidly?
What is the maximum rate of change ofTat 2,3,1 ?
SOLUTION.
(a) 222100 zyxT
4910023122002,3,1200
4915023132002,3,1200
495023112002,3,1200
22222222
22222222
22222222
zz
yy
xx
TzyxzT
TzyxyT
TzyxxT
By Definition 7.2, 50 150 100 50
1,3, 2 3 249 49 49 49
T i j k i j k . A unit
vectoruhaving the direction of kjir is
kjirr
u
3
11 .
By Theorem 7.1,
349
10010015050
349
1
3
1
49
100
49
150
49
502,3,12,3,1
kjikjiuTTDu
(b) The maximum rate of change of Tat 2,3,1 occurs in the direction of the
gradient, that is, in the direction of the vector kji 23 . By Theorem 7.2, the
maximum rate of increase ofTat 2,3,1 is
2500 22500 10000
1,3, 2 3.82401 2401 2401
T .
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7-5
7.2 EXTREMA OF FUNCTIONS
DEFINITION 7.3. Iff is A function of two variables is said to have a local maximumat
ba,if there is a region R containing
ba,such that
bafyxf ,, for all other pairs
yx, inR. It follows that if baf , is a local maximum, then 0, bafx and 0, bafy .
DEFINITION 7.4. Iff is A function of two variables is said to have a local minimumat
dc, if there is a region R containing dc, such that dcfyxf ,, for all other pairs
yx, inR. It follows that if dcf , is a local maximum, then 0, dcfx and 0, dcfy .
NOTE.
The local maxima and minima are called the extremaoff.
NOTE.
The converse of Theorem 7.3 is false, that is, if 0,, bafbaf yx , it does notnecessarily follow that ba, is the local extrema off. If 0,, bafbaf yx , then ba, is
called the critical point off. Critical points that are not the local extrema are called the saddle
points.
THEOREM 7.3. Iff has first partial derivatives and ba, is the local extrema off,
then 0,, bafbaf yx .
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7-6
TEST FOR EXTREMA
Let f be a function of two variables which has continuous second partial derivatives on a
regionR and let
2
,,,, yxfyxfyxfyxG xyyyxx
for all yx, in R. If ba, is in R and 0, bafx , 0, bafy , then the following
statements hold.
(i) If 0, baG and 0, bafxx , then baf , is a local maximum off.
(ii) If 0, baG and 0, bafxx , then baf , is a local minimum off.
(iii) If 0, baG , then baf , is not an extremum, that is, ba, is a saddle point.
(iv) If 0, baG , then no conclusion can be made from this test.
ABSOLUTE EXTREMA OF FUNCTIONS
If a function fof two variables is continuous on a closed region R, then f has an absolute
maximum baf , and an absolute minimum dcf , for some ba, and dc, inR. This
means that
bafyxfdcf ,,,
for all yx, inR.
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7-7
EXAMPLE 7.3. Find the extrema offif 242, 32 yxyxyxf .
SOLUTION. 242, 32 yxyxyxf .
234,
44,
yxyxf
yxyxf
y
x
By Theorem 7.3, if ba, is the local extrema off, then 0,, bafbaf yx .
4
3034,
044,
22 y
xyxyxf
yxyxyxf
y
x
Solving these simultaneously gives us the pairs of critical points 0,0 and 34,34 .
Use the Test for Extrema to find the extrema off. The second partial derivatives off
are 4, yxfxx , yyxfyy 6, and 4, yxfxy .
At 0,0 : 40,0 xxf , 00,0 yyf and 40,0 xyf .
016404
,,,,
2
2
yxfyxfyxfyxG xyyyxx
0,0f is not an extremum, that is, 0,0 is a saddle point
At 34,34 : 434,34 xxf , 834,34 yyf and 434,34 xyf .
016484
,,,,
2
2
yxfyxfyxfyxG xyyyxx
034,34 G and 034,34 xxf ,
34,34f is a local minimum off.
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7-8
EXAMPLE 7.4. A rectangular box with no top and having a volume of 12 m3 is to be
constructed. The cost per m2 of the material to be used is $4 for the bottom, $3 for two of the
opposite sides and $2 for the remaining pair of opposite sides. Find the dimensions of the box
that will minimize the cost.
SOLUTION. {DIY}
Volume 12
xyz m
3
xyz
12
Cost : 4 3 2 2 2 4 6 4C xy xz yz xy xz yz
12 12 72 484 6 4 4C xy x y xy
xy xy y x
2
484
xC y
x ;
2
724
yC x
y
To determine the possible extrema, we must find the simultaneous solutions of
2
484 0
xC y
x and
2
724 0
yC x
y .
These equation imply that 2 12x y and 2 18xy . Substituting 212y x in 2 18xy
gives us 318 144x . Solving for x we obtain 2x . Since 212y x , the
corresponding value ofy is 3. The Test for Extrema can be used to show that these
values ofx andy determine a minimum value of C. Finally,12
2zxy
.
The minimum cost occurs if the dimensions of the box are 3 m 2 m 2 m as
shown below.
x
y
z
2 m
3 m
2 m
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7-9
7.3 LAGRANGE MULTIPLIERS {SKIP P 9-12; READ P13-19: R3 SURFACES }
NOTE.
is called the Lagrange multiplier. Theorem 7.4 provides candidates for extreme points.
SUMMARY.
To find the extrema offsubject to the constraint 0, yxg , take the following steps.
STEP 1 : Define a new function yxgyxfyxL ,,, .
( is independent ofx andy)
STEP 2 : Find the critical points for the function yxL , by solving the system
0, yxLx
0, yxLy
0, yxL
.
THEOREM 7.4. Iffhas a local extrema at ba, subject to the constraint 0, yxg ,
then there exists a value c such that at cbayx ,,,, ,
0, yxLx
0, yxLy
0, yxL
.
Here, yxgyxfyxL ,,, .
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7-10
EXAMPLE 7.5. Find the absolute extrema of yyxyxf 223, on the region
4|, 222 yxyxE .
SOLUTION.
(i) Extrema at 4|,222
yxyxE {This forms the constraint}
12,
6,
yyxf
xyxf
y
x
By Theorem 7.3, if ba, is the local extrema off, then 0,, bafbaf yx . Hence,
2
1012,
006,
yyyxf
xxyxf
y
x
21,0 is a critical point in 4|, 222 yxyxE .
Innerregion
ofE
BoundaryofE
y
x
y
x
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7-11
(ii) Extrema at 4|, 222 yxyxE
To find the extrema of f at the boundary, use Theorem 7.4. Let
04, 22 yxyxg . Hence,
43, 2222 yxyyxyxL 026, xxyxLx (1)
0212, yyyxLy (2)
04, 22 yxyxL
. (3)
(1) y : 026 xyxy
(2) x : 022 xyxxy )
04 xxy
014 yx 0 x or 41y .
Substitute 0x and 41y in (3), we obtain
2
04022
y
y
463
1663
044122
x
x
The critical points are 2,0 , 2,0 , 41,463 and 41,463 .
Findfat these critical points to identify the absolute maximum and minimum.
2 2
0, 1 2 3 0 1 2 1 2 1 4 1 2 0.25f
622032,0 22 f
222032,0 22 f
125.124141463341,463 22 f
125.124141463341,463 22 f
The absolute maximum is 12.125 and the absolute minimum is 0.25 .
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7-12
EXAMPLE 7.6. The temperature at any given yx, of the curve 1124 22 yx is
given by xyxT 2244 22 . Find the points on the curve that give the hottest and coldest
temperature.
SOLUTION.
To find the extrema of f at the boundary, use Theorem 7.4. Let
xyxyxf 2244, 22 and 01124, 22 yxyxg . Hence,
11242244, 2222 yxxyxyxL
0828, xxyxLx (1)
02448, yyyxLy (2)
01124, 22 yxyxL
. (3)
(1) 3y : 024624 xyyxy
(2) x : 02448 xyxy
)
0624 yxy
0624 xy 41 x or 0y .
Substitute 41x and 0y in (3), we obtain
41
011241422
y
y
21
01012422
y
x
The critical points are 41,41 , 41,41 , 0,21 and 0,21 .
Findfat these critical points to identify the absolute maximum and minimum.
25.2412412441441,41 22 f
25.2412412441441,41 22 f
02120242140,21 22 f
22120242140,21 22 f
The hottest temperature occurs at 41,41 and the lowest temperature occurs
at 0,21 .
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7-13
7.4 SURFACES IN R3
PLANES
A plane consists of a set of points that satisfies the equation 0 dczbyax where
dcba ,,, are constant coefficients.
EXAMPLE 7.7.
x x
x
y y
y
z z
z
z=1plane
yzplane
x = 0
xyplane
z= 0
xzplane
y = 0
z=y plane
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7-14
QUADRICSURFACES
The equation 2 2 2 0Ax By Cz Dxy Exz Fyz Gx Hy Iz J in an xyz-coordinate
system is called a second-degree equation in x, y and z. The graphs of such equations are
called quadric surfacesor sometimes quadrics. Six common quadric surfaces (cylinder,sphere, cone, ellipsoid, paraboloid and hyperboloid) are discussed in this section.
CYLINDER
Cylindrical surface is generated by a straight line which moves along a fixed curve and
remains parallel to a fixed straight line. A cylinder whose cross section is an ellipse, parabola,
or hyperbola is called an elliptic cylinder (2 2
2 21
x y
a b ), parabolic cylinder ( 2 2 0x rz ), or
hyperbolic cylinder (2 2
2 21
x y
a b ) respectively.
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7-15
EXAMPLE 7.8.
(a) 422 yx
(b) 2yz
x
y
z
0
2
x
y
1
11
z
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7-16
SPHERE
The surface area of a sphere is given by the formula 2222 rczbyax with
center cba ,, and radius r.
EXAMPLE 7.9.
(a) 2222 221 zyx
(b) 2222 azyx
x
y
z
x
y
z
(1, -2, 0)
a
(0, 0, 0)
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ELLIPTICPARABOLOID
The surface area of an elliptic paraboloid is given by the formula czb
y
a
x
2
2
2
2
where 0c .
EXAMPLE 7.12.
HYPERBOLOID OFONE SHEET
The surface area of hyperboloid of one sheet is given by the formula 12
2
2
2
2
2
c
z
b
y
a
xwhere
0,, cba .
EXAMPLE 7.13.
x
y
z
x
y
z
c
ab
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HYPERBOLOID OFTWO SHEETS
The surface area of hyperboloid of one sheet is given by the formula 12
2
2
2
2
2
c
z
b
y
a
x
where 0,, cba .
EXAMPLE 7.14.
x
y
z
c
c