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i. Level curve : is the line of constant f- IX. y ) =L
horizontal Curves
plane known
slice of flx.y.tt/=C .
in ×
,y , as contours
variable flc y Gives
z .
Section curve : vertical slice taken
by freezing one
of the ,
e.g , , 21=0 .
curves in yZ or XZ or
Xy plane
3. Examples :
fix y ) x 't
sing
=
i ,
.
for fixed y ,
we
get a
parabola
.
for fixed X ,
we
get a sin curve
z
×
y
sin JXHYZ sint
z f = =
r
JX2-1YZ Z = cos
Xy
When
XY
= IT , 2- = -1
×y= , z=o
21T I 2=+1
XY
=
's
4 Z= X -
3×42
1. 2 partial derivative
I. In 3D ( or more )
diagram ,
there are more than I
slope you can draw from 1
point
slope :
steepness
hmm
; d£ ,
where ds is a small
displacement in the direction
vector
quantity
local about
gradient : vector
hmm
quantity that contains all the
information the
slope
e.
g two
slopes
2.
Meaning partial derivative : is the
slope in a
specific
mmmm
direction
, 3- .
Derivative with
respect to × , and treat
y as a constant
JF1X 1YI flxth.gl -
f- IX. y )
=
Iim
jx x→o h
ff
fx=j×
f- xx =
JX2
•
Derivative with
respect to
y , and treat
× as a constant
JF1X 1YI fix .
yth ) -
f- IX. y )
=
Iim
jx x→o h
ff
fy=
try
f- =
'
yy jy
•
fxy =
Ey I # I
f- yx =
# III )
¥§×j ¥¥×
" "
4 Clairaut 's theorem : For f- IX1 , . . .
,
Xn ) : IR → R if the mixed partial derivative and
,
.
i
exist and are continuous at a
point then
If ff
=
txitxj Jxjtxi
If If
e.
g =
txty tyd×
Questions :
Find the derivatives
fx.fxx.fy.fyy.fi/y ,
fyx :
-3×42
'
a) fix , y) = ×
f- × = 3×2 -
3yd
f- xx =
6X
f- y = -
bxy
6X
fyy
-
=
fxy =
-
by
f- yx = -
by
=
fxy
It f- ix. y ) =
cosy +
sinlxy )
f- ×
=
ycoslxy )
'
f×× =
-
y sinlxy )
fy =
-
sing +
xcoslxy )
fyy X'
=
cosy sinlxy )
-
-
f- xy costly ) Xysinlxy )
-
=
fyx =
costly ) -
xysinlxy ) =fxy
y.t /=X2ytyztZZX,findfx.fyandfz.fx=2XytZZfy=x2+z
c) f- ix.
directions
stops in × ,
y ,
Z
fz =
Y1-2XZ
I -3 Differential and Directional derivatives
Differentials :
When variable
i. . we have a
single ,
gradient df =
% DX
direction
• Consider the rate of change of f- IX.
y) in an
arbitrary
in as result
AY
and a
make small AX in y
changes × ,
simultaneous
suppose that we
f- + If
so Af =
f- IXTAX ,
ythy )
-
f IX.
y)
add then subtract
f- f IX1Y-1AY ) fix ,ytAy ) f
ytAy ) y)
= IXTAX -
+ -
ix. -
,
, =
-
f IXTAX ,
Y1-4YI -
f- IX. ytAy )
µ×
f- IX1Y-1AY ) -
flay)
+
AX
µy
Ay
=
¥ DX t
Ity dy
•
The
change df is
given by the sum of contributions
df =
If ,
DX
tatty dy + t£dZ
z .
df =
If DX
tatty dy + t£dZ
•
small
changes df , DX ,
dy.dz a re called differentials
•
df is the total differential of the function f- IX1Y 1ZI .
without
any approximation
3. Total differentials : n
of a function f- IX1 1XA , . . . Xn )
df =
II. DX ,
+ II. dxa +
. . .
+ In dXn
'
Questions I Find the total differential of f- ix. y ) X. y t
costly
: =
'
f- × =
2Xy
-
ysinxy
f- y =
3×25 -
Xsinxy
[ 3×42 xsinxy ]dy
'
So df =
[ sexy -
ysinxyldx + -
4 Exact differential :
Any total differential df adx b-
dy cdz be
+ t can
integrated
.
. =
mum
to find the
original function .
c=fx , b- =fy , c=fz
conserved
physical meaning
: the system is
e.g Energy
df adx b- dy t cdz where a # fx to # fy c # fz
Inexact differential : +
- = . ,
,
cannot define the
global function
the system is not conserved
physical meaning
:
e.
g Friction involves
Show that df Xdy exact differential
2 + ydx is
=
a
Of
a
=
Y i. f =
xy + Aly )
If × i.
f BIX)
xy +
= =
ay
i. f- IX. y)
=
Xy + C
i. It is a exact differential since f- IX.
y)
is consistent
> show that df =
Xdy + Zydx is a inexact differential
Jf
Jx
=
by i.
f- =
3xy + A ly )
If
X f- xy + 131×1
Jy
= . =
. .
Two function are not consistent , so there is no
integral fix , y )
inexact differential
5. Test for exactness :
By < Iairaut 's theorem
df AIX y) DX
Blxiyldy
suppose that : = , +
•
If it is exact ,
A- IX.
y)
=
¥ =f× , Blxiy ) =¥y =fy
By clairaut 's theorem
Ay =
fxy =fy× = Bx
.
Therefore , df = cdx + b- dy
If Fy =
II ,
it is exact
.
If df = a DX tbdy-c.dz
We need to check all pairs :
If =
II =
fxy
, ffz =
¥ =f×z
1¥ =
Ey =fyz
The claim of exactness :
I =
lfx.fy.fi/--Jf
Test of exactness is equivalent to : JX1 Ff ) =o
Gradient and Vector calculus :
i. Determine the
slope of a
function , f- IX. y ) in any arbitrary direction :
•
df =
# DX 1-
¥y dy a
dot
=
( DX ,
dy ) .
1¥ , Ity ) product
direction
gradient of f- IX1Y ) , If
=
I. ds .
I ¥ ,¥y ) dxitdyi
small distance
has
dG
d" "'
dy
level
( dx ) 't Idyt= Ids )
'
DX
>
curves are
orthogonal
mm
to
. So
ddfs =
I. If the
gradient 1
by definition)
2 .
.
Gradient of flx.gl , If :
If =
III. %) a vector ix. g) or II ) or
xityf
1¥ ,¥n )
"
For Jf : IR ,
If =
, ,
. . .
- vector differential operator .
J =
1¥ , #) F -
Laplacian
called del
e.
g t.IN/--fxIJfIx+fyIJfIyI-fzIJfIz--YIzt#ya-fIza-- It .
Jtf
3 .
If and the surface f- IX. y ) :
• If =
In IJF -
magnitude
direction
df
rate of in direction I
Is : is the
change
• a
I. HIIJFI I HI initial
¥ cost IFFI case
- - =
= =
=
111111 cost =
cost
when f- 0 II and in parallel ) ¥ =
IFFI
df
maximum If points the
ds
up slope
=
, , ,
when f- = 1 I and In are
orthogonal ) ,
¥5 =o
,
¥ minimum
,
the level curves are at
right angle to
of
" '
4 Find the rate
of of f- IX. y ) X' y at 10,11 direction of itaf
change
= -1 + × in the the vector
y
'
f- × =
I1-2XY
sub X=o ,
y =L , f- × = I
f- y 4y
>
+
'
2y×
=
sub ×=o ,
y =/ . f- y
=
4
i. It =
IN
Itai
unit vector I = =
III
15 )
] 12 -122
'
the rate
of change
= I. If =
¥11 ) I 4) .
=
€
-
Y
The T X' e At
temperature metal plate direction does
5 is 12,11
on a ix.
y)
= . the
point in what
?
the
temperature increase most
rapidly
-
y
Tx =
2Xe
sub X=2 .
y=I , T× = ¥
Ty =
-
HEY
-
¥
sub X=2 y I Ty
-_ =
,
i. IT =
EH )
so the direction of greatest increase : I =
.IT/--fzH/
+ µ,
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