Single Math B part 2 (first-year math)

1. Partial differentiation
IRHB chapter 51

1. 1 Graphs of functions

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,