Class notes

Oefenzittingen Wiskundige Modellen @GroepT

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Course
Wiskundige Modellen (T1AWM1)

Institution
Katholieke Universiteit Leuven (KU Leuven)

Dit document bevat enerzijds, zo goed als alle oefeningen in de selectie voor het vak, en anderzijds een mooi overzicht van bepaalde belangrijke formules / stramienen om te volgen bij het maken van oefeningen uit ieder hoofdstuk. Bij de oefeningen zelf staan hier en daar wat annotaties om verduidel...

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Uploaded on May 25, 2024
Number of pages 18
Written in 2022/2023
Type Class notes
Professor(s) Koen eneman
Contains All classes

oefeningen
wiskunde
wiskundige modellen
industrieel ingenieur
ingenieur
integralen
afgeleiden

Institution
Katholieke Universiteit Leuven (KU Leuven)
Education
Industriële Wetenschappen
Course
Wiskundige Modellen (T1AWM1)

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1
. Functies meerdere
veranderlijken
1 . f(x y) ,
= x3 -

2xy + 3y2
al f( -
2, 3) = -

8 -

2( -
2 .

3) + 3(9)
= 31

a f(z ) ,
=
()" -
2 .

z - -

+ 3 .

(2)
-
=
+
E

2) f(x y + u) f(x y) X 2x(y + n) 3(y u) x 3x
-

+ + 2xy
-
, ,
=
-
+ -

K
K

-

-
2xy -
axn + 3(y* + 2yn + 4) -xy
-
3yh
=

U

2xn + 3y2
- + 6y4 + 342 -
3y
-
2

=- =
-
2x + 6y + 3n
K

3
. Bepaal domf(x Y) :
,

(n((16 yz(x* + y 4)) (n((x 16)(x y2 4)
*

a) f(x , y) = -
x" - - = + ya -
+

↳ In (x) als x 0 -e man niet !
↳ 4 < X+ <16 is het domein

b) f(x y) ,
= 6 -

2x -

34

↳ 6-2x 3y)/ -
0

2 Particle
.
Afgeleiden
1 2 15te + 29 Orde
.
Particle afgeleide
x2
.

b) f(x y) ,
=
- x + Y
-xY
X + Y
a) f(x , y) =
3xy-Siny
2x(x + y) x() Y(x + y) xy()
=
CSS
-
-

Als er wordt
gevrage
-

(x + y)2 xz :

(x + y)2
bereuen 1 24 orde :
2x + 2xy -
x xy + y -
xy X
+

= -

> oou
gemengde doen !
-

x+ y2 x*+ y

(Gyn
2xy + 2xy +

Dit is een soort 24 Orde
X3
=
+ 2 xy + y2 = 1

x2 + 2xy + y2 X

x2
xX d) f(x y)
+

Gf/by =
- ,
= e

x+ 2xy + y2

1
x3 + x y -
x - x+ xy + xY
=

x + 2xy + yz

=
1

Let
=
10 0,

O
exy y x2
=

10 0) +
2xy2
+
,
+ 2xy .
·
2xy = 2xe + .
2xe

= 2xe
+
Y(2xy + 1)
3
.
& in (1 1) ,
voor 2(x y) ,
=
xarctan
(

() (y) ()(1 m ) ()(x y m 1) 2x(x + y ax 1)
= + =
+ -
2x
+
m
- + +
(x" 1)2
+ =
+
+ y -
2y +

P(1 1) , invullen geeft :

3It

, . bereuen da
8 voor z = xY-34 9
. a) Adhv Particle afgeleiden :

d2(x y)
↳
,

= GY
U(X y) ,
= x2 -
ex
- -
&
) e
U
totale -x
= 2x . ex + x?) - .

differential
= 2xydx + (x2 3)dy -

*
= e (2x y) -

vian door (1 ,
2 , 2(1 , 2) = xe
↓ du =
(e
**
(2x y))dx
- +
(xe x)dy
*

enz-2(n 2)
↓
,
=

( x)) -

,,
1 2
,
+
z(4 41)) -

,1 2)
,
= X-e"x

↳ Lo
2(1 2) ,
= -
4
-2xy 2) Rechtstreens :

df(x z) , y,
=

G+ +
*In =
2 .
1 .
2 : 4
,1 , 2
= -2

y2d(8xy +23) y 23d(3x2 y2)
d8xy 23 3x 8x
-

=
(= ) z =
4(x 1) -
-

a(y 2) -
-
4
3x2Y2 (3x2y2)2
() z = 4x 4 -
2y + 4 4

8(y
- -

23dx xy2d(23)]
-

()
d(8x 23) +
= + x23b(yz) +
0 =
4x =
2y -
2 -
4

d(3x + 2) =
3(y2d(x)) + x zdy + x ydz]
186 . Bereuen dy/dx : 18 .
f(x , y) = 1 -
x -
1 -
y
= 1

In-InF ()1 1
y
InE-Incy
1
-

x =
-

ax-2Y
-

= 2 Es =

(x) ) 2y)a
X -

Y
+ = -

eerst x en y apart,
=) InV-ax = Inty-2y dezedan afleiden

(
(
-
# -
2)ax =
-

2)dy

( 2)dx ( 2)dy
1 -
x
= - =
-

192 .
dy/dx en dy/dx' bepalen :

-
x + cos(x + y) = 0

&
Es
↑

3
. Vervolg 101
,
12
,
13
,
16
,
20b
,
21

102 .
Richtingsafgeleide :

F(X ,
y, 2) =
Xy2 -
y2 in P(2 ,
1
,
1) en S(v .
1
, -1)

= 2 =
(2) =
: Welke

Fi =
richting snelste toename ?

: x2-2y
-

=

(5 . , 5 =
[Y2 ,
xz -
2y , xy]i
Schrijven
E
mag ook 20 :

Fi
=Y
=
x y
8f =
(y2 x2 2y xY) deze vector
- volgens
-

, ,

das Of (2 ,
1 1) =
(1 0, 2) Stijlste helling
F
, ,

-
Fi (2 ,
1
, 1) =

met grootte =
1+ 2 =
55

&
=Pos
12 .

Berenenen voor T = X-XY + y met
Y =
PSinO

**
2

xTy t = (3x y)(oso -
+ ( -
x + 3y2)sino
X Y

·
-

↳ - ↓
GTo (3xy)(-Psino) + -x + yos
B O
* Sino 00

, 2x-y et
Berenen
13
. in to voor U =

u↑
& Y

2
=

=
z

3e
-
t
t
+ 1 =

[ ↳
Y

Ed
=
= 4x +
2
dus = o
= 27
= cost dus
lo
=
S
-

2
dus t=
= -3

= -

Y+ 3x2
dus =
-

= = 27 2 + ( 3)( ) + ( -( -3) = 6

16 .
Toon aan dat U(X ) ,
E
met so voldoet an :

( ( ( + () + =

= ( + " met
X

1 1
el
Y
cos
en si

A O

=

(cos + sind)
=
( *" coso 2/cssin) + + sinc

= metpin enc
= /) sino-2(sinoco
=

( Sino-2(sin + cos
RL =
(*) coso &(csin) + + Sinc + Sino-A/sinc) 1205 +

=

(*) coso 1 since + +
1 Sino 12. cos28 +

=
( *) (coso Sino) () (Sintcos + +

"M 1

=
( () = +

zob .
Bereuen een
uitdrunning die de gevraagde Partiele afgeleiden bevat,

O .
b . v .

volgende implicite definitie :

G Bereuen de numeriee waarden in (ab , = 2 ,

voor f(a ,
b, c) = a coshb-ab + c -
5 = 0 = )df = 0

If (ad(b)
2
= a d(coshb) + coshb -
d(a)) -
+ bd(a)) + d() = 0

= a 2 Sinhb db .
+ 29 coshbd(a) -
adb + bd(a) + b(c) = 0

=
(a 2. sinhb- a) db + 12acoshb -
b)da + d = 0

& da2
E E
= 2
(2acoshb-b) da 1dC -
4 .
0 + 2
db = ·

-asinb-a -
a "Sinb-a

/12 ,
0
, 2)
=
dc =

-
40 + 2
=
2

, ba (a 'Sinhb-a) =
b-zacoshb
ba (a'Sinnb-a) + b'a(2asinhb +
acosh-1 = -co
ba (4 0-2) .
+ 2(4 0 .
+ 4 -
1 . 2 = 2-4-1 .
2

-

2b"a + 8 = 6

ba = 7

21 1"Orde Partiele
.
afgeleide van z als 2 impliciet gedef . Wordt dr :

arctan (xy2)
E maw
ten
=
zoen
-

() Xyz = tan
[1
(z)xyz +
an = g
-

1

=> y2dx + x2dy + xydz = 0

() Xydz = -
yadx -

x2dy

)dz =
- -
()dz = -

Edx zdy -

d ↳
= =-
. Optimalisatie
3 1, 2
,
3 ,
5 ,
6 ,8

1
. Stationaire Punten van F(x , y) = 5x2 + 4x + x y + y"

I1 )
=

=
Of

&
10x

4y3 +
: (10x + 2xy , 4y3

+ 2xy2

2x y
=

+ 4
--
=
+ 2x

0()x(10 + 2y2) = 0()x = 0

0()4y3 +
y + 4)

4 =
= 0

0()y3 = - 1()y = -
1

=> (0,-1) is Stationair

deth 10 ...):
I (01)
=
10 +

Y*12x2 +
2424xY

2x10 ,
=

-2)
170

012
=
144 10 =

>
-
(0 -1) ,

welke
is extremum

soort ?
!

a
Ly,
-

= 10 +

=
Lonaal Minimum

. Lagrange multiplicatoren
2
f(x ,
y
,
x) =
f(x y) ,
-
t .

g(x 4) ,

=>
((x ,
y ,
+) = x + 4y -
2x + 8y -
+ (x + 2y -
7)

= 2x - 2 -

+, = 84 + 8 - 2 , =
-

(x + 24 -
7)

- Allen O (want we stellen XC(x Y x) =)
aan
gelijn Stellen : , ,

2x -
2 -
+ = 0 (= + = 2x -
2

.
8 -
8 -
24 = 0 ()4y -
4 -
X = 0E)2X -
2 = 4y -
4( = ) X =
2y -
1

x(x + 2y 7) 0( 2y

04 3
- = -
1 + 2y -
7 =

=
P(3 4),

(4
=
= 4)

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