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Oefenzittingen Wiskundige Modellen @GroepT

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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 verduidelijking te bieden.

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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

Subjects

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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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