aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

This commit is contained in:
Christoph J. Scherr 2024-09-23 16:31:50 +02:00
parent f8f061d93c
commit 36cc8ebd4e
Signed by: cscherrNT
GPG Key ID: 8E2B45BC51A27EA7
5 changed files with 154 additions and 3 deletions

Binary file not shown.

View File

@ -24,13 +24,13 @@ h(x,y) &= sqrt(y^2+x^2) ; (delta h)/(delta x) = x/sqrt(y^2+x^2) ; (delta h)/(del
&= sqrt(y^2+x^2) + x^2/sqrt(y^2+x^2) \ &= sqrt(y^2+x^2) + x^2/sqrt(y^2+x^2) \
&= (y^2+x^2)/sqrt(y^2+x^2) + x^2/sqrt(y^2+x^2) \ &= (y^2+x^2)/sqrt(y^2+x^2) + x^2/sqrt(y^2+x^2) \
&= (y^2+2x^2)/sqrt(y^2+x^2) checkmark\ &= (y^2+2x^2)/sqrt(y^2+x^2) checkmark\
\ \
(delta f)/(delta y) &= g' dot h + g dot h' \ (delta f)/(delta y) &= g' dot h + g dot h' \
&= 0 dot dots + x dot y/sqrt(y^2+x^2) \ &= 0 dot dots + x dot y/sqrt(y^2+x^2) \
&= (x y)/sqrt(y^2+x^2) checkmark\ &= (x y)/sqrt(y^2+x^2) checkmark\
$ $
Hier ist die Lösung mal wieder falsch 😞, aber ich habe es mit wolfram alpha Hier ist die Lösung mal wieder falsch 😞, aber ich habe es mit wolfram alpha
nachgerechnet. nachgerechnet.
#pagebreak() #pagebreak()
@ -58,7 +58,7 @@ $
=> p(V,t) &= N dot R dot T dot -1/V^2 => p(V,t) &= N dot R dot T dot -1/V^2
$ $
$p$ ist stetig für $RR^2 \\ {v=0}$ und somit für diesen Bereich auch stetig $p$ ist stetig für $RR^2 \\ {v=0}$ und somit für diesen Bereich auch stetig
differenzierbar. differenzierbar.
=== Exercise 3 @Exercise[5, 3] === Exercise 3 @Exercise[5, 3]

90
src/exercise/6.typ Normal file
View File

@ -0,0 +1,90 @@
== Exercise Sheet 6 - Differentials
=== Exercise 1 @Exercise[6, 1]
#block(
fill: luma(230),
inset: 8pt,
radius: 4pt,
[
Determine the directional derivative of $f(x, y) = cos(x + y) · sin(x y)$ at $(0, 0)$ in direction $(2, 1)$.
])
$
(delta f)/(delta x) &= u' dot v + u dot v' \
&= -sin(x+y) dot sin(x-y) + cos(x+y) dot cos(x-y) \
(delta f)/(delta y) &= u' dot v + u dot v' \
&= -sin(x+y) dot sin(x-y) - cos(x+y) dot cos(x-y) \
\
v &= (2,1) dot 1/(norm((2,1))) = (2/sqrt(5) , 1/sqrt(5)) \
f_v (0,0) &= lr(angle.l nabla f(1,1) , v angle.r) \
&= lr(angle.l (?,?) , (2,1) angle.r) =^(???) 1/sqrt(5)
$
#pagebreak()
=== Exercise 2 @Exercise[6, 2]
#block(
fill: luma(230),
inset: 8pt,
radius: 4pt,
[
Determine the tangential plane for the function
$
f(x,y) = x sqrt(1/(y^2+1)) + cos(pi dot x y)
$
at $P = (a,b) = (1,1)$.
])
$
(delta f)/(delta x) &= sqrt(1/(y^2+1)) - sin(pi dot x y) \
(delta f)/(delta y) &= x dot 1/(2(y^2+1)) dot 2y - sin(pi dot x y) \
\
z &= f(a,b) + (delta f)/(delta x) dot (x-a) + (delta f)/(delta y) dot (y-b) \
&= sqrt(1/2) + cos(pi) + (sqrt(1/2) - sin(pi))(x-1) + (1 dot 1/(2 dot 2) dot 2 - sin(pi))(y-1) \
&= sqrt(1/2) + cos(pi) + sqrt(1/2)(x-1) + 1/2(y-1) \
&= sqrt(1/2) - 1 + (x-1)/sqrt(2) + (y-1)/2 \
$
Die Lösung hier hat $z = 1/sqrt(2) - 1 + (x-1)/sqrt(2) - (y-1)/2^(3/2)$ mit
$nabla f(1,1) = (1/sqrt(2) , -2^(-3/2))$ und ich habe keine Ahung
warum. Actually doch, hab falsch abgeleitet.
$
(delta f)/(delta x) &= sqrt(1/(y^2+1)) - sin(pi dot x y) dot pi y \
(delta f)/(delta y) &= (x dot 1/sqrt(y^2+1))' - sin(pi dot x y) dot pi x \
&= [x dot (-1)/(2 (y^2 + 1)^(3/2)) dot 2y] - sin(pi dot x y) dot pi x \
&= [x dot (-1)/((y^2 + 1)^(3/2)) dot y] - sin(pi dot x y) dot pi x \
&= -x ( [1/(y^2+1)]^(2/3) dot y + sin(pi dot x y) dot pi x) \
\
z &= f(a,b) + (delta f)/(delta x) dot (x-a) + (delta f)/(delta y) dot (y-b) \
&= (1/sqrt(2) - 1) + (sqrt(1/2) - sin(pi))(x-1) - ([1/2]^(2/3) + sin(pi))(y-1) \
&= (1/sqrt(2) - 1) + sqrt(1/2)(x-1) - [1/2]^(2/3) (y-1) checkmark
$
#pagebreak()
=== Exercise 3 @Exercise[6, 3]
#block(
fill: luma(230),
inset: 8pt,
radius: 4pt,
[
Determine the total derivatives of the functions
#set enum(numbering: "(a)")
+ $ f(x,y) = (x^2-y^2)/(1+x^2+y^2) $
+ $ f(x,y) = ln(x^2+y^2+1) $
])
#set enum(numbering: "(a)")
+ $
u(x,y) &= x^2 - y^2 quad => quad (delta u)/(delta x) = 2x and (delta u)/(delta y) = -2y \
v(x,y) &= 1 + x^2 + y^2 quad => quad (delta v)/(delta x) = 2x and (delta v)/(delta y) = 2y \
\
=> f(x,y) &= u(x,y)/v(x,y) \
(delta f)/(delta x) &= (u dot v' - u' dot v)/v^2 \
&= ((x^2-y^2)(2y) - (-2y)(1+x^2+y^2))/(1+x^2+y^2)^2 \
&= ((x^2-y^2)2y + 2y(1+x^2+y^2))/(1+x^2+y^2)^2 \
dots "falsch??"
$
Lösung sus, also weiter gehts. Rest kb gehabt, aber Lösung geschaut.

57
src/exercise/7.typ Normal file
View File

@ -0,0 +1,57 @@
== Exercise Sheet 7 - Extrene Values & Optimization
=== Exercise 1 @Exercise[7, 1]
#block(
fill: luma(230),
inset: 8pt,
radius: 4pt,
[
Investigate the local extrema of the functions
#set enum(numbering: "(a)")
+ $ f(x,y) = 7-x^2-4x-y^2+2y $
+ $ f(x,y) = x^4 + 3y^6 $
+ $ f(x,y) = e^(-x^2-y^2) $
])
#set enum(numbering: "(a)")
+ $
nabla f(x,y) &= (-2x-4 , -2y+2) \
nabla f(x,y) &=^! (0,0) "Notwendiges Kriterium" \
upright("I"): -2x-4 &= 0 \
"II" : -2y+2 &= 0 \
=> x = -2 space&space y = 1 \
H(-2,1) &= mat(
-2 , 0 ;
0 , -2 ;
) \
det(H) "negativ" &=> "Lokales Maximum" \
=> "Minimum in" (-2, 1): f(-2,1) &= 7-4+8-1+2 = 12 checkmark \
"Lösung mal wieder falsch... Kek"
$
+ $
nabla f(x,y) &= ( 4x^3 , 18y^5 ) \
nabla f(x,y) &=^! (0,0) "Notwendiges Kriterium" \
upright("I") : 4x^3 &= 0 \
upright("II") : 18y^5 &= 0 \
=> x &= y = 0 \
H(x,y) &= mat(
4x^3 , 0 ;
0 , 90y^4 ;
) \
=> H(0,0) &= mat(0,0;0,0) "indeterminiert" \
f &"ist immer positiv, daher muss das extrema ein Minimum sein." \
=> P(0,0) &"ist ein lokales Minimum mit" f(0,0) = 0
$
+ $
nabla f(x,y) &= (e^(-x^2-y^2) dot (-2 x) , e^(-x^2-y^2) dot (-2 y) ) \
=> x &= y = 0 "(Notwendige Bedingung)" \
=> H(x,y) &= mat(
2x^2-1 , 4x y ;
4x y , 2y^2-1 ;
) dot e^(-x^2-y^2) "Achtung, Produktregel"\
H(0,0) &= mat(
-1,0;
0, -1
) => "Maximum" \
f(0,0) &= 1 checkmark
$

View File

@ -7,3 +7,7 @@
#include "4.typ" #include "4.typ"
#pagebreak() #pagebreak()
#include "5.typ" #include "5.typ"
#pagebreak()
#include "6.typ"
#pagebreak()
#include "7.typ"