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src/exercise/6.typ

@@ -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.

src/exercise/7.typ

@@ -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
+
+$

src/exercise/index.typ

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