cry

src/exercise/5.typ

@@ -17,13 +17,103 @@ $
 
 $
 g(x,y) &= x ; (delta g)/(delta x) = 1 ; (delta g)/(delta y) = 0 \
-h(x,y) &= sqrt(y^2+x^2) ; (delta h)/(delta x) = sqrt(y^2+x^2) dot 2x ; (delta h)/(delta y) = sqrt(y^2+x^2) dot 2y \
+h(x,y) &= sqrt(y^2+x^2) ; (delta h)/(delta x) = x/sqrt(y^2+x^2) ; (delta h)/(delta y) = y/sqrt(y^2+x^2) \
 \
-(delta f)/(delta x) &= g dot h' + g' dot h = x dot sqrt(y^2+x^2) dot 2x + 1 dot sqrt(x^2+y^2) \
+(delta f)/(delta x) &= g' dot h + g dot h' \
-&= sqrt(x^2+y^2) (2x^2 + 1) \
+&= 1 dot sqrt(y^2+x^2) + x dot x/sqrt(y^2+x^2) \
-
+&= sqrt(y^2+x^2) + x^2/sqrt(y^2+x^2) \
-(delta f)/(delta y) &= g dot h' + g' dot h = x dot sqrt(x^2+y^2) dot 2y + 0 dot dots \
+&= (y^2+x^2)/sqrt(y^2+x^2) + x^2/sqrt(y^2+x^2) \
-&= x dot sqrt(x^2+y^2) dot 2y
+&= (y^2+2x^2)/sqrt(y^2+x^2) checkmark\
-
+\ 
-\ "dödö falsch"
+(delta f)/(delta y) &= g' dot h + g dot h' \
+&= 0 dot dots + x dot y/sqrt(y^2+x^2) \
+&= (x y)/sqrt(y^2+x^2) checkmark\
 $
+
+Hier ist die Lösung mal wieder falsch 😞, aber ich habe es mit wolfram alpha 
+nachgerechnet.
+
+#pagebreak()
+=== Exercise 2 @Exercise[5, 2]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+	The function
+$
+	p(V,T) = N dot (R dot T)/V
+$
+	describes the pressure $p$ as a function of volume $V$, temperature $T$, the amount of substance
+	($N$ ) and the ideal gas constant $R$. Where does the function p have partial derivatives with
+	respect to $V$ and $T$ , respectively? Is $p$ also continuously (partially) differentiable?
+])
+
+$
+	(delta p)/(delta T) &= N dot R/V checkmark \
+	\
+	p(V,T) &= N dot R dot T dot v(V,T) \
+	v(V,T) &= 1/V = V^(-1) \
+	(delta v)/(delta V) &= -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 
+differenzierbar.
+
+=== Exercise 3 @Exercise[5, 3]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+	Where are the following function twice partially differentiable? Determine the partial derivatives
+	up to (including) order two.
+#set enum(numbering: "(a)")
++ $ f(x,y) = (2x-3y)^4 $
++ $ f(x,y) = ln(x^4+y^2) $
+])
+
+#set enum(numbering: "(a)")
++ $
+	g(x,y) &= x^4 => g'(x,y) = 4x^3 => g''(x,y) = 12x^2 \
+	dots
+$
+
+Boah ne, die aufgabe ist mir zu doof
+
+#pagebreak()
+=== Exercise 4 @Exercise[5, 4]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+	Show that the function
+
+$
+	f(x,y) = cases(
+		x^5/(x^4+y^4) & "if" (x,y) != (0,0),
+		0			  & "if" (x,y) = (0,0),
+	)
+$
+	is partially differentiable with respect to $y$, but it is not continuously partially differentiable with
+	respect to $y$ in $(0, 0)$.
+])
+
+Die partielle Ableitung mit $y$ in $(0,0)$ existiert:
+
+$
+lim_(y -> infinity) (f(0,y) - f(0,0))/(y-0) = lim_(y -> infinity) (0/y^4)/y = lim_(y -> infinity) 0 = 0
+$
+Aber
+$
+(delta f)/(delta y) &= - x^5/(x^4+y^4)^2 dot (x^4+y^4)' \
+&= - x^5/(x^4+y^4)^2 dot 4y^3 = -(4x^5y^3)/(x^4+y^4)^2 \
+\
+(delta f)/(delta y)(0,y) &= 0 \
+(delta f)/(delta y)(y,y) &= -(4y^8)/((2y^4)^2) = -(4y^8)/(4y^8) = -1 \
+=> "Nicht stetig!"
+$
+
+Somit ist $f$ nach $y$ partiell ableitbar, aber nicht stetig partiell ableitbar.