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Vorschlag von Marco eingebunden
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@ -29,7 +29,10 @@
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\begin{beweis}\leavevmode
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\begin{enumerate}[label=\alph*)]
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\item $l(\gamma) = \int_a^b \|\gamma'(t)\| \mathrm{d} t = \int_a^b 1 \mathrm{d} t = b - a$.
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\item $1 = \|\gamma'(t)\| = \|\gamma'(t)\|^2 = \langle \gamma'(t), \gamma'(t) \rangle$\\
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\item Im Folgenden wird die Aussage nur für $\gamma: [a, b] \rightarrow \mdr^2$ bewiesen.
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Allerdings funktioniert der Beweis im $\mdr^n$ analog. Es muss nur
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die Ableitung angepasst werden.
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$1 = \|\gamma'(t)\| = \|\gamma'(t)\|^2 = \langle \gamma'(t), \gamma'(t) \rangle$\\
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$\begin{aligned}[t]
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\Rightarrow 0 &= \frac{\mathrm{d}}{\mathrm{d}t} \langle \gamma'(t), \gamma'(t) \rangle\\
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&= \frac{\mathrm{d}}{\mathrm{d}t} (\gamma_1'(t)\gamma_1'(t) + \gamma_2'(t)\gamma_2'(t))\\
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