Formuliere den allgemeinen Integranden ${\displaystyle ds}$ in eine für das Problem geeignete Form um.

${\displaystyle min{\stackrel {!}{=}}\int _{t_{0}}^{t_{1}}dt=\int _{1}^{2}{\frac {ds}{v}}}$

Aus den bekannten Variablen für das System kann eine Form für ${\displaystyle v}$ gefunden werden. Dazu wird die Energiebilanz des Systems aufgestellt:

${\displaystyle {\begin{aligned}E_{ges}=E_{pot}-E_{kin}\\=ymg-{\frac {1}{2}}my'^{2}\end{aligned}}}$

Mit ${\displaystyle y_{t_{0}}=0}$ ist ${\displaystyle E_{ges}=0}$ . Nach Umformen ergibt sich also:

${\displaystyle v={\sqrt {2yg}}}$

Und damit:

${\displaystyle {\begin{aligned}\int _{t_{0}}^{t_{1}}dt=\int _{1}^{2}{\frac {ds}{v}}\\=\int {\frac {ds}{\sqrt {2gy}}}\end{aligned}}}$

Mit: ${\displaystyle ds={\sqrt {1+y'^{2}}}dx}$

Ergibt sich:

${\displaystyle min{\stackrel {!}{=}}\int {\sqrt {\frac {1+y'^{2}}{2gy}}}dx}$

Stelle die Eulersche Gleichung auf und löse diese.

• ${\displaystyle min{\stackrel {!}{=}}\int {\sqrt {\frac {1+y'^{2}}{2gy}}}dx={\frac {1}{\sqrt {2g}}}\int {\sqrt {\frac {1+y'^{2}}{y}}}dx}$
• Die Eulersche Gleichung lautet:

${\displaystyle {\frac {\partial f}{\partial y}}-{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}=0}$

Dabei ist:

${\displaystyle {\begin{aligned}{\frac {\partial f}{\partial y}}&=-{\frac {1}{2}}{\sqrt {1+y'^{2}}}y^{-{\frac {3}{2}}}\\{\frac {\partial f}{\partial y'}}&={\frac {y'}{\sqrt {y(1+y'^{2})}}}\end{aligned}}}$

Weiter ist dann:

${\displaystyle {\begin{aligned}{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}&={\frac {d}{dx}}{\frac {y'}{\sqrt {y(1+y'^{2})}}}\end{aligned}}}$

Mit:

${\displaystyle {\begin{aligned}{\frac {\partial }{\partial x}}{\frac {y'}{\sqrt {y(1+y'^{2})}}}=0\\{\frac {\partial }{\partial y}}{\frac {y'}{\sqrt {y(1+y'^{2})}}}\cdot y'=-{\frac {1}{2}}{\frac {y'^{2}}{\sqrt {1+y'^{2}}}}\cdot {\frac {1}{y^{\frac {3}{2}}}}\\{\frac {\partial }{\partial y'}}{\frac {y'}{\sqrt {y(1+y'^{2})}}}\cdot y''={\frac {y''}{\sqrt {y(1+y'^{2})}}}-{\frac {y'^{2}y''}{\sqrt {y}}}\cdot {\frac {1}{(1+y'^{2})^{\frac {3}{2}}}}\end{aligned}}}$

Es ergibt sich also insgesamt:

${\displaystyle {\begin{aligned}0={\frac {\partial f}{\partial y}}-{\frac {d}{dx}}{\frac {\partial f}{\partial y'}}\\=-{\frac {1}{2}}{\sqrt {1+y'^{2}}}y^{-{\frac {3}{2}}}+{\frac {1}{2}}{\frac {y'^{2}}{\sqrt {1+y'^{2}}}}\cdot {\frac {1}{y^{\frac {3}{2}}}}-{\frac {y''}{\sqrt {y(1+y'^{2})}}}+{\frac {y'^{2}y''}{\sqrt {y}}}\cdot {\frac {1}{(1+y'^{2})^{\frac {3}{2}}}}&&|\cdot {\sqrt {1+y'^{2}}}\\=-{\frac {1+y'^{2}}{2}}y^{-{\frac {3}{2}}}+{\frac {1}{2}}{\frac {y'^{2}}{y^{\frac {3}{2}}}}-{\frac {y''}{\sqrt {y}}}+{\frac {y'^{2}y''}{\sqrt {y}}}\cdot {\frac {1}{1+y'^{2}}}&&|\cdot y^{\frac {3}{2}}\\=-{\frac {1+y'^{2}}{2}}+{\frac {y'^{2}}{2}}-y''y+{\frac {yy'^{2}y''}{1+y'^{2}}}\\\Longrightarrow 1+y'^{2}=y'^{2}-2yy''+{\frac {2yy'^{2}y''}{1+y'^{2}}}\end{aligned}}}$

• Fehler beim Parsen (Konvertierungsfehler. Der Server („https://wikimedia.org/api/rest_“) hat berichtet: „Cannot get mml. Expected width > 0.“): {\displaystyle {\begin{aligned}\end{aligned}}}