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Frühe Ableitungsregeln nutzen

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In seinen \"Lectiones Geometricae\" geht Isaac Barrow davon aus, dass man sich Kurven (vgl. Abb. unten) aus unendlich kleinen Geradenstrücken (\"linelets'') zusammengesetzt vorstellen kann.

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Sei $MN$ so ein unendlich kleines Geradenstück ($M$ und $N$ also infinitesimal benachbarte Punkte) einer Kurve mit der Gleichung $f(x,y)=0$, so darf man nach Barrow folgern, dass gilt:

$f(x+e,y+a)=f(x,y)=0$

und dass alle auftretenden $e^m,a^n$ mit $m,n>1$ zu Null werden.

Ermitteln Sie mit diesem Ansatz die Tangentensteigung $\\frac{a}{e}$ für die folgende Kreisgleichung, d. h. für die implizite Funktion $f(x,y)=x^2+y^2-\\var{c}^2=0$

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", "advice": "

Bitte einfach die korrekten Antworten der Lücken ansehen.

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Es folgt zunächst:

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[[0]]$=x^2+y^2-\\var{c}^2=0$

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Durch Klammerauflösen und Unterdrücken aller übereinstimmenden Terme erhalten wir daraus:

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[[1]]$+a^2=0$

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Durch Unterdrücken aller $e^m,a^n$ mit $m,n>1$ bleibt schließlich:

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[[2]]$=0$, was uns zur Ableitung $\\frac{a}{e}=$[[3]] führt.

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