Lösung von Teilaufgabe c) 1.: Unterschied zwischen den Versionen

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Version vom 24. Januar 2010, 01:49 Uhr

Inhaltsverzeichnis

Tangente im Punkt Wa( a + 2 / 2 ) an Gfa mit dem Schnittpunkt A (0 / 2012 )

mit:\;

x = 0\;
y = 2012\;
x_0 = a + 2\;
f_a( x_0 ) = f_a( a + 2 ) = 2\;
f^{'}_a( x_0 ) = f^{'}_a( a + 2 ) = m = -1\;


f^{'}_a( a + 2 ) = e^{a + 2 - ( a + 2 )}\cdot ( 1 + a - ( a + 2 ) )
= e^{a + 2 - a - 2 }\cdot ( 1 + a - a - 2 ) )
= e^{0}\cdot ( -1 ) )
= -1\;


Lösung; Tangentengleichung

Tangentengleichung: siehe Formelsammlung Seite 58

y = f^{'}( x_0 )\cdot ( x - x_0 ) + f ( x_0)


y = f^{'}_a( a + 2 )\cdot ( x - ( a + 2 )) + f ( a + 2 )
y = (-1)\cdot ( x - a - 2 ) + 2
y = -x + a + 2 + 2\;
y = -x + a + 4\;
2012 = 0 + a + 4\;\;\;\;\;\;\; | -4
a = 2008\;

Lösung; Fußweg

y = m\cdot x + t
f_a( x_0 ) = f^{'}_a( x_0 )\cdot x_0 + t
f_a( a + 2 ) = f^{'}_a( a + 2 )\cdot x_0 + t
2 = -1\cdot x_1 + t \;\;\;\;\;\; | - ( -1\cdot x_0)
t = 2 - ( -1\cdot x_0 )
t = 2 - ( -1\cdot ( a + 2 ))
t = 2 - ( -a - 2)\;
t = 2 + a + 2 \;
t = a + 4 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; |\;einsetzen\; in\; y = m\cdot x + t


y = m\cdot x + a + 4
2012 = -1\cdot 0 + a + 4
2012 = a + 4 \;
a = 2008\;

Lösung; Clever

\frac{y_2 - y_1}{x_2 - x_1} = f{'}_a ( x )
\frac{2012 - 2}{0 - ( a + 2 )} = -1
\frac{2010}{(-a - 2 )} = -1 \;\;\;\;\;\;\;\;\;\;| \cdot( -a - 2 )
2010 = a + 2\;
2008 = a\;



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