Aus Online Mathematik Brückenkurs 2

Version vom 14:21, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 1.1:5 moved to Solution 1.1:5: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 13:04, 10. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	Suppose that the tangent touches the curve at the point
-	<center> [[Image:1_1_5-1(3).gif]] </center>	+	<math>\left( x_{0} \right.,\left. y_{0} \right)</math>. That point must, first and foremost, lie on the curve and therefore satisfy the equation of the curve, i.e.
-	<center> [[Image:1_1_5-2(3).gif]] </center>	+
-	<center> [[Image:1_1_5-3(3).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	<math>y_{0}=-x_{0}^{2}\quad \quad \quad \left( 1 \right)</math>
		+
		+
		+	If we now write the equation of the tangent as
		+	<math>y=kx+m</math>, the gradient of the tangent,
		+	<math>k</math>, is given by the value of the curve's derivative,
		+	<math>{y}'=-2x</math>, at
		+	<math>x=x_{0}</math>,
		+
		+
		+	<math>k=-2x_{0}\quad \quad \quad \left( 2 \right)</math>
		+
		+	The condition that the tangent goes through the point
		+	<math>\left( x_{0} \right.,\left. y_{0} \right)</math>
		+	gives us that
		+
		+
		+	<math>y_{0}=k\centerdot x_{0}+m\quad \quad \quad \left( 3 \right)</math>
		+
		+
		+	In addition to this, the tangent should also pass through the point
		+	<math>\left( 1 \right.,\left. 1 \right)</math>,
		+
		+
		+	<math>1=k\centerdot 1+m\quad \quad \quad \left( 4 \right)</math>
		+
		+
		+	Equations (1)-(4) constitute an equation system in the unknowns
		+	<math>x_{0},\ y_{0},\ k</math>
		+	and
		+	<math>m</math>.
		+
		+	Because we are looking for
		+	<math>x_{0}\text{ }</math>
		+	and
		+	<math>y_{0}</math>, the first step is to try and eliminate
		+	<math>k</math>
		+	and
		+	<math>m</math>
		+	from the equations.
		+
		+	Equation (2) gives that
		+	<math>k=-\text{2 }x_{0}</math>
		+	and substituting this into equation (4) gives
		+
		+
		+	<math>1=-2x_{0}+m\quad \Leftrightarrow \quad m=2x_{0}+1</math>
		+
		+	With
		+	<math>k</math>
		+	and
		+	<math>m</math>
		+	expressed in terms of
		+	<math>x_{0}</math>
		+	and
		+	<math>y_{0}</math>, (3) becomes an equation that is expressed completely in terms of
		+	<math>x_{0}</math>
		+	and
		+	<math>y_{0}</math>,
		+
		+
		+	<math>y_{0}=-2x_{0}^{2}+2x_{0}+1\quad \quad \quad \left( 3' \right)</math>
		+
		+
		+	This equation, together with (1), is an equation system in
		+	<math>x_{0}</math>
		+	and
		+	<math>y_{0}</math>
		+
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	y_{0}=-x_{0}^{2} \\
		+	y_{0}=-2x_{0}^{2}+2x_{0}+1 \\
		+	\end{array} \right.</math>
		+
		+
		+	Substituting equation (1) into (3') gives us an equation in x0,
		+
		+
		+	<math>-x_{0}^{2}=-2x_{0}^{2}+2x_{0}+1</math>
		+
		+	i.e.
		+
		+
		+	<math>x_{0}^{2}-2x_{0}-1=0</math>
		+
		+
		+	This second-degree equation has solutions
		+
		+
		+	<math>x_{0}=1-\sqrt{2}</math>
		+	and
		+	<math>x_{0}=1+\sqrt{2}</math>
		+
		+
		+	Equation (1) gives the corresponding y-values:
		+
		+
		+	<math>y_{0}=-3+2\sqrt{2}</math>
		+	and
		+	<math>y_{0}=-3-2\sqrt{2}</math>
		+
		+
		+	Thus, the answers are the points
		+	<math>\left( 1-\sqrt{2} \right.,\left. -3+2\sqrt{2} \right)</math>
		+	and
		+	<math>\left( 1+\sqrt{2} \right.,\left. -3-2\sqrt{2} \right)</math>.
		+
	[[Image:1_1_5_3.gif|center]]		[[Image:1_1_5_3.gif|center]]

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Version vom 13:04, 10. Okt. 2008

Suppose that the tangent touches the curve at the point \displaystyle \left( x_{0} \right.,\left. y_{0} \right). That point must, first and foremost, lie on the curve and therefore satisfy the equation of the curve, i.e.

\displaystyle y_{0}=-x_{0}^{2}\quad \quad \quad \left( 1 \right)

If we now write the equation of the tangent as \displaystyle y=kx+m, the gradient of the tangent, \displaystyle k, is given by the value of the curve's derivative, \displaystyle {y}'=-2x, at \displaystyle x=x_{0},

\displaystyle k=-2x_{0}\quad \quad \quad \left( 2 \right)

The condition that the tangent goes through the point \displaystyle \left( x_{0} \right.,\left. y_{0} \right) gives us that

\displaystyle y_{0}=k\centerdot x_{0}+m\quad \quad \quad \left( 3 \right)

In addition to this, the tangent should also pass through the point \displaystyle \left( 1 \right.,\left. 1 \right),

\displaystyle 1=k\centerdot 1+m\quad \quad \quad \left( 4 \right)

Equations (1)-(4) constitute an equation system in the unknowns \displaystyle x_{0},\ y_{0},\ k and \displaystyle m.

Because we are looking for \displaystyle x_{0}\text{ } and \displaystyle y_{0}, the first step is to try and eliminate \displaystyle k and \displaystyle m from the equations.

Equation (2) gives that \displaystyle k=-\text{2 }x_{0} and substituting this into equation (4) gives

\displaystyle 1=-2x_{0}+m\quad \Leftrightarrow \quad m=2x_{0}+1

With \displaystyle k and \displaystyle m expressed in terms of \displaystyle x_{0} and \displaystyle y_{0}, (3) becomes an equation that is expressed completely in terms of \displaystyle x_{0} and \displaystyle y_{0},

\displaystyle y_{0}=-2x_{0}^{2}+2x_{0}+1\quad \quad \quad \left( 3' \right)

This equation, together with (1), is an equation system in \displaystyle x_{0} and \displaystyle y_{0}

\displaystyle \left\{ \begin{array}{*{35}l} y_{0}=-x_{0}^{2} \\ y_{0}=-2x_{0}^{2}+2x_{0}+1 \\ \end{array} \right.

Substituting equation (1) into (3') gives us an equation in x0,

\displaystyle -x_{0}^{2}=-2x_{0}^{2}+2x_{0}+1

i.e.

\displaystyle x_{0}^{2}-2x_{0}-1=0

This second-degree equation has solutions

\displaystyle x_{0}=1-\sqrt{2} and \displaystyle x_{0}=1+\sqrt{2}

Equation (1) gives the corresponding y-values:

\displaystyle y_{0}=-3+2\sqrt{2} and \displaystyle y_{0}=-3-2\sqrt{2}

Thus, the answers are the points \displaystyle \left( 1-\sqrt{2} \right.,\left. -3+2\sqrt{2} \right) and \displaystyle \left( 1+\sqrt{2} \right.,\left. -3-2\sqrt{2} \right).