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Lösung 1.1:5

Aus Online Mathematik Brückenkurs 2

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

Version vom 13:04, 10. Okt. 2008

Suppose that the tangent touches the curve at the point x0y0 . That point must, first and foremost, lie on the curve and therefore satisfy the equation of the curve, i.e.


y0=x201 


If we now write the equation of the tangent as y=kx+m, the gradient of the tangent, k, is given by the value of the curve's derivative, y=2x, at x=x0,


k=2x02 

The condition that the tangent goes through the point x0y0  gives us that


y0=kx0+m3 


In addition to this, the tangent should also pass through the point 11 ,


1=k1+m4 


Equations (1)-(4) constitute an equation system in the unknowns x0 y0 k and m.

Because we are looking for x0 and y0, the first step is to try and eliminate k and m from the equations.

Equation (2) gives that k=2 x0 and substituting this into equation (4) gives


1=2x0+mm=2x0+1

With k and m expressed in terms of x0 and y0, (3) becomes an equation that is expressed completely in terms of x0 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).