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Lösung 1.1:5

Aus Online Mathematik Brückenkurs 2

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Suppose that the tangent touches the curve at the point
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Suppose that the tangent touches the curve at the point <math>(x_0,y_0)</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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<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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{{Displayed math||<math>y_0 = -x_0^2\,\textrm{.}</math>|(1)}}
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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 <math>y=kx+m</math>, the slope of the tangent, ''k'', is given by the value of the curve's derivative, <math>y^{\,\prime} = -2x</math>, at <math>x=x_0</math>,
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{{Displayed math||<math>k = -2x_0\,\textrm{.}</math>|(2)}}
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If we now write the equation of the tangent as
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The condition that the tangent goes through the point <math>(x_0,y_0)</math> gives us that
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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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{{Displayed math||<math>y_{0} = k\cdot x_0 + m\,\textrm{.}</math>|(3)}}
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<math>k=-2x_{0}\quad \quad \quad \left( 2 \right)</math>
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In addition to this, the tangent should also pass through the point (1,1),
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The condition that the tangent goes through the point
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{{Displayed math||<math>1 = k\cdot 1 + m\,\textrm{.}</math>|(4)}}
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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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 +
Equations (1)-(4) constitute a system of equations in the unknowns <math>x_0</math>, <math>y_{0}</math>, <math>k</math> and <math>m</math>.
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<math>y_{0}=k\centerdot x_{0}+m\quad \quad \quad \left( 3 \right)</math>
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Because we are looking for <math>x_0</math> and <math>y_0</math>, the first step is to try and eliminate ''k'' and ''m'' from the equations.
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Equation (2) gives that <math>k = -2 x_0</math> and substituting this into equation (4) gives
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In addition to this, the tangent should also pass through the point
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{{Displayed math||<math>1 = -2x_0 + m\quad\Leftrightarrow\quad m = 2x_0+1\,\textrm{.}</math>}}
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<math>\left( 1 \right.,\left. 1 \right)</math>,
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 +
With ''k'' and ''m'' 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>
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and <math>y_0</math>,
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<math>1=k\centerdot 1+m\quad \quad \quad \left( 4 \right)</math>
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{{Displayed math||<math>y_0 = -2x_0^2 + 2x_0 + 1\,\textrm{.}</math>|(3')}}
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This equation, together with (1), is a system of equations in <math>x_0</math> and <math>y_0</math>,
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Equations (1)-(4) constitute an equation system in the unknowns
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{{Displayed math||<math>\left\{\begin{align}
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<math>x_{0},\ y_{0},\ k</math>
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y_{0} &= -x_0^{2}\,,\\[5pt]
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and
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y_{0} &= -2x_0^2 + 2x_0 + 1\,\textrm{.}
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<math>m</math>.
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\end{align}\right.</math>}}
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Because we are looking for
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Substituting equation (1) into (3') gives us an equation in <math>x_0</math>,
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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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{{Displayed math||<math>-x_0^2 = -2x_0^2 + 2x_0 + 1\,,</math>}}
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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.
i.e.
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{{Displayed math||<math>x_0^2 - 2x_0 - 1 = 0\,\textrm{.}</math>}}
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<math>x_{0}^{2}-2x_{0}-1=0</math>
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This quadratic equation has solutions
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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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{{Displayed math||<math>x_0 = 1-\sqrt{2}\qquad\text{and}\qquad x_0 = 1+\sqrt{2}\,\textrm{.}</math>}}
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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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Equation (1) gives the corresponding y-values,
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Thus, the answers are the points
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{{Displayed math||<math>y_0 = -3+2\sqrt{2}\qquad\text{and}\qquad y_0 = -3-2\sqrt{2}\,\textrm{.}</math>}}
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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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Thus, the answers are the points <math>(1-\sqrt{2},-3+2\sqrt{2})</math> and
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<math>(1+\sqrt{2},-3-2\sqrt{2})\,</math>.
[[Image:1_1_5_3.gif|center]]
[[Image:1_1_5_3.gif|center]]

Version vom 13:32, 14. 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=−x20. (1)

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

k=−2x0. (2)

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

y0=kx0+m. (3)

In addition to this, the tangent should also pass through the point (1,1),

1=k1+m. (4)

Equations (1)-(4) constitute a system of equations 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=−2x0 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 y0,

y0=−2x20+2x0+1. (3')

This equation, together with (1), is a system of equations in x0 and y0,

y0y0=−x20=−2x20+2x0+1.

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

−x20=−2x20+2x0+1

i.e.

x20−2x0−1=0.

This quadratic equation has solutions

x0=1−2andx0=1+2. 

Equation (1) gives the corresponding y-values,

y0=−3+22andy0=−3−22. 

Thus, the answers are the points (1−2−3+22)  and (1+2−3−22) .

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:5“