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Lösung 1.1:4

Aus Online Mathematik Brückenkurs 2

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K (Lösning 1.1:4 moved to Solution 1.1:4: Robot: moved page)
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{{NAVCONTENT_START}}
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If we write the equation of the tangent as
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<center> [[Image:1_1_4-1(3).gif]] </center>
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<center> [[Image:1_1_4-2(3).gif]] </center>
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<center> [[Image:1_1_4-3(3).gif]] </center>
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<math>y=kx+m</math>
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{{NAVCONTENT_STOP}}
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we know that the tangent's gradient
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<math>k</math>
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is equal to the derivative of
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<math>y=x^{\text{2}}\text{ }</math>
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at the point
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<math>x=\text{1}</math>, and since
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<math>{y}'=\text{2}x</math>, so
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<math>k={y}'\left( 1 \right)=2\centerdot 1=2</math>
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We can determine the constant
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<math>m\text{ }</math>
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with the condition that the tangent should go through the grazing point
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<math>\left( 1 \right.,\left. 1 \right)</math>, i.e. the point
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<math>\left( 1 \right.,\left. 1 \right)</math>
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should satisfy the equation of the tangent
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<math>1=2\centerdot 1+m</math>
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which gives that
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<math>m=-\text{1}</math>
[[Image:1_1_4_1.gif|center]]
[[Image:1_1_4_1.gif|center]]
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The normal to the curve
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<math>y=x^{\text{2}}\text{ }</math>
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at the point
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<math>\left( 1 \right.,\left. 1 \right)</math>
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is the straight line which is perpendicular to the tangent at the same point.
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Because two straight lines which are perpendicular to each other have gradients which satisfy
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<math>k_{1}\centerdot k_{2}=-1</math>, the normal must have a gradient which is equal to
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<math>-\frac{1}{k}=-\frac{1}{2}</math>
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The equation of the normal can therefore be written as
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<math>y=-\frac{1}{2}x+n</math>
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where
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<math>n</math>
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is some constant.
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Since the normal must pass through the line
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<math>\left( 1 \right.,\left. 1 \right)</math>, we can determine the constant
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<math>n</math>
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if we substitute the point into the equation of the normal,
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<math>1=-\frac{1}{2}\centerdot +n</math>
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and this gives
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<math>n=\frac{3}{2}</math>.
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[[Image:1_1_4-3(3).gif|center]]
[[Image:1_1_4-3(3).gif|center]]

Version vom 12:12, 10. Okt. 2008

If we write the equation of the tangent as


y=kx+m


we know that the tangent's gradient k is equal to the derivative of y=x2 at the point x=1, and since

y=2x, so


k=y1=21=2 

We can determine the constant m with the condition that the tangent should go through the grazing point 11 , i.e. the point 11  should satisfy the equation of the tangent


1=21+m


which gives that m=−1



The normal to the curve y=x2 at the point 11  is the straight line which is perpendicular to the tangent at the same point.

Because two straight lines which are perpendicular to each other have gradients which satisfy k1k2=−1, the normal must have a gradient which is equal to


−k1=−21


The equation of the normal can therefore be written as


y=−21x+n


where n is some constant.

Since the normal must pass through the line 11 , we can determine the constant n if we substitute the point into the equation of the normal,


1=−21+n


and this gives n=23.


Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:4“