Lösung 1.1:4
Aus Online Mathematik Brückenkurs 2
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If we write the equation of the tangent as | If we write the equation of the tangent as | ||
+ | {{Displayed math||<math>y=kx+m</math>}} | ||
- | <math>y= | + | we know that the tangent's slope ''k'' is equal to the derivative of <math>y = x^2</math> at the point <math>x=1\,</math>, and since <math>y^{\,\prime} = 2x\,</math>, so |
+ | {{Displayed math||<math>k = y^{\,\prime}(1) = 2\cdot 1 = 2\,\textrm{.}</math>}} | ||
- | + | We can determine the constant ''m'' with the condition that the tangent should go through the grazing point (1,1), i.e. the point (1,1) should satisfy the equation of the tangent | |
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- | + | {{Displayed math||<math>1 = 2\cdot 1 + m</math>}} | |
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+ | which gives that <math>m=-1</math>. | ||
[[Image:1_1_4_1.gif|center]] | [[Image:1_1_4_1.gif|center]] | ||
+ | The normal to the curve <math>y=x^2</math> at the point (1,1) 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 slopes which satisfy <math>k_{1}\cdot k_{2} = -1\,</math>, the normal must have a slope which is equal to | |
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- | Because two straight lines which are perpendicular to each other have | + | |
- | <math>k_{1}\ | + | |
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+ | {{Displayed math||<math>-\frac{1}{k} = -\frac{1}{2}\,\textrm{.}</math>}} | ||
The equation of the normal can therefore be written as | The equation of the normal can therefore be written as | ||
+ | {{Displayed math||<math>y=-\frac{1}{2}x+n</math>}} | ||
- | + | where ''n'' is some constant. | |
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- | where | + | |
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- | is some constant. | + | |
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- | + | Since the normal must pass through the line (1,1), we can determine the constant ''n'' if we substitute the point into the equation of the normal, | |
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+ | {{Displayed math||<math>1=-\frac{1}{2}\cdot + n</math>}} | ||
+ | and this gives <math>n=\tfrac{3}{2}\,</math>. | ||
[[Image:1_1_4-3(3).gif|center]] | [[Image:1_1_4-3(3).gif|center]] |
Version vom 13:11, 14. Okt. 2008
If we write the equation of the tangent as
we know that the tangent's slope k is equal to the derivative of =2x
![]() ![]() |
We can determine the constant m with the condition that the tangent should go through the grazing point (1,1), i.e. the point (1,1) should satisfy the equation of the tangent
![]() |
which gives that
The normal to the curve
Because two straight lines which are perpendicular to each other have slopes which satisfy k2=−1
The equation of the normal can therefore be written as
where n is some constant.
Since the normal must pass through the line (1,1), we can determine the constant n if we substitute the point into the equation of the normal,
![]() |
and this gives