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kakt00s
Forumaš(ica)
Pridružen/a: 17. 10. 2007. (12:19:40)
Postovi: (183)16
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Lokacija: :ɐɾıɔɐʞoן
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Saf
Forumski umjetnik
Pridružen/a: 10. 06. 2005. (21:55:28)
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Lokacija: Zagreb
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 Postano: 19:31 ned, 11. 4. 2010 Naslov: |
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Ok, pada mi na pamet ili eliminacija nekog izraza ili sirovo izlučivanje, npr iz prve y^2 izrazit preko z i x, pa uvrstiti u drugu... Iz toga što dobiješ izraziš recimo z preko x i ponovno uvrstiš u jednadžbu za y... sada kažeš x=t, a ostalo kako ide...
Sve to ok u teoriji, ali ako idemo tom tehnikom, malo je zeznuto izrazit z preko x...
Evo primjer:
[code:1]Algebraically, we are looking for the points (x,y,z) that make both of the equations true:
x2z - xy2 = 4
2xz - y2 = 0
In some cases this may the simplest way to describe algebraically a curve in space: as the set of common solutions of two or more equations. But in this case we can use another method, which may be more useful: giving the coordinates of the points on the curve by expressions in some common variable (called a "parameter") which may or may not be one of the coordinates. In this case we can express y and z,and of course x itself, in terms of x on each of the two green curves, so we can "parametrize" the intersection curves by x: From the second equation we get y2 = 2xz, and substituting into the first equations gives x2z - x(2xz) = 4, or z = -4/x2 -- from which we can see immediately that the z-values will be negative. From the equation y2 = 2xz we can see that the x-values will also be negative; and substituting into this equation our expression for z in terms of x shows that y = ±sqrt(-8/x). So we have the desired parametrizations of the intersection curves:
x = x, y = ±sqrt(-8/x), z = -4/x2[/code:1]
Ok, pada mi na pamet ili eliminacija nekog izraza ili sirovo izlučivanje, npr iz prve y^2 izrazit preko z i x, pa uvrstiti u drugu... Iz toga što dobiješ izraziš recimo z preko x i ponovno uvrstiš u jednadžbu za y... sada kažeš x=t, a ostalo kako ide...
Sve to ok u teoriji, ali ako idemo tom tehnikom, malo je zeznuto izrazit z preko x...
Evo primjer:
| Kod: | Algebraically, we are looking for the points (x,y,z) that make both of the equations true:
x2z - xy2 = 4
2xz - y2 = 0
In some cases this may the simplest way to describe algebraically a curve in space: as the set of common solutions of two or more equations. But in this case we can use another method, which may be more useful: giving the coordinates of the points on the curve by expressions in some common variable (called a "parameter") which may or may not be one of the coordinates. In this case we can express y and z,and of course x itself, in terms of x on each of the two green curves, so we can "parametrize" the intersection curves by x: From the second equation we get y2 = 2xz, and substituting into the first equations gives x2z - x(2xz) = 4, or z = -4/x2 -- from which we can see immediately that the z-values will be negative. From the equation y2 = 2xz we can see that the x-values will also be negative; and substituting into this equation our expression for z in terms of x shows that y = ±sqrt(-8/x). So we have the desired parametrizations of the intersection curves:
x = x, y = ±sqrt(-8/x), z = -4/x2 |
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tierra
Forumaš(ica)
Pridružen/a: 07. 11. 2007. (12:46:15)
Postovi: (4D)16
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Lokacija: zg
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matovillka
Forumaš(ica)
Pridružen/a: 27. 03. 2009. (20:47:50)
Postovi: (3E)16
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tammy
Forumaš(ica)
Pridružen/a: 02. 07. 2007. (20:37:10)
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matovillka
Forumaš(ica)
Pridružen/a: 27. 03. 2009. (20:47:50)
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chinchi
Forumaš(ica)
Pridružen/a: 25. 02. 2005. (00:15:30)
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Gost
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chinchi
Forumaš(ica)
Pridružen/a: 25. 02. 2005. (00:15:30)
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matovillka
Forumaš(ica)
Pridružen/a: 27. 03. 2009. (20:47:50)
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kakt00s
Forumaš(ica)
Pridružen/a: 17. 10. 2007. (12:19:40)
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Lokacija: :ɐɾıɔɐʞoן
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mali_zeleni
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Pridružen/a: 30. 08. 2006. (12:12:05)
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chinchi
Forumaš(ica)
Pridružen/a: 25. 02. 2005. (00:15:30)
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kakt00s
Forumaš(ica)
Pridružen/a: 17. 10. 2007. (12:19:40)
Postovi: (183)16
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Lokacija: :ɐɾıɔɐʞoן
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tierra
Forumaš(ica)
Pridružen/a: 07. 11. 2007. (12:46:15)
Postovi: (4D)16
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Lokacija: zg
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