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- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
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How to use it?
- Canonical form:
- sqrt(x+1)-11/(10*x)=0
- 24xy+24y+8xy2 +8y2
- 6x1^2+5x2^2+7x3^2-4x1x2+4x1x3=0
- y(x^2+1)
- Plot:
- 27*x=(3*y+x)^3
- z=/sqrt(3*x+2*y+2)
- y^2=4*x
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x^4*y-x*y^3=-2
-
Identical expressions
- 6x1^ two +5x two ^ two +7x3^2-4x1x2+4x1x3= zero
- 6x1 squared plus 5x2 squared plus 7x3 squared minus 4x1x2 plus 4x1x3 equally 0
- 6x1 to the power of two plus 5x two to the power of two plus 7x3 squared minus 4x1x2 plus 4x1x3 equally zero
- 6x12+5x22+7x32-4x1x2+4x1x3=0
- 6x1²+5x2²+7x3²-4x1x2+4x1x3=0
- 6x1 to the power of 2+5x2 to the power of 2+7x3 to the power of 2-4x1x2+4x1x3=0
- 6x1^2+5x2^2+7x3^2-4x1x2+4x1x3=O
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Similar expressions
- 6x1^2-5x2^2+7x3^2-4x1x2+4x1x3=0
- 6x1^2+5x2^2+7x3^2-4x1x2-4x1x3=0
- 6x1^2+5x2^2-7x3^2-4x1x2+4x1x3=0
- 6x1^2+5x2^2+7x3^2+4x1x2+4x1x3=0
6x1^2+5x2^2+7x3^2-4x1x2+4x1x3=0 canonical form
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2 2 2 5*x2 + 6*x1 + 7*x3 - 4*x1*x2 + 4*x1*x3 = 0
$$6 x_{1}^{2} - 4 x_{1} x_{2} + 4 x_{1} x_{3} + 5 x_{2}^{2} + 7 x_{3}^{2} = 0$$
6*x1^2 - 4*x1*x2 + 4*x1*x3 + 5*x2^2 + 7*x3^2 = 0
Invariants method
Given equation of the surface of 2-order:
$$6 x_{1}^{2} - 4 x_{1} x_{2} + 4 x_{1} x_{3} + 5 x_{2}^{2} + 7 x_{3}^{2} = 0$$
This equation looks like:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + 2 a_{14} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{14} = 0$$
$$a_{22} = 5$$
$$a_{23} = -2$$
$$a_{24} = 0$$
$$a_{33} = 6$$
$$a_{34} = 0$$
$$a_{44} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 18$$
$$I_{3} = \left|\begin{matrix}7 & 0 & 2\\0 & 5 & -2\\2 & -2 & 6\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}7 & 0 & 2 & 0\\0 & 5 & -2 & 0\\2 & -2 & 6 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}7 - \lambda & 0 & 2\\0 & 5 - \lambda & -2\\2 & -2 & 6 - \lambda\end{matrix}\right|$$
$$I_{1} = 18$$
$$I_{2} = 99$$
$$I_{3} = 162$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 18 \lambda^{2} - 99 \lambda + 162$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - 18 \lambda^{2} + 99 \lambda - 162 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = 3$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x1^{2} + 6 \tilde x2^{2} + 9 \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{3}\right)^{2}}\right) = 0$$
this equation is fora type imaginary cone
- reduced to canonical form
$$6 x_{1}^{2} - 4 x_{1} x_{2} + 4 x_{1} x_{3} + 5 x_{2}^{2} + 7 x_{3}^{2} = 0$$
This equation looks like:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + 2 a_{14} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = 7$$
$$a_{12} = 0$$
$$a_{13} = 2$$
$$a_{14} = 0$$
$$a_{22} = 5$$
$$a_{23} = -2$$
$$a_{24} = 0$$
$$a_{33} = 6$$
$$a_{34} = 0$$
$$a_{44} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
|a11 a12| |a22 a23| |a11 a13|
I2 = | | + | | + | |
|a12 a22| |a23 a33| |a13 a33|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
|a11 a14| |a22 a24| |a33 a34|
K2 = | | + | | + | |
|a14 a44| |a24 a44| |a34 a44| |a11 a12 a14| |a22 a23 a24| |a11 a13 a14|
| | | | | |
K3 = |a12 a22 a24| + |a23 a33 a34| + |a13 a33 a34|
| | | | | |
|a14 a24 a44| |a24 a34 a44| |a14 a34 a44|substitute coefficients
$$I_{1} = 18$$
|7 0| |5 -2| |7 2|
I2 = | | + | | + | |
|0 5| |-2 6 | |2 6|$$I_{3} = \left|\begin{matrix}7 & 0 & 2\\0 & 5 & -2\\2 & -2 & 6\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}7 & 0 & 2 & 0\\0 & 5 & -2 & 0\\2 & -2 & 6 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}7 - \lambda & 0 & 2\\0 & 5 - \lambda & -2\\2 & -2 & 6 - \lambda\end{matrix}\right|$$
|7 0| |5 0| |6 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| |7 0 0| |5 -2 0| |7 2 0|
| | | | | |
K3 = |0 5 0| + |-2 6 0| + |2 6 0|
| | | | | |
|0 0 0| |0 0 0| |0 0 0|$$I_{1} = 18$$
$$I_{2} = 99$$
$$I_{3} = 162$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 18 \lambda^{2} - 99 \lambda + 162$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
I3 != 0
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - 18 \lambda^{2} + 99 \lambda - 162 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = 3$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x1^{2} + 6 \tilde x2^{2} + 9 \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{3}\right)^{2}}\right) = 0$$
this equation is fora type imaginary cone
- reduced to canonical form