Mister Exam
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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:
- y-x
- x^2=2y^2+3z^2
- 3x^2+4xy+3z^2+8xz+8yz=0
- 15x^2+6y^2-10z^2+90x-72y+60z+231=0
- Plot:
- z=3/sqrt(x^2+y-9)
-
(-x^2)*y^3+(x^2+y^2-1)^3=0
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x^2+4y^2
- tan(sin(e−1x))−x=0
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Identical expressions
- 15x^ two +6y^ two -1 zero z^ two +90x-72y+60z+ two hundred and thirty-one =0
- 15x squared plus 6y squared minus 10z squared plus 90x minus 72y plus 60z plus 231 equally 0
- 15x to the power of two plus 6y to the power of two minus 1 zero z to the power of two plus 90x minus 72y plus 60z plus two hundred and thirty minus one equally 0
- 15x2+6y2-10z2+90x-72y+60z+231=0
- 15x²+6y²-10z²+90x-72y+60z+231=0
- 15x to the power of 2+6y to the power of 2-10z to the power of 2+90x-72y+60z+231=0
- 15x^2+6y^2-10z^2+90x-72y+60z+231=O
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Similar expressions
- 15x^2+6y^2-10z^2+90x+72y+60z+231=0
- 15x^2+6y^2-10z^2-90x-72y+60z+231=0
- 15x^2+6y^2+10z^2+90x-72y+60z+231=0
- 15x^2-6y^2-10z^2+90x-72y+60z+231=0
- 15x^2+6y^2-10z^2+90x-72y+60z-231=0
- 15x^2+6y^2-10z^2+90x-72y-60z+231=0
15x^2+6y^2-10z^2+90x-72y+60z+231=0 canonical form
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The solution
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2 2 2 231 - 72*y - 10*z + 6*y + 15*x + 60*z + 90*x = 0
$$15 x^{2} + 90 x + 6 y^{2} - 72 y - 10 z^{2} + 60 z + 231 = 0$$
15*x^2 + 90*x + 6*y^2 - 72*y - 10*z^2 + 60*z + 231 = 0
Invariants method
Given equation of the surface of 2-order:
$$15 x^{2} + 90 x + 6 y^{2} - 72 y - 10 z^{2} + 60 z + 231 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 15$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 45$$
$$a_{22} = 6$$
$$a_{23} = 0$$
$$a_{24} = -36$$
$$a_{33} = -10$$
$$a_{34} = 30$$
$$a_{44} = 231$$
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
$$I_{1} = 11$$
$$I_{3} = \left|\begin{matrix}15 & 0 & 0\\0 & 6 & 0\\0 & 0 & -10\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}15 & 0 & 0 & 45\\0 & 6 & 0 & -36\\0 & 0 & -10 & 30\\45 & -36 & 30 & 231\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}15 - \lambda & 0 & 0\\0 & 6 - \lambda & 0\\0 & 0 & - \lambda - 10\end{matrix}\right|$$
$$I_{1} = 11$$
$$I_{2} = -120$$
$$I_{3} = -900$$
$$I_{4} = 27000$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 11 \lambda^{2} + 120 \lambda - 900$$
$$K_{2} = -1680$$
$$K_{3} = -45000$$
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
$$\lambda^{3} - 11 \lambda^{2} - 120 \lambda + 900 = 0$$
$$\lambda_{1} = 15$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = -10$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$15 \tilde x^{2} + 6 \tilde y^{2} - 10 \tilde z^{2} - 30 = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\frac{1}{10} \sqrt{10}}{\frac{1}{30} \sqrt{30}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{15} \sqrt{15}}{\frac{1}{30} \sqrt{30}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{1}{30} \sqrt{30}}\right)^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form
$$15 x^{2} + 90 x + 6 y^{2} - 72 y - 10 z^{2} + 60 z + 231 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 15$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 45$$
$$a_{22} = 6$$
$$a_{23} = 0$$
$$a_{24} = -36$$
$$a_{33} = -10$$
$$a_{34} = 30$$
$$a_{44} = 231$$
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
$$I_{1} = 11$$
|15 0| |6 0 | |15 0 |
I2 = | | + | | + | |
|0 6| |0 -10| |0 -10|$$I_{3} = \left|\begin{matrix}15 & 0 & 0\\0 & 6 & 0\\0 & 0 & -10\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}15 & 0 & 0 & 45\\0 & 6 & 0 & -36\\0 & 0 & -10 & 30\\45 & -36 & 30 & 231\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}15 - \lambda & 0 & 0\\0 & 6 - \lambda & 0\\0 & 0 & - \lambda - 10\end{matrix}\right|$$
|15 45 | | 6 -36| |-10 30 |
K2 = | | + | | + | |
|45 231| |-36 231| |30 231| |15 0 45 | | 6 0 -36| |15 0 45 |
| | | | | |
K3 = |0 6 -36| + | 0 -10 30 | + |0 -10 30 |
| | | | | |
|45 -36 231| |-36 30 231| |45 30 231|$$I_{1} = 11$$
$$I_{2} = -120$$
$$I_{3} = -900$$
$$I_{4} = 27000$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 11 \lambda^{2} + 120 \lambda - 900$$
$$K_{2} = -1680$$
$$K_{3} = -45000$$
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
$$\lambda^{3} - 11 \lambda^{2} - 120 \lambda + 900 = 0$$
$$\lambda_{1} = 15$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = -10$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$15 \tilde x^{2} + 6 \tilde y^{2} - 10 \tilde z^{2} - 30 = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\frac{1}{10} \sqrt{10}}{\frac{1}{30} \sqrt{30}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{15} \sqrt{15}}{\frac{1}{30} \sqrt{30}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{1}{30} \sqrt{30}}\right)^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form