Mister Exam
Other calculators
- 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:
- 4x^2-y^2+8x+4y+4=0
- 3*x^2-7*y^2+3*z^2+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=0
- 2xy+2xz-2yz
- 16x^2-24xy+9y^2+25x-50y+50=0
- Plot:
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e^y/x=x*y+1
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|arctg(y/x)|=pi/12
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x^2+y^2
- tan(x*factorial(y))=1
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Identical expressions
- three *x^ two - seven *y^ two + three *z^ two + eight *x*y- eight *x*z- eight *y*z+ ten *x- fourteen *y- six *z- eight = zero
- 3 multiply by x squared minus 7 multiply by y squared plus 3 multiply by z squared plus 8 multiply by x multiply by y minus 8 multiply by x multiply by z minus 8 multiply by y multiply by z plus 10 multiply by x minus 14 multiply by y minus 6 multiply by z minus 8 equally 0
- three multiply by x to the power of two minus seven multiply by y to the power of two plus three multiply by z to the power of two plus eight multiply by x multiply by y minus eight multiply by x multiply by z minus eight multiply by y multiply by z plus ten multiply by x minus fourteen multiply by y minus six multiply by z minus eight equally zero
- 3*x2-7*y2+3*z2+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=0
- 3*x²-7*y²+3*z²+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=0
- 3*x to the power of 2-7*y to the power of 2+3*z to the power of 2+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=0
- 3x^2-7y^2+3z^2+8xy-8xz-8yz+10x-14y-6z-8=0
- 3x2-7y2+3z2+8xy-8xz-8yz+10x-14y-6z-8=0
- 3*x^2-7*y^2+3*z^2+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=O
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Similar expressions
- 3*x^2-7*y^2+3*z^2-8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=0
- 3*x^2+7*y^2+3*z^2+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=0
- 3*x^2-7*y^2+3*z^2+8*x*y-8*x*z+8*y*z+10*x-14*y-6*z-8=0
- 3*x^2-7*y^2+3*z^2+8*x*y-8*x*z-8*y*z+10*x+14*y-6*z-8=0
- 3*x^2-7*y^2+3*z^2+8*x*y+8*x*z-8*y*z+10*x-14*y-6*z-8=0
- 3*x^2-7*y^2-3*z^2+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=0
- 3*x^2-7*y^2+3*z^2+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z+8=0
- 3*x^2-7*y^2+3*z^2+8*x*y-8*x*z-8*y*z+10*x-14*y+6*z-8=0
- 3*x^2-7*y^2+3*z^2+8*x*y-8*x*z-8*y*z-10*x-14*y-6*z-8=0
3*x^2-7*y^2+3*z^2+8*x*y-8*x*z-8*y*z+10*x-14*y-6*z-8=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 2 -8 - 14*y - 7*y - 6*z + 3*x + 3*z + 10*x - 8*x*z - 8*y*z + 8*x*y = 0
$$3 x^{2} + 8 x y - 8 x z + 10 x - 7 y^{2} - 8 y z - 14 y + 3 z^{2} - 6 z - 8 = 0$$
3*x^2 + 8*x*y - 8*x*z + 10*x - 7*y^2 - 8*y*z - 14*y + 3*z^2 - 6*z - 8 = 0
Invariants method
Given equation of the surface of 2-order:
$$3 x^{2} + 8 x y - 8 x z + 10 x - 7 y^{2} - 8 y z - 14 y + 3 z^{2} - 6 z - 8 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 4$$
$$a_{13} = -4$$
$$a_{14} = 5$$
$$a_{22} = -7$$
$$a_{23} = -4$$
$$a_{24} = -7$$
$$a_{33} = 3$$
$$a_{34} = -3$$
$$a_{44} = -8$$
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
$$I_{1} = -1$$
$$I_{3} = \left|\begin{matrix}3 & 4 & -4\\4 & -7 & -4\\-4 & -4 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & 4 & -4 & 5\\4 & -7 & -4 & -7\\-4 & -4 & 3 & -3\\5 & -7 & -3 & -8\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 4 & -4\\4 & - \lambda - 7 & -4\\-4 & -4 & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = -1$$
$$I_{2} = -81$$
$$I_{3} = 81$$
$$I_{4} = 81$$
$$I{\left(\lambda \right)} = - \lambda^{3} - \lambda^{2} + 81 \lambda + 81$$
$$K_{2} = -75$$
$$K_{3} = 162$$
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
$$\lambda^{3} + \lambda^{2} - 81 \lambda - 81 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = -9$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} - \tilde y^{2} - 9 \tilde z^{2} + 1 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{1}{3}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(1^{-1}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{3}\right)^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form
$$3 x^{2} + 8 x y - 8 x z + 10 x - 7 y^{2} - 8 y z - 14 y + 3 z^{2} - 6 z - 8 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 4$$
$$a_{13} = -4$$
$$a_{14} = 5$$
$$a_{22} = -7$$
$$a_{23} = -4$$
$$a_{24} = -7$$
$$a_{33} = 3$$
$$a_{34} = -3$$
$$a_{44} = -8$$
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
$$I_{1} = -1$$
|3 4 | |-7 -4| |3 -4|
I2 = | | + | | + | |
|4 -7| |-4 3 | |-4 3 |$$I_{3} = \left|\begin{matrix}3 & 4 & -4\\4 & -7 & -4\\-4 & -4 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & 4 & -4 & 5\\4 & -7 & -4 & -7\\-4 & -4 & 3 & -3\\5 & -7 & -3 & -8\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 4 & -4\\4 & - \lambda - 7 & -4\\-4 & -4 & 3 - \lambda\end{matrix}\right|$$
|3 5 | |-7 -7| |3 -3|
K2 = | | + | | + | |
|5 -8| |-7 -8| |-3 -8| |3 4 5 | |-7 -4 -7| |3 -4 5 |
| | | | | |
K3 = |4 -7 -7| + |-4 3 -3| + |-4 3 -3|
| | | | | |
|5 -7 -8| |-7 -3 -8| |5 -3 -8|$$I_{1} = -1$$
$$I_{2} = -81$$
$$I_{3} = 81$$
$$I_{4} = 81$$
$$I{\left(\lambda \right)} = - \lambda^{3} - \lambda^{2} + 81 \lambda + 81$$
$$K_{2} = -75$$
$$K_{3} = 162$$
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
$$\lambda^{3} + \lambda^{2} - 81 \lambda - 81 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = -9$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} - \tilde y^{2} - 9 \tilde z^{2} + 1 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{1}{3}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(1^{-1}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{3}\right)^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form