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:
- x^2+y^2+z^2=2x+2y-1
- y^2-x-4y+5=0
- -31x^2-24xy+14y^2=0
- 9x^2+4xy+6y^2+40x+20y-50=0
- Plot:
- y=-2*sqrt(|x|)+4
- x^2=y
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x^3(y-x)=x^3+1
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x^2/4+y^2/2=16
- Factor polynomial:
- x^2+y^2+z^2
- Limit of the function:
- x^2+y^2+z^2
- Integral of d{x}:
- x^2+y^2+z^2
- 2x+2y-1
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Identical expressions
- x^ two +y^ two +z^ two =2x+2y- one
- x squared plus y squared plus z squared equally 2x plus 2y minus 1
- x to the power of two plus y to the power of two plus z to the power of two equally 2x plus 2y minus one
- x2+y2+z2=2x+2y-1
- x²+y²+z²=2x+2y-1
- x to the power of 2+y to the power of 2+z to the power of 2=2x+2y-1
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Similar expressions
- x^2+y^2-z^2=2x+2y-1
- x^2+y^2+z^2=2x-2y-1
- x^2+y^2+z^2=2x+2y+1
- x^2-y^2+z^2=2x+2y-1
x^2+y^2+z^2=2x+2y-1 canonical form
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The solution
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2 2 2 1 + x + y + z - 2*x - 2*y = 0
$$x^{2} - 2 x + y^{2} - 2 y + z^{2} + 1 = 0$$
x^2 - 2*x + y^2 - 2*y + z^2 + 1 = 0
Invariants method
Given equation of the surface of 2-order:
$$x^{2} - 2 x + y^{2} - 2 y + z^{2} + 1 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -1$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = -1$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 1$$
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
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & -1\\0 & 1 & 0 & -1\\0 & 0 & 1 & 0\\-1 & -1 & 0 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = 3$$
$$I_{3} = 1$$
$$I_{4} = -1$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - 3 \lambda + 1$$
$$K_{2} = 1$$
$$K_{3} = -1$$
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
$$\lambda^{3} - 3 \lambda^{2} + 3 \lambda - 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = 1$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + \tilde y^{2} + \tilde z^{2} - 1 = 0$$
$$\frac{\tilde z^{2}}{\left(1^{-1}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(1^{-1}\right)^{2}} + \frac{\tilde y^{2}}{\left(1^{-1}\right)^{2}}\right) = 1$$
this equation is fora type ellipsoid
- reduced to canonical form
$$x^{2} - 2 x + y^{2} - 2 y + z^{2} + 1 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -1$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = -1$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 1$$
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
$$I_{1} = 3$$
|1 0| |1 0| |1 0|
I2 = | | + | | + | |
|0 1| |0 1| |0 1|$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & -1\\0 & 1 & 0 & -1\\0 & 0 & 1 & 0\\-1 & -1 & 0 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|$$
|1 -1| |1 -1| |1 0|
K2 = | | + | | + | |
|-1 1 | |-1 1 | |0 1| |1 0 -1| |1 0 -1| |1 0 -1|
| | | | | |
K3 = |0 1 -1| + |0 1 0 | + |0 1 0 |
| | | | | |
|-1 -1 1 | |-1 0 1 | |-1 0 1 |$$I_{1} = 3$$
$$I_{2} = 3$$
$$I_{3} = 1$$
$$I_{4} = -1$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - 3 \lambda + 1$$
$$K_{2} = 1$$
$$K_{3} = -1$$
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
$$\lambda^{3} - 3 \lambda^{2} + 3 \lambda - 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = 1$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + \tilde y^{2} + \tilde z^{2} - 1 = 0$$
$$\frac{\tilde z^{2}}{\left(1^{-1}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(1^{-1}\right)^{2}} + \frac{\tilde y^{2}}{\left(1^{-1}\right)^{2}}\right) = 1$$
this equation is fora type ellipsoid
- reduced to canonical form