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:
- y^2+z^2-3z=0
- -x^2+2y^2-2xy-2x-2y+2=0
- 3x^2+y^2-12x-2y+4=0
- (x^2)/6-(y^2)/4-(z^2)/1=0
- Plot:
- y-(|x|)=(1/2)
-
y=1.6((4x/y+6)(x/y+1)/(4x/y+1))^(1/2)
-
x^2/4+y^2/9=1
-
x^2/36-y^2=1
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Identical expressions
- y^ two +z^ two -3z= zero
- y squared plus z squared minus 3z equally 0
- y to the power of two plus z to the power of two minus 3z equally zero
- y2+z2-3z=0
- y²+z²-3z=0
- y to the power of 2+z to the power of 2-3z=0
- y^2+z^2-3z=O
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Similar expressions
- y^2-z^2-3z=0
- y^2+z^2+3z=0
y^2+z^2-3z=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Invariants method
Given equation of the surface of 2-order:
$$y^{2} + z^{2} - 3 z = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = - \frac{3}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & - \frac{3}{2}\\0 & 0 & - \frac{3}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} - \lambda$$
$$K_{2} = - \frac{9}{4}$$
$$K_{3} = - \frac{9}{4}$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 2 \lambda^{2} + \lambda = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\tilde x^{2} + \tilde y^{2} - \frac{9}{4} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{2}{3}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{3}}\right)^{2}} = 1$$
this equation is fora type elliptical cylinder
- reduced to canonical form
$$y^{2} + z^{2} - 3 z = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = - \frac{3}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 2$$
|0 0| |1 0| |0 0|
I2 = | | + | | + | |
|0 1| |0 1| |0 1|$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & - \frac{3}{2}\\0 & 0 & - \frac{3}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|$$
|0 0| |1 0| | 1 -3/2|
K2 = | | + | | + | |
|0 0| |0 0| |-3/2 0 | |0 0 0| |1 0 0 | |0 0 0 |
| | | | | |
K3 = |0 1 0| + |0 1 -3/2| + |0 1 -3/2|
| | | | | |
|0 0 0| |0 -3/2 0 | |0 -3/2 0 |$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} - \lambda$$
$$K_{2} = - \frac{9}{4}$$
$$K_{3} = - \frac{9}{4}$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 2 \lambda^{2} + \lambda = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\tilde x^{2} + \tilde y^{2} - \frac{9}{4} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{2}{3}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{3}}\right)^{2}} = 1$$
this equation is fora type elliptical cylinder
- reduced to canonical form