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
-
How to use it?
- Canonical form:
- 0x^2+0xy+5y^2-3x-4y+62=0
- -2x-y-4z+26=0
- 4z-yz-3z=12
- x^2-y^2-1=0
- Plot:
- (|y|)=(|3*x^2-4|)
- (|y|)=(|2*x-5|)
- (|y|)=(x-3)+2
- (|y|)=(5/2)*(|x|)-1/((|x|)-(5/2)*x*x)
- Integral of d{x}:
-
12
- Derivative of:
- 12
- Limit of the function:
-
12
-
Identical expressions
- 4z-yz-3z= twelve
- 4z minus yz minus 3z equally 12
- 4z minus yz minus 3z equally twelve
-
Similar expressions
- 4z-yz+3z=12
- 4z+yz-3z=12
4z-yz-3z=12 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:
$$- y z + z - 12 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = - \frac{1}{2}$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = \frac{1}{2}$$
$$a_{44} = -12$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 0 & - \frac{1}{2}\\0 & - \frac{1}{2} & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 0 & - \frac{1}{2} & 0\\0 & - \frac{1}{2} & 0 & \frac{1}{2}\\0 & 0 & \frac{1}{2} & -12\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & - \lambda & - \frac{1}{2}\\0 & - \frac{1}{2} & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = - \frac{1}{4}$$
$$K_{3} = 3$$
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
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2} - 12 = 0$$
$$\frac{\tilde x^{2}}{24} - \frac{\tilde y^{2}}{24} = -1$$
this equation is fora type hyperbolic cylinder
- reduced to canonical form
$$- y z + z - 12 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = - \frac{1}{2}$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = \frac{1}{2}$$
$$a_{44} = -12$$
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
$$I_{1} = 0$$
|0 0| | 0 -1/2| |0 0|
I2 = | | + | | + | |
|0 0| |-1/2 0 | |0 0|$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 0 & - \frac{1}{2}\\0 & - \frac{1}{2} & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 0 & - \frac{1}{2} & 0\\0 & - \frac{1}{2} & 0 & \frac{1}{2}\\0 & 0 & \frac{1}{2} & -12\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & - \lambda & - \frac{1}{2}\\0 & - \frac{1}{2} & - \lambda\end{matrix}\right|$$
|0 0 | |0 0 | | 0 1/2|
K2 = | | + | | + | |
|0 -12| |0 -12| |1/2 -12| |0 0 0 | | 0 -1/2 0 | |0 0 0 |
| | | | | |
K3 = |0 0 0 | + |-1/2 0 1/2| + |0 0 1/2|
| | | | | |
|0 0 -12| | 0 1/2 -12| |0 1/2 -12|$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = - \frac{1}{4}$$
$$K_{3} = 3$$
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
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2} - 12 = 0$$
$$\frac{\tilde x^{2}}{24} - \frac{\tilde y^{2}}{24} = -1$$
this equation is fora type hyperbolic cylinder
- reduced to canonical form