If u =xyz, v =xy+yz+zx and w =x+y+z find jacobian of u,v,w on x,y,z.
Answers
The jacobian of u,v,w on x,y,z is (x-y) (y-z) (z-x).
Given:
u =xyz, v =xy+yz+zx and w =x+y+z
To find:
Jacobian of u,v,w on x,y,z.
Solution:
u =xyz, v =xy+yz+zx and w =x+y+z
∂u/∂x=yz
∂u/∂y=xz
∂u/∂z=xy
∂w/∂x=1
∂w/∂y=1
∂w/∂z=1
∂v/∂x=y+z
∂v/∂y=x+z
∂v/∂z=x+y
∂(u,v,w)/∂(x,y,z) = ∂u/∂x ∂u/∂y ∂u/∂z = yz xz xy
∂v/∂x ∂v/∂y ∂v/∂z y+z x+z x+y
∂w/∂x ∂w/∂y ∂w/∂z 1 1 1
= yz(x+z-x-y)-xz(y+z-x-y)+xy(y+z-x-z)
=yz(z-y)-xz(z-x) +xy(y-x)
=Yz^2-y^2z-xz^2+x^2z+xy^2-x^y
=(x-y) (y-z) (z-x)
- A matrix with a first-order partial vector function derivative, which can have any form, is referred to as a Jacobian matrix.
- The determinant of the jacobian matrix is called the jacobian.
- Each partial derivative of a vector function will be contained in the matrix.
- The transformation of coordinates is where Jacobian is most frequently used.
- It discusses differentiation as it relates to coordinate transformation.
- The components of this Jacobian matrix, which is derived from the state matrix, are used to calculate the results of sensitivity tests.
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The Jacobian matrix for u, v, and w with respect to x, y, and z is:
J = | yz xz xy |
| y+z x+z x+y |
| 1 1 1 |
The Jacobian matrix J is defined as:
J = | ∂u/∂x ∂u/∂y ∂u/∂z |
| ∂v/∂x ∂v/∂y ∂v/∂z |
| ∂w/∂x ∂w/∂y ∂w/∂z |
Here,
u = xyz
v = xy + yz + zx
w = x + y + z
Taking partial derivatives of u, v, and w with respect to x, y, and z,
we get:
∂u/∂x = yz, ∂u/∂y = xz, ∂u/∂z = xy
∂v/∂x = y + z, ∂v/∂y = x + z, ∂v/∂z = x + y
∂w/∂x = 1, ∂w/∂y = 1, ∂w/∂z = 1
Substituting these partial derivatives into the Jacobian matrix, we get:
J = | yz xz xy |
| y+z x+z x+y |
| 1 1 1 |
Therefore, the Jacobian matrix for u, v, and w with respect to x, y, and z is:
J = | yz xz xy |
| y+z x+z x+y |
| 1 1 1 |
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