Jacobian matrix with 2 variables

In summary, the jacobian matrix for the system z=x^2+x^3 and z=y+sin(x) is given by J = [2x 3x^2; cos(x) 1] and the determinant is 2x - 3x^2cos(x).
  • #1
feelau
61
0
jacobian matrix with 2 variables please help!

Homework Statement


so we have z=x^2+x^3 and z=y+sin(x). Find the jacobian matrix of this system. Find the determinant of this jacobian.

The Attempt at a Solution


The determinant part is easy, the only problem is trying to set this up. I'm not sure where to start. I think I need to take partial derivatives...do we do like dz/dx and dz/dy and just put them into matrix? please help
 
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  • #2
Yes, exactly. The partial derivatives of the first function, with respect to x and y, are the first row of the matrix, the first derivatives of the second function, with respect to x and y, are the second row of the matrix.

Just out of curiosity, how is the determinant part "easy" if you don't know what you are to take the determinant of?
 
  • #3
what do you mean? the determinant doesn't necessarily have to be a certain numerical value, just show what the answer is symbolically(that's what I believe atleast heh). so if I did this right the determinant should be just 2x+3x^2
 
  • #4
Are you sure you wrote the problem out right? both equations are for z?
 
  • #5
feelau said:
what do you mean? the determinant doesn't necessarily have to be a certain numerical value, just show what the answer is symbolically(that's what I believe atleast heh). so if I did this right the determinant should be just 2x+3x^2

Yes, that is correct. I said nothing about "a certain numerical value". I'm still wondering how you got 2x+ 3x2 for the determinant of the Jacobean matrix if you did not know what that matrix was! Oh, and I assumed that the two "z"s actually referred to different variables. That was what joob was complaining about. Normally, a Jacobean matrix reflects a change in variables. changing x,y to u, v or z,w would make sense. Changing x,y to z, z would not!
 
  • #6
[tex]
z_1=x^2+x^3
\medskip
z_2=y+sin(x)
[/tex]

Jacobian is:
[tex]
J=\left[ \begin{array}{cc}
\frac{dz_1}{dx} & \frac{dz_1}{dy} \\
\frac{dz_2}{dx} & \frac{dz_2}{dy}
\end{array}
\right]
[/tex]

...which is
[tex]
J=\left[ \begin{array}{cc}
2x & 3x^2 \\
cos(x) & 1
\end{array}
\right]
[/tex]

and determinant:

[tex]
2x -3x^2cos(x)
[/tex]
 

1. What is a Jacobian matrix with 2 variables?

A Jacobian matrix is a square matrix that contains partial derivatives of a vector-valued function. In the context of 2 variables, it represents the rate of change of a function with respect to each variable at a specific point.

2. How is a Jacobian matrix calculated?

To calculate a Jacobian matrix with 2 variables, you first find the partial derivative of the function with respect to each variable. These partial derivatives are then arranged in a matrix, with the derivative with respect to the first variable in the first row and the derivative with respect to the second variable in the second row.

3. What is the significance of a Jacobian matrix in mathematics?

A Jacobian matrix is important in mathematics because it helps in determining the behavior of a vector-valued function at a specific point. It also helps in solving optimization problems and in transforming variables in multivariate calculus.

4. How is a Jacobian matrix used in physics?

In physics, a Jacobian matrix is used to calculate the velocity and acceleration of a system described by a vector-valued function. It also plays a role in determining the stability of a system and in solving problems involving multiple variables and their derivatives.

5. Can a Jacobian matrix be used for functions with more than 2 variables?

Yes, a Jacobian matrix can be used for functions with any number of variables. In fact, it is commonly used in multivariate calculus for functions with 3 or more variables. The number of rows and columns in the matrix will depend on the number of variables in the function.

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