Question

Use Jacobian determinants to test the existence of functional dependence between the paired functions. (a) \(y_{1}=3 x_{1}^{2}+x_{2}\) \(y_{2}=9 x_{1}^{4}+6 x_{1}^{2}\left(x_{2}+4\right)+x_{2}\left(x_{2}+8\right)+12\) (b) \(y_{1}=3 x_{1}^{2}+2 x_{2}^{2}\) \(y_{2}=5 x_{1}+1\)

Short answer

(a) Functionally dependent. (b) Not functionally dependent for \(x_{2} \, \neq \, 0\).

Step-by-step solution

Determine Partial Derivatives for Part (a)

Given the functions: \[y_{1} = 3x_{1}^{2} + x_{2}\] \[y_{2} = 9x_{1}^{4} + 6x_{1}^{2}(x_{2} + 4) + x_{2}(x_{2} + 8) + 12\]Let's find the partial derivatives for each function:For \(y_{1}\), the partial derivatives are:\[\frac{\partial y_{1}}{\partial x_{1}} = 6x_{1}, \quad \frac{\partial y_{1}}{\partial x_{2}} = 1\]For \(y_{2}\), the partial derivatives are:\[\frac{\partial y_{2}}{\partial x_{1}} = 36x_{1}^{3} + 12x_{1}(x_{2} + 4), \quad \frac{\partial y_{2}}{\partial x_{2}} = 6x_{1}^{2} + 2x_{2} + 8\]

Calculate Jacobian for Part (a)

The Jacobian determinant \(J\) is calculated as follows:\[J = \begin{vmatrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{1}}{\partial x_{2}} \ \frac{\partial y_{2}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{2}} \\end{vmatrix}.\]Substituting the partial derivatives:\[J = \begin{vmatrix} 6x_{1} & 1 \ 36x_{1}^{3} + 12x_{1}(x_{2} + 4) & 6x_{1}^{2} + 2x_{2} + 8 \\end{vmatrix}\]Calculating this determinant:\[J = (6x_{1})(6x_{1}^{2} + 2x_{2} + 8) - (1)(36x_{1}^{3} + 12x_{1}(x_{2} + 4))\]After simplification, this reduces to:\[J = 0\]

Analyze Part (a)

Since the Jacobian determinant \(J = 0\), it indicates that there is a functional dependence between \(y_{1}\) and \(y_{2}\) when expressed using \(x_{1}\) and \(x_{2}\).

Determine Partial Derivatives for Part (b)

Given the functions:\[y_{1} = 3x_{1}^{2} + 2x_{2}^{2}\]\[y_{2} = 5x_{1} + 1\]Let's find the partial derivatives for each function:For \(y_{1}\), the partial derivatives are:\[\frac{\partial y_{1}}{\partial x_{1}} = 6x_{1}, \quad \frac{\partial y_{1}}{\partial x_{2}} = 4x_{2}\]For \(y_{2}\), the partial derivatives are constant:\[\frac{\partial y_{2}}{\partial x_{1}} = 5, \quad \frac{\partial y_{2}}{\partial x_{2}} = 0\]

Calculate Jacobian for Part (b)

Now, compute the Jacobian determinant \(J\):\[J = \begin{vmatrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{1}}{\partial x_{2}} \ \frac{\partial y_{2}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{2}} \\end{vmatrix}.\]Substituting the partial derivatives:\[J = \begin{vmatrix} 6x_{1} & 4x_{2} \ 5 & 0 \\end{vmatrix}\]Calculating this determinant:\[J = (6x_{1})(0) - (4x_{2})(5) = -20x_{2}\]

Analyze Part (b)

For part (b), since the Jacobian \(J = -20x_{2}\) and is not identically zero (it depends on \(x_{2}\)), the functions \(y_{1}\) and \(y_{2}\) are not functionally dependent for values of \(x_{2} \, eq \, 0\).

Key concepts

Functional Dependence

In mathematics, the concept of functional dependence is crucial when examining relationships between variables. Imagine having functions like the ones in our exercise: they link input variables to outputs using specific rules. Functional dependence occurs if one function can be expressed as a combination of other functions. It is analogous to having a relationship between ingredients in a recipe, where changing one ingredient necessitates altering others in specific ways.

To test for functional dependence in the paired functions in the exercise, we use Jacobian determinants. If the Jacobian determinant of the functions is zero, it suggests that the variables are connected or dependent in some complex way. This dependency indicates a hidden relationship where one function's existence relies on another. It's like detecting a secret recipe hidden in multiple dishes.

Partial Derivatives

Understanding partial derivatives can seem daunting initially, but they simplify how we look at changes in functions. Partial derivatives focus on one variable at a time and examine how changes to that specific variable impact the function. Imagine tweaking the temperature setting on a thermostat while keeping all other conditions constant. The partial derivative is like measuring how the room temperature changes solely because of adjustments to the thermostat, not considering any other factor.

In our exercise, partial derivatives such as \(\frac{\partial y_{1}}{\partial x_{1}}\) or \(\frac{\partial y_{2}}{\partial x_{2}}\) are calculated. They show how changes to \(x_1\) or \(x_2\) directly impact the corresponding function \(y_1\) or \(y_2\). By isolating each variable's effect, we gain clearer insight into the function's behavior.

Determinant Calculation

The determinant calculation in mathematics offers valuable insights, especially when evaluating a matrix. In the context of this exercise, determinants play a key role in understanding the functional relationships between the given functions.

The Jacobian determinant is a special type of determinant constructed from the partial derivatives of the functions. Its purpose is to test for dependency. For instance, if our Jacobian \(J\) equals zero, it confirms functional dependence among our functions. Calculating determinants involves simple arithmetic operations like multiplication and addition in a structured format akin to a puzzle, where each number adds its unique impact on the final outcome.

In the solutions, we computed two determinants: one for each part of the problem. For example, we saw that for part (a), the Jacobian determinant became zero, indicating dependence. Meanwhile, the determinant in part (b) resulted in an expression \(-20x_2\), suggesting independence unless \(x_2\) is zero.