Systems of Multivariable Equations
 Table of Contents

# Systems of Multivariable Equations

Suppose that we have two multivariable functions, $F(x, y, z, w) = 0$ and $G(x, y, z, w) = 0$. We thus have two equations with four variables or "unknowns", so we have solutions to this system. Where solutions exist, we will be able to partial differentiate the system in order to get partial derivatives of these solutions.

Before we look at some examples, we must first get a bit of notation out of the way. In the example above, if we have the following system of multivariable equations:

(1)
\begin{align} \left\{\begin{matrix} F(x, y, z, w) = 0\\ G(x, y, z, w) = 0 \end{matrix}\right. \end{align}

Then we have exactly two equations. If we wanted to compute $\frac{\partial x}{\partial z}$ then some ambiguity would arise. The variable $x$ is implied to be a dependent variable, while the variable $z$ is implied to be an independent variable in this example. The question to ask is whether the variables $y$ or $w$ are dependent or independent. Since we have two equations here, then one of the other variables (either $y$ or $w$) is a dependent variable, while the other is an independent variable.

The notation $\left ( \frac{\partial x}{\partial z} \right)_y$ implies that $y$ is also an independent variable, while the notation $\left ( \frac{\partial x}{\partial z} \right)_w$ implies that $w$ is also an independent variable. In general, we denote additional independent variables as subscripts when it is unclear from the context.

Now suppose that we wanted to compute $\left ( \frac{\partial x}{\partial z} \right )_w$. Then this implies that $x$ and $y$ are dependent variables and $z$ and $w$ are independent variables. If we partial differentiate the function $F(x, y, z, w) = 0$ with respect to $z$ and holding $w$ as fixed and noting that $\frac{\partial z}{\partial z} = 1$ and $\frac{\partial w}{\partial z} = 0$) then:

(2)
\begin{align} \quad F_1 \frac{\partial x}{\partial z} + F_2 \frac{\partial y}{\partial z} + F_3 \frac{\partial z}{\partial z} + F_4 \frac{\partial w}{\partial z} = F_1 \frac{\partial x}{\partial z} + F_2 \frac{\partial y}{\partial z} + F_3 = 0 \end{align}

Similarly, if we partial differentiate the function $G(x, y, z, w) = 0$ with respect to $z$ and holding $w$ as fixed then:

(3)
\begin{align} \quad G_1 \frac{\partial x}{\partial z} + G_2 \frac{\partial y}{\partial z} + G_3 \frac{\partial z}{\partial z} + G_4 \frac{\partial w}{\partial z} = G_1 \frac{\partial x}{\partial z} + G_2 \frac{\partial y}{\partial z} + G_3 = 0 \end{align}

Thus we obtain the following system of equations:

(4)
\begin{align} \left\{\begin{matrix} F_1 \frac{\partial x}{\partial z} + F_2 \frac{\partial y}{\partial z} = -F_3 \\ G_1 \frac{\partial x}{\partial z} + G_2 \frac{\partial y}{\partial z} = - G_3 \end{matrix}\right. \end{align}

This system is linear (thinking of $\frac{\partial x}{\partial z}$ and $\frac{\partial y}{\partial z}$ as the "variables"). Now if you have taken a course on Linear Algebra, you might have heard of Cramer's Rule. We can apply Cramer's rule to find the partial derivative $\left ( \frac{\partial x}{\partial z} \right)_w$

(5)
\begin{align} \left ( \frac{\partial x}{\partial z} \right )_w = \frac{\det \begin{vmatrix} -F_3 & F_2\\ -G_3 & G_2 \end{vmatrix}}{\det \begin{vmatrix} F_1 & F_2\\ G_1 & G_2 \end{vmatrix}} = \frac{-F_3G_2 + F_2G_3}{F_1G_2 - F_2G_1} = -\frac{F_3G_2 - F_2G_3}{F_1G_2 - F_2G_1} \end{align}

Similarly we can obtain $\left ( \frac{\partial x}{\partial z} \right)_y$ as:

(6)
\begin{align} \left ( \frac{\partial x}{\partial z} \right )_y = \frac{\det \begin{vmatrix} F_1 & -F_3\\ G_1 & -G_3 \end{vmatrix}}{\det \begin{vmatrix} F_1 & F_2\\ G_1 & G_2 \end{vmatrix}} = \frac{-F_1G_3 +F_3G_1}{F_1G_2 - F_2G_1} = -\frac{F_1G_3 - F_3G_1}{F_1G_2 - F_2G_1} \end{align}

We will see the significance of these partial derivatives soon.

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