The Implicit Function Theorem Examples 2

# The Implicit Function Theorem Examples 2

Be sure to review The Implicit Function Theorem page before looking at the examples given below.

## Example 1

Consider the system $\left\{\begin{matrix} xe^y + uz - \cos v = 2\\ u \cos y + x^2 v - yz^2 = 1 \end{matrix}\right.$. Show that this system can be solved for $u$ and $v$ in terms of $x$, $y$ and $z$ near the point $P$ where $(x, y, z) = (2, 0, 1)$ and $(u, v) = (1, 0)$. Also compute $\left ( \frac{\partial u}{\partial z} \right )_{x, y}$ at $P$.

Let $F(x, y, z, u, v) = xe^y + uz - \cos v - 2 = 0$ and let $G(x, y, z, u, v) = u \cos y + x^2 v - yz^2 - 1 = 0$. The problem implies that $u$, $v$ are dependent variables and $x$, $y$, and $z$ are independent variables. Consider the Jacobian:

(1)
\begin{align} \quad \frac{\partial (F, G)}{\partial (u, v)} = \begin{vmatrix} \frac{\partial F}{\partial u} & \frac{\partial F}{\partial v}\\ \frac{\partial G}{\partial u} & \frac{\partial G}{\partial v}\\ \end{vmatrix} = \begin{vmatrix} z & \sin v\\ \cos y & x^2\\ \end{vmatrix} = x^2 z - \sin v \cos y \end{align}

If we evaluate this point for $(x, y, z) = (2, 0, 1)$ and $(u, v) = (1, 0)$ then we have that:

(2)
\begin{align} \quad = 2^2 (1) - \sin (0) \cos (1) = 4 \neq 0 \end{align}

Since $\frac{\partial (F, G)}{\partial (u, v)}$ is nonzero at $P$ then we can solve the system for $u$ and $v$ in terms of $x$, $y$, and $z$ near $P$.

For the second part of the question, we have that:

(3)
\begin{align} \quad \left ( \frac{\partial u}{\partial z} \right )_{x, y} = - \frac{\frac{\partial (F, G)}{\partial (z,v)}}{\frac{\partial (F, G)}{\partial (u, v)}} = -\frac{1}{4} \frac{\partial (F, G)}{\partial (z, v)} = -\frac{1}{4} \begin{vmatrix} \frac{\partial F}{\partial z} & \frac{\partial F}{\partial v}\\ \frac{\partial G}{\partial z} & \frac{\partial G}{\partial v}\\ \end{vmatrix} = -\frac{1}{4} \begin{vmatrix} u & \sin v\\ -2yz & x^2\\ \end{vmatrix} = -\frac{1}{4} (ux^2 + 2yz \sin v) \end{align}

Evaluating the derivative at $P$ gives us that:

(4)
\begin{align} \quad \left ( \frac{\partial u}{\partial z} \right )_{x, y} = -\frac{1}{4} (4 + 0) = -1 \end{align}