Science:Math Exam Resources/Courses/MATH200/December 2012/Question 03 (b)/Solution 1

What to do

The linear approximation of a function $z = f (x, y)$ in a point $(p_{x}, p_{y})$ is

$z = f (x, y) \approx f (p_{x}, p_{y}) + (x - p_{x}) \frac{\partial z}{\partial x} (p_{x}, p_{y}) + (y - p_{y}) \frac{\partial z}{\partial y} (p_{x}, p_{y})$

Find $f (- 1, 1)$

We have that $(p_{x}, p_{y}) = (- 1, 1)$ . To find $f (- 1, 1)$ , we must solve the equation $x y z + x + y^{2} + z^{3} = 0$ , where $(x, y) = (- 1, 1)$ for the variable $z$ .

$\begin{aligned} (- 1) 1 z + (- 1) + 1^{2} + z^{3} & = 0 \\ - z - 1 + 1 + z^{3} & = 0 \\ z (z^{2} - 1) & = 0 \\ z (z - 1) (z + 1) & = 0 \end{aligned}$

Hence, $z_{1} = - 1, z_{2} = 0, z_{3} = 1$ are solution to this equations. With the knowledge,that $f (- 1, 1) < 0$ , the value must be $z = f (- 1, 1) = - 1$

Find $\frac{\partial z}{\partial x} (- 1, 1)$

We already know from part a), that $\frac{\partial z}{\partial x} = - \frac{y z + 1}{x y + 3 z^{2}}$ . Pluging in the point $(x, y, z) = (- 1, 1, - 1)$ gives $\frac{\partial z}{\partial x} (- 1, 1) = 0$

Find $\frac{\partial z}{\partial y} (- 1, 1)$

We know that $\frac{\partial z}{\partial y} = - \frac{F_{y}}{F_{z}} = - \frac{x z + 2 y}{x y + 3 z^{2}}$ Pluging in the point $(x, y, z) = (- 1, 1, - 1)$ gives $\frac{\partial z}{\partial y} (- 1, 1) = - \frac{3}{2}$

The linear approximation

Hence the linear approximation is $f (x, y) \approx - 1 - \frac{3}{2} (y - 1)$