Find symmetric equaihous of the normadLine to the Surface2xy=0 at point (-3,-2,12

Name: Find symmetric equaihous of the normadLine to the Surface2xy=0 at poi
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Description: Find symmetric equaihous of the normadLine to the Surface2xy=0 at point (-3,-2,12

Answered: Find Symmettric equahóus of the normal…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter3: Functions And Graphs

Section3.6: Quadratic Functions

Problem 36E

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Question

Find symmetric equaihous of the normad
Line to the Surface
2xy
=0 at point (-3,-2,12

Expert Solution

Step 1

Given that, the function is $F (x, y, z) = 2 x y - z$ and the point is $(- 3, - 2, 12)$ .

Obtain the value of $F_{x} (x, y, z), F_{y} (x, y, z), F_{z} (x, y, z)$ .

$\begin{array}{rcl} F_{x} (x, y, z) & = & \frac{\partial}{\partial x} (2 x y - z) \\ = & 2 y \\ F_{y} (x, y, z) & = & \frac{\partial}{\partial x} (2 x y - z) \\ = & 2 x \\ F_{z} (x, y, z) & = & \frac{\partial}{\partial x} (2 x y - z) \\ = & - 1 \end{array}$

Obtain the value of $F_{x} (- 3, - 2, 12), F_{y} (- 3, - 2, 12), F_{z} (- 3, - 2, 12)$ .

$\begin{array}{rcl} F_{x} (- 3, - 2, 12) & = & 2 (- 2) \\ = & - 4 \\ F_{y} (- 3, - 2, 12) & = & 2 (- 3) \\ = & - 6 \\ F_{z} (- 3, - 2, 12) & = & - 1 \end{array}$

Therefore, $\nabla F = <- 4, - 6, - 1>$ .

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