Answered: 1 V3' V3 V3 f(x, y, г) — 1 + 4хуz; P(1,…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Gradients in three dimensions Consider the following functions ƒ, points P, and unit vectors u.
a. Compute the gradient of ƒ and evaluate it at P.
b. Find the unit vector in the direction of maximum increase of ƒ at P.
c. Find the rate of change of the function in the direction of maximum
increase at P.
d. Find the directional derivative at P in the direction of the given vector.

1
V3' V3 V3
f(x, y, г) — 1 + 4хуz; P(1, —1, — 1):

Expert Solution

Step 1

(a)

Consider the given function $f (x, y, z) = 1 + 4 x y z$

Now partially differentiate

$\begin{array}{rcl} f_{x} (x, y, z) & = & 0 + 4 y z \\ = & 4 y z \end{array}$

Now partially differentiate

$\begin{array}{rcl} f_{y} (x, y, z) & = & 0 + 4 y z \\ = & 4 x z \end{array}$

Now partially differentiate

$\begin{array}{rcl} f_{z} (x, y, z) & = & 0 + 4 y x \\ = & 4 y x \end{array}$

Step 2

The gradient is $\nabla f (x, y, z) = <4 y z, 4 x z, 4 y x>$

At the point $P (1, - 1, - 1)$

$\begin{array}{rcl} \nabla f (1, - 1, - 1) & = & <4 (- 1) (- 1), 4 (1) (- 1), 4 (- 1) (1)> \\ = & <4, - 4, - 4> \end{array}$

(b)

The unit vector in the direction of maximum increase of $f$ is $\nabla f (x, y, z)$ at $P$

$\begin{array}{rcl} U_{a} & = & \frac{\nabla f (1, - 1, - 1)}{|\nabla f (1, - 1, - 1)|} \\ = & \frac{<4, - 4, - 4>}{\sqrt{4^{2} + {(- 4)}^{2} + {(- 4)}^{2}}} \\ = & \frac{<4, - 4, - 4>}{4 \sqrt{3}} \\ = & <\frac{1}{\sqrt{3}}, \frac{- 1}{\sqrt{3}}, \frac{- 1}{\sqrt{3}}> \end{array}$

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