Short Answer

Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.
$f (x, y, z) = x y^{2} z^{3}, P = (0, 0, 0), v = ⟨ 1, - 2, - 1 ⟩$

$D_{u} f (0, 0, 0) = 0$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

$The directional derivative of a function f (x, y, z) at a point P = (x_{0}, y_{0}, z_{0}) in the direction of a unit vector u = ⟨ a, b, c ⟩ is given by D_{u} f (x_{0}, y_{0}, z_{0}) = lim_{h \to 0} \frac{f (x_{0} + a h, y_{0} + b h, z_{0} + c h) - f (x_{0}, y_{0}, z_{0})}{h} Here f (x, y, z) = x y^{2} z^{3}, and P = (0, 0, 0) .$

Step 2. Solution

$Also the vector is v = ⟨ 1, - 2, - 1 ⟩ The unit vector " u " in the direction of " v " is given by \begin{array}{l} u = \frac{1}{∥ v ∥} v \\ = \frac{1}{\sqrt{1^{2} + (- 2)^{2} + (- 1)^{2}}} ⟨ 1, - 2, - 1 ⟩ \\ = \frac{1}{\sqrt{6}} ⟨ 1, - 2, - 1 ⟩ \\ Hence the directional derivative is: \\ \begin{array}{l} D_{u} f (0, 0, 0) = lim_{h \to 0} \frac{f (\frac{h}{\sqrt{6}}, - \frac{2 h}{\sqrt{6}}, - \frac{h}{\sqrt{6}}) - f (0, 0, 0)}{h} \\ = lim_{h \to 0} \frac{\frac{h}{\sqrt{6}} {(\frac{- 2 h}{\sqrt{6}})}^{2} {(\frac{- h}{\sqrt{6}})}^{3} - 0}{h} \\ = lim_{h \to 0} \frac{- 4 h^{5}}{6^{3}} \\ = 0 \\ H e n c e D_{u} f (0, 0, 0) = 0 \end{array} \end{array}$

Find Study Materials for

Create Study Materials

Select your language

Calculus

Answers without the blur.

Short Answer

Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.
$f (x, y, z) = x y^{2} z^{3}, P = (0, 0, 0), v = ⟨ 1, - 2, - 1 ⟩$

Step by Step Solution

Step 1. Given Information

Step 2. Solution

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Calculus

Answers without the blur.

Short Answer

Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors. f(x,y,z)=xy2z3,P=(0,0,0),v=⟨1,−2,−1⟩

Step by Step Solution

Step 1. Given Information

Step 2. Solution

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Calculus

Pure Maths

Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors.
$f (x, y, z) = x y^{2} z^{3}, P = (0, 0, 0), v = ⟨ 1, - 2, - 1 ⟩$