What is the Correct Directional Derivative for Vector w in the Given Scenario?

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SUMMARY

The correct directional derivative for the vector w in the given scenario is calculated as -√5 when considering a unit vector. The discussion highlights the importance of using the norm of the vector, with the expression D representing the magnitude of the gradient of the function f at that point. Different authors, such as Munkres in "Analysis on Manifolds," may define directional derivatives differently, allowing for non-zero vectors. Clarity on the definition of directional derivative is essential for accurate computation.

PREREQUISITES
  • Understanding of directional derivatives in multivariable calculus
  • Familiarity with vector norms and unit vectors
  • Knowledge of gradient vectors and their significance
  • Basic concepts from "Analysis on Manifolds" by Munkres
NEXT STEPS
  • Study the definition of directional derivatives in various calculus textbooks
  • Learn about the properties of gradient vectors and their applications
  • Explore the differences in definitions of directional derivatives across different mathematical authors
  • Practice calculating directional derivatives using various vectors and functions
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Students and educators in mathematics, particularly those focusing on multivariable calculus, as well as professionals working with vector calculus in physics and engineering.

DottZakapa
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i compute the partial derivative, the vector that i have to use the one in the text or
w=(2/(5^(1/2)), 1/(5^(1/2)))

using the last one i get minus square root of five , if i don't divide by the norm the answer should be B.
i don't understand what D means
 
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DottZakapa said:
View attachment 255446
i compute the partial derivative, the vector that i have to use the one in the text or
w=(2/(5^(1/2)), 1/(5^(1/2)))

using the last one i get minus square root of five , if i don't divide by the norm the answer should be B.
i don't understand what D means

A direction derivative should be taken wrt a unit vector, so ##-\sqrt{5}## looks right. Unless ##\frac{\partial f}{\partial \vec v}## means something else?

The expression in ##D## is the magnitude of the gradient of ##f## at that point. Sometimes ##||u||## is used instead of ##|u|## for the "norm" or magnitiude of vector.
 
PeroK said:
A direction derivative should be taken wrt a unit vector, so ##-\sqrt{5}## looks right. Unless ##\frac{\partial f}{\partial \vec v}## means something else?

The expression in ##D## is the magnitude of the gradient of ##f## at that point. Sometimes ##||u||## is used instead of ##|u|## for the "norm" or magnitiude of vector.

There are authors who do not ask for this. For example Munkres in "Analysis on manifolds" who defines the directional derivative for any non-zero vector.

Also, I believe it is better to use ##\Vert v \Vert## (Use \Vert v \Vert as tex command) instead of ##|| v ||## (formats a bit more nicely imo).
 
so according to the text what shall i consider ?
 
DottZakapa said:
so according to the text what shall i consider ?

Depends on your definition of directional derivative. So please give us the definition your book provides.

If it is w.r.t. a unit vector it is ##-\sqrt{5}##, otherwise ##-5## is correct.
 
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