Dot product between grad f and an arbitrary vector

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SUMMARY

The discussion clarifies the relationship between the dot product of the gradient of a function \( f: \mathbb{R}^n \to \mathbb{R} \) and an arbitrary vector \( v \) at a point \( x \). It establishes that the dot product \( \nabla f \cdot v \) is equivalent to the directional derivative \( D_v f \) multiplied by the magnitude of \( v \). The directional derivative represents the rate of change of \( f \) in the direction of \( v \), confirming that the dot product with a unit vector gives the derivative in that direction.

PREREQUISITES
  • Understanding of gradient notation and vector calculus
  • Familiarity with directional derivatives
  • Knowledge of the concept of dot products in linear algebra
  • Basic understanding of functions from \( \mathbb{R}^n \) to \( \mathbb{R} \)
NEXT STEPS
  • Study the properties of directional derivatives in vector calculus
  • Learn about the implications of the gradient in optimization problems
  • Explore the relationship between gradients and level curves
  • Investigate applications of the dot product in physics and engineering
USEFUL FOR

Mathematicians, physics students, and anyone studying multivariable calculus or optimization techniques will benefit from this discussion.

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Given a function f: R^n -> R, a point x in R^n, and an arbitrary vector v in R^n - is the dot product between grad f and v (evaluated at x) the same as df/dv?

If yes, it would be great if someone were to demonstrate a proof.

If no, what should be the correct interpretation of the dot product?
 
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The first thing you will have to do is define df/dv! I know, for example, Dvf as the directional derivative Defennder refers to- the rate of change of f in the direction of v which is independent of the length of v. The dot product of grad f with an arbitrary unit vector is the derivative in that direction. The dot product of grad f with an arbitrary vector is the derivative in that direction multiplied by the length of the vector.
 
Thanks for the links! It makes sense now.
 

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