Finding the Gradient Vector for a Given Point on a Surface

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Discussion Overview

The discussion revolves around finding the gradient vector of a surface defined by the equation f(x,y) at a specific point (x,y,z) and determining the direction in which the directional derivative Duf is equal to zero. The scope includes mathematical reasoning and technical explanation related to calculus and vector analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how to find the gradient vector for a given point on a surface where the directional derivative is zero.
  • Another participant explains that the directional derivative can be expressed in terms of a unit vector n and its components, suggesting that one can solve for the angle of n by setting the directional derivative to zero.
  • A third participant confirms the previous explanation, stating that the directional derivative at a point is the gradient of f dotted with the unit vector in the direction of interest, and notes the complexity of solving for where this derivative equals zero due to the three components of the direction vector.
  • A later post reiterates the initial question about finding the gradient vector and distinguishes between the gradient vector and the direction for which the directional derivative is zero, suggesting that the unit vector should be chosen such that the dot product with the gradient is zero.

Areas of Agreement / Disagreement

Participants express different perspectives on the relationship between the gradient vector and the direction for which the directional derivative is zero, indicating a lack of consensus on the best approach to the problem.

Contextual Notes

The discussion includes assumptions about the definitions of gradient and directional derivatives, and the complexity of solving for the direction vector is acknowledged but not resolved.

reddawg
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Say you are given the equation of a surface f(x,y) and a point (x,y,z) on the surface.

How would one find the gradient vector in which the directional derivative Duf is equal to zero.
 
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Given the equation z = f(x,y) and the point (x0,y0,z0) you want to find the direction along which the directional derivative is zero.
The directional derivative as a function of direction (the latter given by a unity vector n, with components n_x and n_y) can be written as
$$\frac{\partial f}{\partial n} = \frac{\partial f}{\partial x} n_x + \frac{\partial f}{\partial y} n_y$$
You can express n as a function of the angle with the coordinate axis, at which point you can equate the expression above to zero, and try to solve it for the angle of n.
 
That's exactly correct. The directional derivative of some function f(x,y,z) at a point x0, y0, z0 in the r direction is just the gradient of f at that point dotted with the unit vector in the r direction. Ie, df/dr = ∇f(x0, y0, z0) \bullet r. You'll want to solve for where df/dr = 0, which might be a little tricky since r has three components. Good luck!
 
reddawg said:
Say you are given the equation of a surface f(x,y) and a point (x,y,z) on the surface.

How would one find the gradient vector in which the directional derivative Duf is equal to zero.

The gradient vector and the direction in which ##D_u(f)=0## are two different things. Which do you want? You have ##\nabla f =\langle f_x, f_y\rangle## and a unit vector ##\hat u =
\langle a,b\rangle##. Choose ##\vec u## such that ##\nabla f \cdot \hat u## is zero.
 

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