Directional Derivative for f(x)

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Homework Help Overview

The discussion centers around the concept of the directional derivative, particularly in the context of functions of a single variable, such as f(x) = sin(x). Participants explore whether it is meaningful to define a directional derivative in one dimension and how unit vectors might apply in this scenario.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants question the validity of taking a directional derivative for a single-variable function, suggesting that varying x does not correspond to a direction in the same way as in higher dimensions. Others propose that while gradients can be defined, the concept of directional derivatives may not apply meaningfully.

Discussion Status

The discussion is active, with participants expressing differing opinions on the applicability of directional derivatives in one dimension. Some provide examples to illustrate their points, while others challenge the assumptions underlying the original question. No consensus has been reached, but various interpretations are being explored.

Contextual Notes

Participants note that in one dimension, there are only two possible directions (left and right), which complicates the notion of directional derivatives compared to higher dimensions. The original poster's use of a unit vector in a one-dimensional context is also questioned.

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Homework Statement


Is it possible to take the directional derivative of a function of a single variable?
For example if f(x)=sinx and the unit vector specifying the direction is u=<cos pi/4, sin pi/4>,
then could you say that duf = .5*(sqrt2)*cosx ?
 
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I vote no. The question looks meaningless to me. If they had said f(x,y)=sin(x) then I might agree with that.
 
Sure, you can do it in 1D. Say your function is ##f(x)=-x^2## Then ##\nabla f = -2x\vec i##. Now suppose, for example, this function represents the temperature in a 1D rod along the x-axis. At ##x=2## we have ##\nabla f(2) = -4\vec i##. Notice that the gradient points the direction of maximum increase in temperature (##-\vec i##) and its magnitude gives the rate of change in that direction.
 
LCKurtz said:
Sure, you can do it in 1D. Say your function is ##f(x)=-x^2## Then ##\nabla f = -2x\vec i##. Now suppose, for example, this function represents the temperature in a 1D rod along the x-axis. At ##x=2## we have ##\nabla f(2) = -4\vec i##. Notice that the gradient points the direction of maximum increase in temperature (##-\vec i##) and its magnitude gives the rate of change in that direction.

True, but what can you do with "the unit vector specifying the direction is u=<cos pi/4, sin pi/4>"?
 
Maybe you can take the gradient of a function of a single variable but just not the directional derivative? It seems to me now that the directional derivative wouldn't make any sense for f(x) because you can't vary x in a "direction". Varying x in the negative or positive direction still gives you the same result for the derivative, I think.

But since duf=gradf . u I think they might both have to go.
 
There are only two directions in one dimension- "left" and "right" (as opposed to an infinite number of directions in two or more dimensions).

You can think of the two directional derivatives of f(x) at x= a as f'(a) (to the right) and -f'(a) (to the left).
 
Dick said:
True, but what can you do with "the unit vector specifying the direction is u=<cos pi/4, sin pi/4>"?

Nothing, of course. I just answered the question I think he meant to ask. :rolleyes:
 

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