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I Directional Derivative demonstration

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  1. Jan 1, 2018 #1
    I find directional derivatives confusing. For example if there is a change in a direction and if this direction have both x and y components should not the change be calculated as square root of squares, i.e the pythogores theorem? Would you please provide a simple demonstration showing the directional derivative in terms of partial derivatives with respect to both x and y.

    Thank you.
     
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  3. Jan 1, 2018 #2

    PeroK

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    The directional derivative is the derivative in a given direction. Whatever that direction, you could imagine rotating your function so that the direction lies along the x-axis. In which case, the directional derivative is just the partial derivative with respect to x.

    Otherwise, the directional derivative is just the relevant proportions of the various partial derivatives. For example if the unit vector in the given direction is ##\vec{u} = (a, b)##, then:

    ##D_{\vec{u}}f = a f_x + bf_y##

    There's a proof of that here:

    http://tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx

    Does that help?
     
  4. Jan 1, 2018 #3
    That's not much help. I hope to see a visual example. How we demonstrate decomposition of a vector in a plane by projecting its components onto x and y-axes. Then we put one's tail to other's head to sum them. That example is easy to demonstrate. But how can we do this with directional derivatives?

    Thank you.
     
  5. Jan 1, 2018 #4
    This is confusing for me because there is a magnitude in x direction and there is a magnitude in y-direction. These magnitudes are magnitudes of rate of changes. These two directions are perpendicular to each other. So should not the Pythaogoras theorem be applied?
     
  6. Jan 1, 2018 #5

    PeroK

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    "Confusing" or "surprising"? The proof shows why ##a## and ##b## come out, rather than ##\sqrt{a^2 + b^2}##. You can't just apply Pythagoras. This is about the rate of change of a function, not the magnitude of a vector with two components.
     
  7. Jan 1, 2018 #6
    But isn't there any geometric explanation or interpretation or even an animation?

    Thank you.
     
  8. Jan 1, 2018 #7

    PeroK

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    Well, I guess you can google for an animation. I can't immediately see a neat geometric explanation - not that that means there isn't one. The proof comes from the chain rule for derivatives, rather than Pythagoras.

    Interesting!
     
  9. Jan 1, 2018 #8
    Yes, I had started before you wrote this and I encountered with this. This is a strange tool. Have you ever used before? A Computable Document does not make any sense to me.

    One of the important key phrase is "Visualize directional derivatives".

    http://demonstrations.wolfram.com/download-cdf-player.html

    Thank you.
     
  10. Jan 31, 2018 #9

    Stephen Tashi

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    Perhaps you aren't noticing that the pythagorean theorem is implicitly used in the definition of a directional derivative.

    Suppose you are in a room where the floor is "flat", but not level. You walk across the room from one corner to the diagonally opposite corner of the room and go uphill. You would go uphill by the same amount as if you had walked beside two of the walls to reach the opposite corner.

    (length of x-wall)(slope of x-wall) + (length of y-wall)(slope of y-wall) = (length of diagonal)(slope of diagonal).

    Symbollicaly:

    ## x s_x + y s_y = d s_d##
    ## s_d = (x s_x + y s_y)/ d = (x/d)s_x + (y/d)s_y ##

    The factor ##d## does equal ##sqrt(x^2 + y^2)##.

    Computing the directional derivative involves using a vector that defines a direction. In the case at hand, the unit vector that points diagonally across the room is ##(x/d, y/d)##.

    If you were thinking the Pythagorean theorem should be applied to ##s_x## and ##s_y##, you can see by the above example, that there is no reason to do that.
     
  11. Feb 11, 2018 #10

    FactChecker

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    Suppose you have tiny Δx and Δy in a particular direction. Then you also have a tiny vector in that direction of length Δs = √(Δx2 + Δy2). But Δs is not directly useful for finding the change of a function, f(x,y). In fact, Δs gives us no information regarding direction at all, so it is useless as a directional derivative. Since Δs came from Δx and Δy, which indicate direction, the way to estimate the change of f in a particular direction is to use the partials wrt x and y times Δx and Δy, respectively.

    On the other hand, suppose you parameterize a line in the desired direction using the variable s to get f(s). Then df/ds is exact. ∂f/∂x ∂x/∂s ds + ∂f/∂y ∂y/∂s ds should be a close approximation for small deltas, but that depends on the nature of the function f.
     
    Last edited: Feb 11, 2018
  12. Jun 4, 2018 #11
    This really made it click for me. THANK YOU!!
     
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