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I Proving grad operator yields perpendicular vector to contour

  1. Apr 14, 2016 #1
  2. jcsd
  3. Apr 14, 2016 #2

    stevendaryl

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    If you have a function [itex]f(x,y,z)[/itex], and you have a constant [itex]c[/itex], then the set of points satisfying [itex]f(x,y,z) = c[/itex] is a two-dimensional surface. So now we let [itex]\vec{r}(t)[/itex] be any path that stays on that surface. In terms of components, if we write [itex]\vec{r}(t) = (x(t), y(t), z(t))[/itex], then to say that [itex]\vec{r}(t)[/itex] stays on the surface [itex]f(x,y,z)=c[/itex] just means that for all [itex]t[/itex], we have:

    [itex]f(x(t), y(t), z(t)) = c[/itex]

    So far, this is just true by assumption. We're assuming that [itex]x(t), y(t), z(t)[/itex] are three functions such that [itex]f(x(t), y(t), z(t)) = c[/itex].

    Since [itex]f(x(t), y(t), z(t)) = c[/itex], then it immediately follows that:

    [itex]\frac{d}{dt} f(x(t), y(t), z(t)) = \frac{dc}{dt} = 0[/itex]

    All they're doing is defining [itex]f(x(t), y(t), z(t)) = g(t)[/itex]. That's just the definition of [itex]g[/itex]. So it immediately follows that [itex]\frac{d}{dt} g(t) = 0[/itex].
     
  4. Apr 14, 2016 #3
    Thank you very much, that make a lot of sense, I understand now!
     
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