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Is this derivation valid?

  1. Jun 28, 2013 #1
    For directional derivatives:

    Let [itex]\hat{u}=<a,b,c>[/itex] be the direction.

    Thus, [itex]\frac{∂\hat{u}}{∂x}=\frac{\sqrt{a^2+b^2+c^2}}{a}[/itex] and so on. So,

    [itex]\frac{∂x}{∂\hat{u}}=\frac{a}{\sqrt{a^2+b^2+c^2}}=a[/itex]

    Thus,

    [itex]\frac{∂F}{∂\hat{u}}=\frac{∂F}{∂x}a+\frac{∂F}{∂y}b+\frac{∂F}{∂z}c=∇F \bullet \hat{u}[/itex].
     
    Last edited: Jun 28, 2013
  2. jcsd
  3. Jun 28, 2013 #2

    Mark44

    Staff: Mentor

    Are a, b, and c constants? If so, the derivative of ##\hat{u}## would be the zero vector.
     
  4. Jun 28, 2013 #3
    I probably should have defined it better; [itex]\hat{u}[/itex] is a unit vector in the direction that we are trying to find the derivative in. The linear derivatives of [itex]\hat{u}[/itex] are defined as derivatives of a line in that direction.
     
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