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Direction Derivatives

  1. Nov 5, 2007 #1
    1) http://www.geocities.com/asdfasdf23135/advcal7.JPG

    Now I don't understand why the stuff in blue are equal (completely lost...), can someone please explain why?

  2. jcsd
  3. Nov 5, 2007 #2


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    It's a definition! They are equal because they are defined that way! (That is, of course, just the ordinary definition of derivative put in terms of vectors.
  4. Nov 5, 2007 #3
    But how can you simplify d/dt (f(a+tu)) |t=0 to get
    lim [f(a+tu)-f(a)] / t ?

    I don't understand how I can calculate d/dt (f(a+tu)) |t=0, what is this equal to in terms of limits and how can I get there?
  5. Nov 5, 2007 #4
    Hm, let me try to explain...

    You can tell that the expression on the left is essentially just the limit definition of the derivative that you learned in Calc 1, except for the fact that now it's a function that takes a vector that we're taking the derivative of. The derivative is being evaluated at the point a and in the direction of the vector u. The parameter t lets you "walk" a little bit away from a in the direction of u on the function (analogous to making a step [itex]\Delta x[/itex] for a single-variable function).

    Then, the right hand side is just writing this same concept in a different form. It takes the derivative of the function f along our "walk" (with respect to the parameter t). Then, where do we want to find the derivative? At the location a, of course! So we have to evaluate our derivative df/dt when t=0 (when we're still at a).

    In terms of actually calculating something, can't you just substitute a+tu into the function and take the derivative with respect to t?

    [tex]\vec{f}(x,y) = x \hat{i} + y^2 \hat{j}[/tex]

    Let [tex]\vec{a} = \langle a_1, a_2 \rangle = \langle 1, 1 \rangle[/tex], [tex]\vec{u} = \langle u_1, u_2 \rangle = \langle -1, 2 \rangle[/tex]

    [tex]\vec{f}(\vec{a} + t\vec{u}) = \vec{f}(a_1 + t u_1, a_2 + t u_2) = \vec{f}(1-t, 1+2t) = (1-t) \hat{i} + (1+2t)^2 \hat{j}[/tex]

    [tex]\frac{d\vec{f}(\vec{a} + t\vec{u})}{dt} \bigg|_{t=0} = -1 \hat{i} + 2 (1+2t)^2 \hat{j} \bigg|_{t=0} = -\hat{i} + 2\hat{j}[/tex]

    Can someone confirm this is correct? It's been a while since I've done this. The notation may not be spot on either.
    Last edited: Nov 5, 2007
  6. Nov 6, 2007 #5
    "...the expression on the left is essentially just the limit definition of the derivative that you learned in Calc 1..."

    But what I've learnt in calculus 1 is that,
    f '(a)= lim [f(a+h) - f(a)] / h

    So shouldn't
    lim [f(a+tu)-f(a)] / t be equal to f '(a), instaed of d/dt (f(a+tu)) |t=0?
  7. Nov 6, 2007 #6


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    ?? d/dt (f(a+ tu))|t=0 IS f '(a)!
  8. Nov 7, 2007 #7
    Why? For d/dt (f(a+ tu))|t=0, shouldn't you calculate the derivative FIRST and then evaluate at t=0?
  9. Nov 7, 2007 #8


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    Yes, which is exactly what f'(a) means! If that isn't how you would calculate f'(a) then how would you?
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