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Problem about the derivative of an unknown function

  1. Jan 28, 2017 #1
    1. The problem statement, all variables and given/known data
    $$f:\mathbb{R^2}\to\mathbb{R}$$ a differentiable function in the origin so:

    $$f(t,t) =t^3+t$$ and $$f(t,-2t)=2t$$

    Calculate $$D_vf(0,0)$$

    $$v=(1,3)$$

    2. Relevant equations
    3. The attempt at a solution


    I have no idea on how to approach this problem.
    I know that because f is differentiable we have

    $$D_vf(0,0)= Df(0,0)v$$

    So I should be able to determine the partial derivatives. But how can I do it?

    My biggest obstacle is not knowing the value of $$f(0,0)$$ (and therefore not being able to take limit definition of partial derivative)
     
  2. jcsd
  3. Jan 28, 2017 #2

    LCKurtz

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    Well, for what it's worth, ##f(0,0)## = 0 from either of your equations given about ##f##, putting ##t=0##.
     
  4. Jan 28, 2017 #3

    LCKurtz

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    After looking a bit more carefully at your problem, here's another hint. If you have ##f(x,y)## and you want to know the rate of change along a path ##\vec R(t) = \langle x(t), y(t)\rangle##, have you seen the formula$$
    \frac{df}{dt} = \nabla f \cdot \frac{d \vec R}{dt}\text{ ?}$$
     
  5. Jan 28, 2017 #4
    Is what you mean equivalent to the derivative of a composition of functions?

    I think I got it. What I did was to differentiate both the equations given obtaining:

    $$\frac{df}{dx}(t,t) + \frac{df}{dy}(t,t) = 3t^2+1$$

    $$\frac{df}{dx}(t,-2t) -2 \frac{df}{dy}(t,-2t) = 2$$

    Then making t=0:
    $$\frac{df}{dx}(0,0) + \frac{df}{dy}(0,0) = 1$$

    $$\frac{df}{dx}(0,0) -2 \frac{df}{dy}(0,0) = 2$$

    Solving this system we obtain the partial derivatives of f in the point (0,0).
     
    Last edited: Jan 28, 2017
  6. Jan 28, 2017 #5

    LCKurtz

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    Aren't you missing a plus sign that I inserted above? Other than that it should work.
     
  7. Jan 28, 2017 #6
    Yes, thanks!
     
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