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Test question marked wrong, even after going to office hours, but I may be right?

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose f is a smooth function of x and y, and at the point (1,1), [itex]\partial[/itex]f/[itex]\partial[/itex]x = 1, [itex]\partial[/itex]f/[itex]\partial[/itex]y = -1.

    Find the unit vector u such that Duf(1,1) = 0

    2. Relevant equations
    Duf(1,1) = [itex]\nabla[/itex]f(1,1) * u = 0


    3. The attempt at a solution

    My answer was merely u= k, which is the unit vector k of course (I don't know how to put the symbol on this. Work it out, I don't see how it is wrong. I will tell you why they said that I was wrong, and I will give my argument after an answer or two.
     
  2. jcsd
  3. Oct 24, 2011 #2

    ehild

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    f(x,y) is defined in the (x,y) plane. The problem asked a vector u also in the plane.

    ehild
     
  4. Oct 24, 2011 #3
    Okay, but that doesn't make any sense to me. Why would a function of x and y, or f(x,y), only be defined in the x y plane? If you had a function such as f(x,y) = (1/2)x^2 - (1/2)y^2, that wouldn't just be defined in the x,y plane, and the partial derivatives at (1,1) are still 1 and -1 respectively.

    Analogously, if you had a function of x, you wouldn't say that you have a function that is only defined in terms of the x axis.

    Edit: okay, I made a slight boo-boo in terminology. I do admit that. I meant, as stated below, to say what the graph looked like, which I expressed in a colloquial (and technically wrong) manner.
     
    Last edited: Oct 24, 2011
  5. Oct 24, 2011 #4

    Mark44

    Staff: Mentor

    "Defined in the x-y plane" refers to the domain of the function, not what the graph looks like, which requires three dimensions.
     
  6. Oct 24, 2011 #5
    Okay, I understand, but still: why is my answer wrong? The derivative in this direction 0, so how exactly was I supposed to know that my answer had to lie in the xy plane? I know that the gradient vectors would lie in the xy plane, but you can have something that lies in space that is perpendicular to these gradient vectors. What in the problem made this clear?
     
  7. Oct 24, 2011 #6

    HallsofIvy

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    The problem said "f is a smooth function of x and y, and at the point (1,1)". You can't just arbitrarily add more dimensions. Why not give a fourth or fifth dimension?
     
  8. Oct 24, 2011 #7
    I see what you are saying, but I interpreted (1,1) as x = 1, y = 1 for the function (though it says point). If it is a function of two variables, you would have a point (x, y, f(x,y)) not just (x,y), right? Otherwise, it wouldn't be a function of two variables, but a relation.

    I hope I'm not going insane... does anyone else see the problem that I am having?

    Edit: I am starting to see why I am wrong, but no one has been able to provide me with any insight that I didn't already have/has already been repeated to me for this problem. I didn't arbitrarily add another dimension to the problem; I did it because I took the cross product of the gradient vector and the vector at point (1,1) due to its underlying geometric significance (each respective component is perpendicular to its cross product).

    My question boils down to this: If you have a function of two variable f(x,y) , why is the direction ( u )of the directional derivative restricted to the xy plane (2 dimensions) if the function itself can be graphed in 3 dimensions?
     
    Last edited: Oct 24, 2011
  9. Oct 24, 2011 #8

    vela

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    The directional derivative is defined as
    [tex]D_\vec{u} f(\vec{x}) = \lim_{h \to 0} \frac{f(\vec{x}+h\vec{u})-f(\vec{x})}{h}[/tex]In this problem, the domain of f is the xy-plane, so [itex]\vec{x}+h\vec{u}[/itex] has to lie in the xy-plane. That means, in particular, [itex]\vec{u}[/itex] has to also lie in the xy-plane.
     
  10. Oct 24, 2011 #9
    Okay, finally! Thank you. I guess that's what I get for abandoning the limit definition of the directional derivative in favor of the dot product.
     
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