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Simple Differential Equation Question

  1. Nov 22, 2009 #1
    So I think that this question is easier than I am making it, but it's been awhile since I have had to solve a Diff EQ. And I am not even sure that the question even requires me to solve one:


    Instead of using arrows to represent vector functions, we sometimes use families of curves called field lines. A curve y = y(x) is a field line of the vector function F(x,y) if at each point (xo, yo) on the curve, F(xo, yo) is tangent to the curve.

    (a) Show that the field lines y = y(x) of a vector function F(x,y) = iFx(x,y) + jFy(x,y) are solutions of the differential equation



    (b) Determine the field lines of the function v(x,y) = iy + jx


    I know that the definition of field lines is very similar, if not identical, to that of the curves in an integral field.

    But I am really not sure, for part (a), how I am supposed to start. I am trying to show that y = y(x) is a solution to the diff eq, but I am only given the 'properties' of y(x). So I am a little confused as to how to start the math.

    Any hints?

  2. jcsd
  3. Nov 22, 2009 #2


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    Right. You don't have to solve the differential equation. The slope of the tangent to the curve y=y(x) is dy/dx, yes? What's the slope of the vector F(x,y)?
  4. Nov 23, 2009 #3
    Sorry Dick; I think it is going over my head :redface: When you ask what the slope of F(x,y) is, it is a little ambiguous to me. Do you mean what is the gradient?

    grad( F(x,y)) = grad(iFx(x,y) + jFy(x,y) )

    Wow. This brings to surface another of my mathematical deficiencies (what else is new); I have to go look up how to take the gradient of a vector function now. For some reason the fact that it is not a scalar function is throwing me for a loop.

    But, that aside, is that what you are referring to? Its gradient?
  5. Nov 23, 2009 #4


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    No, you don't have to take the gradient. The slope of a vector is just the y component over the x component, isn't it?
    Last edited: Nov 24, 2009
  6. Nov 24, 2009 #5
    Oh. I guess it is :redface:

    But since that I have brought it up and I am supposed to be working right now instead of being on PF, I cannot think of a better time to ask another question:

    1.) I am having a hard time wrapping my head around what the gradient of a vector function is. I am assuming it should be analogous to that of a scalar function. Would it just be the direction in the vector field that 'sees' sharpest change in vector magnitude?

    2.) How would you find the slope of a vector that is 3D? Or is that the ambiguous question? It has more than 1? Or is it a meaningless question?

  7. Nov 24, 2009 #6


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    The 'gradient' of a vector is a tensor. It's basically a collection of vectors, one for each scalar component of the vector. The thing that would see the sharpest change in the vector magnitude isn't that, it's just grad(|F|) - a different animal. And, yes, a vector in 3d doesn't have a unique 'slope'. You have to define the x and y axes of a plane containing it to measure a 'slope'.
  8. Nov 24, 2009 #7


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    The "gradient of a vector function" is not defined (unless you want to think of it as the linear transformation represented by the matrix having each row the gradient a component of the vector and I don't think that's what you are talking about). Dick was just saying that we can think of a vector (in two dimensions) as the direction of a line through the origin and he was asking about the slope of that line.

    The problem is that while direction of a line (perhaps determined by a vector) in 2 dimensions can be given as a single angle, the direction of a line in 3 dimensions cannot- although you could do it with two angles. The "direction cosines" the cosines of the angles the line makes with lines parallel to the axes. One can show that, if we call those angles, the angles a line makes with lines parallel to the x, y, and z axes, [itex]\theta[/itex], [itex]\phi[/itex], and [itex]\psi[/itex] respectively, then [math]cos^2(\theta)+ cos^2(\phi)+ cos^2(\psi)= 1[/itex] so that, knowing two of the angles, you can calculate the third. In fact, if you represent the line by the vector having those direction cosines as components, [itex]cos(\theta)\vec{i}+ cos(\phi)\vec{j}+ cos(\psi)\vec{k}[/itex] you just have a unit vector pointing in the same direction as the line.

    The corresponding thing in two dimensions would be [itex]cos(\theta)[/itex] and [itex]cos(\phi)[/itex], the cosines of the angles the line makes with the x and y axes respectively. But, of course, looking at the right triangle formed by that line and the two axes, those two angles are "complimentary": [itex]\phi= \pi/2- \theta[/itex] and so [itex]cos(\phi)= sin(\theta)[/itex]. You only need the single angle [itex]\theta[/itex].
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