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Partial derivative of coordinates

  1. Apr 10, 2010 #1
    Hello, I have a simple question. Lets say we have a mechanical system with 2 degrees of freedom. Say the generalized coordinates (which are independent from each other in terms of constraints) are [tex]x[/tex] and [tex]y[/tex]. When we solve Lagrange-Euler equations for this system, we get time evolutions of coordinates as functions of time: [tex]x(t)[/tex] and [tex]y(t)[/tex]. I am wondering what the partial derivative of x with respect to y would be. this would not be 0, right?
     
  2. jcsd
  3. Apr 10, 2010 #2

    clem

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    It is zero if they are independent coordinates.
     
  4. Apr 10, 2010 #3
    But mathematically, I can find a relation between [tex]x(t)[/tex] and [tex]y(t)[/tex], then I could compute the partial derivative, right? Would that still be zero?
     
  5. Apr 10, 2010 #4

    clem

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    Yes. The total derivative dx/dy might not be zero, but the partial will.
    [tex]\frac{\partial x}{\partial y}[/tex] means the change in x keeping x fixed.
     
    Last edited: Apr 10, 2010
  6. Apr 10, 2010 #5
    clem, thanks much for the explanations. But I am not sure what you mean by [tex]\frac{\partial x}{\partial y}[/tex] is "the change in x keeping x fixed.", do you mean the change in x keeping t fixed? Could you elaborate on that please?
     
  7. Apr 11, 2010 #6

    clem

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    The meaning of a partial derivative like [tex]\frac{\partial x}{\partial y}[/tex] is that all independent variables other than y are held fixed in the differentiation. Therefor x can't change in that derivative, so [tex]\frac{\partial x}{\partial y}[/tex] is zero by definition. The way I worded it was sort of funny non sequitor )obvious statement).
     
  8. Apr 11, 2010 #7
    Ok, I think I understand what you mean. But, I thought [tex]t[/tex] is the independent variable in the mechanics problems. (Or am i wrong?)

    Let me give an example:
    Say we are analyzing the movement of a point mass, which is thrown from earth's surface (assumed planar for the problem) with a velocity [tex]v[/tex], whose vertical component is [tex]v_y[/tex] (In +y direction) and horizontal component is [tex]v_x[/tex]. The gravitational acceleration, [tex]g[/tex], acts on the mass vertically to pull it in -y direction. [tex]x(t)[/tex] and [tex]y(t)[/tex] denote the horizontal and vertical components of position of the mass as a function of time. The solution for [tex]x(t)[/tex] is
    [tex]x(t) = v_x \cdot t[/tex]
    The solution for [tex]y(t)[/tex] is
    [tex]y(t) = v_y \cdot t - \frac{1}{2} g t^2[/tex]
    I am assuming that we only account for the time until the mass hits the ground, [tex]0 \leq t \leq 2\frac{v_y}{g}[/tex]. Having these, I will write [tex]y(t)[/tex] in terms of [tex]x(t)[/tex] as:
    [tex]y(x) = \frac{v_y}{v_x}\cdot x - \frac{1}{2}g (\frac{x}{v_x})^2[/tex]
    which I found simply by replacing [tex]t = \frac{x}{v_x}[/tex], which treats [tex]t[/tex] as a function of [tex]x[/tex], over the course of motion. Now I can compute the partial derivative of [tex]y[/tex] with respect to [tex]x[/tex] as:
    [tex]\frac{\partial y}{\partial x} = \frac{v_y}{v_x} - g \frac{x}{v_x^2}[/tex]
    which is not identically equal to [tex]0[/tex]. Did i do something wrong, or is this actually the total derivative of [tex]y[/tex] wrt [tex]x[/tex]?
     
  9. Apr 12, 2010 #8

    clem

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    What you have calculated is the total derivative.
    `Partial' means something is held fixed. What have you held fixed?
     
  10. Apr 12, 2010 #9
    Alright, I also thought that it should be the total derivative (as I was asking at the end of my last post). Ok then let me put it this way with another example, which I think should better illustrate my problem in understanding: Lets say I have three variables: x,y,z, which are parametrized with respect to t. Lets assume that [tex]t[/tex] is the ultimate independent variable, like time. [tex]x(t)[/tex], [tex]y(t)[/tex] and [tex]z(t)[/tex] given as:
    [tex]x(t) = t[/tex]
    [tex]y(t) = t^2[/tex]
    [tex]z(t) = t^2 + 4 \cdot t[/tex]
    now, I can write [tex]z[/tex] as:
    [tex]z = y + 4 \cdot x[/tex]
    then, partial derivative of [tex]z[/tex] with respect to [tex]x[/tex] and wrt [tex]y[/tex] would be:
    [tex]\frac{\partial z}{\partial x} = 4[/tex]
    [tex]\frac{\partial z}{\partial y} = 1[/tex]

    Now, I can also write [tex]z[/tex] as:
    [tex]z = x^2 + 4 \cdot x[/tex]
    then partial derivatives change:
    [tex]\frac{\partial z}{\partial x} = 2 \cdot x + 4[/tex]
    [tex]\frac{\partial z}{\partial y} = 0[/tex]

    Now, I can also write these using the chain rule:
    [tex]\frac{\partial z}{\partial x} = \frac{\partial z}{\partial t} \cdot \frac{\partial t}{\partial x}[/tex]
    and
    [tex]\frac{\partial z}{\partial x} = \underbrace{(2\cdot t + 4)}_{\frac{\partial z}{\partial t}} \cdot 1[/tex]

    and similarly for partial derivative of z wrt y
    [tex]\frac{\partial z}{\partial y} = \frac{\partial z}{\partial t} \cdot \frac{\partial t}{\partial y}[/tex]
    and
    [tex]\frac{\partial z}{\partial y} = (2\cdot t + 4) \cdot \underbrace{\frac{1}{2\sqrt{y}}}_{\frac{\partial t}{\partial y}}[/tex]

    The partial derivatives come out differently. Am I making a mistake here? Thanks a lot.
     
  11. Apr 12, 2010 #10

    SpectraCat

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    I think you made some mistakes in the quoted equations ... please check it over and correct them, or if what is there is really what you intended to write, then please indicate that. Then I can comment on what errors you might be making.
     
  12. Apr 12, 2010 #11

    Cyosis

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    The problem is that you're mixing up all kinds of derivatives. Partial, total and a type of derivative you invented yourself (the one in between partial and total).

    You have the functions x y and z that depend on t therefore your t depends on x y and z.

    The first derivative you take is some kind of mixed derivative. You arbitrarily take into account the x dependence of t, but not of t^2. It is like saying the derivative of of x^2+x is 1. If you want to take the x y z dependence of t into account you need to do it for all ts. That is what you did in your second dz/dx. However taking those dependencies into account means that you are not computing a partial derivative but a total derivative. Your third derivation (the chainrule one) is again not a partial derivative but a total derivative equal to your second attempt at calculating dz/dx. Why? Because again you take the x/t dependence into account.

    As has been said before if you want to calculate the partial derivative you will have to keep all coordinates, except the coordinate you're differentiating with respect to, fixed.

    partial derivative:
    [tex]
    \frac{\partial z}{\partial x}=\frac{\partial }{\partial x} (t^2+4t)=0
    [/tex]

    Total derivative:
    [tex]
    \frac{d z}{d x}=\frac{\partial z}{\partial t} \frac{d t}{dx}
    [/tex]
     
  13. Apr 13, 2010 #12
    Ok, I think I understand that I am inventing a new type of derivative there, which does not make physical sense. (Thanks for bearing with me)

    I have one more question. Lets assume we have three variables y, x, and t such that t is again independent variable. x and y are functions of t:
    [tex]y = t^2 + t[/tex]
    [tex]x = t[/tex]
    Thus, partial derivative of y wrt x is 0. (right?) But, assuming that t is also a function of x, if I write the chain rule (which I got from Stewart's calculus book):
    [tex]\frac{\partial y}{\partial x} = \frac{\partial y}{\partial t} \cdot \frac{\partial t}{\partial x} = 2t + 1[/tex].
    This looks like the total derivative of y wrt x to me. Am I using the definition wrongly?
     
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