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Derivative of a parametrized vector on a nonfixed basis

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the parameter derivative of the vector function v(u, v) in, say, polar coordinates, i.e, this: http://en.wikipedia.org/wiki/Vector...tive_of_a_vector_function_with_nonfixed_bases but deriving with respect to the parameter u or v instead of the time.


    2. Relevant equations



    3. The attempt at a solution
    I'd say the solution would be to do as in the link with the parameter u in the place of t, but I'm having problems at finding the u-derivative of the basis vectors. Should I apply the chain rule and then write it as de1/du = de1/dρ * dρ/du + de1/dθ * dθ/du? (with e1 being a basis vector and ρ and θ being the polar coordinates)

    What are then the derivatives of the polar coordinates respect to the parameter u? The derivative of the components of v(u, v) along those coordinates? (I mean, the derivatives of P(u,v) and Θ(u, v), given that v(u,v) = P(u, v)e1 + Θ(u, v)e2)
     
    Last edited: Jan 29, 2014
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  3. Jan 31, 2014 #2

    BvU

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    Is there a way for you to clarify the problem statement a little?
    v(u,v) is rather difficult to interpret.
    Filling in 'all variables and given/known data' and what you were handed as ammo under 'Relevant equations' would be a good start...

    The impression I get now is that the link you provide might be overly complicating the issue.
     
  4. Jan 31, 2014 #3
    Ok, suppose you have a curve defined by a vector parametrized through the variable $u$. You derive it with respect to that parameter. What happens to the base vectors? If you derive with respect to time you have to derive them as well (given that they are not fixed). What if you derive respect to a parameter? Do you consider them constant?
     
  5. Jan 31, 2014 #4

    BvU

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    Still unclear to me. Why would base vectors be dependent on parameters ?
    Perhaps the litteral wording of the problem you are trying to solve (under 1. ) is helpful ?
     
  6. Jan 31, 2014 #5
    No. You consider them functions of u, or, in polar coordinates, functions of θ (which is a function of u). You need to express the changes in both unit vectors in terms of the changes in θ, and then use the chain rule to multiply by the derivative of θ with respect to u.
     
  7. Feb 1, 2014 #6
    Thanks.

    I understand you mean this: dr/du = dr/dρ*dρ/du + dr/dθ*dθ/du (being r the vector)

    However I cannot see how the dependences of the polar coordinates on the parameters are defined, i.e ρ(u) and θ(u). I have functions that relate the components of the vector aong those coordinates, that is r = a(u)eρ + b(u)eθ. Are those the dependences?
     
  8. Feb 1, 2014 #7

    BvU

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    Yes.
     
  9. Feb 1, 2014 #8
    Not exactly. First of all, in polar coordinates, all position vectors drawn from the origin have only a radial component: [itex]\vec{r}=ρ\vec{e}_r(θ)[/itex]. The coordinates ρ and θ are regarded as function of u. So...
    [tex]\frac{d\vec{r}}{du}=\frac{dρ}{du}\vec{e}_r(θ)+ρ\frac{d\vec{e}_r(θ)}{dθ}\frac{dθ}{du}[/tex]
    But, [itex]\frac{d\vec{e}_r(θ)}{dθ}=\vec{e}_θ(θ)[/itex]
    Therefore,[tex]\frac{d\vec{r}}{du}=\frac{dρ}{du}\vec{e}_r(θ)+ρ\vec{e}_θ(θ)\frac{dθ}{du}[/tex]
    Note also that, in polar coordinates, vectors other than position vectors will typically have components in both coordinate directions.
     
  10. Feb 1, 2014 #9
    I still don't know how to obtain [tex]\frac{dθ}{du}[/tex] and [tex]\frac{dρ}{du}[/tex]
     
  11. Feb 1, 2014 #10
    You are having trouble understanding how the trajectory of a particle can be parameterized in terms of a parameter u, correct? If ρ=ρ(u) and θ=θ(u) then you can eliminate u from these equations to get ρ as a function of θ along the trajectory, or vice versa. So expressing ρ and θ in terms of u doesn't change the trajectory. It just makes it easier to work with. Now, if θ is changing along the trajectory, so are the unit vectors in the ρ and θ directions. So, if you want to find out how a position vector r is changing along the particle trajectory, you need to take into account how the unit vectors are changing along the trajectory.
     
  12. Feb 1, 2014 #11

    BvU

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    Would you do me (us) the favour and render the full problem statement? I follow this thread with interest and I wonder about the level at which all this takes place. Is there really a moving coordinate system? Or is it only about parametrisation and derivatives?
     
  13. Feb 2, 2014 #12
    We have a cone described by a vector function on cylindrical coordinates: [itex]\vec{r} (u, v) = u\vec{\hat{e}_z} + utg(α)\vec{\hat{e}_ρ} + v\vec{\hat{e}_θ}[/itex], where α is the angle of the cone.

    Find [itex]d\vec{r}[/itex]
     
  14. Feb 2, 2014 #13
    Ah. I finally see what this is all about now. You have a cone (2D surface) suspended somewhere in 3D space. The parameters u and v represent a grid of coordinates that you have inscribed onto the surface of the cone. You are expressing the locations of all the points on the surface of the cone in 3D space parametrically in terms of the inscribed coordinates u and v by writing: ρ=ρ(u,v), θ=θ(u,v), and z=z(u,v). The game is: you tell me the inscribed coordinates u and v of a point on the cone, and I will tell you the polar coordinates of the point, as well as the equation for the position vector [itex]\vec{r}[/itex]drawn to the point from the origin of the 3D spatial polar coordinate system. Your job is to determine the equation for a differential position vector [itex]d\vec{r}[/itex] within the surface of the cone joining the point (u,v) to the neighboring point (u+du, v+dv). This differential position vector must be expressed in terms of du and dv. Note that, in this problem statement, we are assuming that the functions ρ=ρ(u,v), θ=θ(u,v), and z=z(u,v) are specified in advance and that you do not have to determine them.

    I can tell you right now that there is something a little wrong with the problem statement, because in polar coordinates, a position vector [itex]\vec{r}[/itex] drawn from the origin to any point is space does not have a component in the θ direction. The equation in your problem statement for the position vector does feature a θ component. But, we are going to let that slide for now, in favor of working out the general methodology for solving a problem like this one.

    So we are going to write:
    [tex]d\vec{r}=\frac{∂\vec{r}}{∂u}du+\frac{∂\vec{r}}{∂v}dv[/tex]
    Your job now is to evaluate the two partial derivatives in this equation, taking into account that [itex]\vec{\hat{e}_r}[/itex] and [itex]\vec{\hat{e}_θ}[/itex] are both functions of θ, and θ is a function of u and v. Do you want to take a shot at deriving those partial derivatives, or would you like me to help a little more?

    Chet
     
  15. Feb 4, 2014 #14
    The whole thing is useless if the vector is wrong... I can manage to calculate the derivatives, but I need to know how to write the position vector for the cone. Should I just ditch the θ component?
     
  16. Feb 4, 2014 #15
    I don't know how to advise you on this. It's a judgement call. My inclination would be to do it both ways just for the practice (with the understanding that, in reality, the θ component of r should be zero). It also appears that you are supposed to let θ and v represent the same variable.
     
    Last edited: Feb 4, 2014
  17. Feb 5, 2014 #16

    BvU

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    Stayed subscribed out of curiosity. Felt a little lost initially. This looks like something to practice math in a course of Lagrangian mechanics, maybe towards general relativity, don't know (looked at your other posts). Something for advanced physics?

    And I have a hard time even imagining what kind of cone is described. (No further hints, like g(α) is a constant or something?) In such a case I start with simpler versions (like: g=1) and feel a lot more comfortable. I see a light cone. My abstraction level is such that I need to transform to x,y,z , or -- since v = Θ (in my ancient version of cylindrical coordinates we always used ##\phi## because that was also used for spherical coordinates), to x, z.

    You can see I'm preconditioned, because I don't think I want to set t=1. Pity, because then we would have a cone on its tip, standing still. And life would be so easy: u = z, ρ = z, ##\phi## = ##\phi## (getting [itex]\varphi[/itex] to look halfway decent turns out to be a nasty bugger in this forum).

    I could go on like this, but I'd rather someone shoots at this bla to remove the nonsense. Perhaps we can then continue with what's left standing.

    ps Can you help me understand why ##\vec{r} (u, v) ## is not ##\vec{r} (u, v, t, α) ## ?
     
  18. Feb 5, 2014 #17
    I think what you are really asking is what this kind of development is used for. This approach was originally developed by Gauss to describe the metrical properties and geometry of 2D curved surfaces. It involves inscribing a set of coordinates onto the surface, and representing mathematically a position vector from an arbitrary origin in 3D space to points on the surface. The 3D coordinates of the points on the surface are represented parametrically in terms of the two coordinates within the surface. Next, one calculates a differential position vector within the surface between two arbitrary closely neighboring points. This then establishes the basis for describing the metrical properties of the curved surface.

    Of course, this was extended to provide the mathematics for General Relativity of curved 4D manifolds. However, there are other important applications. In materials science, we use this to describe the kinematics of large deformations, usually by expressing the spatial coordinates of a material point within a body at time t in terms of the spatial coordinates of the same material point at time t=0. This allows use to figure out how much stretching occurs in various directions between material points. This deformational kinematics is expressed in terms of the Cauchy Green tensor and/or the finite strain tensor.

    Chet
     
  19. Feb 5, 2014 #18
    The cone is described by a picture, that vector is my try at a position vector. There are not t nor g(α), when i wrote tg(α) I actually meant tan(α), I should have written it like that.

    I need it for this: https://www.physicsforums.com/showthread.php?t=734065
     
  20. Feb 5, 2014 #19

    BvU

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    Ok, that makes it a lot clearer - and simpler. No general relativity. Have to go now, but perhaps chet or tsny are online.
     
  21. Feb 5, 2014 #20
    Oh. So the θ component shouldn't have been in there in the first place. It should have read:

    [itex]\vec{r} (u, v) = u\vec{\hat{e}_z} + utan(α)\vec{\hat{e}_ρ(θ)}[/itex]

    This would be the equation for the cone in the figure. Of course, θ = v.
     
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