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How can I linearize ?

  1. Jun 4, 2008 #1
    Is there anyone that can help me about linearization of the equations around x=teta=0 ? It is a bit urgent. I am waiting answer from the calculus experts.
     

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    Last edited: Jun 4, 2008
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  3. Jun 4, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi e135193! Welcome to PF! :smile:

    hmm … (c cosθ, c sinθ) is the point at angle θ on the circle of radius c round the origin.

    r1 is the distance from that point, and r2 is the distance from the opposite point, (-c cosθ, -c sinθ).

    but I don't understand what you mean by "linearization of the equations around x = θ = 0" :confused:
     
  4. Jun 4, 2008 #3
    As you see equations are non-linear. I need equations like in the picture below. I heard that it can be made by using taylor series.
     

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  5. Jun 4, 2008 #4

    tiny-tim

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    Hi e135193! :smile:

    (btw, anyone who contributes to a thread gets an email whenever someone else replies, so there's no need to use PMs. :smile: )

    This may help: √(1 + x) = (1 + x/2 + …); sinx = x + … ; cosx = 1 - x²/2 + …

    But what I don't understand is that you have three variables, x y and θ, for a 2-dimensional situation. :confused:
     
  6. Jun 4, 2008 #5
    Thanks for your interest to help me.

    I'm sorry about the insufficient information. The "y" and "c" terms are constants.
    Then, there are only 2 variables, x and θ.

    Can you help me further from that because I am not good at mathematics.
     
    Last edited: Jun 4, 2008
  7. Jun 4, 2008 #6
    Revised Question...

    This is the clear definiton of my problem.. Does Anyone Know the solution ?
     

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  8. Jun 4, 2008 #7

    HallsofIvy

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    I once walked into a test on differential equations that asked for the "linearization" of a non-linear de and I had no idea what "linearization" meant! Turns out it's easy.

    You have
    [tex]r_1= \sqrt{(c cos(\theta)- x)^2+ (y- c sin(\theta)^2}[/tex]
    [tex]r_2= \sqrt{(c cos(\theta)+ x)^2+ (y+ c sin(\theta)^2}[/tex]
    and you say that c and y are constant. So you want linear functions for r1 and r2 in terms of x and [itex]\theta[/itex]. You don't really need the full "Taylor's series", you just need the tangent line equations (which are the order one Taylor's Polynomials).

    In general, if r= f(x, [itex]\theta[/itex]), then the "tangent plane" approximation, about x= 0, [itex]\theta= 0[/itex], is given by
    [tex]r= f(0,0)+ \frac{\partial f}{\partial x}(0,0) x+ \frac{\partial f}{\partial \theta}(0,0)\theta[/tex]
    Find the partial derivatives of your functions and plug into that.
     
  9. Jun 4, 2008 #8
    thank you so much for your effort. I'm gonna try to do but its not gonna be easy for me. But the point I smash is that I can not plug that functions into the expression you define above. Can you please a bit help :)))
     
    Last edited: Jun 4, 2008
  10. Jun 4, 2008 #9

    tiny-tim

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    Hi e135193! :smile:

    ok, y and c are constants.

    So r is a function of x and θ.

    What is the derivative ∂r/∂x? It's the rate at which r increases if you increase x (but leave θ fixed).

    So if you only go a very small distance (in x), then r depends linearly on x, and the factor is ∂r/∂x. Similarly for ∂r/∂θ.

    Work out ∂r/∂x and ∂r/∂θ, and then find their values at (x=0,θ=0). :smile:
     
  11. Jun 4, 2008 #10
    hi ,tiny-tim.

    Thanks for your effort again but I said that I was not good at mathematics.


    I can not take the partial derivative :) of that function.

    could you please explain a bit further.


    kind regards.
     
  12. Jun 4, 2008 #11

    tiny-tim

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    ok, let's start with an example: ∂/∂x of √(x² + px + q) … can you do that? :smile:

    (if not, just d/dx of √(x² + px + q)? :smile: )
     
  13. Jun 4, 2008 #12
    if I said no, would you angry of me ? :smile:
     
  14. Jun 4, 2008 #13

    tiny-tim

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    ok, just try d/dx of x² + px + q.

    You can do that, can't you?

    Do you then know how to use substitution to get d/dx of √(x² + px + q)? :smile:

    hmm … gotta go to bed now … :zzz:
     
  15. Jun 4, 2008 #14
    hi tiny-tim

    I'm begging you please. It's very important for me. I have to finish this before the sun rises :smile:.

    Again I'm saying I'm not good at mathematics.


    There is someone who waits for answer. :smile:
     
  16. Jun 5, 2008 #15

    tiny-tim

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    Hi e135193! :smile:
    But I don't know your time-zone! :confused:

    And what part of …
    … did you not understand? :wink:

    Look, if you really can't do d/dx of x² + px + q, then you'd better ask your teacher/TA for special help.

    That sort of d/dx is something you should be able to do in your head by now, like seven times eight equals fifty-six.

    But if you can do it, show us what it is, and we'll help you through the next steps. :smile:
     
  17. Jun 5, 2008 #16

    HallsofIvy

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    If you do not know how to differentiate at all (which is what you seem to be saying) then you should not trying to do a problem like this. This kind of problem is typically given in a calculus course about a year after learning things like how to differentiate x2+ px+ q.
     
  18. Jun 5, 2008 #17
    Hi tiny-tim,


    Finally, I have calculated equations but I don't have any idea about whether they are correct or nor. Atleast could you check it please ?
     

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  19. Jun 5, 2008 #18

    tiny-tim

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    Hi e135193! :smile:

    I'm sorry, but i have no idea how you got that …

    are they supposed to be ∂r1/∂x and ∂r2/∂x, or what?

    and where did those √(c² + y²) some from?

    We'd better start with:

    what is d/dx of √(x² + px + q)? :smile:
     
  20. Jun 5, 2008 #19
    equations I found is the solution of my equations not the one you suggested me to solve

    The equation I found is that the linearized one of the non-linear equations.

    I just asked that is that true ? could you check it ?
     

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  21. Jun 5, 2008 #20

    tiny-tim

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    oh … now I understand.

    Your -cx/√(c² + y²) and your + √(c² + y²) are correct,

    but I've no idea how you got the cos(c) and the sin(y) in the middle (presumably from your ∂/∂θ).

    Do the ∂/∂θ again, and this time show your working, so we can see where you're going wrong.
     
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