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Lineary DE in Kinematics

  1. Jan 14, 2005 #1
    Linear DE in Kinematics

    Hey
    I would like to know, how it is possible to solve the following
    differentialequations

    [tex]\ddot{x}(t)+\alpha \dot{x}(t)^2=0[/tex]

    and the one that really gives troubles

    [tex]\ddot{y}(t)+\alpha\dot{y}(t)^2=-g[/tex]

    when given that [tex]x(0)=0[/tex], [tex]y(0)=0[/tex], [tex]\dot{y}(0)=v_0\sin \theta[/tex] and [tex]\dot{x}(0)=v_0\cos \theta[/tex]. Where [tex]\alpha=k/m[/tex]. The problem in equation 2 is, that i can't even solve the integral

    [tex]\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}[/tex]

    Doing the substitution here doesn't get get me anywhere. Maybe someone have made this problem before, it's motion in 2-dimensions, with
    airresistance. The 2 equations comes from

    [tex]m\vec{a}=\vec{F}_g - \vec{F}_{airr}[/tex]
     
    Last edited by a moderator: Jan 14, 2005
  2. jcsd
  3. Jan 14, 2005 #2

    dextercioby

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    HINT:not [\tex],but [/tex]

    Daniel.
     
  4. Jan 14, 2005 #3

    dextercioby

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    1.Your equations are not linear.
    2.Write them using the velocity.E.g.
    [tex] m\frac{dv(t)}{dt}-kv^{2}(t)=-mg [/tex]

    Can u solve it,now??

    Daniel.
     
    Last edited: Jan 14, 2005
  5. Jan 14, 2005 #4
    That doesn't help, I started by writing it in that way, but then rewrited it here. The problem is the integral

    [tex]\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}[/tex]

    How can you solve this. By using the substitution, you end up with an even more complicated integral, and partial integration doesn't help. Can you that problem? I hope you can help me. Thanks in advance.
     
  6. Jan 14, 2005 #5

    dextercioby

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    Can't u solve this integral??

    [tex] \int \frac{dv}{-mg+kv^{2}} [/tex]

    Daniel.

    PS.If u can't,what are u doing solving ODE-s??
     
  7. Jan 14, 2005 #6

    arildno

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    Check up an arcus tangens and artanh (inverse of hyperbolic tangens)
     
  8. Jan 14, 2005 #7

    dextercioby

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    Let's not go there,Arildno... :tongue2: Maybe separation into simple fractions...?? :rolleyes:

    Daniel.
     
  9. Jan 14, 2005 #8
    No I can't solve this integral:

    [tex] \int \frac{dv}{-mg+kv^{2}} [/tex]

    Because when I substitude lets say

    [tex]u=-mg+kv^{2}[/tex]

    then

    [tex]v=\sqrt{\frac{u+mg}{k}}[/tex]

    and

    [tex]\frac{du}{dv}=2kv \Leftrightarrow dv=\frac{du}{2kv}[/tex]

    But when I use the partial integration technique, I end up with a even more complicated integral.... Try solve it and you will see. Perhaps I should check up an arcus tangens and artanh?
     
  10. Jan 14, 2005 #9

    dextercioby

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    Do u have any experience doing integrals??In this case,u can decompose the fraction/integrand into 2 simple fractions.

    Daniel.

    PS.Do you know this integration technique??
     
  11. Jan 14, 2005 #10
    If you want to use an inverse hyperbolic/trig integral, don't subsitute like this. What you need to do is make the denominator into the form 1 - av^2 (where a is a constant (can you see how to do that?), then you can look up what this integral is. It's not arctan, as d/dx (arctan x) = 1/(1 + x^2), but I can't remember which one it is.
     
  12. Jan 16, 2005 #11
    looking up integrals :bugeye: ? just draw a right triangle... make pythagoras proud :rofl: [/hint] :devil:

    if you use trig substitution, you'll get to [tex]\int\sec\theta\ d\theta = \ln | \sec\theta + \tan\theta |+C[/tex] which will be the same as if you were to go by seperation.
     
    Last edited: Jan 16, 2005
  13. Jan 17, 2005 #12
    Well, I havn't been online for 2 days now, but I have solved the ODE. I just jused arctan Nylex says.
     
  14. Jan 17, 2005 #13

    dextercioby

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    WHAT??U mean "arctanh" as from "arcus tangens hyperbolicus",right???
    As i said,partial fraction would have dunnit much easier.

    Daniel.
     
  15. Jan 17, 2005 #14
    No, I used arctan;
    [tex]\frac{d}{dx}\arctan x=\frac{1}{1+x^2}[/tex]
    And yes I sepretated into simple fractions.
     
    Last edited by a moderator: Jan 17, 2005
  16. Jan 17, 2005 #15

    dextercioby

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    Then your result was WRONG,WRONG!!!!!!You should have used "arctanh".

    Daniel.

    PS.Are u sure it was the integral discussed above,the one with one minus at the denominator??
     
  17. Jan 17, 2005 #16

    dextercioby

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    Please,separate into simple fractions
    [tex] \frac{1}{1+x^{2}} [/tex]

    Daniel.
     
  18. Jan 17, 2005 #17
    I have the calculations in Mathcad, how can I upload a file in here, so that you can see them..
     
  19. Jan 17, 2005 #18

    dextercioby

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    If it's not more than 50KB,you can post them as an attachement.OOPS,can u transfer them into another format??Attachment manager doesn't support "mathcad/matlab" files,sorry.

    Try to post the general idea,at least.

    Valid extensions:
    "Valid file extensions: bmp doc gif jpe jpeg jpg pdf png psd txt zip"

    Daniel.
     
  20. Jan 17, 2005 #19
    Here is how i solve it;

    [tex]m\frac{d}{dt}v_y(t)+kv_y(t)^2=-mg[/tex]

    [tex]\frac{d}{dt}v=-g-\alpha v^2[/tex]

    where \alpha = k/m, then

    [tex]\frac{dv}{-g-\alpha v ^2}=dt [/tex]

    [tex]\int\frac{1}{-g-\alpha v^2}dv=t + c[/tex]

    [tex]\frac{1}{-g}\int\frac{1}{1+\frac{\alpha}{g}v^2}dv=t + c[/tex]

    [tex]\frac{1}{-g}\int\frac{1}{1+\left(\sqrt{\frac{\alpha}{g}}v\right)^2}dv=t + c[/tex]

    Then i substitude

    [tex] u=\sqrt{\frac{\alpha}{g}}v[/tex]

    [tex] \frac{1}{-g\sqrt{\frac{\alpha}{g}}}\int\frac{1}{1+u^2}du[/tex]

    therefor

    [tex] \frac{-1}{\sqrt{g\alpha}}\arctan\left(\sqrt{\frac{\alpha}{g}}v\right)=t+c[/tex]

    Where does it go wrong :(?
     
    Last edited by a moderator: Jan 17, 2005
  21. Jan 17, 2005 #20

    dextercioby

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    I thought so.The initial equation is wrong.Gravity and aerodynamic force have opposite signs...


    Daniel.

    P.S.Redo calculations.
     
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