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Do you understand what linearization means?

  1. Feb 4, 2008 #1
    do you understand what "linearization" means?

    1. The problem statement, all variables and given/known data

    Linearization
    4x'' + 3 cos(x-y)y'' +2 y^2 sin(x-y) +3g sin (x) = 0

    2. Relevant equations

    initial condition x(0) = y(0) = 0


    3. The attempt at a solution

    pls tell me the relevant steps to solve this problem.thanx
     
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Feb 4, 2008 #2

    HallsofIvy

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    Do you understand what "linearization" means? (I ask because I finally realized how simple it was the day after a test!)

    Just replace any non-linear functions of the dependent variable by linear approximations.

    It would help, of course, to separate dependence on the different variables (am I correct that both x and y are dependent variables, depending on a third variable and the derivatives are with respect to that third variable?): using the trig sum formulas cos(x- y)= cos(x)cos(y)+ sin(x)sin(y) and sin(x-y)= sin(x)cos(y)- cos(x)sin(y). Since you will want to approximate around x= y= 0, the linear approximation to sin(x) is x near x= 0 and to cos(x) is 1 around x= 0. (You can see that by looking at their Taylor's series.) Of coure, the only "linear" approximation to y2 or xy around y= 0 is 0.
     
    Last edited: Feb 4, 2008
  4. Feb 4, 2008 #3
    x'' means diferentiation twice.however, i think it should be solve by using taylor series.Am i correct??But i still can do it......
     
  5. Feb 4, 2008 #4

    HallsofIvy

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    My only point about Taylor's series is that by dropping all the terms of degree 2 or above, you get the "linearization".
    sin(x)= x- x3/3!+ ... and so the "linearization" of sin(x), around x= 0, is x.
    cos(x)= 1- x2/2!+... so the "linearization" of cos(x), around 0, is 1.
     
  6. Feb 4, 2008 #5
    sorry,there is another equation given which is 3 cos(x-y) y'' + 2y'' - 2y' ^2 sin(x-y) +g sin(y) =0 .

    but,the answer for this question is 4x'' + 3y'' + 3g x = 0 and 3x'' + 2y'' + gy = 0.

    pls help......
     
  7. Feb 4, 2008 #6

    Ben Niehoff

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    If you can't be bothered to type in complete sentences, and clearly state your point of confusion, why should somebody take the time to explain anything to you?

    I really don't get why people do this.
     
  8. Feb 4, 2008 #7
    1. The problem statement, all variables and given/known data

    Linearization

    4x'' + 3 cos(x-y)y'' +2 y^2 sin(x-y) +3g sin (x) = 0

    3 cos(x-y) y'' + 2y'' - 2y' ^2 sin(x-y) +g sin(y) =0



    2. Relevant equations

    initial condition x(0) = y(0) = 0


    3. The attempt at a solution

    the answer for this question is 4x'' + 3y'' + 3g x = 0 and 3x'' + 2y'' + gy = 0.

    pls tell me the relevant steps to solve this problem.thanx
     
  9. Feb 5, 2008 #8
    pls help.........thanx
     
  10. Feb 5, 2008 #9

    HallsofIvy

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    ?? Are you saying that you started with two equations:
    4x'' + 3 cos(x-y)y'' +2 y^2 sin(x-y) +3g sin (x) = 0
    and
    3 cos(x-y) y'' + 2y'' - 2y' ^2 sin(x-y) +g sin(y) =0 .

    And it didn't occur to you to tell us that?
     
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