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Homework Help: Vectors and differentiation

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data
    The vector a depends on a parameter t, i.e. [itex]a=a(t)=a_x(t)i +a_y(t)j +a_z(t)k[/itex]..
    it satisfies the equation [itex]da/dt= j [/itex][tex]\times[/tex][itex] a[/itex]
    show that [itex]d^2a_x/dt^2 =-a_x[/itex] , [itex]da_y/dt=0[/itex] and [itex]d^2a_z/dt^2 =-a_z[/itex].

    For the vector a, find its value for t=pi if at t=0 [itex]a(0)=i+j[/itex] and [itex]da/dt(0)=0k [/itex]


    2. Relevant equations
    [itex] a.b = mod(a)mod(b)cos\theta [/itex]
    [itex] a [/itex] X [itex] b = mod(a)mod(b)sin\theta[/tex] [tex]\hat{n}[/tex] [/itex]


    3. The attempt at a solution
    i have absolutely no idea how to start...
     
  2. jcsd
  3. Jan 10, 2010 #2

    Dick

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    To start you have to figure out what da/dt=jxa means in terms of the components of a, [ax,ay,az]. Can you find the cross product of the vector j with a?
     
  4. Jan 10, 2010 #3
    Have you written the second asked-to-be-shown equation correctly? I think it must be modified as follows: [itex]da_y/dt=c[/itex] (c being a constant).

    To not get into the trouble of Latex, you can scan a photo of the printed question and put it on the forum.

    AB
     
    Last edited: Jan 10, 2010
  5. Jan 10, 2010 #4
    [itex]da_y/dt=0[/itex] follows from [itex]da/dt= j \times a[/itex]
     
  6. Jan 10, 2010 #5
    Oh, yes! I straightly put [itex]d^2a_y/{dt^{2}}=0[/itex] without looking at the first derivative. Thanks...
     
  7. Jan 10, 2010 #6
    [itex]j[/itex] X [itex]a[/itex] will be [itex]a_z(t)i+0j-a_x(t).[/itex]

    so,
    [tex]\stackrel{da_y}{dt}[/tex]=0 , [tex]\stackrel{da_x}{dt}[/tex]=[itex]a_z (t)[/itex] and [tex]\stackrel{da_z}{dt}[/tex]= [itex]-a_x(t) [/itex]
     
  8. Jan 10, 2010 #7

    Dick

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    That's a good start. Can you continue from there?
     
  9. Jan 10, 2010 #8
    I need to differentiate [itex]a_z (t)[/itex] with respect to t... can i just say it's [itex]-a_x [/itex] ?
     
    Last edited: Jan 10, 2010
  10. Jan 10, 2010 #9

    Dick

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    Well, no. da_z/dt isn't just a_z. Your last result says it's -a_x. Try looking at the second derivative part. You want to show e.g. d/dt(da_z/dt))=(-a_z). How would that work?
     
  11. Jan 10, 2010 #10
    ahh i meant to write 'is it just -a_x'

    can you integrate both sides? so da_z/dt is -a_x t , which is a function of t??
     
  12. Jan 10, 2010 #11

    Dick

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    a_x isn't a constant. You can't integrate it by multiplying it by t. Just differentiate da_z/dt, that will give you the second derivative, right?
     
  13. Jan 10, 2010 #12
    ok...but how do i differentiate -a_x(t) ?

    ohhh, it's just -(the x -component of j x a)?
     
  14. Jan 10, 2010 #13

    Dick

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    Right. da/dt=jxa tells you how to differentiate the components of a.
     
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