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Calculate the length of the given curve part 2

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\vec r (t) = \vec u +t \vec v for -4 \le t \le 6[/itex] where u and v are constant vectors and v not equal 0

    Do i proceed with the usual calculation of differentiating r wrt t, find the magnitude and then integrate between limits? Looks too simple for that. Thanks
     
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  3. Nov 4, 2011 #2

    Simon Bridge

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    http://www.ucl.ac.uk/Mathematics/geomath/level2/fvec/fv4.html

    but in this case it indeed looks like:

    [tex]\frac{d\vec r}{dt}=\vec v[/tex]

    ... think what differentiation means. How [itex]\vec r[/itex] changes with time is it grows in the [itex]\vec v[/itex] direction and the rate of that growth is the magnitude.

    The intergral will be the area between the path mapped out by [itex]\vec r(t)[/itex] and ... something.

    Of course, I'm only guessing that you have been asked to differentiate and integrate that function ... you didn't actually say.

    If we consider r to be a displacement, then [itex]\vec u = \vec r(0)[/itex] and [itex]\vec v[/itex] is, indeed, the velocity.
    It is not clear what the area under the displacement-time graph represents.
    It's easier to think about if r is a velocity, u is initial velocity, and v is the acceleration.
     
    Last edited: Nov 4, 2011
  4. Nov 4, 2011 #3
    It says I have to calculate the length of the curve for the given problem.
     
  5. Nov 4, 2011 #4

    LCKurtz

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    It is very simple. Just do it.
     
  6. Nov 4, 2011 #5
    I get [itex]10 \vec v[/itex] by differentiating r(t) wrt t , getting its magnitude which is v and then integrating between the limits wrt t...
     
  7. Nov 4, 2011 #6

    LCKurtz

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    That is confusing. You don't get [itex]10\vec v[/itex] as the derivative if that is what you mean. And if you mean that is the answer it isn't correct either because the answer must be a scalar.
     
  8. Nov 4, 2011 #7
    Well yes I realise the answer is a scalar because I know the vector v is a constant as stated in the problem hence 10 times a constant?

    r(t)= u + tv.
    dr(t)= v.

    Magnitude = (v^2)^(1/2) =v. Then integrate this between the limits from -4 to 6....
     
  9. Nov 4, 2011 #8

    LCKurtz

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    You need to use different notation for a vector [itex]\vec v[/itex] and its magnitude [itex]|\vec v| = v[/itex]. Otherwise it is difficult to tell whether you are using the vector or its magnitude. You are using v for both above.
     
  10. Nov 4, 2011 #9
    [itex]d \vec r (t)= \vec v[/itex]. Therefore the magnitude is [itex]| \vec v| = v[/itex]. Then

    [itex]\int_{-4}^{6} v dt = 10v[/itex]...?

    That is what I did . Correct?
     
  11. Nov 4, 2011 #10

    LCKurtz

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    Yes. That is much better and it is correct.
     
  12. Nov 5, 2011 #11

    Simon Bridge

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    Neato :)
    I'm a big fan of understanding the problem.

    If r(t) is a position vector, then the thingy's motion has speed |v| in the direction v/|v| and passes through position u at t=0. Since v is a constant vector, the path is a straight line. We want to know how far the thingy has traveled between t=-4 and t=6 units.

    Since it's linear, that's just the final position subtracted from the initial position - or the speed multiplied by the elapsed time.

    Of course, that's probably not mathematically rigorous.
     
  13. Nov 5, 2011 #12

    Simon Bridge

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    Neato :)
    I'm a big fan of understanding the problem.

    If r(t) is a position vector, then the thingy's motion has speed |v| in the direction v/|v| and passes through position u at t=0. Since v is a constant vector, the path is a straight line. We want to know how far the thingy has traveled between t=-4 and t=6 units.

    Since it's linear, that's just the final position subtracted from the initial position - or the speed multiplied by the elapsed time.

    Of course, that's probably not mathematically rigorous.
     
  14. Nov 5, 2011 #13
    good insight! Thanks
     
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