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Calculate the length of the given curve

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]2sin(t) \vec i +5t \vec j -2cos(t) \vec k for -5 <= t<=5[/itex]


    3. The attempt at a solution

    [itex]let \vec r (t)= 2sin(t) \vec i +5t \vec j -2cos(t) \vec k[/itex]

    Then [itex] d \vec r(t) = 2cos(t) \vec i +5 \vec j + 2sin(t) \vec k[/itex]

    [itex]|| d\vec r(t) ||= \sqrt( (2cos(t) \vec i)^2+(5 \vec j)^2+ (2 sin(t) \vec k )^2)[/itex]....?
     
  2. jcsd
  3. Nov 3, 2011 #2

    Mark44

    Staff: Mentor

    All those i's, j's, and k's are getting in your way.
    Let r(t) = <2sin(t), 5t, -2cos(t)>
    dr/dt = <2cos(t), 5, 2sin(t)>
    How do you calculate the magnitude of a vector?
     
    Last edited: Nov 3, 2011
  4. Nov 3, 2011 #3
    It will be the square root of the sum of the squares

    [itex]||r'(t)||= \sqrt{2cos(t)^2+5^2+(2sin(t)^2)}=\sqrt{29}[/itex] where sin^2(t)+cos^2(t)=1...?
     
    Last edited by a moderator: Nov 3, 2011
  5. Nov 3, 2011 #4

    I like Serena

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    Good! :smile:

    Do you have a formula for the length of a curve?
     
  6. Nov 4, 2011 #5
    The length of the curve is the integral of this value with respect to t from -5 to 5...?

    [itex]= |\sqrt{29}t|^{5}_{-5}[/itex].....?

    If the above is correct, how do you justify removing the basis i,j and k suddenly to proceed with the calculation?
     
    Last edited: Nov 4, 2011
  7. Nov 4, 2011 #6

    I like Serena

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    That's it.

    The formula for the length of a vector is [itex]||x \vec i + y \vec j + z \vec k|| = \sqrt{x^2 + y^2 + z^2}[/itex]

    And if you really want to include i, j, and k, consider that [itex]\vec i \cdot \vec i = 1[/itex] and that [itex]\vec i \cdot \vec j= 0[/itex]
     
  8. Nov 4, 2011 #7
    So based on your last post this is corect since the dot product of i by itself, j by itself and k by itself is one....

    Thanks
     
  9. Nov 4, 2011 #8

    I like Serena

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    Yes, it is correct, but it is very unusual to write it like this.
     
  10. Nov 4, 2011 #9
    Ok, but it seems strange to suddenly drop the basis. Thanks anyhow!
     
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