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A question on vectors.

  1. Jul 24, 2006 #1


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    my question is from klppner's text an introduction to mechanics, on page 26, they suppose a vector A(t) has a constant magnitude A, he states that the direction dA(t)/dt is always perpendicular to A(t), now they define [tex]\triangle A=A(t+\triangle t)-A(t)[/tex], he states that [tex]|\triangle A|=2Asin(\frac{\triangle\theta}{2})[/tex] i dont understand how did they arrive at the last equation, and dont understand which angle does he refer to. he states that it's defined in a sketch in the book but i cant find where does he state this?

    thanks in advance.
  2. jcsd
  3. Jul 24, 2006 #2
    I don't have the book with me now, but I think [tex]\Delta\theta[/tex] is the angle between the [tex]A(t)[/tex] and [tex]A(t+\Delta t)[/tex]. The equation becomes more accurate as the angle becomes smaller.
  4. Jul 24, 2006 #3


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    [itex]\Delta \theta[/itex] is the angle between[itex]A(t+\triangle t)[/itex] and A(t). Since both vectors are of magnitude A, the law of cosines gives

    [tex]\Delta A^2 = A^2+A^2-2AAcos(\Delta \theta)[/tex]

    And [itex] cos(2x) = 1 - 2sin^2(x)[/itex], so...
  5. Jul 24, 2006 #4


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    Ok, Draw a vector for A(t), and then draw another for [tex]A(t+\triangle t)[/tex] with the same length and starting from the same point in space. The angle [tex]\triangle\theta[/tex] is acute angle between these two vectors and [tex]\triangle A[/tex] is the vector that forms the third side of the isosceles triangle having A(t) and [tex]A(t+\triangle t)[/tex] for sides (and it terminates in the same place [tex]A(t+\triangle t)[/tex] does). Construct the altitude of this triangle and consider one of the resulting identical right triangles with one leg being the altitude, one leg is of length [tex]\frac{1}{2}\left| \triangle A\right| [/tex], and the hypotenuse is of length A. Use trig.
  6. Jul 24, 2006 #5
    [tex]|\Delta A| = \sqrt{(A(t+\Delta t)-A(t))\cdot(A(t+\Delta t)-A(t))}[/tex]

    [tex] = \sqrt{|A(t+\Delta t)|^2 - 2A(t+\Delta t)\cdot A(t) + |A(t)|^2}
    = \sqrt{A^2 - 2A^2\cos {\Delta \theta} + A^2}[/tex]

    [tex] = \sqrt{2A^2 - 2A^2\cos{\Delta \theta}} = \sqrt{2}A\sqrt{1-\cos{\Delta \theta}}.[/tex]

    Now use the identity [itex] \sin^2{\phi} = \frac{1}{2}(1-\cos{2\phi})[/itex] to change that into

    [tex] \sqrt{2}A\sqrt{2\sin^2{\frac{\Delta \theta}{2}}}[/tex]

    [tex] = 2A\sin \frac{\Delta \theta}{2}.[/tex]

    Here, [itex]\Delta \theta[/itex] is the angle between [itex]A(t+\Delta t)[/itex] and [itex]A(t)[/itex] and [itex]A[/itex] is the (constant) magnitude of the vector function.
  7. Jul 24, 2006 #6


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    thank you for your help.
  8. Jul 24, 2006 #7


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    by the way, neutrino, you havent bought the booog, have you?
    ive seen in amazon that it's quite coasts a lot.
  9. Jul 24, 2006 #8
    What's that? Some new variant for the word book?
  10. Jul 25, 2006 #9


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    no just big funny typing error.
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