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Prove that v(t) is any vector that depends on time

  1. Sep 18, 2007 #1
    Prove that v(t) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then [tex]\dot{v}[/tex](t) is orthogonal to v(t). Prove the converse that if [tex]\dot{v}[/tex](t) is orthogonal to v(t), then lv(t)l is constant. Hint: consder the derivative of v^2.

    This is a very hand result. It explains why, in 2-D polars, d[tex]\hat{r}[/tex]/dt has to be in the direction of [tex]\hat{\phi}[/tex] and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since teh acceleration is perpendicular to the velocity.



    How to I approach proving this? I'm also unclear on Orthogonal, which I'll look up now but if someone could explain this to compliment what I'm off to figure out, I'd appreciate it a great deal.
    Thanks!
     
  2. jcsd
  3. Sep 18, 2007 #2

    learningphysics

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    The trick is to use this equation [tex]\vec{v}\cdot\vec{v} = |\vec{v}|^2[/tex]

    Try taking the derivative of both sides... see what happens.
     
  4. Sep 18, 2007 #3
    How did you know that?
     
  5. Sep 18, 2007 #4

    learningphysics

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    Just kind of guessed... They want orthogonality... Orthogonality means perpendicular... so I thought it probably had to do with dot products. I looked up the derivative of the dot product, and checked that this idea worked.

    When you have a problem that has to do with orthogonality... or angles between vectors... and the magnitudes of the vectors... you're probably dealing with dot product or cross product...
     
  6. Sep 18, 2007 #5
    haha. You's brilliant eh?

    I'll let you know how it turns out
     
  7. Sep 18, 2007 #6

    learningphysics

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    lol Nope. I wish. It's just experience. Just seen these types of problems so many times. So I kind of know what to expect.

    cool.
     
  8. Sep 18, 2007 #7
    I forget, do I treat v as a variable or does a vector differentiate differently?
     
  9. Sep 18, 2007 #8

    learningphysics

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    look up the rule for taking the derivative of a dot product.

    http://mathworld.wolfram.com/DotProduct.html

    The [tex]|\vec{v}|[/tex] is just a scalar... so just differentiate that normally, like an ordinary variable.
     
  10. Sep 19, 2007 #9
    [tex]\vec{v}[/tex].[tex]\frac{d\vec{v}}{dt}[/tex] + [tex]\vec{v}[/tex].[tex]\frac{d\vec{v}}{dt}[/tex] = 2l[tex]\vec{v}[/tex]l

    Hows that look so far?
     
  11. Sep 19, 2007 #10
    That's not actually equal is it?
     
  12. Sep 19, 2007 #11

    learningphysics

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    Left side looks good. Right side you need to multiply that by [tex]\frac{d|\vec{v}|}{dt}[/tex], using the chain rule. because you're taking the derviative wrt t.
     
  13. Sep 19, 2007 #12
    Right... oops. (embarrasing)

    that gives me
    2[tex]\vec{v}[/tex][tex]\vec{a}[/tex] = 2l[tex]\vec{v}[/tex]l [tex]\vec{a}[/tex]

    and v and a are perpendicular?
     
  14. Sep 19, 2007 #13

    learningphysics

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    careful. left side looks good, but not the right side. [tex]\vec{a} = \frac{d\vec{v}}{dt}[/tex]. But [tex]\frac{d|\vec{v}|}{dt}[/tex] isn't the same as [tex]\frac{d\vec{v}}{dt}[/tex] (ie think of the velocity keeping the same magnitude but changing directions.)

    You know the magnitude of the velocity is constant for this question, hence [tex]\frac{d|\vec{v}|}{dt}[/tex] is 0.
     
  15. Sep 19, 2007 #14
    Right, the slope of a scalar.
    So, 2va = 0?
     
  16. Sep 19, 2007 #15

    learningphysics

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    yeah, [tex]2\vec{v}\cdot\vec{a} = 0[/tex]...hence [tex]\vec{v}\cdot\vec{a} = 0[/tex] if the dot product of two vectors is 0, then they are orthogonal/perpendicular.
     
  17. Sep 19, 2007 #16
    To prove the converse that if [tex]\dot{v}[/tex](t) is orthogonal to v(t), (that v(t) is constant), do i start in the same manner?
     
  18. Sep 19, 2007 #17
    a dv/dt + v da/dt = 0

    this leaves me with the derivative of acceleration though...
     
  19. Sep 19, 2007 #18

    learningphysics

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    Do the same thing as the first part... except backwards...

    basically it's just the reverse of the steps starting with the end...

    start with [tex]\vec{v}\cdot\vec{a} = 0[/tex]
     
  20. Sep 19, 2007 #19
    integrate instead of derive?
     
  21. Sep 19, 2007 #20

    learningphysics

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    yes, exactly.
     
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