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Dot Product, Derivative, and Vector Valued Functions

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that, if [tex]\vec{r}[/tex](t) is a differentiable vector valued function, then so is ||[tex]\vec{r}[/tex](t)||, and [tex]\vec{r}[/tex](t) [tex]\bullet[/tex] [tex]\vec{r'}[/tex](t) = ||[tex]\vec{r}[/tex](t)|| ||[tex]\vec{r}[/tex](t)||'


    2. Relevant equations
    I know how to do a dot product, but what bothers me is the fact that the question involves the derivative of the magnitude of r. But the magnitude of a vector is a scalar, the derivative of which would be zero.

    If it's a typo and really supposed to be the magnitude of the derivative, I'm still lost as to how to relate the dot product of the vectors with their magnitudes??


    3. The attempt at a solution
    I know that the unit tangent vector involves r' and ||r'||, but I'm not sure that this helps.
     
  2. jcsd
  3. Nov 7, 2009 #2

    lanedance

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    Homework Helper

    scalars can still be time dependent, and clearly have non-zero derivatives... the magnitude just represents the length of r, if the length is changing the derivative will be non-zero


    [tex] |\vec{r}| = (\vec{r(t)} \bullet \vec{r(t)})^{1/2} [/tex]
     
  4. Nov 7, 2009 #3
    Okay, so if it really is the derivative of the magnitude, I can understand that. But, what does [tex]\
    (\vec{r(t)} \bullet \vec{r'(t)})^{1/2}
    [/tex] equal?

    If you replace r with something like f(x)+g(x) and r' with f'(x)+g'(x) and do the dot product, you get [tex]\sqrt{f(x)f'(x)+g(x)g'(x)}[/tex]. But how would that help?


    I'm still not clear...
     
  5. Nov 7, 2009 #4

    lanedance

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    Homework Helper

    how about starting wth
    [tex] |\vec{r(t)}|^2 = |\vec{r}||\vec{r}|= (\vec{r(t)} \bullet \vec{r(t)}) [/tex]

    if i rememebr correctly, the derivative product rulle works just as normal, but if you're unsure, try writing everything out in component form

    [tex] \vec{r(t)} =(x(t), y(t), z(t))[/tex]
     
  6. Nov 7, 2009 #5
    Thanks. I think I've got it now... it just took some time staring at the problem! Thanks again.
     
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