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Homework Help: Acceleration & arc length

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that d^2 R / dt^2 is NOT a scalar which is a multiple of d^2 R / ds^2 where R is a vector, s is arc length

    2. Relevant equations and 3. The attempt at a solution

    I was thinking maybe it has something to do with the fact k = |d^2 R / ds^2|
    a = d^2 R / dt^2 = d|v|/dt * T + k |v|^2 N
    so perhaps intuitively it can't be a multiple??? I'm not really sure how to go about this.....
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2
    Parameterize R as a function of s and note that s is a function of t. Then apply the chain rule.
  4. Feb 22, 2009 #3
    How does parameterizing show that for the second derivative?

    Also, how would you do it when you aren't given a function?

    s = int[dR/dt]dt from 0 to t ... but I don't see how you can really go from there
    Last edited: Feb 22, 2009
  5. Feb 22, 2009 #4
    Let R be a function of s and s be a function of t. Then the second derivative of R with respect to s is [itex]\frac{d^2R}{ds^2}[/itex]. What is the second derivative of R with respect to t (Just use the chain rule)? Is it a scalar multiple of [itex]\frac{d^2R}{ds^2}[/itex]?
  6. Feb 22, 2009 #5
    That's what I'm asking. I don't know how to properly show it.

    s = int[dR/dt]dt from 0 to t
    Thus, s(t)

    Given d^2 R / ds^2
    = d^2 R / ds^2 * ds^2 / dt^2 = d^2 R / dt^2 by the chain rule
    Is that what you mean?
  7. Feb 22, 2009 #6
    That formula is not true. The chain rule applies to a single derivative only, not to second derivatives. You can, however, apply the chain rule to
    [tex]\frac{d^2 R}{dt^2} = \frac{d}{dt}\left(\frac{dR}{dt}\right)[/tex]
    to find it in terms of dR/ds. Remember R is a function of s and s is a function of t, so you have R(s(t)).
    Last edited: Feb 22, 2009
  8. Feb 22, 2009 #7
    Doesn't that imply that it is a multiple, when the whole point is to show that it is not?

    That is why I am still confused. I understand using the chain rule to show:
    d/ds (dR/ds) = d^2 R / ds^2
    d/dt (dR/dt) = d^2 R / dt^2
    But not how it is used to show that the second isn't a multiple of the first.
  9. Feb 22, 2009 #8
    d/dt is not a number, it is a differential operator. Note that dR/dt = (dR/ds)(ds/dt) by the chain rule. You then have to find (d/dt)[(dR/ds)(ds/dt)] for which you will have to use the product rule.
  10. Feb 22, 2009 #9
    Thank you. It makes A LOT more sense now.
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