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Limits involving absolute values

  1. Mar 26, 2015 #1
    This is actually a physics problem, but since my question is really about the math involved, I decided to post it in the calculus subforum.
    I'm supposed to get from the term:
    $$\lim_{\Delta t → 0} |\vec{v}_r (t + \Delta t)| \frac{\sin \Delta \theta}{\Delta t}$$
    To:
    $$v_r (t) \frac{d\theta}{dt}$$
    ##\theta## is a function of ##t## (so ##\Delta \theta## approaches zero as ##\Delta t## approaches zero), and ##v_r## is the scalar component (not the magnitude) of the vector ##\vec{v}_r##, in a particular direction.
    Computing the limit as ##\Delta t## approaches zero:
    $$\lim_{\Delta t → 0} |\vec{v}_r (t + \Delta t)| \frac{\sin \Delta \theta}{\Delta t} = |\vec{v}_r (t)| \frac{d\theta}{dt} = |v_r (t)| \frac{d\theta}{dt}$$
    How do I get rid of those absolute value bars?
     
  2. jcsd
  3. Mar 26, 2015 #2

    Mark44

    Staff: Mentor

    Then by this logic ##\lim_{\Delta t}\frac{\Delta y}{\Delta t} = \frac 0 0##, assuming as you did that y is a function of t.

    What they're probably doing here is using the idea that, if u is close to zero, then ##\sin(u) \approx u##.
    Is ##v_r## the magnitude of ##\vec{v_r}##? If so ##v_r## would be a nonnegative scalar.
     
  4. Mar 26, 2015 #3
    ##v_r## is not the magnitude of ##\vec{v}_r##. It is the scalar component of ##\vec{v}_r## in a particular direction. It can be positive, negative, or zero.
    Its absolute value is the magnitude of ##\vec{v}_r##.
     
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