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Homework Help: Basic Calculus III- Arc Length Parameter and Length- Getting a negative length

  1. Aug 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the Arc Length Parameter along the curve from the point where t = 0 by evaluating the integral:

    s = |v(τ)| dτ from 0 to t​

    Then find the length of the indicated portion of the curve.

    2. Relevant equations

    The vector I am using for this:

    r(t) = (etcos t)i + (etsin t)j + etk, -ln (4) ≤ t ≤ 0

    3. The attempt at a solution

    I got the correct answer for the Arc Length Parameter, so no algebra or calculus mistakes:

    s(t) = √ ([(eτcos τ - eτsin τ)2 + (eτcos τ + eτsin τ)2 + (eτ)2]) dτ from 0 to t​

    After multiplying, canceling, and applying the trig identity cos2u + sin2u = 1, I have

    s(t) = √ (3e) dτ from 0 to t​

    which is

    √(3) et - √(3)​

    That is what is in the back of the book.

    However, the second part is giving me problems. I plug in -ln4, and I end up with the right number, but negative. How can there be a negative length?

    So I plug -ln4 in for t, and I get (I suspect that this is where I went wrong):

    √(3) (et - 1)

    √(3) (e-ln 4 - 1)

    √(3) (eln(.25) - 1)

    √(3) (1/4 - 1)

    √(3) (- 3/4)


    Which is off by a sign.

    Why am I getting this wrong?


    EDIT- I know it is easy to just say "take the absolute value," but that won't get me any understanding. You see, I did the exact same thing for the previous problem that I've done in this current one , except that the interval was 0 ≤ t ≤ π/2, and just plugging in π/2 for t gave me the correct answer. Why does this not work in this problem?

    Clearly, the obvious difference is that the interval for my vector is -ln 4 ≤ t ≤ 0, which is kind of "reverse" from my previous problem. Basically I'm looking to understand what's going on here, obviously. I'm sure it's obvious.
    Last edited: Aug 21, 2009
  2. jcsd
  3. Aug 21, 2009 #2


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    You are integrating a positive function from t=0 to t=(-ln(4)). Of course the result is negative. The arc length parameter, s(t), as you've defined it is negative for t<0 and positive for t>0. The arc length distance from t=0 is indeed |s(t)|.
  4. Aug 21, 2009 #3
    So, to be on the safe side, would I be mathematically justified in putting the absolute value on s(t) in all cases from now on?
  5. Aug 21, 2009 #4


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    Depends on what you are doing. You should always think about something before you just do it. If you want the arclength distance between two points t1<t2 you want s(t2)-s(t1). Definitely not |s(t2)|-|s(t1)|. You see why, right?
  6. Aug 21, 2009 #5
    Yes, I believe so. If I plug a negative number in for s(t1) and a positive for s(t2) I get two different answers.
  7. Aug 21, 2009 #6


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    Sounds like you get the point. s(t) is signed arclength distance from t=0. Just like the x coordinate is the signed distance from x=0 along the x-axis.
  8. Aug 21, 2009 #7
    Thanks Dick. Now I'm off to partial derivatives.
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