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) = ∫ √ (3e2τ) 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) -3√(3)/4 Which is off by a sign. Why am I getting this wrong? Thanks 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.