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

[Battlemage!]

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.

[Dick]

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)|.

[Battlemage!]

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)|.

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?

[Dick]

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?

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?

[Battlemage!]

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?

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.

[Dick]

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.

[Battlemage!]

Thanks Dick. Now I'm off to partial derivatives.