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

1. Aug 21, 2009

### 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) = √ (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)

-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.

Last edited: Aug 21, 2009
2. Aug 21, 2009

### 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)|.

3. Aug 21, 2009

### Battlemage!

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?

4. Aug 21, 2009

### Dick

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?

5. Aug 21, 2009

### Battlemage!

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.

6. Aug 21, 2009

### 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.

7. Aug 21, 2009

### Battlemage!

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