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

In summary, the conversation discusses finding the Arc Length Parameter and the length of a portion of a curve using an integral. The first part is solved correctly, but the second part results in a negative length. It is determined that this is due to the interval being reversed. It is concluded that the absolute value should be taken in certain cases when calculating the arc length distance. The concept of signed distance is also mentioned.
  • #1
Battlemage!
294
45

Homework Statement



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.

Homework Equations



The vector I am using for this:


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

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.
 
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  • #2
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
Dick said:
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?
 
  • #4
Battlemage! said:
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?
 
  • #5
Dick said:
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.
 
  • #6
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
Thanks Dick. Now I'm off to partial derivatives.
 

1. What is an arc length parameter in calculus?

An arc length parameter is a mathematical concept used to measure the distance along a curve. It is often denoted by "s" and is defined as the integral of the magnitude of the derivative of the curve with respect to the independent variable.

2. How is arc length parameter calculated?

Arc length parameter is calculated by taking the integral of the square root of the sum of the squares of the derivatives of the curve with respect to the independent variable. This can be written as ∫√(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 dt, where x, y, and z are the coordinates of the curve and t is the independent variable.

3. Why would the arc length parameter result in a negative length?

The arc length parameter can result in a negative length when the curve being measured has a decreasing slope. This means that the derivative of the curve is negative and when squared, it becomes a positive value. When this value is integrated, it results in a negative length.

4. Is a negative arc length parameter valid?

Yes, a negative arc length parameter can be a valid measurement. It simply means that the distance along the curve is in the opposite direction of the direction of the curve. This can happen when the curve is decreasing or when the measurement starts at a point further along the curve and ends at a point closer to the origin.

5. How can a negative arc length parameter be interpreted?

A negative arc length parameter can be interpreted as the distance traveled in the opposite direction of the curve. It can also be thought of as the difference between the starting point and the endpoint, with the negative sign indicating the direction of travel along the curve.

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