Expert: Paul Klarreich - 3/31/2006

Question
Assume x=f(t), y=g(t), and arc lenth=ds, then
ds=sqrt[(f')^2+(g')^2]dt.

If y=h(x), then ds=sqrt[1+{g'/f')^2][f'] and
ds=sqrt[1+(y')^2]dx.

What happens when f' is negative or f' is zero?
When I wrote the second line I assumed f' was positive and f' was not zero.

Answer
Hi, Ron,

You wrote:
Subject:  Arc Length
Question:  Assume x=f(t), y=g(t), and arc lenth=ds, then
ds=sqrt[(f')^2+(g')^2]dt.

If y=h(x), then ds=sqrt[1+{g'/f')^2][f'] and
ds=sqrt[1+(y')^2]dx.

What happens when f' is negative or f' is zero?
When I wrote the second line I assumed f' was positive and f' was not zero.

WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


Let's try the classic example:

x = cos t,  y = sin t,  integrate for t = 0 to 2 pi.

Arc length is:
{2pi
|    sqrt( f'^2 + g'^2) dt
}0

f'(t) = - sin t,  g'(t) = cos t

f'^2 + g'^2 = 1

Arc length is:

{2pi
|    dt =
}0

t from 0 to 2 pi = 2 pi - 0 = 2 pi.

Right! That's the circumference of a unit circle.

Now  x^2 + y^2 = 1, and:

y = sqrt(1 - x^2)
    1     - 2x
y' = - -------------
    2 sqrt(1 - x^2)

          x
y' = - -------------
      sqrt(1 - x^2)

                   x^2
1 + y'^2 = 1 + ------------
                1 - x^2
    1
= --------
 1 - x^2

Arc length is:

{x=1
|   (1 - x^2)^-1/2  dx
}x=-1

We'll do this by a trig substitution:

x = sin t
dx = cos t dt
(1 - x^2) = cos^2 t

{    cos t dt
|  -------------
}    cos t

{   
| dt = t
}   

= arcsin x  from x=-1 to x=1

= arcsin(1) - arcsin(-1) = pi/2 - (-pi/2) = pi,  just as before.

Oops -- not quite just as before.  Something has gone wrong.

It appears that when we integrated, it was not good enough to do x=-1 to x=1.  We really obtained only the upper half of the semicircle.

So the answer to your question(s) will be:

What happens when f' is negative or f' is zero?

Something could go wrong.  It will be best if you identify regions where the derivatives are negative and positive and integrate those separately.