Length of a curve?

By Olof Sisask (P3033) on Sunday, October 22, 2000 - 06:18 pm :

Hi,

This is my first post - hope it works! The other day I was thinking about how you could find the "length" of a curve, ie the distance it covers. The only thing that occured to me would be to estimate the curve with small triangles, and to find the length of each little segment of the curve by considering it to be the hypotenuse of the triangle, and using Pythagoras thus: length = sqrt[(dx)² + (dy)² ]. The next step would be to make each triangle infinitesimally small, and sum infinitely many of them (by integration?). Am I on the right path? Is there some other really neat method? Hope this is legible :).

Thank you,
Olof Sisask

By Dan Goodman (Dfmg2) on Sunday, October 22, 2000 - 06:30 pm :

You've got the idea basically right, the way it's usually done is this: Let the curve be

(x (t), y (t))

, i.e. the

x

and

y

coordinates are functions of some parameter

t

. Let's also suppose that we're trying to find the length of the curve between

t = a

and

t = b

. As you rightly pointed out,

{dL}^{2} = {dx}^{2} + {dy}^{2}

. Divide both sides by

{dt}^{2}

to get

(dL / dt)^{2} = (dx / dt)^{2} + (dy / dt)^{2}

. Take the square root of both sides and integrate between

a

and

b

, to get

L (a, b) = \int_{a}^{b} dL / dt dt = \int_{a}^{b} \sqrt{(} (dx / dt)^{2} + (dy / dt)^{2}) dt

. This is sometimes quite a nasty integral to do unfortunately. You can use this to prove (for instance) that the circumference of a circle is

2 π r

, using the equation

x (t) = r \cos (t)

and

y (t) = r \sin (t)

for

0 \leq t \leq 2 π

, you get

dx / dt = - r \sin (t)

dy / dt = r \cos (t)

, therefore

(dx / dt)^{2} + (dy / dt)^{2} = r^{2} (\cos^{2} (t) + \sin^{2} (t)) = r^{2}

. Therefore

L (0, 2 π) = \int_{0}^{2 π} r dt = [r . t]_{0}^{2 π} = 2 π r

By Olof Sisask (P3033) on Sunday, October 22, 2000 - 07:39 pm :

Thanks a lot Dan!

Olof