Question

Is the length of a vector defined function from a to b, the integral of magnitudes of its tangents?

Answer 1

spose, f(t) = <sin(t),cos(t),t> what is the length of the resulting curve in the interval [a,b] ?

Answer 2

Yes because\[\int\limits_{a}^{b}\sqrt{(dx/dt)^2+(dy/dt)^2}dt\]
is arc length and the integrand is the magnitude of the parametric tangents, right?

Answer 3

if it is in R^2 then I would say yes; but does the extra dimension casue havok?

Answer 4

im not dyslexic, but i type for those who are :)

Answer 5

in 3d space you have twist as well as turn

Answer 6

I don't see why it would, most formulas for n-dimensions have obvious extensions in higher dimensions.

Answer 7

or am i confusing this with a line integral?

Answer 8

wouldn't it just be\[\int\limits_{a}^{b}\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}dt\]

Answer 9

I think you are, but I'm not positive.

Answer 10

example 6:

http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx

Answer 11

Well you asked about arc length, and the formula given for the line integral is the integral of the function times the magnitude of the tangent vectors (arc length), so as far as the length of the function itself is concerned it looks the same to me.

Answer 12

thats what i tend to think as well, but then I see stuff like this:

http://www.youtube.com/watch?v=fjEvsinvtnw

where it has: \(\int_{a}^{b}\ f(x,y)\sqrt{dx^2+dy^2}\ dt\) and have to wonder what im missing

Answer 13

I think someone's stealing my wi-fi, so that video is still downloading, but I'll watch it and get back to you on that.

Answer 14

k

Answer 15

Seems the same to me as the example in Paul's notes. I just see regular arc length formula times f(x,y) as the integrand. Once again, the arc length itself is the magnitude of the tangent vectors. I don't really see a contradiction.
Anyway, I'm late for something so we can pick this up later if you like.