## Length of Curve

For the discussion of math. Duh.

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RyanK
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### Length of Curve

Is there any connection between the length of a function on a given interval and it's integral or derivative? I know Derivatives and Integrals are related but is length related too?

Qoppa
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### Re: Length of Curve

Yes. The arc length of a function f(x) from a to b is given by
$s=\int_a^b \sqrt{1+f'(x)^2}dx$

Code: Select all

_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

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### Re: Length of Curve

Great question. It's usually answered somewhere in the second half of second-semester calculus. You can determine the answer by breaking up the interval into lots of little pieces and approximating the function with straight lines on each piece (that is, connect the point on the graph at the start of a little interval to the point at the end with a straight line). This works because, as the little intervals get smaller and smaller (and thus more numerous), the total length of the straight lines gets closer and closer to the actual length of the function. If you write out the formula for the total length of the line segments as a summation, and then use the mean value theorem, you can make it look like a Riemann sum for the function Qoppa provided. That means the limit as the intervals get small is just the integral of that function.
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Sir_Elderberry
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### Re: Length of Curve

Qoppa wrote:Yes. The arc length of a function f(x) from a to b is given by
$s=\int_a^b \sqrt{1+f'(x)^2}dx$

For further information, note that if the function is defined parametrically, with x and y functions of t, then:

$s=\int_a^b \sqrt{x'(t)^2+y'(t)^2}dx$
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Qoppa
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### Re: Length of Curve

Except then you have to integrate with respect to [imath]t[/imath], not [imath]x[/imath]

Code: Select all

_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

Sir_Elderberry
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### Re: Length of Curve

Qoppa wrote:Except then you have to integrate with respect to [imath]t[/imath], not [imath]x[/imath]

Oh yeah. That too. Sorry.
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qinwamascot
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### Re: Length of Curve

A derivation of that is possible with only the fundamental theorem of calculus. the Pythagorean theorem, and a little knowledge of parametric curves. I won't do it here, but you'll almost definitely see it later. For a curve defined in polar coordinates the formula is:

$s=\int_a^b \sqrt{r^2+r'(θ)^2}dθ$
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thornahawk
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### Re: Length of Curve

Seeing that everybody's giving their arc-length expressions, I always preferred the Argand (complex) form to parametric equations:

$s=\int_{a}^b \left|z^\prime\left(t\right)\right|\mathrm{d}t$

for a plane curve represented as [imath]z(t)=f(t)+i\,g(t)[/imath], [imath]f(t)[/imath] and [imath]g(t)[/imath] real functions of [imath]t[/imath].

The formula mentioned by Elderberry is quite easily extended to curves in three or higher dimensions; take the derivatives of the component functions (or in vector notation, the vector-valued function) with respect to the parameter before integrating.

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stephentyrone
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### Re: Length of Curve

thornahawk wrote:Seeing that everybody's giving their arc-length expressions, I always preferred the Argand (complex) form to parametric equations

I always prefered the classical differential geometry approach: parametrize the curve by arc length. Now the arc length is given by:
$s = b - a$
Even better, normalize your parameter first. Now the arc length is given by:
$s = 1$
Of course, the rest of the expressions you're interested in might have gotten a lot more complicated...
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Yesila
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### Re: Length of Curve

There's also the put a string down "on top of" the curve then lift the string pull straight and measure.... Incidentally if you want to put the string down correctly you'll need to do some derivatives to make sure that the tangent lines to the string at each point correspond with the tangent lines to the graph, similar with 2nd derivatives... and 3 rd... etc.