Forum index » Arc Length Problem

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speedychaos4

Post subject: Arc Length Problem

Posted: Fri Apr 09, 2010 6:20 am

Joined: Mon Mar 22, 2010 2:42 am

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Joined: Mon Mar 22, 2010 2:42 am
Posts: 42

How to evaluate the length of arc of y=x^4/4 from on [0,a]? thanks.

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origin415

Post subject: Re: Arc Length Problem

Posted: Fri Apr 09, 2010 8:17 am

Joined: Fri Oct 23, 2009 11:42 pm

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Joined: Fri Oct 23, 2009 11:42 pm
Posts: 61

The formula for arc length is

$\int_0^a \sqrt{(1 + (\frac{dy}{dx})^2} dx$

Sorry I edited it to simplify as you were posting!

Last edited by origin415 on Fri Apr 09, 2010 8:20 am, edited 2 times in total.

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speedychaos4

Post subject: Re: Arc Length Problem

Posted: Fri Apr 09, 2010 8:19 am

Joined: Mon Mar 22, 2010 2:42 am

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Joined: Mon Mar 22, 2010 2:42 am
Posts: 42

origin415 wrote:

First you want to parametrize x and y, so
x(t) = t
y(t) = t^4/4

then the formula for arc length is

$\int_0^a \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$

I'm ok with this formula...but I'm stuck with the integral...so i want to ask for help, thanks.

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origin415

Post subject: Re: Arc Length Problem

Posted: Fri Apr 09, 2010 8:30 am

Joined: Fri Oct 23, 2009 11:42 pm

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Joined: Fri Oct 23, 2009 11:42 pm
Posts: 61

Well you can't actually do that integral, so don't try it (http://www.wolframalpha.com/input/?i=In ... %29%2Cx%29). What are the multiple choice answers? Maybe you can estimate the answer if the answers are not too close to one another.

If a is small, than $\sqrt{1+x^6}$ is close to 1, if a is large, its close to $x^3$ .

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joeleitz

Post subject: Re: Arc Length Problem

Posted: Mon Apr 19, 2010 6:43 pm

Joined: Mon Apr 19, 2010 6:33 pm

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Joined: Mon Apr 19, 2010 6:33 pm
Posts: 4
Location: Washington State

Nooo.. Every time I estimate based on multiple choice I get it wrong!

I swear that unlike all other humans, my first guess is my worst.

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