Arc Length Problem

Forum for the GRE subject test in mathematics.
speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

Arc Length Problem

Postby speedychaos4 » Fri Apr 09, 2010 6:20 am

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

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

Re: Arc Length Problem

Postby origin415 » Fri Apr 09, 2010 8:17 am

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.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

Re: Arc Length Problem

Postby speedychaos4 » Fri Apr 09, 2010 8:19 am

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.

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

Re: Arc Length Problem

Postby origin415 » Fri Apr 09, 2010 8:30 am

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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