Can anyone explain a method on how to go about this problem? I don't care for the answer; I'm just perplexed on how to begin the problem in a successful manner. Thanks!

edit here's syntax:http://www5a.wolframalpha.com/Calcul...3&w=171.&h=30.

Printable View

- April 24th 2013, 11:05 AMSteelers72Exact length of a curve?
Can anyone explain a method on how to go about this problem? I don't care for the answer; I'm just perplexed on how to begin the problem in a successful manner. Thanks!

edit here's syntax:http://www5a.wolframalpha.com/Calcul...3&w=171.&h=30. - April 24th 2013, 11:12 AMHallsofIvyRe: Exact length of a curve?
Where did you get this problem? It is pretty standard Calculus problem- are you not taking Calculus and have not learned this?

If y= f(x), the arclength, from x= a to x= b, is given by . - April 24th 2013, 11:21 AMSteelers72Re: Exact length of a curve?
Thanks! Oh, so you have to find the derivative of y and plug into the length formula you've provided right?

My instructor hasn't covered this (yet?) and gave this as one of a bunch of questions to do for next week. Had no clue what formula to use but thanks for clearing that up! - April 24th 2013, 11:40 AMSteelers72Re: Exact length of a curve?
By the way, how would you go about finding the intervals? I'm guessing set y=0? Not sure.

I've gotten this far:

http://www4a.wolframalpha.com/Calcul...6&w=204.&h=62. - April 25th 2013, 02:33 PMSteelers72Re: Exact length of a curve?
Original problem:y=sqrt(x-x^2)+arcsin(sqrt(x))

Sorry, this is the derivative of y I got to plug into the Length formula:

sqrt(-(-1+x) x)/x

Does anyone know what the bounds are and how to find them? Do we set the original y equation = to 0?

I think it may be 0 to 1 but have no clue as to why. - April 25th 2013, 04:18 PMProve ItRe: Exact length of a curve?
- April 25th 2013, 05:12 PMSteelers72Re: Exact length of a curve?
Thanks for the work through! I was confused as to how to get those endpoints but your work through helps a lot too.

Edit: never mind. They were 0 to 1