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.

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- Apr 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. - Apr 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 $\displaystyle \int_a^b \sqrt{1+ f'^2(x)} dx$. - Apr 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! - Apr 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. - Apr 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. - Apr 25th 2013, 04:18 PMProve ItRe: Exact length of a curve?
$\displaystyle \displaystyle \begin{align*} f(x) &= \sqrt{ x - x^2 } + \arcsin{ \left( \sqrt{x} \right) } \\ \\ f'(x) &= \frac{1 - 2x}{2\,\sqrt{ x - x^2 }} + \frac{1}{2\,\sqrt{x} \, \sqrt{ 1 - x } } \\ &= \frac{ 1 - 2x }{ 2 \, \sqrt{ x \left( 1 - x \right) } } + \frac{1}{2 \, \sqrt{ x \left( 1 - x \right) } } \\ &= \frac{2 - 2x}{2\,\sqrt{x \left( 1 - x \right) }} \\ &= \frac{1 -x}{\sqrt{x \left( 1 - x \right) }} \\ &= \frac{ \sqrt{x \left( 1 - x \right) } }{x} \end{align*}$

So now plugging into the formula for the arclength (I can't see your images so I'm not sure what your bounds should be)...

$\displaystyle \displaystyle \begin{align*} l &= \int_a^b{ \sqrt{ 1 + \left[ f'(x) \right] ^2 } \, dx } \\ &= \int_a^b{ \sqrt{1 + \left[ \frac{\sqrt{x\left( 1 - x \right) }}{x} \right] ^2 } \, dx } \\ &= \int_a^b{ \sqrt{ 1 + \frac{x \left( 1 - x \right) }{x^2} } \,dx } \\ &= \int_a^b{ \sqrt{ 1 + \frac{1 - x}{x} }\,dx } \\ &= \int_a^b{ \sqrt{ \frac{1}{x} } \,dx } \\ &= \int_a^b{ x^{-\frac{1}{2} } \,dx} \\ &= \left[ 2x^{\frac{1}{2}} \right] _a^b \\ &= 2\,\sqrt{b} - 2\,\sqrt{a} \end{align*}$

So now plug in your endpoints... - Apr 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