arc length

**silencecloak** · April 2nd 2009, 02:21 PM

$y = sin^-1(x)+\sqrt{1-x^2}$ starting at point (0,1)

I have worked this down and obtained the integral

$\int \frac{2}{(x+1)}dx$

but I do not know how to set up my limits on the integral. I know the arcsin function is bound between -1 and 1 on the x axis and -pi/2 and pi/2 on the y axis

**skeeter** · April 2nd 2009, 03:10 PM

Originally Posted by silencecloak

$y = sin^-1(x)+\sqrt{1-x^2}$ starting at point (0,1) ... and ending where?

I have worked this down and obtained the integral

$\int \frac{2}{(x+1)}dx$

but I do not know how to set up my limits on the integral. I know the arcsin function is bound between -1 and 1 on the x axis and -pi/2 and pi/2 on the y axis

.

**silencecloak** · April 2nd 2009, 03:14 PM

Originally Posted by skeeter

.

It doesnt tell me where it ends

"Find the arc length function for the curve below with starting point (0, 1)."

**skeeter** · April 2nd 2009, 03:45 PM

Originally Posted by silencecloak

It doesnt tell me where it ends

"Find the arc length function for the curve below with starting point (0, 1)."

the water clears ... the question is asking for a function.

note that I reworked and corrected your arc length integral ...

$S(x) = \sqrt{2} \int_0^x \frac{1}{\sqrt{t+1}} \, dt$

$S(x) = 2\sqrt{2}\left[\sqrt{t+1}\right]_0^x$

$S(x) = 2\sqrt{2}\left[\sqrt{x+1} - 1\right]$

**silencecloak** · April 2nd 2009, 07:41 PM

how did you know to start the integral at 0?

**skeeter** · April 3rd 2009, 05:07 AM

Originally Posted by silencecloak

how did you know to start the integral at 0?

what's the given starting point?

**silencecloak** · April 3rd 2009, 09:04 AM

Originally Posted by skeeter

what's the given starting point?

Mmmm ok, didn't know it was that straightforward

Gracias Senor

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