# arc length

• Apr 2nd 2009, 02:21 PM
silencecloak
arc length
$y = sin^-1(x)+\sqrt{1-x^2}$ 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
• Apr 2nd 2009, 03:10 PM
skeeter
Quote:

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

.
• Apr 2nd 2009, 03:14 PM
silencecloak
Quote:

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)."
• Apr 2nd 2009, 03:45 PM
skeeter
Quote:

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 ...

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

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

$S(x) = 2\sqrt{2}\left[\sqrt{x+1} - 1\right]$
• Apr 2nd 2009, 07:41 PM
silencecloak
how did you know to start the integral at 0?
• Apr 3rd 2009, 05:07 AM
skeeter
Quote:

Originally Posted by silencecloak
how did you know to start the integral at 0?

what's the given starting point?
• Apr 3rd 2009, 09:04 AM
silencecloak
Quote:

Originally Posted by skeeter
what's the given starting point?

Mmmm ok, didn't know it was that straightforward

Gracias Senor