1. ## 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

2. 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
.

3. 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)."

4. 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 ...

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

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

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

5. how did you know to start the integral at 0?

6. Originally Posted by silencecloak
how did you know to start the integral at 0?
what's the given starting point?

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

Mmmm ok, didn't know it was that straightforward

Gracias Senor