Question

arc length of polar curve... im stuck (attaching below)

Answer 1

okay the problem is find the length of the curve of the parabolic segment: \[r = 2/(1-\cos \theta), \pi/2 \le \theta \le \pi\]

Answer 2

and i got stuck at \[\int\limits_{\pi/2}^{\pi} \frac{ 2\sqrt{1+\sin^2\theta} }{ 1-\cos }\]

Answer 3

doesnt look right

Answer 4

wolfram gives
http://www.wolframalpha.com/input/?i=arc+length+r+%3D+2%2F%281%E2%88%92cos%CE%B8%29%2C+pi%2F2%3C%5Ctheta%3Cpi

Answer 5

\[s = \int \limits_{\frac{\pi}{2}}^{\pi} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} d\theta\]

Answer 6

\[s = \int \limits_{\frac{\pi}{2}}^{\pi} \sqrt{\left(\frac{1}{2-\cos\theta}\right)^2 + \left(\left(\frac{2}{1-\cos\theta}\right)'\right)^2} d\theta\]

Answer 7

oh crap, okay i found what i did wrong haha forgot to square the denom of the derivative

Answer 8

lol okay :)
did u simplify the given equation :

\(\dfrac{2}{1 - \cos \theta} = \dfrac{2}{2\sin^2(\theta/2)} = \csc^2(\theta/2)\)

?

Answer 9

^that may help...

Answer 10

what did you use to convert that? idt ive seen that b4... is that half angle formula?

Answer 11

yup !

1-cos = 2sin^2

Answer 12

\[s = \int \limits_{\frac{\pi}{2}}^{\pi} \sqrt{\left(\frac{2}{1-\cos\theta}\right)^2 + \left(\left(\frac{2}{1-\cos\theta}\right)'\right)^2} d\theta \]

is same as



\[s = \int \limits_{\frac{\pi}{2}}^{\pi} \sqrt{\left(csc^2(\theta/2)\right)^2 + \left(\left( \csc^2(\theta/2) \right)'\right)^2} d\theta \]