Arc length of cardioid using line integral techniques

Oct 2014
32
0
Canada
I've written out the question and my work thus far.
If someone tell me if I'm on the right track or not, it would help a lot.

image.jpg
 

romsek

MHF Helper
Nov 2013
6,744
3,037
California
you seem to be doing it the hard way

$r=r(\theta)$

$ds=\sqrt{r^2 + r_{\theta}^2} ~d\theta$

$L=\int ds$

$r(\theta)=\cos(\theta)+1$

$r_{\theta}=-\sin(\theta)$

$ds=\sqrt{(1+\cos(\theta))^2+\sin^2(\theta)}=$

$\sqrt{1+2\cos(\theta)+\cos^2(\theta)+\sin^2( \theta)}=$

$\sqrt{2+2\cos(\theta)}$

so

$L = \displaystyle{\int_0^{2\pi}}\sqrt{2+2\cos(\theta)}~d\theta=8$
 
Dec 2013
2,002
757
Colombia
Yes, you are on the right track. Your curve is parametrised by \(\displaystyle \theta\), so you just need to build the integral \(\displaystyle \int \limits_C \mathrm d s = \int \limits_C \big| \big|\mathbf{ r}'(\theta) \big| \big| \, \mathrm d \theta\).
 
Oct 2014
32
0
Canada
So I've hit a wall again, I'm not sure how to finish simplifying the magnitude so I can integrate it.
I feel like I'm not doing this right.
image.jpg
 

romsek

MHF Helper
Nov 2013
6,744
3,037
California
So I've hit a wall again, I'm not sure how to finish simplifying the magnitude so I can integrate it.
I feel like I'm not doing this right.
View attachment 35837
maybe this is more help, there's a fair bit of trig simplification you can do. Look for factors of

$\cos^2(\theta)+\sin^2(\theta)$ which reduce to 1

Clipboard01.jpg
 
Dec 2013
2,002
757
Colombia
You have \(\displaystyle cos^4 x + 2\cos^2 x \sin^2 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2\) and \(\displaystyle 2\cos x\sin^2 x + 2 \cos ^3 x = 2\cos x(\sin^2 x + \cos^2 x)\).
 
Oct 2014
32
0
Canada
Thanks everyone! I finally got through it, I appreciate your help!