Thread: Polar length of curve

1. Polar length of curve

I am having some trouble with this problem. I have the formula, but I am getting trouble with the simplification. Can i please have some help?

Find the length of the curve $\displaystyle r= a \ cos^2 \ (\frac{\theta}{2}$), for 0 less than or equal theta less than or equal pi/2.

2. You say you "have the formula" so I presume you know that the arclength is given by $\displaystyle \int \sqrt{r^2+ (dr/d\theta)^2} d\theta$.

Here, $\displaystyle r= a cos^2(\theta/2)$ so $\displaystyle dr/d\theta= -a sin(\theta/2)cos(\theta)$ and so $\displaystyle r^2+ (dr/d\theta)^2= a^2 cos^4(\theta/2)+ a^2 sin^2(\theta/2)cos^2(\theta/2)$$\displaystyle = a^2 cos^2(\theta/2)(cos^2(\theta/2)+ sin^2(\theta/2))= a^2 cos^2(\theta/2). I don't see anything difficult about that! 3. Originally Posted by HallsofIvy You say you "have the formula" so I presume you know that the arclength is given by \displaystyle \int \sqrt{r^2+ (dr/d\theta)^2} d\theta. Here, \displaystyle r= a cos^2(\theta/2) so \displaystyle dr/d\theta= -a sin(\theta/2)cos(\theta) and so \displaystyle r^2+ (dr/d\theta)^2= a^2 cos^4(\theta/2)+ a^2 sin^2(\theta/2)cos^2(\theta/2)$$\displaystyle = a^2 cos^2(\theta/2)(cos^2(\theta/2)+ sin^2(\theta/2))= a^2 cos^2(\theta/2)$.

I don't see anything difficult about that!
Its not difficult.. but it looks like i am not getting that differential to work out...? I am getting some other thing.. can u help me with the steps to get to the value u have for dr/dtheta?