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Math Help - Polar length of curve

  1. #1
    Ife
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    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 r= a \ cos^2 \ (\frac{\theta}{2}), for 0 less than or equal theta less than or equal pi/2.
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  2. #2
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    You say you "have the formula" so I presume you know that the arclength is given by \int \sqrt{r^2+ (dr/d\theta)^2} d\theta.

    Here, r= a cos^2(\theta/2) so dr/d\theta= -a sin(\theta/2)cos(\theta) and so r^2+ (dr/d\theta)^2= a^2 cos^4(\theta/2)+ a^2 sin^2(\theta/2)cos^2(\theta/2) = 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!
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  3. #3
    Ife
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    Quote Originally Posted by HallsofIvy View Post
    You say you "have the formula" so I presume you know that the arclength is given by \int \sqrt{r^2+ (dr/d\theta)^2} d\theta.

    Here, r= a cos^2(\theta/2) so dr/d\theta= -a sin(\theta/2)cos(\theta) and so r^2+ (dr/d\theta)^2= a^2 cos^4(\theta/2)+ a^2 sin^2(\theta/2)cos^2(\theta/2) = 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?
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