graph the curve and find its length, $r=\cos^2(\frac {\theta}{2}) $

Question

graph the curve and find it's length, $r=\cos^2(\frac {\theta}{2}) $

I graphed it and found that it was a cardioid (or a sideways heart). I am getting stuck on the arc length.

this is what I have:

$$r=(\frac 12(1+\cos\theta) $$

$$\frac {dr}{d\theta}= -\frac 12\sin\theta $$

what I get for under the square root is :

$$\frac 14+\frac 12\cos\theta+\frac 14\cos^2\theta+ \frac 14\sin^2\theta $$

I ended up getting stuck with $\frac12+\frac12\cos\theta $

But I don't think this is right, where did I go wrong?

Answer 1

Things look fine. By the double-angle identity you have already used, the square root is $|\cos(\theta/2)|$. Integrate. Note the absolute value sign.

Answer 2

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\begin{align} \color{#66f}{\Large L}&=2\int_{0}^{\pi}\root{\pars{\dd r}^{2} + r^{2}\pars{\dd\theta}^{2}} =2\int_{0}^{\pi}\root{\bracks{\totald{{\rm r}\pars{\theta}}{\theta}}^{2} + {\rm r}^{2}\pars{\theta}}\,\dd\theta \\[3mm]&=2\int_{0}^{\pi} \root{{1 \over 4}\,\sin^{2}\pars{\theta} + \cos^{4}\pars{\theta \over 2}} \,\dd\theta \\[3mm]&=2\int_{0}^{\pi}\root{{1 \over 4}\,\sin^{2}\pars{\theta} + \bracks{1 + \cos\pars{\theta} \over 2}^{2}}\,\dd\theta =\int_{0}^{\pi}\root{2\bracks{1 + \cos\pars{\theta}}}\,\dd\theta \\[3mm]&=\int_{0}^{\pi}\root{4\cos^{2}\pars{\theta \over 2}}\,\dd\theta =2\int_{0}^{\pi}\verts{\cos\pars{\theta \over 2}}\,\dd\theta =4\int_{0}^{\pi/2}\cos\pars{\theta}\,\dd\theta = \color{#66f}{\Large 4} \end{align}

Answer 3

$L=\int_{0}^{2\pi}\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta = \int_{0}^{2\pi}\sqrt{\frac{1}{4}(1+2cos\theta+cos^2\theta)+\frac{1}{4}sin^2\theta}d\theta = \int_{0}^{2\pi}\sqrt{\frac{1}{4}(2+2cos\theta)}d\theta = \int_{0}^{2\pi}\sqrt{\frac{1}{4}(2+2(2cos^2\frac{\theta}{2}-1))}d\theta = \int_{0}^{2\pi}\sqrt{cos^2\frac{\theta}{2}}d\theta = \int_{0}^{2\pi}|\cos\frac{\theta}{2}|d\theta = \int_{0}^{\pi}\cos\frac{\theta}{2}d\theta - \int_{\pi}^{2\pi}\cos\frac{\theta}{2}d\theta = [2sin\frac{\theta}{2}]_{0}^{\pi} - [2sin\frac{\theta}{2}]_{\pi}^{2\pi} = (2-0)-(0-2)=4$