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How can I find the arc length of the polar curve: $r(t)=1+\cos(t)$ using the hint: $\sqrt{1-\cos(t)}\cdot \sqrt{1+\cos(t)} =\sin(t)$ for $0\leq t\leq \pi$ ?

I think the main thing I'm wondering is the factorization, since I'm pretty sure I can use the the formula: $$L=\int_0^{\pi} \sqrt{(dr/dt)^2 +r^2} dt$$ To find the arc length of the upper half of the cardioid and then just multiply it by 2?

So I'm not sure how I can use the hint when I got $$\int_0^{\pi}\sqrt{(-\sin(t))^2 +(1+\cos(t))^2} dt.$$

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  • $\begingroup$ As regards the hint, see my P.S. $\endgroup$ – Robert Z Sep 3 '16 at 19:03
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The length $L$ of the cardioid is given by \begin{align*}L&=2\int_0^{\pi} \sqrt{(dr/dt)^2 +r^2} dt\\ &= 2\int_0^{\pi}\sqrt{(-\sin(t))^2 +(1+\cos(t))^2} dt \\&=2\int_0^{\pi}\sqrt{\sin^2(t) +1+2\cos(t)+\cos^2(t)} dt \\&=2\sqrt{2}\int_0^{\pi}\sqrt{1+\cos(t)} dt =4\int_0^{\pi}\cos(t/2) dt=8[\sin(t/2)]_0^{\pi}=8. \end{align*}

P.S. As regards the hint, note that $$2\sqrt{2}\int_0^{\pi}\sqrt{1+\cos(t)} dt = 2\sqrt{2}\int_0^{\pi}\frac{\sin(t)}{\sqrt{1-\cos(t)}} dt= 4\sqrt{2}\int_0^{\pi}\frac{d(1-\cos(t))}{2\sqrt{1-\cos(t)}}\\ =4\sqrt{2}[\sqrt{1-\cos(t)}]_0^{\pi}=8.$$

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  • $\begingroup$ As the OP says, don't you want to multiply your answer by 2 to get the length of the whole cardioid? $\endgroup$ – user84413 Sep 3 '16 at 21:43
  • $\begingroup$ @user84413 That's OK. $\endgroup$ – Robert Z Sep 3 '16 at 22:12

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