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Find the curve length when $\alpha(t)=(a(t-\sin t),a(1-\cos t))$ where $a>0$ and $t\in [0,2\pi]$

So I am using: $$\int_{a}^{b}\|\alpha'(t)\|\,\mathrm dt$$

$\alpha'(t)=(a-a\cos t,a\sin t)$

$$\|\alpha'(t)\|=\sqrt{(a-a\cos t)^2+(a\sin t)^2}=\sqrt{a^2-2a^2 \cos t+a^2\cos^2t+a^2\sin^2t}=\sqrt{a^2-2a^2 \cos t+a^2(\cos^2t+\sin^2t)}=\sqrt{2a^2-2a^2 \cos t}=\sqrt{2a^2(1-\cos t)}=\sqrt{2a^2(2\sin^2(t/2))}=\sqrt{4a^2(\sin^2(t/2))}=2a\sin(t/2)$$

So we have $$\int_{0}^{2\pi}2a\sin(t/2)\,\mathrm dt=2a\int_{0}^{\pi}\sin(t/2)\,\mathrm dt=2a(-2\cos(t/2))|^{\pi}_{0}=4a$$

But the answer is $8a$

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  • $\begingroup$ You have an error on the third line, second equality: $a^{2}(\cos^{2}t+\sin^{2}t) = a^{2}$, not $0$ $\endgroup$
    – pwerth
    Jul 18, 2017 at 16:12
  • $\begingroup$ @pwerth I added it to the $2a^2$ $\endgroup$
    – gbox
    Jul 18, 2017 at 16:20
  • $\begingroup$ ah yes I see, my fault $\endgroup$
    – pwerth
    Jul 18, 2017 at 16:21
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    $\begingroup$ Although it didn't hurt you here, in general you should be very careful to remember that $\sqrt{x^2} = |x|$. Omitting the absolute values often leads to wrong answers! $\endgroup$ Jul 18, 2017 at 16:25
  • $\begingroup$ @TedShifrin How does it effect? I need to take the integral only where the function is nonnegative? in sin for example between 0 to $\pi$ what about $\pi$ to $2\pi$? $\endgroup$
    – gbox
    Jul 18, 2017 at 16:27

1 Answer 1

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There's a typo in the integral calculation. $$\int_0^{2\pi}2a\sin\frac t2\,dt=2a\int_0^{\mathbf 2\pi}\sin\frac t2\,dt=2a\left[-2\cos\frac t2\right]^{2\pi}_0=8a$$

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  • $\begingroup$ There are case that I need to look when the function is positive? in this case I do not as it is $0\leq sin\frac{t}{2}\leq 1$? $\endgroup$
    – gbox
    Jul 18, 2017 at 16:24
  • $\begingroup$ @gbox See Ted Shifrin's comment. Note that arc length integrands are always positive. $\endgroup$ Jul 18, 2017 at 16:26
  • $\begingroup$ So I just need to take the absolute value of the function? $\endgroup$
    – gbox
    Jul 18, 2017 at 16:27

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