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I am trying to find the arc length for the parametric equations $x=\cos^3t,\,y=\sin^3t$, for $t\in[0,\,2\pi]$.

This interval of $t$-values traces the curve exactly once, yet if you use the standard formula for the arc length of a parametric curve ($\int_0^{2\pi}\sqrt{\dot{x}^2+\dot{y}^2}dt$), it gives you an answer of $0$ which is obviously wrong.

If, instead of doing this, I simply multiplied the arc length between $t=0$ and $t=\pi/2$ by $4$, I would get the correct answer ($6$). Can someone explain why the normal formula does not work in this instance?

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  • $\begingroup$ Please use MathJax. math.stackexchange.com/help/notation $\endgroup$
    – Paul Frost
    Jun 7, 2022 at 11:54
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    $\begingroup$ You should show your calculation in the question body but, ssince $\dot{x}^2+\dot{y}^2=(3\sin t\cos t)^2$, my guess is you integrated $3\sin t\cos t$ rather than $|3\sin t\cos t|$, which is the actual value of the square root as it's by definition non-negative. $\endgroup$
    – J.G.
    Jun 7, 2022 at 11:56

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if you use the standard formula for the arc length of a parametric curve ($\int_0^{2\pi}\sqrt{\dot{x}^2+\dot{y}^2}dt$), it gives you an answer of $0$ which is obviously wrong.

The integrand is always non-negative, so the integral can only be zero if the integrand is zero, which is not the case.

We can only guess what went wrong, presumably integrating over a singularity or picking the wrong branch of the square root. Like you used $\sqrt{t^2}=t$ which is not correct if $t< 0$. What's correct is $\sqrt{t^2} = |t|$.

$|t|$ as part of the integrand can be treated by handling different cases of $t\leqslant 0$ and $t\geqslant 0$. This means you can split the integral accordingly, so that $|t|$ can be replaced by either $-t$ or $+t$.

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  • $\begingroup$ Well, this is embarassing. After checking my working out I did indeed make a square-root -related mistake. I hadn't encountered this sort of problem before in calculus where you have to consider absolute values. Thank you very much for the help! $\endgroup$ Jun 7, 2022 at 12:55
  • $\begingroup$ @Lily Morgan: As your question was not very specific, all I could do was guessing, so that my answer is not very specific, either... But it's nice that you could locate the bug. $\endgroup$ Jun 7, 2022 at 15:10

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