# Why does the arc length formula not work for this parametric curve?

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?

• Please use MathJax. math.stackexchange.com/help/notation Jun 7, 2022 at 11:54
• 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.
– J.G.
Jun 7, 2022 at 11:56

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.
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 $$+t$$.