The problem I came across asks the following:

Find the arc length of a curve in its parametric form: $$x=a(2\cos{t}-\cos{2t}),$$ $$y=a(2\sin{t}-\sin{2t})$$

To solve this we can use the formula $\int_{t_1}^{t_2}{\sqrt{(x')^2+(y')^2}dx}$.
We can easily find that $$x'=2a(\sin{2t}-\sin{t}),$$ $$y'=-2a(\cos{2t}-\cos{t})$$

The part where I struggle with is finding $t_1$ and $t_2$. This is a wider problem I have when ever I come across these types of problems. How can I find these bounds if I have no idea how this parametric function looks like? How do I know if we can even calculate arc length just by using the previously written formula? (e.g. like with circle, where we find it by finding just one quarter of circle and multiplying it with four) How should I know if I only got some part of arc length and not the whole thing?


1 Answer 1


Regarding a circle with the parameterization $x=\cos t$, $y=\sin t$, rather than finding the length of one quarter of the circle and multiplying by $4$, another way to find the length is the recognize that the parameterization is periodic with period $2 \pi$. You can find the arc length by going exactly once around the circle, integrating from $0$ to $2 \pi$.

Similarly, with your example, you can recognize from the formulas that the parameterization is also periodic with period $2 \pi$: the $\cos 2t$ and $\sin 2t$ terms have a smaller period of $\pi$, but the $\cos t$ and $\sin t$ have period $2 \pi$, and so the smallest period that all of those terms have in common is $2 \pi$. So for this problem also you can find the arc length by integrating from $0$ to $2 \pi$.

  • $\begingroup$ So basically I should try to find the smallest period that all the terms in both $x$ and $y$ have in common? $\endgroup$
    – bb_823
    Commented Jun 6, 2023 at 18:16
  • 1
    $\begingroup$ For periodic parameterizations, yes. In general not all parameterizations have a period, and for such a problem you would have to do something different. $\endgroup$
    – Lee Mosher
    Commented Jun 6, 2023 at 18:19

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