# Calculate arc length. How to know the limits of integration?

The task: $$x(t) = \frac{\cos(3t)}{t^3}; \ y(t)=\frac{\sin(3t)}{t^3};\ t\ge \frac{1}{\sqrt3}$$

I have this condition. It is necessary to calculate the arc length of the curve. I have to use this formula

$$l = \int_{t1}^{t2}{\sqrt{(x'(t))^2+(y'(t))^2}\ dt}$$

But I dont know how to find another limit of integration (first $\boldsymbol{- t\geq 1/\sqrt{3}}$). I ask for your help! Thank you in advance!

• Probably its an improper integral, and the upper limit should be $+\infty$. – Bernard Mar 26 '17 at 14:17
• just +infinity. – user392395 Mar 26 '17 at 14:24
• I do not think that the second limit = + infinity, because we have not yet had a topic with improper integrals :D Even if the graph is plotted, then no more arc is built. But what exactly is the second integration limit to take? – Frip Mar 26 '17 at 14:37

Do you have access to $\textit{Mathematica}$?

If so, try this out:

• what is it? say me pls – Frip Mar 26 '17 at 15:04
• The parametric plot shows that the graph of your function is a spiral moving counter-clockwise towards the origin on the $xy$-plane. But you didn't need the graph to do this problem. Whenever you see $t\ge \frac{1}{\sqrt3}$, it usually means the bound for your integral is $\frac{1}{\sqrt3}\leq t < \infty$. – Mee Seong Im Mar 26 '17 at 15:25
• what web-site is it? – Frip Mar 27 '17 at 10:47
• Frip, this is not a website. I use a mathematics software called $\textit{Mathematica}$ to type those commands, and then executed them via holding down the Shift key and then Enter after each command line. It was faster for me to do the calculations via a computer than on paper. – Mee Seong Im Mar 27 '17 at 16:32

This problem can solved very nicely in the complex plane. Let us take

$$z=\frac{e^{i3t}}{t^3}$$

and note that the arc length is given by

$$s=\int |\dot z| dt$$

(See, for example, Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.)

We can then calculate

$$\dot z=(3t^3 i -3t^2)e^{i3t}/t^6 \text{ and } |\dot z|=\frac{3}{t^4}\sqrt{t^2+1}$$

The indefinite integral is given nicely by

$$\int |\dot z| dt=\frac{(t^2+1)^{3/2}}{t^3}$$

so that finally

$$\int_{1/\sqrt(3)}^\infty |\dot z| dt=\frac{(t^2+1)^{3/2}}{t^3} \Big{\vert}_{1/\sqrt(3)}^\infty=8-1=7$$

(Sorry, the limits on the vertical bar in MathJax are not well done.)