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!
On
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.)
Do you have access to $\textit{Mathematica}$?
If so, try this out: