Calculate the length of the curve: $X(t)=(t^2\cos(t), t^2\sin(t))$, where $-\pi \le t \le \pi$.
So using the following formula - $L(X(t)) = \int^{-\pi}_{\pi}|X(t)'|$ I have found that $L(X(t)) = 0$.
What is the meaning of this? what does it mean that a curve is of length $0$?
EDIT
Here is my calculation - $X'(t)=(2t\cos(t) -t^2\sin(t),2t\sin(t) + t^2\cos(t))$.
Therefore - $ \int^{-\pi}_{\pi}|X(t)'| = \int\sqrt{(2t\cos(t) -t^2\sin(t))^2 + (2t\sin(t) + t^2\cos(t))^2}dt$
Let's look at first only on the expression in the brackets - $(2t\cos(t) -t^2\sin(t))^2 + (2t\sin(t) + t^2\cos(t))^2 = 4t^2\cos^2(t) + t^4\sin^2(t) -4t^3\sin(t)\cos(t)+4t^2\sin^2(t)+t^4cos^2(t)+4t^3\sin(t)\cos(t) = 4t^2\cos^2(t)+4t^2\sin^2(t)+t^4\sin^2(t)+t^4\cos^2(t)=4t^2+t^4$
Going back to our integral -
$ \int\sqrt{(2t\cos(t) -t^2\sin(t))^2 + (2t\sin(t) + t^2\cos(t))^2}dt = \int\sqrt{4t^2+t^4}dt= \int t\sqrt{4+t^2}dt$.
Now we'll define $s = 4+t^2$ and so $dt = \frac{ds}{2t}$, and going back to the integral -
$\int t\sqrt{4+t^2}dt = \int \frac{1}{2}\sqrt{s}ds = \frac{1}{2} \frac{2}{3}s^{\frac{3}{2}}=\frac{1}{3}(4+t^2)^{\frac{3}{2}} = 0 $ (The limits doesn't even matter).
If a curve connects two different points, no matter what the curve is, the length is nonzero (this is one of the properties of a metric). If you do find a curve of length $0$, that would mean all the points on the curve are actually just one point, and so it's not really a curve so much as it is a point. Here, however, this indicates a mistake, as your curve clearly connects at least two different points. The mistake is likely that you forgot to take the absolute value.