How could I calculate T such that the length of curve C is equal to 1?

Question

Where $C$ is parameterized by $s(t) := (tcos(t),tsin(t), \frac{2\sqrt{2}}{3}t^\frac{3}{2})$ with $0 \leq t \leq T$.

Answer 1

Note that the length $S$ of a parametrized curve $s(t)$ is given by $$ S=\int_C \sqrt{x'(t)^2+y'(t)^2+z'(t)^2}dt$$ with the integration being along the curve $C$ and $s(t)=(x(t),y(t),z(t))$, given by your expressions above. In your case we have (after some math and given that t>0): $$S=\int_0^T \sqrt{(1+t)^2} dt=\int_0^T (1+t)dt=T+T^2/2$$ which means that if you want $S=1$ we get $T=-1\pm \sqrt{3}$.

Since T is also positive, we finally get $T=-1+ \sqrt{3}\simeq 0.732051$