Why does parameterizing a curve with its length yield $|g(T,T)|=1$?

Question

Let $l(t)=\int^t_{t_0}|T|(t')dt'$, where $T$ is a tangent vector to some curve $C(t)$.

Why does setting this function as a parameterization of the curve $C$, hence letting $l(t)=\psi C(t)$, imply $|T|^2=|g(T,T)|=1$? I can't seem to figure it out.

Thanks.

Answer 1

Instead of letting $l(t)=\psi C(t)$, what was meant was having $l(t)=t$. Then the result follows simply from $t=\int^t_{t_0}1dt'+t_0$.