Consider the curve $C\in\mathbb R^3$ defined in Cartesian coordinates, by the equations $$x=\sqrt{4-y^2-z^2}\qquad y=z+1$$ from $P_0=(\sqrt3,1,0)$ to $P_1=\left(\frac{\sqrt6}2,\frac32,\frac12\right)$. Define a parametrization of the curve $C$.
My answer was $$(x,y,z)=(\sqrt{4-(t+1)^2-t^2},t+1,t)\qquad t\in\mathbb R$$ However the solution is $t\in[0,\frac{1}{2}]$, which I can't understand why…
You haven't taken into account $C$'s endpoints. With the formula you found, you must now find $t$ such that $C(t)$ gives $P_0$ and $P_1$, and those values will be the bounds on $t$ for $C$. In this case, looking at $z$ immediately gives us the correct bounds of $\left[0,\frac12\right]$.