Does the curve $c(t)=\langle \sqrt{1-t^2}\cos t,\sqrt{1-t^2}\sin t,t\rangle$ lie on unit sphere?

Question

Given curve $c(t)=\langle \sqrt{1-t^2}\cos t,\sqrt{1-t^2}\sin t,t\rangle$ and $|t|\le 1$ does the curve lie on a sphere which has radius of $1$ and is centered at $(0,0,0)$?

I thought that: $$ x=\sqrt{1-t^2}\cos t\implies\cos t=\frac{x}{\sqrt{1-t^2}}\\ y=\sqrt{1-t^2}\sin t\implies\sin t=\frac{y}{\sqrt{1-t^2}} $$ but I have no idea what to do with $z$ coordinate.

Answer 1

Hint The equation for a unit sphere of radius $1$ centered at the origin has the Cartesian equation:

$$x^2+y^2+z^2=1$$

Can you show that your parameterization satisfies this equation?

Answer 2

$$|c(t)|=\sqrt{(\sqrt{1-t^2})^2+t^2}=1$$ using that $\cos^2{t}+\sin^2{t}=1$.

So $c(t)$ lies on the unit sphere.