How to change the parametric equations of a hypersurface in $V_N$ to another form... This exercise was given in the first pages of Synge & Schild Tensor Calculus.
The parametric equations of a hypersurface in $V_N$ are
$x^1=a\cos{u}$,
$x^2 = a\sin{u^1}\cos{u^2}$,
$x^3 = a\sin{u^1}\sin{u^2}\cos{u^3}$,
$\cdot\hspace{0.2cm}\cdot\hspace{0.2cm}\cdot\hspace{0.2cm}\cdot\hspace{0.2cm}\cdot\hspace{0.2cm}\cdot$
$x^{N-1}= a\sin{u^1}\sin{u^2}\sin{u^3}\dots\sin{u^{N-2}}\cos{u^{N-1}}$,
$x^N=a\sin{u^1}\sin{u^2}\sin{u^3}\dots\sin{u^{N-2}}\sin{u^{N-1}}$,
where $a$ is a constant. Find the single equation of the hypersurface in the form $$F(x^1,x^2,x^3,\dots,x^N)=0.$$
The number of parameters is one less than the number of equations, so how to eliminate them to give the equation in the form above?
Sorry if this is too easy. I just can't see it.
 A: $f^i(u^1,u^2,...,u^{N-1})=x^i$
I think I'll attempt to make a collection of answers for the exercises. From what I could discern, finding the length of the parametric when treating it as a vector instead of a coordinate allows for a dramatic simplification of these terms. 
Factoring,
We have $(x^1)^2+(x^2)^2+ ... +(x^N)^2$=
$a^2[{cos}^2(u^1)+{sin}^2(u^1){cos}^2(u^2)+{sin}^2(u^1){sin}^2(u^2){cos}^2(u^3)+ ...]\\$
Within the brackets, in all of the terms except the first, a $sin^2(u^1)$ can be factored out, getting:
${cos}^2(u^1)+{sin}^2(u^1)[{cos}^2(u^2)+{sin}^2(u^2){cos}^2(u^3)+ ... ]\\$
You can keep repeating this with every $sin^2(u^i)$, nesting this process within each previous one, until you get to the innermost expression:
$\\{sin}^2(u^{N-1})+{cos}^2(u^{N-1})$ which reduces to one, resulting in each outer expression doing the same until you have:
$(x^1)^2+(x^2)^2+ ... +(x^N)^2=a^2$, showing the every value of the parametric is $a$ distance from the origin.
We can designate the function F to be
$F(x^1,x^2,...x^N)=\sum_{i=1}^{N} (x^i)^2-a^2$ or 
$\\(x^i)^2-a^2$ for brevity. This means that every point not on the parametric is less than or greater than $a^2$ in distance squared, so plugging in their coordinates into $F$ would give a value other than $0$. 
(by now, it's clear that $f$ is a hypersphere due to its constant distance from the origin)
Finally, we see that  $F(\frac12a,0,...,0)=-\frac34a^2$ and $F(0,0,...,2a)=3a^2$
One vector is greater in magnitude than the function, and one is less. Therfor the two points lie in opposite regions divided by $f$.
