let $g:R\rightarrow R^3$ be a curve such that for every $\textbf{t}$ the vector $g(\textbf{t})$ is orthogonal to the tanget line to the curve at the point
prove that the curve lies on a sphere centered at $(0,0)$.
help please heres what i did :
$$ g(t) = (x(t),y(t),z(t)) \\ g'(t) = (x' , y' , z') \\ <g(t),g'(t)> = 0 \\ xx' + yy' + zz' = 0 $$
now how to continue ? through integrals ?
Once you have $x(t)x'(t)+y(t)y'(t)+z(t)z'(t)=0$, youn only have to look for a Primitive.
$$0=\int_0^t 0ds=\int_0^t(x(s)x'(s)+y(s)y'(s)+z(s)z'(s))dt=\frac{1}{2}\left(x(t)^2+y(t)^2+z(t)^2\right)-C$$ where $C=\frac{1}{2}\left(x(0)^2+y(0)^2+z(0)^2\right)$ is some real constant.
If $C$ where negative, your curve wiil lie in the circle $x^2+y^2+z^2=2C$.