# Parametric curve on cylinder surface

Let $r(t)=(x(t),y(t),z(t)),t\geq0$ be a parametric curve with $r(0)$ lies on cylinder surface $x^2+2y^2=C$. Let the tangent vector of $r$ is $r'(t)=\left( 2y(t)(z(t)-1), -x(t)(z(t)-1), x(t)y(t)\right)$. Would you help me to show that :

(a) The curve always lies on ylinder surface $x^2+2y^2=C$.

(b) The curve $r(t)$ is periodic (we can find $T_0\neq0$ such that $r(T_0)=r(0)$).If we make the C smaller then the parametric curve $r(t)$ more closer to the origin (We can make a Neighboorhood that contain this parametric curve)

My effort:

(a) Let $V(x,y,z)=x^2+2y^2$. If $V(x,y,z)=C$ then $\frac{d}{dt} V(x,y,z)=0$. Since $\frac{d}{dt} V(x,y,z)=(2x,4y,0)\cdot (x'(t),y'(t),z'(t))=2x(2y(z-1))+4y(-x(z-1))=0$, then $r(0)$ would be parpendicular with normal of cylinder surface. Hence the tangent vector must be on the tangent plane of cylinder. So $r(t)$ must lie on cylinder surface.

(b) From $z'=xy$, I analyze the sign of $z'$ (in 1st quadrant z'>0 so the z component of $r(t)$ increasing and etc.) and conclude that if $r(t)$ never goes unbounded when move to another octan ( But I can't guarante $r(t)$ accros another octan.). I also consider the case when $(x=0, y>0), (x=0, y<0,z>1), (x>0, y=0,z>1$and so on) and draw the vector $r'(t)$.

Thank you so much of your help.

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We can reparameterize $S=\{(\sqrt{C}\cos u,\frac{\sqrt{C}}{\sqrt{2}}\sin u, v): u,v\in \mathbb{R}\}$ since $(\sqrt{C}\cos u)^2+2\left(\frac{\sqrt{C}}{\sqrt{2}}\sin u\right)^2=C$. Let $r(t)= (x(t),y(t),z(t))$ and $r(0)=(x_0,y_0,z_0)$. Define $V(x,y,z)=x^2+2y^2$. Since $V(x,y,z)=C$ then $\frac{dV}{dt}=0$. But, by chain rule we get $0=\frac{dV}{dt}=\nabla{V}\cdot(x',y',z')$ so the tangent vector of the parametrized curve that intersect $S$ in a point always parpendicular with $\nabla{V}$. Since $r(0)$ be in $S$ and $\nabla{V}$ parpendicular with the tangent plane of $S$ at $r(0)$ , then $r'(0)$ be on the tangent plane of $S$ at $r(0)$. By this argument, we can conclude that $r(t)$ must be on $S$. Since $S=\{(\sqrt{C}\cos u,\frac{\sqrt{C}}{\sqrt{2}}\sin u, v): u,v\in \mathbb{R}\}$ then $x(t)=\sqrt{C}\cos (t-t_0)$ and $y(t)=\frac{\sqrt{C}}{\sqrt{2}}\sin (t-t_0)$ with $t_0$ satisfying $x_0=\sqrt{C}\cos t_0$ and $y_0=-\frac{\sqrt{C}}{\sqrt{2}}\sin t_0$. Since $z'=xy$ then $z'(t)=\frac{C}{2\sqrt{2}}\sin(2t-t_0)$, hence $z(t)=-\frac{C}{4\sqrt{2}}\cos(2t-t_0)$. Since $r(2\pi)=(\sqrt{C}\cos (2\pi-t_0),\frac{\sqrt{C}}{\sqrt{2}}\sin (2\pi-t_0),-\frac{C}{4\sqrt{2}}\cos(2\pi-t_0))=(x_0,y_0,z_0)=r(0)$ then $r(t)$ is periodic.