# Show that the trace of the curve lies on a sphere $S$ and a cylinder $C$

Let $\alpha: [0,2\pi] \rightarrow \mathbb{R}^3$ be the curve defined by $$\alpha(t)=(\cos^2(t)-1/2,\sin(t)\cos(t),\sin(t)).$$ Problem: Show that the trace of the curve lies on a sphere $S$ and on a cylinder $C$ with the axis of the cylinder as the $z$-axis.

Approach 1: I have found that the curvature $\kappa(t)=2\sqrt{\dfrac{3\cos(2t)+7}{(\cos(2t)+3)^3}}$. However, this is not constant.

Approach 2: Also, letting $\alpha(t)=(x(t),y(t),z(t))$, we have that $x^2(t)+y^2(t)+z^2(t)=1/4+\sin^2(t)$, which again is not constant.

However, it seems like the curve does lie on the unit sphere as I have plotted the curve on Wolfram. I am not sure whether the trace does actually lie on cylinder.

How should I go about approaching the problem and where am I making a mistake?

A curve lies on a sphere if its distance from a fixed point remains constant. If we optimistically assume that the centre of the sphere is the origin, we can just look at $\lVert \alpha(t) \rVert^2$, but it turns out if you do this calculation, it doesn't come out right. Instead, we don't like the look of the $1/2$ in the first coordinate, so try $\lVert \alpha(t)-(-1/2,0,0) \rVert^2$: $$\lVert \alpha(t)+(1/2,0,0) \rVert^2 = (\cos^2{t})^2+\sin^2{t}\cos^2{t}+\sin^2{t} = \cos^4{t} +(1-\cos^2{t})(1+\cos^2{t}) = 1$$ using difference of two squares. Thus $\alpha(t)$ lies on the sphere of radius $1$ with centre $(-1/2,0,0)$ If we hadn't been able to guess, we would have had to try subtracting a general point $(a,b,c)$ and trying to make the answer constant.
The curve will lie on a cylinder with centre on the $z$-axis if its distance from the $z$-axis is constant. The square of this is given by $$(\cos^2{t}-1/2)^2 + \sin^2{t}\cos^2{t} = \cos^4{t}-\cos^2{t}+ \frac{1}{4} + \cos^2{t}-\cos^4{t} = \frac{1}{4},$$ and as the question said, the curve lies on the cylinder centred on the $z$-axis with radius $\sqrt{1/4}=1/2$. And the intersection of the sphere and the cylinder turn out to be this curve: here's a Mathematica plot.