therightwomanwf

2022-07-03

What is the easiest way to see that the path
$\underset{_}{r}:\mathbb{R}\to {\mathbb{R}}^{3}:t↦\left(\mathrm{sin}\phantom{\rule{thinmathspace}{0ex}}t,\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}t,\mathrm{cos}\phantom{\rule{thinmathspace}{0ex}}t\right)$
traces out an ellipse in the plane $y=z$ ?

Lana Schwartz

Expert

Step 1
You are almost there. The rotation by $\pi /4$ will yield the path
$\left({x}^{\prime },{y}^{\prime },{z}^{\prime }\right)=\left(\mathrm{sin}t,\sqrt{2}\mathrm{cos}t,0\right)$
in the new coordinates. How do you eliminate t? Use
${\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1$
${x}^{\prime 2}+\frac{{y}^{\prime 2}}{2}=1$

Palmosigx

Expert

Your path is also contsined in ${x}^{2}+{y}^{2}=1$ , which is a cylinder. So your path is in the intersection of this cylinder and the plane $y=z$ . Geometrically, this is an ellipse, since the plane is not paralel to the axis of the cyclinder, so it cuts all generatrices.