how to parametrize a curve in $\mathbb R^3$?

Question

How to parametrize $x^2+y^2+z^2=4$ and $x+z=2$

$x$, $y$ and $z$ should be function of $t$

I have tried to eliminate $z$ but it doesn't work

Answer 1

Your curve $\gamma$ is the intersection of the sphere $x^2+y^2+z^2=4$ with the plane $x+z=2$, hence a circle. Looking at a figure one immediately sees that $[2{\bf e}_1,2{\bf e_3}]$ is a diameter of the circle, hence ${\bf m}:=(1,0,1)$ is its center, and $\rho:=\sqrt{2}$ its radius. We now need two orthogonal unit vectors spanning the plane of the circle. From the same figure we see that ${\bf a}:=(0,1,0)$, $\>{\bf b}:=\bigl({1\over\sqrt{2}},0,-{1\over\sqrt{2}}\bigr)$ do the job. It follows that a parametric representation of $\gamma$ is given by $$\gamma:\quad t\mapsto{\bf r}(t):={\bf m}+\rho\cos t\>{\bf a}+\rho \sin t\>{\bf b}\>\qquad(0\leq t\leq 2\pi)\ ,$$ or in coordinates: $$\gamma:\qquad t\mapsto\left\{\eqalign{x(t)&:=1+\sin t\cr y(t)&:=\sqrt{2}\cos t\cr z(t)&:=1-\sin t\cr}\right.\qquad(0\leq t\leq 2\pi)\ .$$

Answer 2

From the second equation we get $z=2-x$. Plugging this into the equation of the sphere gives $$4=x^2+y^2+(2-x)^2=y^2+2x^2-4x+4,$$ or $$ 2=y^2+2x^2-4x+2=y^2+2(x-1)^2.\qquad(*) $$ In the $xy$-plane the equation $(*)$ defines an ellipse centered at $(x,y)=(1,0)$ (this is the projection of the curve of intersection into the $xy$-plane). We also see that the semiaxes of that ellipse are parallel to the coordinate axes. If their lengths are $a$ and $b$, then that ellipse has parametrization $$ x=1+a\cos t,\qquad y=b\sin t, $$ with $t$ ranging over the interval $[0,2\pi)$.

Leaving the tasks of finding $a$ and $b$ and writing $z$ as a function of $t$ to you.

Here's how Mathematica renders it.

Answer 3

As ajotatxe says in the comments, $x^2+y^2+z^2=4$ describes a surface (a sphere in fact), not a curve. So we cannot parametrize it with one variable, we need two variables. The natural coordinate system is spherical coordinates in this case, as the surface is a sphere. The two parameters would be $\theta\in[0,\pi]$ and $\phi\in[0,2\pi)$, where $$\begin{cases}x=r\sin(\theta)\cos(\phi)\\y=r\sin(\theta)\sin(\phi)\\z=r\cos(\theta)\end{cases}$$ note that $r$ is the radius of the sphere, which is constant (and in this case is $\sqrt 4=2$). $\theta$ is the inclination angle (in the $z$-direction) and $\phi$ is the azimuth angle, which measure the angle in the $xy$-plane (with the positive $x$ axis being an angle of $0$).

Answer 4

1)$ x^2+y^2+z^2 = 4;$ A sphere center $(0,0,0)$, radius $= 2$.

2) $x+z = 2$ , a plane.

Intersection of the two yields a curve, a circle.

Let $\vec r(t)$ be the curve, parameter $t$.

Normalized normal to the plane:

$\vec n =(1/√2)(1,0,1).$

Centre of the circle $ (1,0,1)$. Radius of the circle: $r_0 := √2$

$\vec r(t)$ lies in the plane, with normal $n$, passing through $(1,0,1):$

$ \vec n \cdot (\vec r - (1,0,1)) = 0;$

$\vec r(t) = (1,0,1) + \vec d(t).$

Direction vector $\vec d(t)$ is perpendicular to $\vec n:$

Two perpendicular vectors are:

$ \vec a:=(1/√2)(-1,0,1)$, and $\vec b:= (0,1,0)$, normalized.

Note:$\vec a$ is perpendicular to $\vec b$.

In analogy to the 2D circle with

$Y -$ and $X -$ axes:

$\vec d(t) = r_0(\sin(t) \vec a + \cos(t) \vec b)$.

$\vec r(t) = (1/√2)r_0\sin(t)(-1,0,1) +r_0 \cos(t)(0,1,0) +(1,0,1)$.

Recall: $r_0= √2$ is the radius of a circle with centre (1,0,1).