Cartesian equation of $r(t) = 4\cos t \textbf{i} + 4\sin t \textbf{j} + 4 \cos t \textbf{k} $

Question

The curve described with $r(t) = 4\cos t \textbf{i} + 4\sin t \textbf{j} + 4 \cos t \textbf{k}$ is an ellipse and I would like to find it's Cartesian equation.

We get $x = z = 4\cos t, y = 4\sin t$. From that we have $t = \arccos\frac{x}{4}$. Then we have:

$y = 4\sin t = 4 \sin (\arccos\frac{x}{4}) \implies y^2 = 16 \sin ^2 (\arccos\frac{x}{4}) = 16 (1 - \cos ^2 (\arccos\frac{x}{4})) = 16 (1 - (\cos \arccos\frac{x}{4})) ^ 2 = 16 ( 1 - (\frac{x}{4})^2) = 16(1 - \frac{x^2}{16})=16 - x^2 \implies y^2 + x^2 = 16.$

That circle is the view of the ellipse when we project it in the $(x, y)$ plane, so we need to plug in $z$ somewhere to get the ellipse, but I'm not sure how to do that.

Thanks!

Answer 1

Notice that the parametrization $r$ defines a $1$-dimensional shape in $\mathbb{R}^3$, so we need two equations to describe it. You already found one, $$y^2+x^2=16$$

To finish, notice that, directly from the expression of $r$, one sees $x=z$.

Thus, the ellipse is given by the equations $$y^2+x^2=16 \\ x=z$$

It is the intersection of a cylinder and a plane.

Answer 2

Well you can add the first equation you find $$\begin{cases} x^2+y^2=16\\ x=z \end{cases}$$ And you get your ellipse $$\Gamma = \{(x,y,z) \in \Bbb R^3: x^2+y^2=16, x=z\}$$

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If you are searching for an equation like $f(x,y,z) = 0$ with $f:\Bbb R^3 \to \Bbb R$ you will not obtain a 3D curve but instead a 3D surface, for example take $f(x,y,z) = z +x^2+y^2-4$ plotting $f(x,y,z) = 0$ results in

Hope this helps!

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P.S. Notice that your curve could be expressed as the zeroes of $F: \Bbb R^3 \to \Bbb R^2$