I'm given the vector valued function (supposedly a circle) $r(t) = (3\cos t, 4\cos t, 5\sin t)$. However, I cannot see immediately how this is a circle. How do I verify that it is?
I also have a related question: how do I find the parametric equation of a circle in 3D in general?
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Another way to see this: notice that your curve lies completely in the plane $4x-3y=0$ and the sphere $x^2+y^2+z^2=25$. It doesn't take much more work to show that the curve is actually the complete intersection of the plane and the sphere, so it must be a circle!
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You can calculate the curvature and torsion of the curve. You will find out that $\kappa= \text{const}, \tau=0$. The only curves with constant curvature & torsion are helixes, and since $\tau=0$ the curve lies in a plane - thus it is a circle.
The most general circle in 3D space is isometric to the simple $r(t)=(R \cos t,R \sin t,0)$. That is there exists an orthogonal matrix $Q \in \mathbb R^{3 \times 3}$ and a vector $b \in \mathbb R^3$ such that $f(x)=Qx+b$ takes the general circle into the simple one.
It might help to write the function in the following way: $$ r(t) = 5\cos t \,\vec v_1 + 5\sin t \,\vec v_2 $$ Where $\vec v_1$ and $\vec v_2$ are the orthogonal unit vectors $$ \vec v_1=\left(\frac35,\frac45,0\right)\qquad \vec v_2 = (0,0,1) $$ For any choice of $\vec v_1$ and $\vec v_2$ that are of length one and perpendicular to each other (that is, orthogonal unit vectors), the above expression gives you a circle of radius $5$. This includes the usual unit circle, where $\vec v_1=(1,0,0)$ and $\vec v_2=(0,1,0)$.
But, supposing that you did not have this stroke of insight, you could always verify that this parametrizes a curve of constant distance origin by noting that for any $t$: $$ [x(t)]^2+[y(t)]^2+[z(t)]^2=5^2 $$ Or by noting that for any $t$, $$ r(t)\cdot r'(t)=0 $$ As Jared notes, if you then notice that the curve lies entirely in a single plane (in this case, $4x-3y=0$) you may deduce that the curve is some piece of a circle.