Question

An exercise gives the parameterized equation of a movement ( r(t)=x(t),y(t),z(t) ) ,and asks to prove that the movement is happening on one plane and that it is circular. How do I do that?

Answer 1

what are the functions:
\(x(t),y(t),z(t)\) ?

Answer 2

x=2+(cost/\sqrt(2))+(sint/\sqrt3), y=2-(cost/\sqrt2)+(sint/\sqrt3), z=1+(sint/\sqrt3)\]

Answer 3

a simple computation, shows that the subsequent condition holds:
\[x + y - 2z = 2\]
please check, by substitution.
Such equation, in the euclidean space, is the cartesian equation of a plane

Answer 4

if we differentiate the coordinates: \(x,y,z\), we get:
\[\left\{ \begin{gathered}
\dot x = \frac{{ - \sin t}}{{\sqrt 2 }} + \frac{{\cos t}}{{\sqrt 3 }} \hfill \\
\hfill \\
\dot y = \frac{{\sin t}}{{\sqrt 2 }} + \frac{{\cos t}}{{\sqrt 3 }} \hfill \\
\hfill \\
\dot z = \frac{{\cos t}}{{\sqrt 3 }} \hfill \\
\end{gathered} \right.\]

Answer 5

so the tangent vector \(T\), is:
\[T = \left( {\frac{{ - \sin t}}{{\sqrt 2 }} + \frac{{\cos t}}{{\sqrt 3 }},\;\;\frac{{\sin t}}{{\sqrt 2 }} + \frac{{\cos t}}{{\sqrt 3 }},\;\;\frac{{\cos t}}{{\sqrt 3 }}} \right)\]
furthermore, we have:
\[{\left( {ds} \right)^2} = {\left( {dx} \right)^2} + {\left( {dy} \right)^2} + {\left( {dz} \right)^2} = {\left( {dt} \right)^2},\quad \Rightarrow ds = dt\]

Answer 6

next, I apply the formulas of \(Differential\;\;Geometry\), and I get:
\[\frac{{dT}}{{ds}} = \frac{{dT}}{{dt}} = \dot T = \left( {\frac{{ - \cos t}}{{\sqrt 2 }} - \frac{{\sin t}}{{\sqrt 3 }},\;\;\frac{{\cos t}}{{\sqrt 2 }} - \frac{{\sin t}}{{\sqrt 3 }},\;\;\frac{{ - \sin t}}{{\sqrt 3 }}} \right)\]

Answer 7

and I can write this:
\[\left\| {\dot T} \right\| = 1,\quad \Rightarrow r = \frac{1}{{\left\| {\dot T} \right\|}} = 1\]
namely the radius of curvature of trajectory, is \(constant\)

Answer 8

Thanks!