CircularPath::usage="A point P moves over the paraboloid z =
2
x
+
2
y
so that its projection to the plane XY runs through the circumference
2
x
+
2
y
= x, z = 0, at a constant angular velocity of ω rad/sec. Parameterize the trajectory of P as a function of time. Deduce that it is a flat curve and calculate its Frenet trihedron, its curvature and its torsion in each P. Obtain the normal and tangential acceleration of P at every moment. References: www2.caminos.upm.es; Personal.";
Development:
Trajectory:
In[]:=
ζ[t_]:=
1
2
Cos[ωt]+
1
2
,
1
2
Sin[ωt],0;
Is it a flat curve?
Let’s calculate your torsion:
In[]:=
χ=
Det[{ζ'[t],ζ''[t],ζ'''[t]}]
2
Norm[ζ'[t]ζ''[t]]
Out[]=
0
It’s a flat curve because its torsion is zero.
Curvature:
In[]:=
κ=
Norm[ζ'[t]ζ''[t]]
3
Norm[ζ'[t]]
//FullSimplify
2
3/2
(
2
Abs[Cos[tω]]
+
2
Abs[Sin[tω]]
)
=
2
3/2
(
2
[Cos[tω]]
+
2
[Sin[tω]]
)
;​​κ=2;
Normal acceleration:
In[]:=
ζ''[t]//FullSimplify

a
n
[t_]:=-
1
2
2
ω
Cos[tω],-
1
2
2
ω
Sin[tω],0;
It can be seen that the magnitude of normal acceleration is:
a
n
=
2
ω
r=
1
2
2
ω
;
Tangential acceleration:

a
t
=r

α
;

α
is the angular acceleration, but ω the magnitude of angular velocity is constant, then the magnitude of the angular acceleration α = 0 , then the magnitude of the tangential acceleration vector is zero
a
t
=0.
Frenet trihedron:
In[]:=
ResourceFunction["TangentVector"][ζ[t],t]
-
2
ωSin[tω]
2
ω
(5+3Cos[2tω])
,
2
2
ωCos[tω]
2
ω
(5+3Cos[2tω])
,0;​​T1[t_]:=-
2
Sin[ωt]
5+3Cos[2ωt]
,
2
2
Cos[ωt]
5+3Cos[2ωt]
,0;
In[]:=
ResourceFunction["NormalVector"][ζ[t],t]//FullSimplify
-
2
2
6
ω
Cos[tω]
2
ω
2
ω
(5+3Cos[2tω])
,-
2
6
ω
Sin[tω]
2
ω
2
ω
(5+3Cos[2tω])
,0;​​N1[t_]:=-
2
2
Cos[ωt]
5+3Cos[2ωt]
,-
2
Sin[ωt]
5+3Cos[2ωt]
,0;
In[]:=
ResourceFunction["BinormalVector"][ζ[t],t]//FullSimplify
0,0,
3
ω
6
ω
;​​B={0,0,1};
Animation (let’s take ω = 1):
In[]:=
SwatchLegend[{Blue,Green,Orange,Gray},{"Tangent vector","Normal vector","Binormal vector","Normal acceleration"}]​​Manipulateζ[t_]:=
1
2
Cos[t]+
1
2
,
1
2
Sin[t],0;T1[t_]:=-
2
Sin[t]
5+3Cos[2t]
,
2
2
Cos[t]
5+3Cos[2t]
,0;N1[t_]:=-
2
2
Cos[t]
5+3Cos[2t]
,-
2
Sin[t]
5+3Cos[2t]
,0;B={0,0,1};​​κ=2;
a
n
=
1
2
;
a
t
=0;χ=0;​​p=Grid{Point,Torsion,FrenetTrihedron(TNB),Curvature,NormalAccelerationAndMagnitude,TangentialAccelerationMagnitude},ζ[t1],0,{T1[t1],N1[t1],{0,0,1}},2,-
1
2
Cos[t1],-
1
2
Sin[t1],0,
1
2
,0,FrameAll,ItemSize->15,ItemStyle->12;​​ShowParametricPlot3D[ζ[t],{t,0,2π},PlotStyle->{Yellow,Thick}]/.Line[pts_,rest___]:>Tube[pts,0.02,rest],Graphics3D[{Yellow,Thick,Ball[ζ[t1],0.05]}],​​Graphics3DBlack,Thick,Ball
1
2
,0,0,0.05,​​Graphics3D[{Green,Point[ζ[t1]],Arrowheads[.03],Arrow[Tube[{ζ[t1],ζ[t1]+N1[t1]}]]}],​​Graphics3D[{Orange,Point[ζ[t1]],Arrowheads[.03],Arrow[Tube[{ζ[t1],ζ[t1]+B}]]}],​​Graphics3D[{Blue,Point[ζ[t1]],Arrowheads[.03],Arrow[Tube[{ζ[t1],ζ[t1]+T1[t1]}]]}],​​Graphics3DGray,Point[ζ[t1]],Arrowheads[.03],ArrowTubeζ[t1],ζ[t1]+-
1
2
Cos[t1],-
1
2
Sin[t1],0,ParametricPlot3D[{uCos[v],uSin[v],
2
u
},{v,0,2π},{u,0,2},PlotStyle->{Green,Opacity[0.2]},PerformanceGoal->"Quality"],AxesLabel->(Style[#,15,Blue]&/@{"X","Y","Z"}),AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRange{{-2,2},{-2,2},{-1,5}},ViewPoint{1.3,-2.4,2.},ImageSize700,Style["Circular path of a point:",Bold,Large],{{t1,0,"Value (t1)"},0,2π,0.00001},Delimiter,{{p,2,"Measurable Parameters"}},ControlPlacementTop
Out[]=
Tangent vector
Normal vector
Binormal vector
Normal acceleration
Out[]=
​
Circular path of a point:
Value (t1)
Measurable Parameters
Point
Torsion
FrenetTrihedronTNB
Curvature
NormalAccelerationAndMagnitude
TangentialAccelerationMagnitude
{0.0769478,0.266509,0}
0
{{-0.300432,-0.953803,0},{0.953803,-0.300432,0},{0,0,1}}
2
{0.423052,-0.266509,0},
1
2

0
Note that normal acceleration (gray vector) always points towards the center of the circumference, which is why it is also known as centripetal acceleration. The tangential velocity coincides with the tangent vector (in blue).