Normal Frames
Normal Frames
Clear[x,y,f,T]
x[t_]:=t
y[t_]:=t^2
f[t_]:={x[t],y[t]}
curve=ParametricPlot[f[t],{t,-3,5}];
T[t_]:=f'[t]/Sqrt[f'[t].f'[t]]
Simplify[T[t]]
tanv=vectorPlot[Table[f[i],{i,5}],Table[f[i]+T[i],{i,5}]];
Show[curve,tanv,AspectRatio->Automatic,PlotRange->{0,40}];Let's look at the vector T'(t)/||T'(t)||.
Simplify[T'[t]/Sqrt[T'[t].T'[t]]]
Yuck! Why bother?
normv=vectorPlot[Table[f[i],{i,5}],
Table[f[i]+(T'[i]/Sqrt[T'[i].T'[i]]),{i,5}]];
Show[curve,tanv,normv,AspectRatio->Automatic];
Simplify[norm[T'[t]/Sqrt[T'[t].T'[t]]]]
Well, well, well. It seems that T'(t)/||T'(t)|| is a unit vector orthogonal to
the tangent vector, which means it is orthogonal (perpendicular) to the curve.
The vector T'(t)/||T'(t)|| is called the unit normal vector, N(t). The unit
normal vector can also be written as
N(t)=T'(t)/(curvature(t) ||F'(t)||). (Homework!)
If the curve is parametrized by arc length, the unit normal vector (as a
function of arc length) is simply
N(s)=(dT(s)/ds)/curvature(s). (Homework!)
Simplify[curvature[f,t]]
Simplify[T'[t]/Sqrt[T'[t].T'[t]]]
Simplify[T'[t]/(curvature[f,t](Sqrt[f'[t].f'[t]]))]
This is called a moving frame.
We have a unit tangent vector T(t) and a unit normal vector N(t). The vector
B(t)=T(t)xN(t) (cross product) is orthogonal to both T and N. It also a unit
vector (verify?). The vector B is called the binormal of the curve. The three
vectors T, N, and B form a right handed coordinate system at each point of the
curve.
Components of Acceleration
Clear[r,T,B,motion,nm]
r[t_]:={t^2,2-3t,t}
motion=ParametricPlot3D[r[t],{t,0,4},AxesLabel->{"x","y","z"}];
T[t_]=r'[t]/Sqrt[r'[t].r'[t]];
nm[t_]=T'[t]/Sqrt[T'[t].T'[t]];
B[t_]=Cross[T[t],nm[t]];
utp=vectorPlot[Table[r[i],{i,4}],Table[r[i]+T[i],{i,4}]];
unp=vectorPlot[Table[r[i],{i,4}],Table[r[i]+nm[i],{i,4}]];
bp=vectorPlot[Table[r[i],{i,4}],Table[r[i]+B[i],{i,4}]];
mvframe=Show[motion,utp,unp,bp];