Normal Frames

We have already seen that T(t) is a unit vector tangent to the curve F(t).

In[88]:=

  Clear[x,y,f,T]
  x[t_]:=t
  y[t_]:=t^2
  f[t_]:={x[t],y[t]}

In[89]:=

  curve=ParametricPlot[f[t],{t,-3,5}];

In[90]:=

  T[t_]:=f'[t]/Sqrt[f'[t].f'[t]]

In[91]:=

  Simplify[T[t]]

Out[91]=

         1              2 t
  {--------------, --------------}
               2               2
   Sqrt[1 + 4 t ]  Sqrt[1 + 4 t ]

In[92]:=

  tanv=vectorPlot[Table[f[i],{i,5}],Table[f[i]+T[i],{i,5}]];

In[93]:=

  Show[curve,tanv,AspectRatio->Automatic,PlotRange->{0,40}];

Let's look at the vector T'(t)/||T'(t)||.

In[94]:=

  Simplify[T'[t]/Sqrt[T'[t].T'[t]]]

Out[94]=

                     2 -2              2
  {-2 t Sqrt[(1 + 4 t )  ] Sqrt[1 + 4 t ], 
   
                 2 -2              2
    Sqrt[(1 + 4 t )  ] Sqrt[1 + 4 t ]}

Yuck! Why bother?

In[95]:=

  normv=vectorPlot[Table[f[i],{i,5}],
  Table[f[i]+(T'[i]/Sqrt[T'[i].T'[i]]),{i,5}]];

In[96]:=

  Show[curve,tanv,normv,AspectRatio->Automatic];

In[97]:=

  Simplify[norm[T'[t]/Sqrt[T'[t].T'[t]]]]

Out[97]=

  1

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!)

In[98]:=

  Simplify[curvature[f,t]]

Out[98]=

                 2 -2
  2 Sqrt[(1 + 4 t )  ]
  --------------------
                 2
     Sqrt[1 + 4 t ]

In[99]:=

  Simplify[T'[t]/Sqrt[T'[t].T'[t]]]

Out[99]=

                     2 -2              2
  {-2 t Sqrt[(1 + 4 t )  ] Sqrt[1 + 4 t ], 
   
                 2 -2              2
    Sqrt[(1 + 4 t )  ] Sqrt[1 + 4 t ]}

In[100]:=

  Simplify[T'[t]/(curvature[f,t](Sqrt[f'[t].f'[t]]))]

Out[100]=

                     2 -2              2
  {-2 t Sqrt[(1 + 4 t )  ] Sqrt[1 + 4 t ], 
   
                 2 -2              2
    Sqrt[(1 + 4 t )  ] Sqrt[1 + 4 t ]}

Components of Acceleration

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.

In[111]:=

  Clear[r,T,B,motion,nm]
  r[t_]:={t^2,2-3t,t}
  motion=ParametricPlot3D[r[t],{t,0,4},AxesLabel->{"x","y","z"}];

In[112]:=

  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]];

In[113]:=

  utp=vectorPlot[Table[r[i],{i,4}],Table[r[i]+T[i],{i,4}]];

In[114]:=

  unp=vectorPlot[Table[r[i],{i,4}],Table[r[i]+nm[i],{i,4}]];

In[115]:=

  bp=vectorPlot[Table[r[i],{i,4}],Table[r[i]+B[i],{i,4}]];

In[116]:=

  mvframe=Show[motion,utp,unp,bp];

This is called a moving frame.

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