Length Function, Parametrizing by Arc Length
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ASIDE: Note then, that ds/dt =||r'(t)|| so ds = ||r'(t)|| dt as in the definition of arc length!
In[41]:=
f[t_]:={Cos[t],Sin[t],t/3}
In[42]:=
fplot=ParametricPlot3D[f[t],{t,-4 Pi,4 Pi}];

In[43]:=
s[t_]:=arclength[f,-4 Pi,t]
Under suitable conditions, this length function has an inverse function, t(s).
In[44]:=
Solve[s[t]==s,t]
t[s_]=t /.%[[1]]
Out[44]=
4 Sqrt[10] Pi - 3 s
{{t -> -(-------------------)}}
Sqrt[10]
Out[45]=
4 Sqrt[10] Pi - 3 s
-(-------------------)
Sqrt[10]
Look at g(s)=f(t(s)), which traces out the same curve over suitable values of s. Since t ranges from -4Pi to 4Pi, s ranges from s(-4Pi) to s(4Pi). When a curve is parametrized in this manner, we say the curve is parametrized by arc length.
In[46]:=
g[s_]:=f[t[s]]
In[47]:=
g[s]
Out[47]=
4 Sqrt[10] Pi - 3 s 4 Sqrt[10] Pi - 3 s
{Cos[-------------------], -Sin[-------------------],
Sqrt[10] Sqrt[10]
-(4 Sqrt[10] Pi - 3 s)
----------------------}
3 Sqrt[10]
In[48]:=
Simplify[g[s]]
Out[48]=
3 s 3 s -4 Pi s
{Cos[--------], Sin[--------], ----- + --------}
Sqrt[10] Sqrt[10] 3 Sqrt[10]
In[49]:=
s[-4 Pi]
Out[49]=
0
Obvious, I hope!
In[50]:=
s[4 Pi]
Out[50]=
8 Sqrt[10] Pi
-------------
3
In[51]:=
gplot=ParametricPlot3D[g[s],{s,s[-4 Pi],s[4 Pi]}];

Is it the same plot?
In[52]:=
Show[GraphicsArray[{fplot,gplot}]];

What's so great about parametrizing a curve by arc length?
In[53]:=
g'[s]
Out[53]=
4 Sqrt[10] Pi - 3 s 4 Sqrt[10] Pi - 3 s
3 Sin[-------------------] 3 Cos[-------------------]
Sqrt[10] Sqrt[10]
{--------------------------, --------------------------,
Sqrt[10] Sqrt[10]
1
--------}
Sqrt[10]
In[54]:=
Simplify[g'[s]]
Out[54]=
3 s 3 s
-3 Sin[--------] 3 Cos[--------]
Sqrt[10] Sqrt[10] 1
{----------------, ---------------, --------}
Sqrt[10] Sqrt[10] Sqrt[10]
In[55]:=
Simplify[norm[g'[s]]]
Out[55]=
1
When a curve is parametrized by arc length, g'(s) always has unit length!
Up to Arc Length