Arc Length

If r(t)=x(t)i+y(t)j+z(t)k from a to b is a continuously differentiable curve, the length of the curve is given by Integral from a to b of ||r'(t)|| with respect to t. The idea behind this is that ||r'(t)|| is the length of a tiny piece of tangent line to the curve. Adding up these tiny pieces gives the arclength. In calculus, adding up tiny pieces is integrating! To make this seem reasonable, let's look at the following:

In[5]:=

  r[t_]:={Cos[t],Sin[t]}

In[6]:=

  ParametricPlot[r[t],{t,0,2 Pi},AspectRatio->Automatic]

  ParametricPlot::ppcom: 

     Function r[t] cannot be compiled; plotting will proceed

       with the uncompiled function.

Out[7]=

  -Graphics-

In[8]:=

  arclength=Integrate[norm2[D[r[t],t]],{t,0,2 Pi}]

Out[8]=

  2 Pi

I hope that was what you expected!

Let's make this a Mathematica function.

In[9]:=

  arclength2[r_,t_,a_,b_]:=
     Integrate[norm2[D[r[t],t]],{t,a,b}]

In[10]:=

  arclength3[r_,t_,a_,b_]:=
     Integrate[norm3[D[r[t],t]],{t,a,b}]

In[11]:=

  r[t_]:={2 Cos[t],2 Sin[t],t^2}

In[12]:=

  ParametricPlot3D[r[t],{t,0,2 Pi}]

  ParametricPlot3D::ppcom: 

     Function r[t] cannot be compiled; plotting will proceed

       with the uncompiled function.

Out[13]=

  -Graphics3D-

In[14]:=

  
  D[r[t],t]

Out[14]=

  {-2 Sin[t], 2 Cos[t], 2 t}

In[15]:=

  norm3[D[r[t],t]]

Out[15]=

          2           2           2
  Sqrt[4 t  + 4 Cos[t]  + 4 Sin[t] ]

In[16]:=

  Simplify[%]

Out[16]=

              2
  Sqrt[4 + 4 t ]

In[17]:=

  Integrate[Sqrt[4+4 t^2],{t,0,2 pi}]

Out[17]=

                    2
  2 pi Sqrt[1 + 4 pi ] + ArcSinh[2 pi]

Here I went through all the steps for you. You could just use:

In[18]:=

  
  arclength3[r,t,0,2 Pi]

Out[18]=

                    2
  2 Pi Sqrt[1 + 4 Pi ] + ArcSinh[2 Pi]

Up to Some Multivariable Calculus Ideas Part II