The ArcLength(C, interval) command computes the arc length of the curve C where the parameter varies over the range specified in interval.

The curve C can be specified as a Vector or as a Vector valued procedure.  This determines the returned object type.

If interval is a range, the function tries to determine a suitable variable name by using the components of C.  To do this, it checks all of the indeterminates of type name in the components of C and removes the ones which are determined to be constants.

If the resulting set has a single entry, that entry is the variable name.  If it has more than one entry, an error is raised.  If interval is an equation, the left side is used as the parameter name.

If either of the endpoints of the range specified by interval is of type complex(float), symbolic integration is not attempted. Instead, numeric integration is used to find the arc length.

If a coordinate system attribute is specified on C, it is interpreted in that coordinate system.  Otherwise, the curve is interpreted as a curve in the current default coordinate system.  If the two are not compatible, an error is raised.

The ArcLength(C, interval, inert) command returns the integral form of the arc length of the curve C over interval.

$\mathrm{with}\left(\mathrm{VectorCalculus}\right)\:$

$\mathrm{ArcLength}\left(⟨r\cos \left(t\right)\&comma;r\sin \left(t\right)⟩\&comma;t=0..2\mathrm{\pi }\right)\mathbf{assuming}0<r$

$2\mathrm{\pi }r$

$\mathrm{ArcLength}\left(\mathrm{PositionVector}\left(\left[\cos \left(t\right)\&comma;\sin \left(t\right)\&comma;t\right]\right)\&comma;0..6\mathrm{\pi }\right)$

$6\mathrm{\pi }\sqrt{2}$

$\mathrm{ArcLength}\left(⟨\cos \left(t\right)\&comma;\sin \left(t\right)\&comma;t⟩\&comma;0..6\mathrm{\pi }\&comma;'\mathrm{inert}'\right)$

${\int }_0^{6\mathrm{\pi }}\sqrt{1+{\sin \left(t\right)}^2+{\cos \left(t\right)}^2}\&DifferentialD;t$

$\mathrm{ArcLength}\left(t\mapsto ⟨t\&comma;t^2⟩\&comma;0..1\right)$

$\frac{\sqrt{5}}{2}+\frac{\mathrm{arcsinh}\left(2\right)}{4}$

$\mathrm{evalf}\left(\right)$

$1.478942857$

$\mathrm{ArcLength}\left(⟨t\&comma;t^2⟩\&comma;0...1.0\right)$

$1.478942858$

$\mathrm{SetCoordinates}\left('\mathrm{polar}'\right)$

$\mathrm{polar}$

$\mathrm{ArcLength}\left(⟨\exp \left(-t\right)\&comma;t⟩\&comma;t=0..\mathrm{\infty }\right)$

$\sqrt{2}$