Show that a curve lying entirely in the [math]xy-[/math]plane has zero torsion.
First of all, let's think about this geometrically. The binormal vector is a unit vector perpendicular to the osculating plane, which for a plane curve is just the [math]xy-[/math]plane. Hence the binormal vector is constant (hmm...maybe something to think about there) and so [math]\vec{B}\,\,'\left(s\right)=\vec{0}[/math] meaning [math]\tau=0[/math].[br][br]Not convinced? We can approach this analytically too. Let [math]\vec{c}\left(t\right)=\left(x\left(t\right),y\left(t\right),0\right)[/math] be a (thrice) differentiable parameterization of a curve lying in the [math]xy-[/math]plane.[br]Then [math]\vec{c}\left(t\right)=\left(x\,'\left(t\right),y\,'\left(t\right),0\right),\vec{c}\,\,''\left(t\right)=\left(x\,''\left(t\right),y\,''\left(t\right),0\right),\text{ and }\vec{c}\,\,'''\left(t\right)=\left(x\,'''\left(t\right),y\,'''\left(t\right),0\right)[/math].[br]If we compute [math]\vec{c}\,\,'\times \vec{c}\,\,''[/math] we get the vector [math]\left(0,0,x\,'\left(t\right)y\,''\left(t\right)-y\,'\left(t\right)x\,''\left(t\right)\right)[/math]. Dotting that with [math]\vec{c}\,\,'''\left(t\right)[/math] we get [math]0[/math].[br][br]