Answer to Question #149787 in Differential Geometry | Topology for Haider

Question

Show that. The curvature and torsion of the straight line is zero.

Answer 1

Consider straight line: $y=cx+d$

Curvature of a twice differentiable function $y=f(x)$ is given by $k=\frac{|y''|}{(1+(y')^2)^\frac{3}{2}}$

$y'=(cx+d)'=c$

$y''=(c)'=0$

$k=\frac{|0|}{(1+c^2)^\frac{3}{2}}=0$

Let $r(t)=\{x_0+p_1t,y_0+p_2t,z_0+p_3 t\}$  be the parametric equation of straight line. Then we can find the torsion using following formula:

$k_2=\frac{x'''(y'z''-y''z')+y'''(x''z'-x'z'')+z'''(x'y''-x''y')}{(y'z''-y''z')^2+(x''z'-x'z'')^2+(x'y''-x''y')^2}$

then:

$k_2=\frac{0\cdot(p_2\cdot0-0\cdot p_3)+0\cdot(0\cdot p_3-p_1\cdot 0)+0\cdot(p_1\cdot 0-0\cdot p_2)}{(p_2\cdot0-0\cdot p_3)^2+(0\cdot p_3-p_1\cdot 0)^2+p_1\cdot 0-0\cdot p_2')^2}= \frac{0}{0}$ i.e. undefined