Question

Let f be a smooth function. Calculate the curvature and the torsion of the curve that is the intersection of x = y and z = f(x).

Answer 1

Solution:

Formula of calculation curvature (for x=y):

$k=\frac{x'z''-z'x''}{(x'^{2}+z'^{2})};$ here $z= f(x); \space and \space z'=\frac{df(x)}{dx}; \space z''=\frac{d^{2}f(x)}{dx^{2}};$

so, $x'=1 \space and \space x''=0;$ $k=\frac{z''}{(1+z'^{2})};$

For torsion:

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

$x''=0\space and\space y=0;$

So, $\tau=0.$

Answer:

$k=\frac{z''}{(1+z'^{2})};$

$\tau=0.$