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).
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."
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