UPSC 2024 Maths Optional Paper 1 Q5e-i — Step-by-Step Solution

5 marks · Section B

Curvature and torsion · Vector Analysis · asked 7× in 14 yrs · Read the full method →

Question

Let $C$ be a plane curve $\bar r(t)=f(t)\hat i+g(t)\hat j$ , where $f$ and $g$ have second-order derivatives. Show that the curvature at a point is

\kappa=\frac{|f'(t)g''(t)-g'(t)f''(t)|}{([f'(t)]^2+[g'(t)]^2)^{3/2}}.

What is the value of torsion $\tau$ at any point of this curve?

Curvature. Embed the plane curve in $\mathbb R^3$ : $\bar r(t)=(f,g,0)$ .

\bar r'=(f',g',0),\quad \bar r''=(f'',g'',0).

\bar r'\times\bar r''=(0,\,0,\,f'g''-g'f''),\quad |\bar r'\times\bar r''|=|f'g''-g'f''|,\quad |\bar r'|=\sqrt{f'^2+g'^2}.

\kappa=\frac{|\bar r'\times\bar r''|}{|\bar r'|^3}=\frac{|f'g''-g'f''|}{(f'^2+g'^2)^{3/2}}.

Torsion. Using $\tau=\dfrac{(\bar r'\times\bar r'')\cdot\bar r'''}{|\bar r'\times\bar r''|^2}$ and $\bar r'''=(f''',g''',0)$ :

(\bar r'\times\bar r'')\cdot\bar r'''=(0,0,f'g''-g'f'')\cdot(f''',g''',0)=0.

Therefore $\tau=0$ at every point. (This is the general fact: a curve lies in a plane iff $\tau\equiv 0$ .)

\boxed{\;\kappa=\frac{|f'g''-g'f''|}{(f'^2+g'^2)^{3/2}};\qquad\tau=0.\;}