UPSC 2018 Maths Optional Paper 1 Q7b — Step-by-Step Solution

12 marks · Section B

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

Question

Find the curvature and torsion of the curve $\vec r=a(u-\sin u)\hat i+a(1-\cos u)\hat j+bu\hat k$ .

Technique

$\kappa=\dfrac{|\vec r\,'\times\vec r\,''|}{|\vec r\,'|^3}$ , $\tau=\dfrac{(\vec r\,'\times\vec r\,'')\cdot\vec r\,'''}{|\vec r\,'\times\vec r\,''|^2}$ .

Solution

Step 1 — Derivatives

\vec r\,'=\big(a(1-\cos u),\ a\sin u,\ b\big),\qquad \vec r\,''=\big(a\sin u,\ a\cos u,\ 0\big),\qquad \vec r\,'''=\big(a\cos u,\ -a\sin u,\ 0\big).

Speed squared:

|\vec r\,'|^2=a^2(1-\cos u)^2+a^2\sin^2u+b^2=a^2\big(1-2\cos u+\cos^2u+\sin^2u\big)+b^2=2a^2(1-\cos u)+b^2.

Step 2 — Cross product and curvature

\vec r\,'\times\vec r\,''= \begin{vmatrix}\hat i&\hat j&\hat k\\ a(1-\cos u)&a\sin u&b\\ a\sin u&a\cos u&0\end{vmatrix} =\big(-ab\cos u,\ ab\sin u,\ a^2(1-\cos u)\cos u-a^2\sin^2u\big).

The $\hat k$ component simplifies: $a^2[(1-\cos u)\cos u-\sin^2u]=a^2[\cos u-\cos^2u-\sin^2u]=a^2(\cos u-1)$ . Then

|\vec r\,'\times\vec r\,''|^2=a^2b^2\cos^2u+a^2b^2\sin^2u+a^4(1-\cos u)^2=a^2b^2+a^4(1-\cos u)^2 =a^2\big[a^2(1-\cos u)^2+b^2\big].

Curvature $\kappa=\dfrac{|\vec r\,'\times\vec r\,''|}{|\vec r\,'|^3}$ :

\boxed{\;\kappa=\frac{a\sqrt{a^2(1-\cos u)^2+b^2}}{\big[2a^2(1-\cos u)+b^2\big]^{3/2}}.\;}

Step 3 — Torsion

\tau=\frac{(\vec r\,'\times\vec r\,'')\cdot\vec r\,'''}{|\vec r\,'\times\vec r\,''|^2}.

Scalar triple product (numerator):

(-ab\cos u,\ ab\sin u,\ a^2(\cos u-1))\cdot(a\cos u,\ -a\sin u,\ 0) =-a^2b\cos^2u-a^2b\sin^2u=-a^2b.

Therefore

Answer

\boxed{\;\tau=\frac{-a^2b}{a^2\big[a^2(1-\cos u)^2+b^2\big]}=\frac{-b}{a^2(1-\cos u)^2+b^2}.\;}