UPSC 2019 Maths Optional Paper 1 Q7b — Step-by-Step Solution

15 marks · Section B

Curvature and torsion · Vector Analysis · asked 6× in 13 yrs · Read the full method →

Question

Find the radius of curvature and radius of torsion of the helix $x=a\cos u,\ y=a\sin u,\ z=au\tan\alpha$ .

Technique

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

Solution

Let $\vec r(u)=\big(a\cos u,\ a\sin u,\ au\tan\alpha\big)$ .

Step 1 — Derivatives

\vec r\,'=(-a\sin u,\ a\cos u,\ a\tan\alpha),\qquad \vec r\,''=(-a\cos u,\ -a\sin u,\ 0),\qquad \vec r\,'''=(a\sin u,\ -a\cos u,\ 0).

Speed:

|\vec r\,'|=\sqrt{a^2\sin^2u+a^2\cos^2u+a^2\tan^2\alpha}=a\sqrt{1+\tan^2\alpha}=a\sec\alpha.

Step 2 — Curvature

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

|\vec r\,'\times\vec r\,''|=a^2\sqrt{\tan^2\alpha\sin^2u+\tan^2\alpha\cos^2u+1}=a^2\sqrt{1+\tan^2\alpha}=a^2\sec\alpha.

Curvature:

\kappa=\frac{|\vec r\,'\times\vec r\,''|}{|\vec r\,'|^3}=\frac{a^2\sec\alpha}{(a\sec\alpha)^3}=\frac{\cos^2\alpha}{a}.

Radius of curvature:

\boxed{\;\rho=\frac1\kappa=\frac{a}{\cos^2\alpha}=a\sec^2\alpha.\;}

Step 3 — Torsion

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

Numerator (scalar triple product):

(a^2\tan\alpha\sin u,\ -a^2\tan\alpha\cos u,\ a^2)\cdot(a\sin u,\ -a\cos u,\ 0) =a^3\tan\alpha\sin^2u+a^3\tan\alpha\cos^2u+0=a^3\tan\alpha.

The two trigonometric terms add (note the matching minus signs in the $\hat j$ components), giving the constant

(\vec r\,'\times\vec r\,'')\cdot\vec r\,'''=a^3\tan\alpha.

Denominator: $|\vec r\,'\times\vec r\,''|^2=a^4\sec^2\alpha$ . Hence

\tau=\frac{a^3\tan\alpha}{a^4\sec^2\alpha}=\frac{\tan\alpha\cos^2\alpha}{a}=\frac{\sin\alpha\cos\alpha}{a}=\frac{\sin2\alpha}{2a}.

Radius of torsion:

Answer

\boxed{\;\sigma=\frac1\tau=\frac{a}{\sin\alpha\cos\alpha}=\frac{2a}{\sin2\alpha}=a\sec^2\alpha\cot\alpha.\;}