UPSC 2014 Maths Optional Paper 1 Q5e — Step-by-Step Solution

10 marks · Section B

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

Question

Find the curvature vector at any point of the curve $\vec r(t)=t\cos t\,\hat i+t\sin t\,\hat j$ , $0\le t\le 2\pi$ . Give its magnitude also.

Technique

Planar curvature formula $\kappa=(x'y''-y'x'')/|\vec r'|^{3}$ ; curvature vector $\vec\kappa=\kappa\hat N$ with $\hat N$ being the inward-pointing unit normal.

Solution

This is a planar curve in the $xy$ -plane (no $\hat k$ -component) — an Archimedean spiral in polar form $r=t$ at angle $\theta=t$ .

Step 1 — First and second derivatives

\vec r'(t)=(\cos t-t\sin t)\hat i+(\sin t+t\cos t)\hat j.

\vec r''(t)=(-2\sin t-t\cos t)\hat i+(2\cos t-t\sin t)\hat j.

Step 2 — Speed

|\vec r'|^{2}=(\cos t-t\sin t)^{2}+(\sin t+t\cos t)^{2}.

Expand: $\cos^{2}t-2t\sin t\cos t+t^{2}\sin^{2}t+\sin^{2}t+2t\sin t\cos t+t^{2}\cos^{2}t=1+t^{2}$ .

So $|\vec r'|=\sqrt{1+t^{2}}$ .

Step 3 — Signed curvature for planar curve

The signed curvature (positive = turning CCW) for a planar curve is

\kappa=\dfrac{x'y''-y'x''}{(x'^{2}+y'^{2})^{3/2}}.

Compute the numerator $x'y''-y'x''$ :

$x'y''=(\cos t-t\sin t)(2\cos t-t\sin t)=2\cos^{2}t-3t\sin t\cos t+t^{2}\sin^{2}t$ .
$y'x''=(\sin t+t\cos t)(-2\sin t-t\cos t)=-2\sin^{2}t-3t\sin t\cos t-t^{2}\cos^{2}t$ .

Difference:

x'y''-y'x''=2(\cos^{2}t+\sin^{2}t)+t^{2}(\sin^{2}t+\cos^{2}t)=2+t^{2}.

So curvature:

\kappa=\dfrac{2+t^{2}}{(1+t^{2})^{3/2}}>0.

Step 4 — Curvature vector

The principal normal $\hat N$ (rotation of $\hat T$ by $+\pi/2$ for positive curvature):

\hat N=\dfrac{(-y',\,x')}{|\vec r'|}=\dfrac{(-(\sin t+t\cos t),\,\cos t-t\sin t)}{\sqrt{1+t^{2}}}.

Curvature vector:

\vec\kappa=\kappa\,\hat N=\dfrac{2+t^{2}}{(1+t^{2})^{3/2}}\cdot\dfrac{(-(\sin t+t\cos t),\,\cos t-t\sin t)}{\sqrt{1+t^{2}}}=\dfrac{2+t^{2}}{(1+t^{2})^{2}}\Bigl(-(\sin t+t\cos t),\,\cos t-t\sin t\Bigr).

\boxed{\;\vec\kappa(t)=\dfrac{2+t^{2}}{(1+t^{2})^{2}}\Bigl[-(\sin t+t\cos t)\,\hat i+(\cos t-t\sin t)\,\hat j\Bigr].\;}

Step 5 — Magnitude

Answer

\boxed{\;|\vec\kappa|=\kappa=\dfrac{2+t^{2}}{(1+t^{2})^{3/2}}.\;}