← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q7c — Step-by-Step Solution

15 marks · Section B

Serret-Frenet formulae · Vector Analysis · asked 2× in 13 yrs · Read the full method →

Question

If the tangent to a curve makes a constant angle θ\theta with a fixed line, then prove that the ratio of radius of torsion to radius of curvature is proportional to tan⁡θ\tan\theta. Further prove that if this ratio is constant, then the tangent makes a constant angle with a fixed direction.

Technique

Frenet–Serret formulas for both directions of the equivalence. Forward: differentiate the constant-angle condition twice. Reverse: construct d⃗=T⃗+(1/k)B⃗\vec d=\vec T+(1/k)\vec B as an explicit constant vector when τ/κ=k\tau/\kappa=k.

Solution

Let T⃗,N⃗,B⃗\vec T,\vec N,\vec B be the Frenet frame, κ\kappa curvature, τ\tau torsion, ρ=1/κ\rho=1/\kappa radius of curvature, σ=1/τ\sigma=1/\tau radius of torsion. Frenet–Serret: T⃗′=κN⃗, N⃗′=−κT⃗+τB⃗, B⃗′=−τN⃗\vec T'=\kappa\vec N,\,\vec N'=-\kappa\vec T+\tau\vec B,\,\vec B'=-\tau\vec N (primes = d/dsd/ds).

Part 1 — Constant angle ⟹ σ/ρ∝tan⁡θ\sigma/\rho\propto\tan\theta

Suppose the tangent makes a constant angle θ\theta with a fixed unit vector d⃗\vec d, i.e., T⃗⋅d⃗=cos⁡θ\vec T\cdot\vec d=\cos\theta (constant).

Differentiate w.r.t. ss:

T⃗′⋅d⃗=0⇒κN⃗⋅d⃗=0⇒N⃗⋅d⃗=0.\vec T'\cdot\vec d=0\Rightarrow\kappa\vec N\cdot\vec d=0\Rightarrow\vec N\cdot\vec d=0.

So d⃗⊥N⃗\vec d\perp\vec N. Therefore d⃗\vec d lies in the (T⃗,B⃗)(\vec T,\vec B)-plane:

d⃗=cos⁡θ T⃗+sθB⃗,sθ2=1−cos⁡2θ=sin⁡2θ  ⇒  sθ=±sin⁡θ.\vec d=\cos\theta\,\vec T+s_\theta\vec B,\quad s_\theta^2=1-\cos^2\theta=\sin^2\theta\;\Rightarrow\;s_\theta=\pm\sin\theta.

Take sθ=sin⁡θs_\theta=\sin\theta (sign convention).

Differentiate N⃗⋅d⃗=0\vec N\cdot\vec d=0:

N⃗′⋅d⃗=0⇒(−κT⃗+τB⃗)⋅d⃗=0.\vec N'\cdot\vec d=0\Rightarrow(-\kappa\vec T+\tau\vec B)\cdot\vec d=0.

Substitute d⃗=cos⁡θ T⃗+sin⁡θB⃗\vec d=\cos\theta\,\vec T+\sin\theta\vec B:

−κcos⁡θ+τsin⁡θ=0⇒τsin⁡θ=κcos⁡θ⇒τκ=cot⁡θ.-\kappa\cos\theta+\tau\sin\theta=0\Rightarrow\tau\sin\theta=\kappa\cos\theta\Rightarrow\frac{\tau}{\kappa}=\cot\theta.

Equivalently, σρ=κτ=tan⁡θ\dfrac{\sigma}{\rho}=\dfrac{\kappa}{\tau}=\tan\theta.

  σρ=tan⁡θ    (proportional to tan⁡θ).  \boxed{\;\frac{\sigma}{\rho}=\tan\theta\;\;(\text{proportional to }\tan\theta).\;}

Part 2 — Constant σ/ρ\sigma/\rho ⟹ tangent makes constant angle with a fixed direction

Suppose σ/ρ=\sigma/\rho= const, equivalently τ/κ=k\tau/\kappa=k for some constant kk. We construct a fixed direction d⃗\vec d with T⃗⋅d⃗=\vec T\cdot\vec d= const.

Try d⃗=T⃗+1kB⃗\vec d=\vec T+\dfrac{1}{k}\vec B (where 1/k=τ/κ1/k=\tau/\kappa).

Show d⃗\vec d is constant (independent of ss): differentiate.

d⃗′=T⃗′+1kB⃗′=κN⃗+1k(−τN⃗)=(κ−τk)N⃗=(κ−κ)N⃗=0⃗.\vec d'=\vec T'+\frac{1}{k}\vec B'=\kappa\vec N+\frac{1}{k}(-\tau\vec N)=\Bigl(\kappa-\frac{\tau}{k}\Bigr)\vec N=\Bigl(\kappa-\kappa\Bigr)\vec N=\vec 0.

(Using τ/k=τ⋅κ/τ=κ\tau/k=\tau\cdot\kappa/\tau=\kappa.)

So d⃗=T⃗+(1/k)B⃗\vec d=\vec T+(1/k)\vec B is constant. ✓

Compute T⃗⋅d⃗\vec T\cdot\vec d:

T⃗⋅d⃗=T⃗⋅T⃗+1kT⃗⋅B⃗=1+0=1.\vec T\cdot\vec d=\vec T\cdot\vec T+\frac{1}{k}\vec T\cdot\vec B=1+0=1.

Magnitude: ∣d⃗∣2=∣T⃗∣2+(1/k)2∣B⃗∣2=1+1/k2|\vec d|^2=|\vec T|^2+(1/k)^2|\vec B|^2=1+1/k^2.

So angle between T⃗\vec T and d⃗\vec d:

cos⁡Θ=T⃗⋅d⃗∣d⃗∣=11+1/k2=k1+k2=\cos\Theta=\frac{\vec T\cdot\vec d}{|\vec d|}=\frac{1}{\sqrt{1+1/k^2}}=\frac{k}{\sqrt{1+k^2}}=

constant.

Therefore the tangent makes a constant angle Θ\Theta with the fixed direction d⃗\vec d. ✓

Answer

  σ/ρ const ⇒ tangent makes constant angle with a fixed direction.  \boxed{\;\sigma/\rho\text{ const }\Rightarrow\text{ tangent makes constant angle with a fixed direction.}\;}
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