← 2017 Paper 1
UPSC 2017 Maths Optional Paper 1 Q7a — Step-by-Step Solution 16 marks · Section B
Curvature and torsion · Vector Analysis · asked 6× in 13 yrs · Read the full method →
Question
Find the curvature vector and its magnitude at any point r ⃗ = ( θ ) \vec r=(\theta) r = ( θ ) r ⃗ = ( a cos θ , a sin θ , a θ ) \vec r=(a\cos\theta,\ a\sin\theta,\ a\theta) r = ( a cos θ , a sin θ , a θ ) x 2 + y 2 − z 2 = a 2 x^2+y^2-z^2=a^2 x 2 + y 2 − z 2 = a 2
Technique
κ ⃗ = d T ⃗ / d s \vec\kappa=d\vec T/ds κ = d T / d s d s / d θ = a 2 ds/d\theta=a\sqrt2 d s / d θ = a 2 λ = − ( r ⃗ ⋅ r ⃗ ′ ) / ∣ r ⃗ ′ ∣ 2 \lambda=-(\vec r\cdot\vec r\,')/|\vec r\,'|^2 λ = − ( r ⋅ r ′ ) /∣ r ′ ∣ 2
Solution
Step 1 — Tangent and arc length
r ⃗ = ( a cos θ , a sin θ , a θ ) , r ⃗ ′ = ( − a sin θ , a cos θ , a ) . \vec r=(a\cos\theta,\ a\sin\theta,\ a\theta),\qquad \vec r\,'=(-a\sin\theta,\ a\cos\theta,\ a). r = ( a cos θ , a sin θ , a θ ) , r ′ = ( − a sin θ , a cos θ , a ) . ∣ r ⃗ ′ ∣ 2 = a 2 sin 2 θ + a 2 cos 2 θ + a 2 = 2 a 2 ⇒ d s d θ = a 2 ( constant ) . |\vec r\,'|^2=a^2\sin^2\theta+a^2\cos^2\theta+a^2=2a^2\ \Rightarrow\ \frac{ds}{d\theta}=a\sqrt2\ (\text{constant}). ∣ r ′ ∣ 2 = a 2 sin 2 θ + a 2 cos 2 θ + a 2 = 2 a 2 ⇒ d θ d s = a 2 ( constant ) . Unit tangent
T ⃗ = r ⃗ ′ ∣ r ⃗ ′ ∣ = 1 2 ( − sin θ , cos θ , 1 ) . \vec T=\frac{\vec r\,'}{|\vec r\,'|}=\frac{1}{\sqrt2}(-\sin\theta,\ \cos\theta,\ 1). T = ∣ r ′ ∣ r ′ = 2 1 ( − sin θ , cos θ , 1 ) . Step 2 — Curvature vector d T ⃗ d s \dfrac{d\vec T}{ds} d s d T
The curvature vector is κ ⃗ = d T ⃗ d s = d T ⃗ / d θ d s / d θ \vec\kappa=\dfrac{d\vec T}{ds}=\dfrac{d\vec T/d\theta}{ds/d\theta} κ = d s d T = d s / d θ d T / d θ
d T ⃗ d θ = 1 2 ( − cos θ , − sin θ , 0 ) , κ ⃗ = 1 a 2 ⋅ 1 2 ( − cos θ , − sin θ , 0 ) . \frac{d\vec T}{d\theta}=\frac{1}{\sqrt2}(-\cos\theta,\ -\sin\theta,\ 0),\qquad
\vec\kappa=\frac{1}{a\sqrt2}\cdot\frac{1}{\sqrt2}(-\cos\theta,\ -\sin\theta,\ 0). d θ d T = 2 1 ( − cos θ , − sin θ , 0 ) , κ = a 2 1 ⋅ 2 1 ( − cos θ , − sin θ , 0 ) . κ ⃗ = d T ⃗ d s = − 1 2 a ( cos θ , sin θ , 0 ) , ∣ κ ⃗ ∣ = κ = 1 2 a . \boxed{\;\vec\kappa=\frac{d\vec T}{ds}=-\frac{1}{2a}\big(\cos\theta,\ \sin\theta,\ 0\big),\qquad |\vec\kappa|=\kappa=\frac{1}{2a}.\;} κ = d s d T = − 2 a 1 ( cos θ , sin θ , 0 ) , ∣ κ ∣ = κ = 2 a 1 . The curvature is the constant 1 2 a \dfrac{1}{2a} 2 a 1
The tangent line at parameter θ \theta θ X ⃗ ( λ ) = r ⃗ + λ r ⃗ ′ \vec X(\lambda)=\vec r+\lambda\vec r\,' X ( λ ) = r + λ r ′ Q Q Q X ⃗ ⋅ r ⃗ ′ = 0 \vec X\cdot\vec r\,'=0 X ⋅ r ′ = 0 Q Q Q
( r ⃗ + λ r ⃗ ′ ) ⋅ r ⃗ ′ = 0 ⇒ λ = − r ⃗ ⋅ r ⃗ ′ ∣ r ⃗ ′ ∣ 2 . (\vec r+\lambda\vec r\,')\cdot\vec r\,'=0\ \Rightarrow\ \lambda=-\frac{\vec r\cdot\vec r\,'}{|\vec r\,'|^2}. ( r + λ r ′ ) ⋅ r ′ = 0 ⇒ λ = − ∣ r ′ ∣ 2 r ⋅ r ′ . r ⃗ ⋅ r ⃗ ′ = ( a cos θ ) ( − a sin θ ) + ( a sin θ ) ( a cos θ ) + ( a θ ) ( a ) = a 2 θ , ∣ r ⃗ ′ ∣ 2 = 2 a 2 , \vec r\cdot\vec r\,'=(a\cos\theta)(-a\sin\theta)+(a\sin\theta)(a\cos\theta)+(a\theta)(a)=a^2\theta,\qquad |\vec r\,'|^2=2a^2, r ⋅ r ′ = ( a cos θ ) ( − a sin θ ) + ( a sin θ ) ( a cos θ ) + ( a θ ) ( a ) = a 2 θ , ∣ r ′ ∣ 2 = 2 a 2 , so λ = − θ 2 \lambda=-\dfrac{\theta}{2} λ = − 2 θ
Q = r ⃗ − θ 2 r ⃗ ′ = ( a cos θ + θ 2 a sin θ , a sin θ − θ 2 a cos θ , a θ − θ 2 a ) , Q=\vec r-\frac{\theta}{2}\vec r\,'=\Big(a\cos\theta+\tfrac{\theta}{2}a\sin\theta,\ a\sin\theta-\tfrac{\theta}{2}a\cos\theta,\ a\theta-\tfrac{\theta}{2}a\Big), Q = r − 2 θ r ′ = ( a cos θ + 2 θ a sin θ , a sin θ − 2 θ a cos θ , a θ − 2 θ a ) , Q = ( a 2 ( 2 cos θ + θ sin θ ) , a 2 ( 2 sin θ − θ cos θ ) , a θ 2 ) . Q=\Big(\tfrac{a}{2}(2\cos\theta+\theta\sin\theta),\ \tfrac{a}{2}(2\sin\theta-\theta\cos\theta),\ \tfrac{a\theta}{2}\Big). Q = ( 2 a ( 2 cos θ + θ sin θ ) , 2 a ( 2 sin θ − θ cos θ ) , 2 a θ ) . Step 4 — Show Q Q Q x 2 + y 2 − z 2 = a 2 x^2+y^2-z^2=a^2 x 2 + y 2 − z 2 = a 2
x 2 + y 2 = a 2 4 [ ( 2 cos θ + θ sin θ ) 2 + ( 2 sin θ − θ cos θ ) 2 ] . x^2+y^2=\frac{a^2}{4}\Big[(2\cos\theta+\theta\sin\theta)^2+(2\sin\theta-\theta\cos\theta)^2\Big]. x 2 + y 2 = 4 a 2 [ ( 2 cos θ + θ sin θ ) 2 + ( 2 sin θ − θ cos θ ) 2 ] . Expand: ( 2 cos θ + θ sin θ ) 2 + ( 2 sin θ − θ cos θ ) 2 = 4 cos 2 θ + 4 θ sin θ cos θ + θ 2 sin 2 θ + 4 sin 2 θ − 4 θ sin θ cos θ + θ 2 cos 2 θ = 4 + θ 2 . (2\cos\theta+\theta\sin\theta)^2+(2\sin\theta-\theta\cos\theta)^2=4\cos^2\theta+4\theta\sin\theta\cos\theta+\theta^2\sin^2\theta+4\sin^2\theta-4\theta\sin\theta\cos\theta+\theta^2\cos^2\theta=4+\theta^2. ( 2 cos θ + θ sin θ ) 2 + ( 2 sin θ − θ cos θ ) 2 = 4 cos 2 θ + 4 θ sin θ cos θ + θ 2 sin 2 θ + 4 sin 2 θ − 4 θ sin θ cos θ + θ 2 cos 2 θ = 4 + θ 2 .
x 2 + y 2 = a 2 4 ( 4 + θ 2 ) = a 2 + a 2 θ 2 4 , z 2 = a 2 θ 2 4 . x^2+y^2=\frac{a^2}{4}(4+\theta^2)=a^2+\frac{a^2\theta^2}{4},\qquad z^2=\frac{a^2\theta^2}{4}. x 2 + y 2 = 4 a 2 ( 4 + θ 2 ) = a 2 + 4 a 2 θ 2 , z 2 = 4 a 2 θ 2 . x 2 + y 2 − z 2 = a 2 + a 2 θ 2 4 − a 2 θ 2 4 = a 2 . x^2+y^2-z^2=a^2+\frac{a^2\theta^2}{4}-\frac{a^2\theta^2}{4}=a^2. x 2 + y 2 − z 2 = a 2 + 4 a 2 θ 2 − 4 a 2 θ 2 = a 2 . Answer
x 2 + y 2 − z 2 = a 2 for all θ , \boxed{\;x^2+y^2-z^2=a^2\quad\text{for all }\theta,\;} x 2 + y 2 − z 2 = a 2 for all θ ,