← 2026 Paper 1
UPSC 2026 Maths Optional Paper 1 Q6c — Step-by-Step Solution 15 marks · Section B
Curvature and torsion · Vector Analysis · asked 7× in 14 yrs · Read the full method →
Question
Determine T ^ , N ^ , B ^ , κ \hat{T}, \hat{N}, \hat{B}, \kappa T ^ , N ^ , B ^ , κ τ \tau τ y 2 = 4 a x y^2 = 4ax y 2 = 4 a x a a a T ^ , N ^ , B ^ , κ \hat{T}, \hat{N}, \hat{B}, \kappa T ^ , N ^ , B ^ , κ τ \tau τ
Technique
Parametrise the parabola as r ⃗ = ( a t 2 , 2 a t , 0 ) \vec r=(at^{2},\,2at,\,0) r = ( a t 2 , 2 a t , 0 ) Serret–Frenet apparatus in parametric form : T ^ = r ⃗ ′ ∣ r ⃗ ′ ∣ \hat T=\dfrac{\vec r\,'}{|\vec r\,'|} T ^ = ∣ r ′ ∣ r ′ B ^ = r ⃗ ′ × r ⃗ ′ ′ ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ \hat B=\dfrac{\vec r\,'\times\vec r\,''}{|\vec r\,'\times\vec r\,''|} B ^ = ∣ r ′ × r ′′ ∣ r ′ × r ′′ N ^ = B ^ × T ^ \hat N=\hat B\times\hat T N ^ = B ^ × T ^ κ = ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ ∣ r ⃗ ′ ∣ 3 \kappa=\dfrac{|\vec r\,'\times\vec r\,''|}{|\vec r\,'|^{3}} κ = ∣ r ′ ∣ 3 ∣ r ′ × r ′′ ∣ τ = [ r ⃗ ′ r ⃗ ′ ′ r ⃗ ′ ′ ′ ] ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ 2 \tau=\dfrac{[\vec r\,'\,\vec r\,''\,\vec r\,''']}{|\vec r\,'\times\vec r\,''|^{2}} τ = ∣ r ′ × r ′′ ∣ 2 [ r ′ r ′′ r ′′′ ] plane curve has τ = 0 \tau=0 τ = 0 , because B ^ \hat B B ^
Solution
Take a > 0 a>0 a > 0 z = 0 z=0 z = 0
Step 1 — Parametrisation.
The standard parametrisation of y 2 = 4 a x y^{2}=4ax y 2 = 4 a x x = a t 2 , y = 2 a t x=at^{2},\ y=2at x = a t 2 , y = 2 a t y 2 = 4 a 2 t 2 = 4 a ⋅ a t 2 = 4 a x y^{2}=4a^{2}t^{2}=4a\cdot at^{2}=4ax y 2 = 4 a 2 t 2 = 4 a ⋅ a t 2 = 4 a x z = 0 z=0 z = 0
r ⃗ ( t ) = a t 2 i ^ + 2 a t j ^ + 0 k ^ . \vec r(t)=at^{2}\,\hat i+2at\,\hat j+0\,\hat k . r ( t ) = a t 2 i ^ + 2 a t j ^ + 0 k ^ . Differentiate:
r ⃗ ′ = 2 a t i ^ + 2 a j ^ , r ⃗ ′ ′ = 2 a i ^ , r ⃗ ′ ′ ′ = 0 ⃗ . \vec r\,'=2at\,\hat i+2a\,\hat j,\qquad \vec r\,''=2a\,\hat i,\qquad \vec r\,'''=\vec 0 . r ′ = 2 a t i ^ + 2 a j ^ , r ′′ = 2 a i ^ , r ′′′ = 0 . Note r ⃗ ′ = 2 a ( t , 1 , 0 ) ≠ 0 ⃗ \vec r\,'=2a(t,1,0)\neq\vec 0 r ′ = 2 a ( t , 1 , 0 ) = 0 t t t t = 0 t=0 t = 0
∣ r ⃗ ′ ∣ = 2 a t 2 + 1 ( = d s d t ) . |\vec r\,'|=2a\sqrt{t^{2}+1}\quad\Bigl(=\tfrac{ds}{dt}\Bigr). ∣ r ′ ∣ = 2 a t 2 + 1 ( = d t d s ) . Step 2 — Unit tangent T ^ \hat T T ^
T ^ = r ⃗ ′ ∣ r ⃗ ′ ∣ = 2 a ( t i ^ + j ^ ) 2 a 1 + t 2 = t i ^ + j ^ 1 + t 2 . \hat T=\frac{\vec r\,'}{|\vec r\,'|}=\frac{2a(t\,\hat i+\hat j)}{2a\sqrt{1+t^{2}}}
=\frac{t\,\hat i+\hat j}{\sqrt{1+t^{2}}}. T ^ = ∣ r ′ ∣ r ′ = 2 a 1 + t 2 2 a ( t i ^ + j ^ ) = 1 + t 2 t i ^ + j ^ . Step 3 — Binormal B ^ \hat B B ^ κ \kappa κ
r ⃗ ′ × r ⃗ ′ ′ = ∣ i ^ j ^ k ^ 2 a t 2 a 0 2 a 0 0 ∣ = i ^ ( 0 − 0 ) − j ^ ( 0 − 0 ) + k ^ ( 0 − 4 a 2 ) = − 4 a 2 k ^ , \vec r\,'\times\vec r\,''=\begin{vmatrix}\hat i&\hat j&\hat k\\ 2at&2a&0\\ 2a&0&0\end{vmatrix}
=\hat i(0-0)-\hat j(0-0)+\hat k(0-4a^{2})=-4a^{2}\,\hat k, r ′ × r ′′ = i ^ 2 a t 2 a j ^ 2 a 0 k ^ 0 0 = i ^ ( 0 − 0 ) − j ^ ( 0 − 0 ) + k ^ ( 0 − 4 a 2 ) = − 4 a 2 k ^ , so ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ = 4 a 2 |\vec r\,'\times\vec r\,''|=4a^{2} ∣ r ′ × r ′′ ∣ = 4 a 2
B ^ = r ⃗ ′ × r ⃗ ′ ′ ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ = − 4 a 2 k ^ 4 a 2 = − k ^ . \hat B=\frac{\vec r\,'\times\vec r\,''}{|\vec r\,'\times\vec r\,''|}=\frac{-4a^{2}\hat k}{4a^{2}}=-\hat k . B ^ = ∣ r ′ × r ′′ ∣ r ′ × r ′′ = 4 a 2 − 4 a 2 k ^ = − k ^ . κ = ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ ∣ r ⃗ ′ ∣ 3 = 4 a 2 ( 2 a 1 + t 2 ) 3 = 4 a 2 8 a 3 ( 1 + t 2 ) 3 / 2 = 1 2 a ( 1 + t 2 ) 3 / 2 . \kappa=\frac{|\vec r\,'\times\vec r\,''|}{|\vec r\,'|^{3}}=\frac{4a^{2}}{\bigl(2a\sqrt{1+t^{2}}\bigr)^{3}}
=\frac{4a^{2}}{8a^{3}(1+t^{2})^{3/2}}=\frac{1}{2a\,(1+t^{2})^{3/2}} . κ = ∣ r ′ ∣ 3 ∣ r ′ × r ′′ ∣ = ( 2 a 1 + t 2 ) 3 4 a 2 = 8 a 3 ( 1 + t 2 ) 3/2 4 a 2 = 2 a ( 1 + t 2 ) 3/2 1 . Step 4 — Principal normal N ^ \hat N N ^
Since ( T ^ , N ^ , B ^ ) (\hat T,\hat N,\hat B) ( T ^ , N ^ , B ^ ) N ^ = B ^ × T ^ \hat N=\hat B\times\hat T N ^ = B ^ × T ^
N ^ = ( − k ^ ) × t i ^ + j ^ 1 + t 2 = − t ( k ^ × i ^ ) − ( k ^ × j ^ ) 1 + t 2 = − t j ^ + i ^ 1 + t 2 = i ^ − t j ^ 1 + t 2 . \hat N=(-\hat k)\times\frac{t\,\hat i+\hat j}{\sqrt{1+t^{2}}}
=\frac{-\,t(\hat k\times\hat i)-(\hat k\times\hat j)}{\sqrt{1+t^{2}}}
=\frac{-t\,\hat j+\hat i}{\sqrt{1+t^{2}}}=\frac{\hat i-t\,\hat j}{\sqrt{1+t^{2}}}. N ^ = ( − k ^ ) × 1 + t 2 t i ^ + j ^ = 1 + t 2 − t ( k ^ × i ^ ) − ( k ^ × j ^ ) = 1 + t 2 − t j ^ + i ^ = 1 + t 2 i ^ − t j ^ . Cross-check by the Frenet definition d T ^ d s = κ N ^ \dfrac{d\hat T}{ds}=\kappa\hat N d s d T ^ = κ N ^
d T ^ d t = d d t [ ( t , 1 , 0 ) 1 + t 2 ] = ( 1 , − t , 0 ) ( 1 + t 2 ) 3 / 2 , d T ^ d s = d T ^ / d t d s / d t = ( 1 , − t , 0 ) 2 a ( 1 + t 2 ) 2 = 1 2 a ( 1 + t 2 ) 3 / 2 ⏟ κ ⋅ ( 1 , − t , 0 ) 1 + t 2 ⏟ N ^ . ✓ \frac{d\hat T}{dt}=\frac{d}{dt}\left[\frac{(t,1,0)}{\sqrt{1+t^{2}}}\right]=\frac{(1,-t,0)}{(1+t^{2})^{3/2}},
\qquad \frac{d\hat T}{ds}=\frac{d\hat T/dt}{ds/dt}=\frac{(1,-t,0)}{2a(1+t^{2})^{2}}
=\underbrace{\frac{1}{2a(1+t^{2})^{3/2}}}_{\kappa}\cdot\underbrace{\frac{(1,-t,0)}{\sqrt{1+t^{2}}}}_{\hat N}. \checkmark d t d T ^ = d t d [ 1 + t 2 ( t , 1 , 0 ) ] = ( 1 + t 2 ) 3/2 ( 1 , − t , 0 ) , d s d T ^ = d s / d t d T ^ / d t = 2 a ( 1 + t 2 ) 2 ( 1 , − t , 0 ) = κ 2 a ( 1 + t 2 ) 3/2 1 ⋅ N ^ 1 + t 2 ( 1 , − t , 0 ) . ✓ Both κ \kappa κ N ^ \hat N N ^
Geometrically N ^ \hat N N ^ t = 1 t=1 t = 1 ( a , 2 a ) (a,2a) ( a , 2 a ) N ^ = 1 2 ( 1 , − 1 , 0 ) \hat N=\tfrac{1}{\sqrt2}(1,-1,0) N ^ = 2 1 ( 1 , − 1 , 0 )
Step 5 — Torsion τ \tau τ
τ = ( r ⃗ ′ × r ⃗ ′ ′ ) ⋅ r ⃗ ′ ′ ′ ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ 2 = ( − 4 a 2 k ^ ) ⋅ 0 ⃗ ( 4 a 2 ) 2 = 0. \tau=\frac{\bigl(\vec r\,'\times\vec r\,''\bigr)\cdot\vec r\,'''}{|\vec r\,'\times\vec r\,''|^{2}}
=\frac{(-4a^{2}\hat k)\cdot\vec 0}{(4a^{2})^{2}}=0 . τ = ∣ r ′ × r ′′ ∣ 2 ( r ′ × r ′′ ) ⋅ r ′′′ = ( 4 a 2 ) 2 ( − 4 a 2 k ^ ) ⋅ 0 = 0. More instructively — and this is the observation the examiner is looking for — the parabola is a plane curve . Its binormal B ^ = − k ^ \hat B=-\hat k B ^ = − k ^ constant vector, so by the third Serret–Frenet formula
d B ^ d s = − τ N ^ = 0 ⃗ ⟹ τ = 0 ( since N ^ ≠ 0 ⃗ ) . \frac{d\hat B}{ds}=-\tau\,\hat N=\vec 0\quad\Longrightarrow\quad \tau=0\ \ (\text{since }\hat N\neq\vec 0). d s d B ^ = − τ N ^ = 0 ⟹ τ = 0 ( since N ^ = 0 ) . Equivalently: torsion measures the rate at which the curve twists out of its osculating plane; here the osculating plane is the fixed plane z = 0 z=0 z = 0 Every plane curve has zero torsion.
Step 6 — Express the results in Cartesian form.
Using t = y 2 a t=\dfrac{y}{2a} t = 2 a y 1 + t 2 = y 2 + 4 a 2 2 a \sqrt{1+t^{2}}=\dfrac{\sqrt{y^{2}+4a^{2}}}{2a} 1 + t 2 = 2 a y 2 + 4 a 2
T ^ = y i ^ + 2 a j ^ y 2 + 4 a 2 , N ^ = 2 a i ^ − y j ^ y 2 + 4 a 2 , B ^ = − k ^ , \hat T=\frac{y\,\hat i+2a\,\hat j}{\sqrt{y^{2}+4a^{2}}},\qquad
\hat N=\frac{2a\,\hat i-y\,\hat j}{\sqrt{y^{2}+4a^{2}}},\qquad \hat B=-\hat k, T ^ = y 2 + 4 a 2 y i ^ + 2 a j ^ , N ^ = y 2 + 4 a 2 2 a i ^ − y j ^ , B ^ = − k ^ , κ = 1 2 a ⋅ 8 a 3 ( y 2 + 4 a 2 ) 3 / 2 = 4 a 2 ( y 2 + 4 a 2 ) 3 / 2 . \kappa=\frac{1}{2a}\cdot\frac{8a^{3}}{(y^{2}+4a^{2})^{3/2}}=\frac{4a^{2}}{\bigl(y^{2}+4a^{2}\bigr)^{3/2}} . κ = 2 a 1 ⋅ ( y 2 + 4 a 2 ) 3/2 8 a 3 = ( y 2 + 4 a 2 ) 3/2 4 a 2 . Since y 2 = 4 a x y^{2}=4ax y 2 = 4 a x y 2 + 4 a 2 = 4 a ( x + a ) y^{2}+4a^{2}=4a(x+a) y 2 + 4 a 2 = 4 a ( x + a )
κ = 4 a 2 [ 4 a ( x + a ) ] 3 / 2 = 4 a 2 8 a 3 / 2 ( x + a ) 3 / 2 = a 2 ( x + a ) 3 / 2 , ρ = 1 κ = 2 ( x + a ) 3 / 2 a . \kappa=\frac{4a^{2}}{\bigl[4a(x+a)\bigr]^{3/2}}=\frac{4a^{2}}{8a^{3/2}(x+a)^{3/2}}=\frac{\sqrt a}{2\,(x+a)^{3/2}},
\qquad \rho=\frac{1}{\kappa}=\frac{2(x+a)^{3/2}}{\sqrt a}. κ = [ 4 a ( x + a ) ] 3/2 4 a 2 = 8 a 3/2 ( x + a ) 3/2 4 a 2 = 2 ( x + a ) 3/2 a , ρ = κ 1 = a 2 ( x + a ) 3/2 . At the vertex x = 0 x=0 x = 0 ρ = 2 a \rho=2a ρ = 2 a
Answer
T ^ = t i ^ + j ^ 1 + t 2 = y i ^ + 2 a j ^ y 2 + 4 a 2 , N ^ = i ^ − t j ^ 1 + t 2 = 2 a i ^ − y j ^ y 2 + 4 a 2 , B ^ = − k ^ , κ = 1 2 a ( 1 + t 2 ) 3 / 2 = 4 a 2 ( y 2 + 4 a 2 ) 3 / 2 = a 2 ( x + a ) 3 / 2 ( ρ = 2 ( x + a ) 3 / 2 a ) , τ = 0 ( the parabola is a plane curve: B ^ is constant ) . \boxed{\;
\begin{aligned}
\hat T&=\frac{t\,\hat i+\hat j}{\sqrt{1+t^{2}}}=\frac{y\,\hat i+2a\,\hat j}{\sqrt{y^{2}+4a^{2}}},\qquad
\hat N=\frac{\hat i-t\,\hat j}{\sqrt{1+t^{2}}}=\frac{2a\,\hat i-y\,\hat j}{\sqrt{y^{2}+4a^{2}}},\qquad
\hat B=-\hat k,\\[2mm]
\kappa&=\frac{1}{2a(1+t^{2})^{3/2}}=\frac{4a^{2}}{(y^{2}+4a^{2})^{3/2}}=\frac{\sqrt a}{2(x+a)^{3/2}}
\quad\Bigl(\rho=\tfrac{2(x+a)^{3/2}}{\sqrt a}\Bigr),\\[2mm]
\tau&=0\quad(\text{the parabola is a plane curve: }\hat B\text{ is constant}).
\end{aligned}\;} T ^ κ τ = 1 + t 2 t i ^ + j ^ = y 2 + 4 a 2 y i ^ + 2 a j ^ , N ^ = 1 + t 2 i ^ − t j ^ = y 2 + 4 a 2 2 a i ^ − y j ^ , B ^ = − k ^ , = 2 a ( 1 + t 2 ) 3/2 1 = ( y 2 + 4 a 2 ) 3/2 4 a 2 = 2 ( x + a ) 3/2 a ( ρ = a 2 ( x + a ) 3/2 ) , = 0 ( the parabola is a plane curve: B ^ is constant ) .