← 2026 Paper 1
UPSC 2026 Maths Optional Paper 1 Q8c-ii — Step-by-Step Solution 10 marks · Section B
Differentiation of a vector function of a scalar variable · Vector Analysis · asked 5× in 14 yrs · Read the full method →
Question
If r ⃗ = a cos t i ^ + a sin t j ^ + b t k ^ \vec{r} = a\cos t\,\hat{i} + a\sin t\,\hat{j} + bt\,\hat{k} r = a cos t i ^ + a sin t j ^ + b t k ^
∣ r ⃗ ′ × r ⃗ ′ ′ ∣ 2 = a 2 ( a 2 + b 2 ) and [ r ⃗ ′ r ⃗ ′ ′ r ⃗ ′ ′ ′ ] = a 2 b . |\vec{r}\,' \times \vec{r}\,''|^2 = a^2(a^2 + b^2) \quad\text{and}\quad [\vec{r}\,'\;\vec{r}\,''\;\vec{r}\,'''] = a^2 b. ∣ r ′ × r ′′ ∣ 2 = a 2 ( a 2 + b 2 ) and [ r ′ r ′′ r ′′′ ] = a 2 b . Technique
Direct differentiation of the circular helix, one cross product and one 3 × 3 3\times3 3 × 3 along the third column , where two of the three entries are zero — that turns a nine-term expansion into a single 2 × 2 2\times2 2 × 2
Solution
Step 1 — The derivatives
r ⃗ = a cos t i ^ + a sin t j ^ + b t k ^ \vec r = a\cos t\,\hat i + a\sin t\,\hat j + bt\,\hat k r = a cos t i ^ + a sin t j ^ + b t k ^ r ⃗ ′ = − a sin t i ^ + a cos t j ^ + b k ^ \vec r\,' = -a\sin t\,\hat i + a\cos t\,\hat j + b\,\hat k r ′ = − a sin t i ^ + a cos t j ^ + b k ^ r ⃗ ′ ′ = − a cos t i ^ − a sin t j ^ + 0 k ^ \vec r\,'' = -a\cos t\,\hat i - a\sin t\,\hat j + 0\,\hat k r ′′ = − a cos t i ^ − a sin t j ^ + 0 k ^ r ⃗ ′ ′ ′ = a sin t i ^ − a cos t j ^ + 0 k ^ \vec r\,''' = \;\;\,a\sin t\,\hat i - a\cos t\,\hat j + 0\,\hat k r ′′′ = a sin t i ^ − a cos t j ^ + 0 k ^ (Useful to note in passing: ∣ r ⃗ ′ ∣ = a 2 sin 2 t + a 2 cos 2 t + b 2 = a 2 + b 2 |\vec r\,'| = \sqrt{a^2\sin^2 t + a^2\cos^2 t + b^2} = \sqrt{a^2+b^2} ∣ r ′ ∣ = a 2 sin 2 t + a 2 cos 2 t + b 2 = a 2 + b 2 t t t s = t a 2 + b 2 s = t\sqrt{a^2+b^2} s = t a 2 + b 2
Step 2 — The cross product r ⃗ ′ × r ⃗ ′ ′ \vec r\,'\times\vec r\,'' r ′ × r ′′
r ⃗ ′ × r ⃗ ′ ′ = ∣ i ^ j ^ k ^ − a sin t a cos t b − a cos t − a sin t 0 ∣ \vec r\,'\times\vec r\,'' = \begin{vmatrix}\hat i & \hat j & \hat k\\[2pt] -a\sin t & a\cos t & b\\[2pt] -a\cos t & -a\sin t & 0\end{vmatrix} r ′ × r ′′ = i ^ − a sin t − a cos t j ^ a cos t − a sin t k ^ b 0 = i ^ [ ( a cos t ) ( 0 ) − b ( − a sin t ) ] − j ^ [ ( − a sin t ) ( 0 ) − b ( − a cos t ) ] + k ^ [ ( − a sin t ) ( − a sin t ) − ( a cos t ) ( − a cos t ) ] = \hat i\Bigl[(a\cos t)(0) - b(-a\sin t)\Bigr] - \hat j\Bigl[(-a\sin t)(0) - b(-a\cos t)\Bigr] + \hat k\Bigl[(-a\sin t)(-a\sin t) - (a\cos t)(-a\cos t)\Bigr] = i ^ [ ( a cos t ) ( 0 ) − b ( − a sin t ) ] − j ^ [ ( − a sin t ) ( 0 ) − b ( − a cos t ) ] + k ^ [ ( − a sin t ) ( − a sin t ) − ( a cos t ) ( − a cos t ) ] = a b sin t i ^ − a b cos t j ^ + a 2 ( sin 2 t + cos 2 t ) k ^ = ab\sin t\,\hat i \;-\; ab\cos t\,\hat j \;+\; a^2\bigl(\sin^2 t + \cos^2 t\bigr)\hat k = ab sin t i ^ − ab cos t j ^ + a 2 ( sin 2 t + cos 2 t ) k ^ r ⃗ ′ × r ⃗ ′ ′ = a b sin t i ^ − a b cos t j ^ + a 2 k ^ \boxed{\;\vec r\,'\times\vec r\,'' = ab\sin t\,\hat i - ab\cos t\,\hat j + a^2\,\hat k\;} r ′ × r ′′ = ab sin t i ^ − ab cos t j ^ + a 2 k ^ Step 3 — First result
∣ r ⃗ ′ × r ⃗ ′ ′ ∣ 2 = a 2 b 2 sin 2 t + a 2 b 2 cos 2 t + a 4 = a 2 b 2 + a 4 = a 2 ( a 2 + b 2 ) . |\vec r\,'\times\vec r\,''|^2 = a^2b^2\sin^2 t + a^2b^2\cos^2 t + a^4 = a^2b^2 + a^4 = a^2\bigl(a^2 + b^2\bigr). ∣ r ′ × r ′′ ∣ 2 = a 2 b 2 sin 2 t + a 2 b 2 cos 2 t + a 4 = a 2 b 2 + a 4 = a 2 ( a 2 + b 2 ) . ∴ ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ 2 = a 2 ( a 2 + b 2 ) . ■ \therefore\quad |\vec r\,'\times\vec r\,''|^2 = a^2(a^2+b^2). \qquad\blacksquare ∴ ∣ r ′ × r ′′ ∣ 2 = a 2 ( a 2 + b 2 ) . ■ Step 4 — Second result: the scalar triple product
[ r ⃗ ′ r ⃗ ′ ′ r ⃗ ′ ′ ′ ] = ( r ⃗ ′ × r ⃗ ′ ′ ) ⋅ r ⃗ ′ ′ ′ = ∣ − a sin t a cos t b − a cos t − a sin t 0 a sin t − a cos t 0 ∣ . [\vec r\,'\;\vec r\,''\;\vec r\,'''] = \bigl(\vec r\,'\times\vec r\,''\bigr)\cdot\vec r\,''' =
\begin{vmatrix} -a\sin t & a\cos t & b\\[2pt] -a\cos t & -a\sin t & 0\\[2pt] a\sin t & -a\cos t & 0\end{vmatrix}. [ r ′ r ′′ r ′′′ ] = ( r ′ × r ′′ ) ⋅ r ′′′ = − a sin t − a cos t a sin t a cos t − a sin t − a cos t b 0 0 . Expand along the third column (only the entry b b b ( − 1 ) 1 + 3 = + 1 (-1)^{1+3}=+1 ( − 1 ) 1 + 3 = + 1
= b ∣ − a cos t − a sin t a sin t − a cos t ∣ = b [ ( − a cos t ) ( − a cos t ) − ( − a sin t ) ( a sin t ) ] = b\begin{vmatrix}-a\cos t & -a\sin t\\ a\sin t & -a\cos t\end{vmatrix}
= b\Bigl[(-a\cos t)(-a\cos t) - (-a\sin t)(a\sin t)\Bigr] = b − a cos t a sin t − a sin t − a cos t = b [ ( − a cos t ) ( − a cos t ) − ( − a sin t ) ( a sin t ) ] = b [ a 2 cos 2 t + a 2 sin 2 t ] = a 2 b . = b\Bigl[a^2\cos^2 t + a^2\sin^2 t\Bigr] = a^2 b . = b [ a 2 cos 2 t + a 2 sin 2 t ] = a 2 b . ∴ [ r ⃗ ′ r ⃗ ′ ′ r ⃗ ′ ′ ′ ] = a 2 b . ■ \therefore\quad [\vec r\,'\;\vec r\,''\;\vec r\,'''] = a^2 b. \qquad\blacksquare ∴ [ r ′ r ′′ r ′′′ ] = a 2 b . ■ (Equivalently, using Step 2 directly: ( a b sin t ) ( a sin t ) + ( − a b cos t ) ( − a cos t ) + a 2 ⋅ 0 = a 2 b ( sin 2 t + cos 2 t ) = a 2 b (ab\sin t)(a\sin t) + (-ab\cos t)(-a\cos t) + a^2\cdot 0 = a^2b(\sin^2t+\cos^2t) = a^2b ( ab sin t ) ( a sin t ) + ( − ab cos t ) ( − a cos t ) + a 2 ⋅ 0 = a 2 b ( sin 2 t + cos 2 t ) = a 2 b r ⃗ ′ ′ ′ = − ( r ⃗ ′ − b k ^ ) \vec r\,''' = -\bigl(\vec r\,' - b\hat k\bigr) r ′′′ = − ( r ′ − b k ^ ) − ( r ⃗ ′ × r ⃗ ′ ′ ) ⋅ r ⃗ ′ + b ( r ⃗ ′ × r ⃗ ′ ′ ) ⋅ k ^ = 0 + b ⋅ a 2 = a 2 b -\bigl(\vec r\,'\times\vec r\,''\bigr)\cdot\vec r\,' + b\bigl(\vec r\,'\times\vec r\,''\bigr)\cdot\hat k = 0 + b\cdot a^2 = a^2b − ( r ′ × r ′′ ) ⋅ r ′ + b ( r ′ × r ′′ ) ⋅ k ^ = 0 + b ⋅ a 2 = a 2 b
Step 5 — Why these two numbers are the ones asked for
The two quantities just computed are exactly the numerators of the curvature and torsion formulae for a curve given by a general parameter:
κ = ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ ∣ r ⃗ ′ ∣ 3 = a a 2 + b 2 ( a 2 + b 2 ) 3 / 2 = a a 2 + b 2 , τ = [ r ⃗ ′ r ⃗ ′ ′ r ⃗ ′ ′ ′ ] ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ 2 = a 2 b a 2 ( a 2 + b 2 ) = b a 2 + b 2 . \kappa = \frac{|\vec r\,'\times\vec r\,''|}{|\vec r\,'|^{3}} = \frac{a\sqrt{a^2+b^2}}{\bigl(a^2+b^2\bigr)^{3/2}} = \frac{a}{a^2+b^2},
\qquad
\tau = \frac{[\vec r\,'\;\vec r\,''\;\vec r\,''']}{|\vec r\,'\times\vec r\,''|^{2}} = \frac{a^2b}{a^2(a^2+b^2)} = \frac{b}{a^2+b^2}. κ = ∣ r ′ ∣ 3 ∣ r ′ × r ′′ ∣ = ( a 2 + b 2 ) 3/2 a a 2 + b 2 = a 2 + b 2 a , τ = ∣ r ′ × r ′′ ∣ 2 [ r ′ r ′′ r ′′′ ] = a 2 ( a 2 + b 2 ) a 2 b = a 2 + b 2 b . Both are constants , independent of t t t κ \kappa κ τ \tau τ is a circular helix). Note also τ κ = b a \dfrac{\tau}{\kappa} = \dfrac{b}{a} κ τ = a b k ^ \hat k k ^
Answer
r ⃗ ′ × r ⃗ ′ ′ = a b sin t i ^ − a b cos t j ^ + a 2 k ^ , ∣ r ⃗ ′ × r ⃗ ′ ′ ∣ 2 = a 2 ( a 2 + b 2 ) , [ r ⃗ ′ r ⃗ ′ ′ r ⃗ ′ ′ ′ ] = a 2 b . \boxed{\;\vec r\,'\times\vec r\,'' = ab\sin t\,\hat i - ab\cos t\,\hat j + a^{2}\hat k,\qquad
|\vec r\,'\times\vec r\,''|^{2} = a^{2}\bigl(a^{2}+b^{2}\bigr),\qquad
[\vec r\,'\;\vec r\,''\;\vec r\,'''] = a^{2}b.\;} r ′ × r ′′ = ab sin t i ^ − ab cos t j ^ + a 2 k ^ , ∣ r ′ × r ′′ ∣ 2 = a 2 ( a 2 + b 2 ) , [ r ′ r ′′ r ′′′ ] = a 2 b . ( hence κ = a a 2 + b 2 , τ = b a 2 + b 2 — both constant . ) \left(\text{hence } \kappa = \frac{a}{a^{2}+b^{2}},\quad \tau = \frac{b}{a^{2}+b^{2}}\ \text{— both constant}.\right) ( hence κ = a 2 + b 2 a , τ = a 2 + b 2 b — both constant . )