← 2024 Paper 1
UPSC 2024 Maths Optional Paper 1 Q5c — Step-by-Step Solution 10 marks · Section B
Differentiation of a vector function of a scalar variable · Vector Analysis · asked 4× in 13 yrs · Read the full method →
Question
At any time t t t
α ˉ = t i ^ + ( t + 1 ) j ^ + ( 2 t + 1 ) k ^ , β ˉ = 2 t i ^ + ( 3 t − 1 ) j ^ + t k ^ , γ ˉ = i ^ + 3 t j ^ + k ^ . \bar\alpha=t\hat i+(t+1)\hat j+(2t+1)\hat k,\quad\bar\beta=2t\hat i+(3t-1)\hat j+t\hat k,\quad\bar\gamma=\hat i+3t\hat j+\hat k. α ˉ = t i ^ + ( t + 1 ) j ^ + ( 2 t + 1 ) k ^ , β ˉ = 2 t i ^ + ( 3 t − 1 ) j ^ + t k ^ , γ ˉ = i ^ + 3 t j ^ + k ^ . What is the rate of change of the vectorial area of the parallelogram with coterminous edges α ˉ \bar\alpha α ˉ γ ˉ \bar\gamma γ ˉ t = 1 t=1 t = 1
Technique
Cross product for vector area; scalar triple product (determinant) for volume; differentiate the resulting polynomials.
Solution
Part 1 — d d t ( α ˉ × γ ˉ ) \dfrac{d}{dt}(\bar\alpha\times\bar\gamma) d t d ( α ˉ × γ ˉ )
α ˉ × γ ˉ = ∣ i ^ j ^ k ^ t t + 1 2 t + 1 1 3 t 1 ∣ . \bar\alpha\times\bar\gamma=\begin{vmatrix}\hat i & \hat j & \hat k\\t & t+1 & 2t+1\\1 & 3t & 1\end{vmatrix}. α ˉ × γ ˉ = i ^ t 1 j ^ t + 1 3 t k ^ 2 t + 1 1 . i ^ -component: ( t + 1 ) ( 1 ) − ( 2 t + 1 ) ( 3 t ) = t + 1 − 6 t 2 − 3 t = − 6 t 2 − 2 t + 1. \hat i\text{-component:}\;(t+1)(1)-(2t+1)(3t)=t+1-6t^2-3t=-6t^2-2t+1. i ^ -component: ( t + 1 ) ( 1 ) − ( 2 t + 1 ) ( 3 t ) = t + 1 − 6 t 2 − 3 t = − 6 t 2 − 2 t + 1. j ^ -component: − [ t ( 1 ) − ( 2 t + 1 ) ( 1 ) ] = − ( t − 2 t − 1 ) = t + 1. \hat j\text{-component:}\;-[t(1)-(2t+1)(1)]=-(t-2t-1)=t+1. j ^ -component: − [ t ( 1 ) − ( 2 t + 1 ) ( 1 )] = − ( t − 2 t − 1 ) = t + 1. k ^ -component: t ( 3 t ) − ( t + 1 ) ( 1 ) = 3 t 2 − t − 1. \hat k\text{-component:}\;t(3t)-(t+1)(1)=3t^2-t-1. k ^ -component: t ( 3 t ) − ( t + 1 ) ( 1 ) = 3 t 2 − t − 1. So A ⃗ ( t ) = ( − 6 t 2 − 2 t + 1 , t + 1 , 3 t 2 − t − 1 ) \vec A(t)=(-6t^2-2t+1,\;t+1,\;3t^2-t-1) A ( t ) = ( − 6 t 2 − 2 t + 1 , t + 1 , 3 t 2 − t − 1 )
d A ⃗ d t = ( − 12 t − 2 , 1 , 6 t − 1 ) . \frac{d\vec A}{dt}=(-12t-2,\;1,\;6t-1). d t d A = ( − 12 t − 2 , 1 , 6 t − 1 ) . Part 2 — d V d t ∣ t = 1 \dfrac{dV}{dt}\Big|_{t=1} d t d V t = 1
V ( t ) = det ( t t + 1 2 t + 1 2 t 3 t − 1 t 1 3 t 1 ) . V(t)=\det\begin{pmatrix}t & t+1 & 2t+1\\2t & 3t-1 & t\\1 & 3t & 1\end{pmatrix}. V ( t ) = det t 2 t 1 t + 1 3 t − 1 3 t 2 t + 1 t 1 . Expanding along the first row (cofactors C 11 , C 12 , C 13 C_{11},C_{12},C_{13} C 11 , C 12 , C 13
C 11 = t [ ( 3 t − 1 ) ( 1 ) − t ( 3 t ) ] = − 3 t 3 + 3 t 2 − t . C_{11}=t[(3t-1)(1)-t(3t)]=-3t^3+3t^2-t. C 11 = t [( 3 t − 1 ) ( 1 ) − t ( 3 t )] = − 3 t 3 + 3 t 2 − t . C 12 = − ( t + 1 ) [ 2 t ( 1 ) − t ( 1 ) ] = − ( t + 1 ) ( t ) = − t 2 − t . C_{12}=-(t+1)[2t(1)-t(1)]=-(t+1)(t)=-t^2-t. C 12 = − ( t + 1 ) [ 2 t ( 1 ) − t ( 1 )] = − ( t + 1 ) ( t ) = − t 2 − t . C 13 = ( 2 t + 1 ) [ 2 t ( 3 t ) − ( 3 t − 1 ) ( 1 ) ] = ( 2 t + 1 ) ( 6 t 2 − 3 t + 1 ) = 12 t 3 − t + 1. C_{13}=(2t+1)[2t(3t)-(3t-1)(1)]=(2t+1)(6t^2-3t+1)=12t^3-t+1. C 13 = ( 2 t + 1 ) [ 2 t ( 3 t ) − ( 3 t − 1 ) ( 1 )] = ( 2 t + 1 ) ( 6 t 2 − 3 t + 1 ) = 12 t 3 − t + 1. V ( t ) = − 3 t 3 + 3 t 2 − t − t 2 − t + 12 t 3 − t + 1 = 9 t 3 + 2 t 2 − 3 t + 1. V(t)=-3t^3+3t^2-t-t^2-t+12t^3-t+1=9t^3+2t^2-3t+1. V ( t ) = − 3 t 3 + 3 t 2 − t − t 2 − t + 12 t 3 − t + 1 = 9 t 3 + 2 t 2 − 3 t + 1. d V d t = 27 t 2 + 4 t − 3 , d V d t ∣ t = 1 = 27 + 4 − 3 = 28. \frac{dV}{dt}=27t^2+4t-3,\qquad\left.\frac{dV}{dt}\right|_{t=1}=27+4-3=28. d t d V = 27 t 2 + 4 t − 3 , d t d V t = 1 = 27 + 4 − 3 = 28. Answer
d A ⃗ d t = ( − 12 t − 2 , 1 , 6 t − 1 ) . \boxed{\;\frac{d\vec A}{dt}=(-12t-2,\;1,\;6t-1).\;} d t d A = ( − 12 t − 2 , 1 , 6 t − 1 ) . d V d t ∣ t = 1 = 28. \boxed{\;\left.\frac{dV}{dt}\right|_{t=1}=28.\;} d t d V t = 1 = 28.