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UPSC 2022 Maths Optional Paper 1 Q5e — Step-by-Step Solution 10 marks · Section B
Curl: definition, physical meaning, computation · Vector Analysis · asked 4× in 13 yrs · Read the full method →
Question
Show that A ⃗ = ( 6 x y + z 3 ) ı ^ + ( 3 x 2 − z ) ȷ ^ + ( 3 x z 2 − y ) k ^ \vec A=(6xy+z^3)\hat\imath+(3x^2-z)\hat\jmath+(3xz^2-y)\hat k A = ( 6 x y + z 3 ) ^ + ( 3 x 2 − z ) ^ + ( 3 x z 2 − y ) k ^ ϕ \phi ϕ A ⃗ = ∇ ϕ \vec A=\nabla\phi A = ∇ ϕ
Technique
Curl test for irrotational; integrate ∂ ϕ / ∂ x = P \partial\phi/\partial x=P ∂ ϕ / ∂ x = P ϕ \phi ϕ ( y , z ) (y,z) ( y , z ) y y y Q Q Q g ( y , z ) g(y,z) g ( y , z ) z z z R R R
Solution
Let A ⃗ = ( P , Q , R ) \vec A=(P,Q,R) A = ( P , Q , R ) P = 6 x y + z 3 P=6xy+z^3 P = 6 x y + z 3 Q = 3 x 2 − z Q=3x^2-z Q = 3 x 2 − z R = 3 x z 2 − y R=3xz^2-y R = 3 x z 2 − y
Step 1 — Compute ∇ × A ⃗ \nabla\times\vec A ∇ × A
∇ × A ⃗ = ( ∂ R ∂ y − ∂ Q ∂ z , ∂ P ∂ z − ∂ R ∂ x , ∂ Q ∂ x − ∂ P ∂ y ) . \nabla\times\vec A=\!\left(\dfrac{\partial R}{\partial y}-\dfrac{\partial Q}{\partial z},\;\dfrac{\partial P}{\partial z}-\dfrac{\partial R}{\partial x},\;\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right). ∇ × A = ( ∂ y ∂ R − ∂ z ∂ Q , ∂ z ∂ P − ∂ x ∂ R , ∂ x ∂ Q − ∂ y ∂ P ) . Compute partials:
∂ R / ∂ y = − 1 \partial R/\partial y=-1 ∂ R / ∂ y = − 1 ∂ Q / ∂ z = − 1 \partial Q/\partial z=-1 ∂ Q / ∂ z = − 1 ∂ P / ∂ z = 3 z 2 \partial P/\partial z=3z^2 ∂ P / ∂ z = 3 z 2 ∂ R / ∂ x = 3 z 2 \partial R/\partial x=3z^2 ∂ R / ∂ x = 3 z 2 ∂ Q / ∂ x = 6 x \partial Q/\partial x=6x ∂ Q / ∂ x = 6 x ∂ P / ∂ y = 6 x \partial P/\partial y=6x ∂ P / ∂ y = 6 x
So:
ı ^ \hat\imath ^ − 1 − ( − 1 ) = 0 -1-(-1)=0 − 1 − ( − 1 ) = 0 ȷ ^ \hat\jmath ^ 3 z 2 − 3 z 2 = 0 3z^2-3z^2=0 3 z 2 − 3 z 2 = 0 k ^ \hat k k ^ 6 x − 6 x = 0 6x-6x=0 6 x − 6 x = 0
∇ × A ⃗ = ( 0 , 0 , 0 ) \nabla\times\vec A=(0,0,0) ∇ × A = ( 0 , 0 , 0 ) A ⃗ \vec A A
Step 2 — Find ϕ \phi ϕ ∇ ϕ = A ⃗ \nabla\phi=\vec A ∇ ϕ = A
∂ ϕ / ∂ x = P = 6 x y + z 3 \partial\phi/\partial x=P=6xy+z^3 ∂ ϕ / ∂ x = P = 6 x y + z 3 x x x
ϕ = 3 x 2 y + x z 3 + g ( y , z ) . \phi=3x^2 y+xz^3+g(y,z). ϕ = 3 x 2 y + x z 3 + g ( y , z ) . ∂ ϕ / ∂ y = 3 x 2 + ∂ g / ∂ y = Q = 3 x 2 − z \partial\phi/\partial y=3x^2+\partial g/\partial y=Q=3x^2-z ∂ ϕ / ∂ y = 3 x 2 + ∂ g / ∂ y = Q = 3 x 2 − z
So ∂ g / ∂ y = − z \partial g/\partial y=-z ∂ g / ∂ y = − z y y y
g ( y , z ) = − y z + h ( z ) . g(y,z)=-yz+h(z). g ( y , z ) = − y z + h ( z ) . ∂ ϕ / ∂ z = 3 x z 2 − y + h ′ ( z ) = R = 3 x z 2 − y \partial\phi/\partial z=3xz^2-y+h'(z)=R=3xz^2-y ∂ ϕ / ∂ z = 3 x z 2 − y + h ′ ( z ) = R = 3 x z 2 − y
So h ′ ( z ) = 0 h'(z)=0 h ′ ( z ) = 0 h ( z ) = C h(z)=C h ( z ) = C
Answer
ϕ = 3 x 2 y + x z 3 − y z + C . \boxed{\;\phi=3x^2 y+xz^3-yz+C.\;} ϕ = 3 x 2 y + x z 3 − y z + C .