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UPSC 2017 Maths Optional Paper 1 Q5d — 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
For what values of the constants a , b a,b a , b c c c V ⃗ = ( x + y + a z ) i ^ + ( b x + 2 y − z ) j ^ + ( − x + c y + 2 z ) k ^ \vec V=(x+y+az)\hat i+(bx+2y-z)\hat j+(-x+cy+2z)\hat k V = ( x + y + a z ) i ^ + ( b x + 2 y − z ) j ^ + ( − x + cy + 2 z ) k ^
Technique
Set ∇ × V ⃗ = 0 ⃗ \nabla\times\vec V=\vec0 ∇ × V = 0
Solution
Step 1 — Irrotational condition ∇ × V ⃗ = 0 ⃗ \nabla\times\vec V=\vec 0 ∇ × V = 0
With V 1 = x + y + a z , V 2 = b x + 2 y − z , V 3 = − x + c y + 2 z V_1=x+y+az,\ V_2=bx+2y-z,\ V_3=-x+cy+2z V 1 = x + y + a z , V 2 = b x + 2 y − z , V 3 = − x + cy + 2 z
∇ × V ⃗ = ( ∂ V 3 ∂ y − ∂ V 2 ∂ z , ∂ V 1 ∂ z − ∂ V 3 ∂ x , ∂ V 2 ∂ x − ∂ V 1 ∂ y ) . \nabla\times\vec V=\left(\frac{\partial V_3}{\partial y}-\frac{\partial V_2}{\partial z},\ \frac{\partial V_1}{\partial z}-\frac{\partial V_3}{\partial x},\ \frac{\partial V_2}{\partial x}-\frac{\partial V_1}{\partial y}\right). ∇ × V = ( ∂ y ∂ V 3 − ∂ z ∂ V 2 , ∂ z ∂ V 1 − ∂ x ∂ V 3 , ∂ x ∂ V 2 − ∂ y ∂ V 1 ) . Compute each component:
i ^ : ∂ V 3 ∂ y − ∂ V 2 ∂ z = c − ( − 1 ) = c + 1 , \hat i:\ \frac{\partial V_3}{\partial y}-\frac{\partial V_2}{\partial z}=c-(-1)=c+1, i ^ : ∂ y ∂ V 3 − ∂ z ∂ V 2 = c − ( − 1 ) = c + 1 , j ^ : ∂ V 1 ∂ z − ∂ V 3 ∂ x = a − ( − 1 ) = a + 1 , \hat j:\ \frac{\partial V_1}{\partial z}-\frac{\partial V_3}{\partial x}=a-(-1)=a+1, j ^ : ∂ z ∂ V 1 − ∂ x ∂ V 3 = a − ( − 1 ) = a + 1 , k ^ : ∂ V 2 ∂ x − ∂ V 1 ∂ y = b − 1. \hat k:\ \frac{\partial V_2}{\partial x}-\frac{\partial V_1}{\partial y}=b-1. k ^ : ∂ x ∂ V 2 − ∂ y ∂ V 1 = b − 1. Setting all three to zero:
a = − 1 , b = 1 , c = − 1. \boxed{\,a=-1,\quad b=1,\quad c=-1.\,} a = − 1 , b = 1 , c = − 1. Step 2 — The field with these values, and its divergence (Cartesian)
V ⃗ = ( x + y − z ) i ^ + ( x + 2 y − z ) j ^ + ( − x − y + 2 z ) k ^ . \vec V=(x+y-z)\hat i+(x+2y-z)\hat j+(-x-y+2z)\hat k. V = ( x + y − z ) i ^ + ( x + 2 y − z ) j ^ + ( − x − y + 2 z ) k ^ . ∇ ⋅ V ⃗ = ∂ ∂ x ( x + y − z ) + ∂ ∂ y ( x + 2 y − z ) + ∂ ∂ z ( − x − y + 2 z ) = 1 + 2 + 2 = 5. \nabla\cdot\vec V=\frac{\partial}{\partial x}(x+y-z)+\frac{\partial}{\partial y}(x+2y-z)+\frac{\partial}{\partial z}(-x-y+2z)=1+2+2=5. ∇ ⋅ V = ∂ x ∂ ( x + y − z ) + ∂ y ∂ ( x + 2 y − z ) + ∂ z ∂ ( − x − y + 2 z ) = 1 + 2 + 2 = 5. The divergence is the scalar constant 5 5 5
Step 3 — Divergence in cylindrical coordinates
Divergence is a scalar invariant; it does not depend on the coordinate system, so its value is still 5 5 5 x = ρ cos ϕ , y = ρ sin ϕ , z = z x=\rho\cos\phi,\ y=\rho\sin\phi,\ z=z x = ρ cos ϕ , y = ρ sin ϕ , z = z
V ρ = V ⃗ ⋅ e ^ ρ , V ϕ = V ⃗ ⋅ e ^ ϕ , V z = V ⃗ ⋅ e ^ z , V_\rho=\vec V\cdot\hat e_\rho,\quad V_\phi=\vec V\cdot\hat e_\phi,\quad V_z=\vec V\cdot\hat e_z, V ρ = V ⋅ e ^ ρ , V ϕ = V ⋅ e ^ ϕ , V z = V ⋅ e ^ z , where e ^ ρ = ( cos ϕ , sin ϕ , 0 ) , e ^ ϕ = ( − sin ϕ , cos ϕ , 0 ) \hat e_\rho=(\cos\phi,\sin\phi,0),\ \hat e_\phi=(-\sin\phi,\cos\phi,0) e ^ ρ = ( cos ϕ , sin ϕ , 0 ) , e ^ ϕ = ( − sin ϕ , cos ϕ , 0 )
V ρ = ρ + ρ sin ϕ cos ϕ − z cos ϕ + ρ cos ϕ sin ϕ − z sin ϕ = ρ + ρ sin 2 ϕ − z ( cos ϕ + sin ϕ ) , V_\rho=\rho+\rho\sin\phi\cos\phi-z\cos\phi+\rho\cos\phi\sin\phi-z\sin\phi=\rho+\rho\sin2\phi-z(\cos\phi+\sin\phi), V ρ = ρ + ρ sin ϕ cos ϕ − z cos ϕ + ρ cos ϕ sin ϕ − z sin ϕ = ρ + ρ sin 2 ϕ − z ( cos ϕ + sin ϕ ) , and applying the cylindrical divergence formula
∇ ⋅ V ⃗ = 1 ρ ∂ ( ρ V ρ ) ∂ ρ + 1 ρ ∂ V ϕ ∂ ϕ + ∂ V z ∂ z \nabla\cdot\vec V=\frac1\rho\frac{\partial(\rho V_\rho)}{\partial\rho}+\frac1\rho\frac{\partial V_\phi}{\partial\phi}+\frac{\partial V_z}{\partial z} ∇ ⋅ V = ρ 1 ∂ ρ ∂ ( ρ V ρ ) + ρ 1 ∂ ϕ ∂ V ϕ + ∂ z ∂ V z reproduces the value
Answer
∇ ⋅ V ⃗ = 5. \boxed{\,\nabla\cdot\vec V=5.\,} ∇ ⋅ V = 5.