← 2023 Paper 1
UPSC 2023 Maths Optional Paper 1 Q8c — Step-by-Step Solution 15 marks · Section B
Vector identities (curl of grad, div of curl, product rules) · Vector Analysis · asked 3× in 13 yrs · Read the full method →
Question
For a scalar point function ϕ \phi ϕ f ⃗ \vec f f ∇ ⋅ ( ϕ f ⃗ ) = ∇ ϕ ⋅ f ⃗ + ϕ ( ∇ ⋅ f ⃗ ) \nabla\cdot(\phi\vec f)=\nabla\phi\cdot\vec f+\phi(\nabla\cdot\vec f) ∇ ⋅ ( ϕ f ) = ∇ ϕ ⋅ f + ϕ ( ∇ ⋅ f ) ∇ ⋅ ( f ( r ) r r ⃗ ) \nabla\cdot\Bigl(\dfrac{f(r)}{r}\vec r\Bigr) ∇ ⋅ ( r f ( r ) r )
Technique
(1) Direct expansion in Cartesian coordinates for the identity. (2) Apply identity with ϕ = f ( r ) / r , f ⃗ = r ⃗ \phi=f(r)/r,\,\vec f=\vec r ϕ = f ( r ) / r , f = r ∇ ( f ( r ) ) = f ′ ( r ) r ^ \nabla(f(r))=f'(r)\hat r ∇ ( f ( r )) = f ′ ( r ) r ^ ∇ ⋅ r ⃗ = 3 \nabla\cdot\vec r=3 ∇ ⋅ r = 3
Solution
Part 1 — Prove the identity
In Cartesian coordinates, f ⃗ = ( f 1 , f 2 , f 3 ) \vec f=(f_1,f_2,f_3) f = ( f 1 , f 2 , f 3 )
∇ ⋅ ( ϕ f ⃗ ) = ∑ i = 1 3 ∂ ( ϕ f i ) ∂ x i = ∑ i = 1 3 ( f i ∂ ϕ ∂ x i + ϕ ∂ f i ∂ x i ) . \nabla\cdot(\phi\vec f)=\sum_{i=1}^3\frac{\partial(\phi f_i)}{\partial x_i}=\sum_{i=1}^3\Bigl(f_i\frac{\partial\phi}{\partial x_i}+\phi\frac{\partial f_i}{\partial x_i}\Bigr). ∇ ⋅ ( ϕ f ) = i = 1 ∑ 3 ∂ x i ∂ ( ϕ f i ) = i = 1 ∑ 3 ( f i ∂ x i ∂ ϕ + ϕ ∂ x i ∂ f i ) . Group:
= ∑ i f i ∂ ϕ ∂ x i + ϕ ∑ i ∂ f i ∂ x i = ∇ ϕ ⋅ f ⃗ + ϕ ( ∇ ⋅ f ⃗ ) . =\sum_i f_i\frac{\partial\phi}{\partial x_i}+\phi\sum_i\frac{\partial f_i}{\partial x_i}=\nabla\phi\cdot\vec f+\phi(\nabla\cdot\vec f). = i ∑ f i ∂ x i ∂ ϕ + ϕ i ∑ ∂ x i ∂ f i = ∇ ϕ ⋅ f + ϕ ( ∇ ⋅ f ) . ∇ ⋅ ( ϕ f ⃗ ) = ∇ ϕ ⋅ f ⃗ + ϕ ( ∇ ⋅ f ⃗ ) . \boxed{\;\nabla\cdot(\phi\vec f)=\nabla\phi\cdot\vec f+\phi(\nabla\cdot\vec f).\;} ∇ ⋅ ( ϕ f ) = ∇ ϕ ⋅ f + ϕ ( ∇ ⋅ f ) . Part 2 — Compute ∇ ⋅ ( f ( r ) r r ⃗ ) \nabla\cdot\bigl(\tfrac{f(r)}{r}\vec r\bigr) ∇ ⋅ ( r f ( r ) r )
Apply the identity with ϕ = f ( r ) / r \phi=f(r)/r ϕ = f ( r ) / r f ⃗ = r ⃗ \vec f=\vec r f = r
∇ ⋅ ( f ( r ) r r ⃗ ) = ∇ ( f ( r ) r ) ⋅ r ⃗ + f ( r ) r ( ∇ ⋅ r ⃗ ) . \nabla\cdot\Bigl(\frac{f(r)}{r}\vec r\Bigr)=\nabla\Bigl(\frac{f(r)}{r}\Bigr)\cdot\vec r+\frac{f(r)}{r}(\nabla\cdot\vec r). ∇ ⋅ ( r f ( r ) r ) = ∇ ( r f ( r ) ) ⋅ r + r f ( r ) ( ∇ ⋅ r ) . Compute ∇ ( f ( r ) / r ) \nabla(f(r)/r) ∇ ( f ( r ) / r ) Let g ( r ) = f ( r ) / r g(r)=f(r)/r g ( r ) = f ( r ) / r ∇ g = g ′ ( r ) ∇ r = g ′ ( r ) r ^ = g ′ ( r ) ( r ⃗ / r ) \nabla g=g'(r)\nabla r=g'(r)\hat r=g'(r)(\vec r/r) ∇ g = g ′ ( r ) ∇ r = g ′ ( r ) r ^ = g ′ ( r ) ( r / r )
g ′ ( r ) = f ′ ( r ) ⋅ r − f ( r ) ⋅ 1 r 2 = r f ′ ( r ) − f ( r ) r 2 g'(r)=\dfrac{f'(r)\cdot r-f(r)\cdot 1}{r^2}=\dfrac{rf'(r)-f(r)}{r^2} g ′ ( r ) = r 2 f ′ ( r ) ⋅ r − f ( r ) ⋅ 1 = r 2 r f ′ ( r ) − f ( r )
So ∇ g = r f ′ ( r ) − f ( r ) r 2 ⋅ r ⃗ r = ( r f ′ ( r ) − f ( r ) ) r ⃗ r 3 \nabla g=\dfrac{rf'(r)-f(r)}{r^2}\cdot\dfrac{\vec r}{r}=\dfrac{(rf'(r)-f(r))\vec r}{r^3} ∇ g = r 2 r f ′ ( r ) − f ( r ) ⋅ r r = r 3 ( r f ′ ( r ) − f ( r )) r
Compute ∇ ⋅ r ⃗ \nabla\cdot\vec r ∇ ⋅ r ∇ ⋅ r ⃗ = ∂ x / ∂ x + ∂ y / ∂ y + ∂ z / ∂ z = 3 \nabla\cdot\vec r=\partial x/\partial x+\partial y/\partial y+\partial z/\partial z=3 ∇ ⋅ r = ∂ x / ∂ x + ∂ y / ∂ y + ∂ z / ∂ z = 3
Substitute:
∇ g ⋅ r ⃗ = ( r f ′ ( r ) − f ( r ) ) r ⃗ ⋅ r ⃗ r 3 = ( r f ′ ( r ) − f ( r ) ) r 2 r 3 = r f ′ ( r ) − f ( r ) r . \nabla g\cdot\vec r=\frac{(rf'(r)-f(r))\vec r\cdot\vec r}{r^3}=\frac{(rf'(r)-f(r))\,r^2}{r^3}=\frac{rf'(r)-f(r)}{r}. ∇ g ⋅ r = r 3 ( r f ′ ( r ) − f ( r )) r ⋅ r = r 3 ( r f ′ ( r ) − f ( r )) r 2 = r r f ′ ( r ) − f ( r ) . f ( r ) r ( ∇ ⋅ r ⃗ ) = 3 f ( r ) r . \dfrac{f(r)}{r}(\nabla\cdot\vec r)=\dfrac{3f(r)}{r}. r f ( r ) ( ∇ ⋅ r ) = r 3 f ( r ) .
Sum:
∇ ⋅ ( f ( r ) r r ⃗ ) = r f ′ ( r ) − f ( r ) r + 3 f ( r ) r = r f ′ ( r ) + 2 f ( r ) r = f ′ ( r ) + 2 f ( r ) r . \nabla\cdot\Bigl(\frac{f(r)}{r}\vec r\Bigr)=\frac{rf'(r)-f(r)}{r}+\frac{3f(r)}{r}=\frac{rf'(r)+2f(r)}{r}=f'(r)+\frac{2f(r)}{r}. ∇ ⋅ ( r f ( r ) r ) = r r f ′ ( r ) − f ( r ) + r 3 f ( r ) = r r f ′ ( r ) + 2 f ( r ) = f ′ ( r ) + r 2 f ( r ) . Answer
∇ ⋅ ( f ( r ) r r ⃗ ) = f ′ ( r ) + 2 f ( r ) r . \boxed{\;\nabla\cdot\Bigl(\frac{f(r)}{r}\vec r\Bigr)=f'(r)+\frac{2f(r)}{r}.\;} ∇ ⋅ ( r f ( r ) r ) = f ′ ( r ) + r 2 f ( r ) .