← 2013 Paper 1

UPSC 2013 Maths Optional Paper 1 Q8a — Step-by-Step Solution

10 marks · Section B

Higher order derivatives; Laplacian · Vector Analysis · asked 2× in 14 yrs · Read the full method →

Question

Calculate ∇2(rn)\nabla^{2}(r^{n}) and find its expression in terms of rr and nn, rr being the distance of any point (x,y,z)(x,y,z) from the origin, nn being a constant and ∇2\nabla^{2} being the Laplace operator.

Technique

Use the radial Laplacian formula for spherically symmetric functions.

Solution

Strategy. rnr^{n} is a function of rr only (spherically symmetric). For any f(r)f(r) in 3D:

∇2f(r)=f′′(r)+2rf′(r).\nabla^{2}f(r)=f''(r)+\frac{2}{r}f'(r).

(This is the radial Laplacian; comes from ∇2=1r2∂∂r ⁣(r2∂∂r)\nabla^{2}=\dfrac{1}{r^{2}}\dfrac{\partial}{\partial r}\!\left(r^{2}\dfrac{\partial}{\partial r}\right) in spherical coordinates.)

Step 1 — Compute derivatives of f(r)=rnf(r)=r^{n}

f′(r)=nrn−1f'(r)=nr^{n-1}, f′′(r)=n(n−1)rn−2f''(r)=n(n-1)r^{n-2}.

Step 2 — Combine

∇2(rn)=n(n−1)rn−2+2r⋅nrn−1=n(n−1)rn−2+2nrn−2=nrn−2[(n−1)+2]=n(n+1)rn−2.\nabla^{2}(r^{n})=n(n-1)r^{n-2}+\frac{2}{r}\cdot nr^{n-1}=n(n-1)r^{n-2}+2nr^{n-2}=nr^{n-2}\bigl[(n-1)+2\bigr]=n(n+1)r^{n-2}.

Answer

  ∇2(rn)=n(n+1) rn−2.  \boxed{\;\nabla^{2}(r^{n})=n(n+1)\,r^{n-2}.\;}
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