- Asked 6/16
- 2073 Magh · 5 marks
Gradient of (a·r)/rⁿ
Short answer
Apply product rule: ∇((a·r) r^-n) = (a·r)(-n r^{-n-2} r) + r^{-n} a = a/r^n - (n/r^{n+2})(a·r) r.
If r be the position vector and a is a constant vector then prove that ∇(a·r / rⁿ) = a/rⁿ − (n/r^(n+2)) (a·r) r.
Let r = xi + yj + zk, r = |r|, and a a constant vector. We use ∇r = r/r (so ∇(rᵐ) = m r^(m−2) r) and ∇(a · r) = a.
Product rule
Write a · r / rⁿ = (a · r) · r⁻ⁿ and use ∇(uv) = u∇v + v∇u:
∇[(a · r) r⁻ⁿ] = (a · r) ∇(r⁻ⁿ) + r⁻ⁿ ∇(a · r)
The two gradients
- ∇(r⁻ⁿ) = −n r^(−n−1) ∇r = −n r^(−n−1) (r/r) = −n r^(−n−2) r.
- ∇(a · r) = ∇(a₁x + a₂y + a₃z) = a₁i + a₂j + a₃k = a.
Combine
∇(a·r / rⁿ) = (a · r)(−n r^(−n−2) r) + r⁻ⁿ a
= a/rⁿ − (n / r^(n+2)) (a · r) r
Hence ∇(a·r / rⁿ) = a/rⁿ − (n/r^(n+2)) (a · r) r.
✓ Answer checked by Claude Opus 5.5 (high effort)
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