 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-3/9a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

x = (r sin θ cos ϕ, r sin θ sin ϕ, r cos θ)

gives the orthogonal decomposition

d x = e_{r} d r + r e_{θ} d θ + r sin θ e_{ϕ} d ϕ,

where

e_{r} e_{θ} e_{ϕ} = (sin θ cos ϕ, sin θ sin ϕ, cos θ), = (cos θ cos ϕ, cos θ sin ϕ, - sin θ), = (- sin ϕ, cos ϕ, 0) .

h_{r} = 1, h_{θ} = r, h_{ϕ} = r sin θ .

Since the total differential satisfies

df = f_{r} d r + f_{θ} d θ + f_{ϕ} d ϕ = d x \cdot \nabla f,

comparison of coefficients gives the gradient

\nabla f = e_{r} \frac{\partial f}{\partial r} + e_{θ} \frac{1}{r} \frac{\partial f}{\partial θ} + e_{ϕ} \frac{1}{r sin θ} \frac{\partial f}{\partial ϕ} .

 New to topics? Read the docs here!