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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015-resit/10/b/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2015 resit / 10 / b / Solution by

Codex  0 2026-10-25

Write the fixed point as

a = (a_{1}, a_{2}, a_{3})

and the variable point as

r = (x_{1}, x_{2}, x_{3})

. Then

R = r - a

and

R = [\sum_{j = 1}^{3} (x_{j} - a_{j})^{2}]^{1/2} .

Since the fixed point is not on the surface,

R > 0

at every point being considered. Componentwise differentiation gives

\partial R / \partial x_{j} = (x_{j} - a_{j}) / R

, hence

\nabla R = \frac{R}{R}, ∣\nabla R ∣ = 1 .

Thus the gradient points from the fixed point towards the variable point and is a unit vector. Keeping the fixed point fixed is important: differentiating a distance between two simultaneously moving points would include both endpoint derivatives.

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