 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/12/d/solution

Write

c = cos ϕ

and

s = sin ϕ

c_{x} = s^{2} / r

s_{x} = - sc / r

c_{y} = - sc / r

and

s_{y} = c^{2} / r

. Apply the product rule to both terms of

f_{x} = c g_{r} - (s / r) g_{ϕ}

and use the chain rule for the derivatives of

g

. Collecting terms gives

f_{xx} = c^{2} g_{rr} - \frac{2 sc}{r} g_{r ϕ} + \frac{2 sc}{r ^{2}} g_{ϕ} + \frac{s ^{2}}{r} g_{r} + \frac{s ^{2}}{r ^{2}} g_{ϕϕ} .

Similarly, differentiating

f_{y} = s g_{r} + (c / r) g_{ϕ}

gives

f_{yy} = s^{2} g_{rr} + \frac{2 sc}{r} g_{r ϕ} - \frac{2 sc}{r ^{2}} g_{ϕ} + \frac{c ^{2}}{r} g_{r} + \frac{c ^{2}}{r ^{2}} g_{ϕϕ} .

On addition, both the mixed partial derivative terms and the first angular derivative terms cancel. Using

s^{2} + c^{2} = 1

\nabla^{2} f = g_{rr} + \frac{1}{r} g_{r} + \frac{1}{r ^{2}} g_{ϕϕ}, r > 0.

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