 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/19/c/solution

For an independent inversion, use the coordinate relations

W = z, θ = arctan \frac{x ^{2} + y ^{2}}{z}, ϕ = arctan \frac{y}{x} .

The positive-octant domain fixes their angle branches unambiguously. Let

s = x^{2} + y^{2} = W tan θ

. Differentiating at fixed Cartesian coordinates gives

θ_{x} = \frac{z x}{s ( z ^{2} + s ^{2} )} = \frac{c o s ^{2} θ c o s ϕ}{W}, θ_{y} = \frac{c o s ^{2} θ s i n ϕ}{W}, θ_{z} = - \frac{s i n θ c o s θ}{W},

and

ϕ_{x} = - y / s^{2} = - sin ϕ / (W tan θ)

ϕ_{y} = x / s^{2} = cos ϕ / (W tan θ)

ϕ_{z} = 0

. Hence the inverse chain rule gives

(\partial_{x})_{y, z} = \frac{cos ^{2} θ cos ϕ}{W} (\partial_{θ})_{W, ϕ} - \frac{sin ϕ}{W tan θ} (\partial_{ϕ})_{W, θ},

(\partial_{y})_{x, z} = \frac{cos ^{2} θ sin ϕ}{W} (\partial_{θ})_{W, ϕ} + \frac{cos ϕ}{W tan θ} (\partial_{ϕ})_{W, θ},

and

(\partial_{z})_{x, y} = (\partial_{W})_{θ, ϕ} - \frac{sin θ cos θ}{W} (\partial_{θ})_{W, ϕ} .

Multiplying these coefficients by the forward coefficient matrix gives the identity. The original PDF correctly holds

x, y

fixed in the final derivative; the transcription's

x, z

subscript there is a typo.

 New to topics? Read the docs here!