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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2/9/a/ii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 9 / a / ii / Solution by

Codex  0 2026-10-20

At fixed

z

, differentiating

2 x^{2} + y^{2} = z^{2}

gives

(\partial x / \partial y)_{z} = - y / (2 x)

. Also,

x^{2} = (z^{2} + u^{2}) /3

gives

(\partial x / \partial u)_{z} = u / (3 x)

. At fixed

y

, eliminate

x^{2}

to write

u^{2} = (z^{2} - 3 y^{2}) /2

, giving

(\partial u / \partial z)_{y} = z / (2 u)

. Finally, the previous part gave

(\partial y / \partial z)_{u} = z / (3 y)

. Thus

(\frac{\partial x}{\partial y})_{z} (\frac{\partial y}{\partial z})_{u} + (\frac{\partial x}{\partial u})_{z} (\frac{\partial u}{\partial z})_{y} = - \frac{y}{2 x} \frac{z}{3 y} + \frac{u}{3 x} \frac{z}{2 u} = 0.

(1)

The held-fixed subscripts matter: substituting a reciprocal partial derivative while silently changing what is held fixed would not preserve these cancellations.

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