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

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

Solve the two algebraic constraints for the squared coordinates:

x^{2} = \frac{z ^{2} + u ^{2}}{3}, y^{2} = \frac{z ^{2} - 2 u ^{2}}{3} .

On any nonsingular real branch, implicit differentiation at fixed

u

gives

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

and

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

. Differentiating

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

with the other original coordinate fixed gives

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

and

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

. Therefore

(\frac{\partial x}{\partial z})_{u} (\frac{\partial z}{\partial x})_{y} + (\frac{\partial y}{\partial z})_{u} (\frac{\partial z}{\partial y})_{x} = \frac{z}{3 x} \frac{2 x}{z} + \frac{z}{3 y} \frac{y}{z} = \frac{2}{3} + \frac{1}{3} = 1.

This is the multivariable chain rule applied to

z (x (z, u), y (z, u)) = z

. These formulas assume the relevant denominators are nonzero and the chosen variables are valid local coordinates.

ii

Solution

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

(\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.

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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