2016 Paper 1 Question 12

(a) With

$u = x^2-y^2, \; v = 2xy$

We can think of $f$ as

$f=f(u,v)$

Then

$\begin{aligned} \frac{\partial f}{\partial x} &= \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} \\ &= \frac{\partial f}{\partial u}\, 2x + \frac{\partial f}{\partial v}\, 2y \end{aligned}$

and

$\begin{aligned} \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u}(-2y) + \frac{\partial f}{\partial v}(2x) \end{aligned}$

Now substitute in to the given equation

$\begin{aligned} y\frac{\partial f}{\partial x} + x\frac{\partial f}{\partial y} &= y 2x \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} 2y^2 + (-2xy)\frac{\partial f}{\partial u} + 2x^2\frac{\partial f}{\partial v} \\ &= 2 (x^2+y^2) \frac{\partial f}{\partial v} \\ &= 0 \end{aligned}$

So for general $x,y$

$\frac{\partial f}{\partial v} = 0$

This means $f$ is independent of $v$ and is a function of $u$ alone. Thus

$f = f(u) = f(x^2-y^2)$

QED

NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

2016 Paper 1 Question 12

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