 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/20/b/solution

On a smooth implicit branch, put

Q = x + z^{2}

and require

Q \neq = 0

. This is exactly the implicit function theorem condition, since the defining equation has vertical partial derivative

3 Q

. Differentiating the cubic relation in

x

and

y

gives

Q z_{x} = - z, Q z_{y} = 1, z_{x} = - \frac{z}{Q}, z_{y} = \frac{1}{Q} .

For a second derivative, the dependence of

Q

z

must also be included:

Q_{x} = 1 + 2 z z_{x}

and

Q_{y} = 2 z z_{y}

. Thus

z_{xx} z_{yy} z_{x y} = - \frac{z _{x}}{Q} + \frac{z ( 1 + 2 z z _{x} )}{Q ^{2}} = \frac{2 x z}{Q ^{3}}, = - \frac{2 z z _{y}}{Q ^{2}} = - \frac{2 z}{Q ^{3}}, = - \frac{z _{y}}{Q} + \frac{2 z ^{2} z _{y}}{Q ^{2}} = \frac{z ^{2} - x}{Q ^{3}} .

The required cancellation follows immediately:

z_{xx} + x z_{yy} = \frac{2 x z}{( x + z ^{2} ) ^{3}} - \frac{2 x z}{( x + z ^{2} ) ^{3}} = 0.

The verification is local to smooth branches with

x + z^{2} \neq = 0

; the implicit relation does not guarantee a differentiable function at its singular points.

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