Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 19

Apply the multivariable chain rule to an arbitrary differentiable scalar function of $x,y,z$. With the subscripts indicating the coordinates held fixed, the first operator isHolding $W,\varphi$ fixed when differentiating with respect to $\theta$ givesFinally, holding $W,\theta$ fixed givesFor example, the coefficient of ${\partial }_x$ in ${\partial }_{\theta }$ is precisely $(\partial x/\partial \theta {)}_{W,\varphi }$; all coefficients follow by differentiating the given coordinate functions.

Suppress the held-fixed subscripts only within this matrix display. Collecting the preceding operator coefficients givesIf $J=\partial (x,y,z)/\partial (W,\theta ,\varphi )$ uses the usual convention of coordinate functions as rows, the displayed coefficient matrix is $J^{\mathsf{T}}$. Its determinant is $W^2{\sec }^2\theta \tan \theta >0$ on the given domain, so the inverse operator transformation exists everywhere there.

For an independent inversion, use the coordinate relationsThe positive-octant domain fixes their angle branches unambiguously. Let $s=\sqrt{x^2+y^2}=W\tan \theta$. Differentiating at fixed Cartesian coordinates givesand ${\varphi }_x=-y/s^2=-\sin \varphi /(W\tan \theta )$, ${\varphi }_y=x/s^2=\cos \varphi /(W\tan \theta )$, ${\varphi }_z=0$. Hence the inverse chain rule givesandMultiplying 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.