Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 14

For the force magnitude in Newton's law of universal gravitation, the total differential isDividing by the positive value $F$ gives the particularly useful relative formFor finite small changes this gives the first-order approximation to the fractional change; higher-order products of increments are omitted.

Set $dF=0$ in the relative total differential and use the two equal $1\%$ mass increases:Thus the separation must increase by $1\%$. In this special case the result is also exact: the mass product increases by $1.01^2$, and preserving the inverse-square force requires $r_{\mathrm{new}}^2=1.01^2{r}^2$, hence $r_{\mathrm{new}}=1.01r$.

The original PDF specifies evaluation at $(1,1,1)$; the local TeX incorrectly gives $(1,1,0)$. Write the scalar field explicitly as $f=c/(x^2+y^2+z^2)$. Its gradient isA directional derivative measures rate per unit distance, so normalize the direction to the unit vector $\mathbf{n}=(1,1,0)/\sqrt{2}$. ThenUsing the unnormalized direction vector would instead differentiate with respect to a path parameter, not distance, and would give a different scale.

The polar-coordinate Jacobian matrix and its Jacobian determinant areFor $r>0$, the corresponding area element is $dxdy=rdrd\theta$. The determinant vanishes at $r=0$, where the polar angle does not give a unique coordinate representation.

Put $D={\sigma }^2>0$ and $\xi =x-x_0$. Logarithmic differentiation of the positive Gaussian function givesDifferentiating once more in space givesMultiplying by $D$ provesThe assumption $\sigma \ne 0$ ensures both the normalization and exponential denominator are defined. This is the heat kernel, with diffusion coefficient $D$ and centre $x_0$.