Introduce the virtual-origin coordinate
$$
\zeta=z+z_0,\qquad z_0=\frac{\mathcal B_0}{\beta},
$$
so that $\mathcal B=\beta\zeta$. Seek $Q=A\zeta^{4/3}$ and $M=C\zeta^{5/3}$. Substitution gives
$$
A=\left(\frac{27E^2\beta}{80}\right)^{1/3},
\qquad
C=\left(\frac{27E\beta^2}{100}\right)^{1/3}.
$$
Therefore the exact similarity solution is
$$
\boxed{Q=A\zeta^{4/3}},\qquad
\boxed{M=C\zeta^{5/3}},
$$
$$
\boxed{W=\frac MQ=\left(\frac{4\beta}{5E}\right)^{1/3}\zeta^{1/3}},
\qquad
\boxed{b=\frac{Q^2}{M}=\frac{3E}{4}\zeta},
$$
and
$$
\boxed{T-T_0=\frac{T_0\mathcal B}{gQ}
=\frac{T_0\beta}{gA}\zeta^{-1/3}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 345 / 4 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-345/4/e/solution