For $C=1/2$,
$$
(W')^2=W^2+2W=(W+1)^2-1,
$$
so
$$
\boxed{W(Z)=\cosh Z-1,
\qquad A(Z)\propto\frac1{\cosh Z-1}.}
$$
The thread thins exponentially for large $Z$, and its tensile stress is proportional to $\sinh Z$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 329 / 2 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-329/2/a/ii/solution