The current in a straight river of constant width \(h\) flows at uniform speed \(\alpha v\) parallel to the river banks, where \(0<\alpha<1\). A boat has to cross from a point \(A\) on one bank to a point \(B\) on the other bank directly opposite to \(A\). The boat moves at constant speed \(v\) relative to the water. When the position of the boat is \((x,y)\), where \(x\) is the perpendicular distance from the opposite bank and \(y\) is the distance downstream from \(AB\), the boat is pointing in a direction which makes an angle \(\theta\) with \(AB\). Determine the velocity vector of the boat in terms of \(v,\theta\) and \(\alpha.\) The pilot of the boat steers in such a way that the boat always points exactly towards \(B\). Show that the velocity vector of the boat is \[ \begin{pmatrix}\dfrac{\mathrm{d}x}{\mathrm{d}t}\\ \tan\theta\dfrac{\mathrm{d}x}{\mathrm{d}t}+x\sec^{2}\theta\dfrac{\mathrm{d}\theta}{\mathrm{d}t} \end{pmatrix}. \] By comparing this with your previous expression deduce that \[ \alpha\frac{\mathrm{d}x}{\mathrm{d}\theta}=-x\sec\theta \] and hence show that \[ (x/h)^{\alpha}=(\sec\theta+\tan\theta)^{-1}. \] Let \(s(t)\) be a new variable defined by \(\tan\theta=\sinh(\alpha s).\) Show that \(x=h\mathrm{e}^{-s},\) and that \[ h\mathrm{e}^{-s}\cosh(\alpha s)\frac{\mathrm{d}s}{\mathrm{d}t}=v. \] Hence show that the time of crossing is \(hv^{-1}(1-\alpha^{2})^{-1}.\)