Take a river $100m$ wide flowing at $10m/s$ ($x$-axis). Take a boat which can go at $8.0m/s$. Clearly this boat cannot go straight across the river without being dragged some distance downstream. What angle $\alpha$ ( relative to the negative $x$-axis) should the boatsperson steer in order to minimise the resultant displacement downstream, ie: maximise the angle $\beta$ (relative to the positive $x$-axis) ?
All velocities are in $ms^{-1}$.
Velocity of river: $$\vec{V}_r=10\vec{i}+0\vec{j}$$
Velocity of boat: $$\vec{V}_b=(-8.0\cos\alpha)\vec{i}+(8.0\sin\alpha)\vec{j}$$
Relative velocity of boat: $$\vec{V}_R=(10-8.0\cos\alpha)\vec{i}+(8.0\sin\alpha)\vec{j}$$
Direction of $\vec{V}_R$: $$\tan\beta=\frac{8.0\sin\alpha}{10-8.0\cos\alpha}$$
So assuming this equation correctly relates $\alpha$ to $\beta$ for given velocities for the river and the boat, what angle of $\alpha$ gives a maximum value for $\beta$ ?
My thoughts are that one should get and expression for $\beta$ in terms of $\alpha$. Differentiate to get $\frac{d\beta}{d\alpha}$. Then solve for $\beta$ where $\frac{d\beta}{d\alpha}=0$. But I am not sure how to differentiate such an expression. Where does one go from here ?
Best Answer
It's clear that $\beta<\pi/2$ because boat cannot go "upstream" or straight across. So $0<\beta<\pi/2$. Now in this range $\tan$ incerases as $\beta$ increases, so to maximize $\beta$ it's enough to maximize $\tan \beta$. Now you can just compute $\frac{d\tan\beta}{d\alpha}$ using quotien rule or whatever you need and find the root.
Actually, resulting derivative has rather nice numerator and you probably get $\cos \alpha = 4/5$ for maximum $\beta$