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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 ?

snulty
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Kantura
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    By the way, you say "shortest route" when you really mean "fastest route". – user247327 Jun 16 '16 at 18:31
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    @user247327 No, the fastest route is not the shortest route, and this question is clearly about the shortest route - the one which results in least downstream displacement. – Joffan Jun 16 '16 at 18:39

2 Answers2

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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 $\beta$

user340297
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To minimise the resultant downstream displacement, the velocity of the boat (relative to the fixed bank) must be perpendicular to the relative velocity, i.e. the velocity of the boat relative to the water.

Therefore you draw a right-angled triangle with the velocity of the water ($10$) as the hypotenuse. The other two sides are $8$, the speed of the boat relative to the water, and $6$ which is the actual observed speed of the boat.

The required direction of motion relative to the negative $x$ axis is therefore given by $$\cos \alpha=\frac{8}{10}=\frac 45$$

David Quinn
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    Intuitively attractive, but you need to justify why perpendicular. Also you mean the absolute velocity relative to the bank (or equivalently the river bed), not the velocity relative to the water, which is all produced by the boatman and cannot be perpendicular to itself. – Joffan Jun 16 '16 at 18:43
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    @Joffan: this is a well-known result and not guess-work on my part. The boatman steers in a direction which will be perpendicular to his resultant velocity. – David Quinn Jun 16 '16 at 20:08
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    I wasn't suggesting it was guesswork, I just think it needs justification, given that it is the more interesting part of your answer rather than the subsequent trig. – Joffan Jun 16 '16 at 20:13
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    I also would like to understand why this angle is 90 degrees. It might explain why I am having trouble with a similar question: https://mathoverflow.net/questions/393801/how-to-calculate-distance-of-shortest-path-when-there-is-a-flow-in-the-river – David Callanan May 26 '21 at 18:28