Irodov 1.9: Aiming Upstream to Minimize Drift – Amit & Deepali Sinha Foundation

Problem Irodov 1.9

A river flows with speed $u$ . A boat’s speed relative to the water, $v$ , is smaller than the flow: $v = u/n$ with $n = 2.0$ . The boat holds a constant heading, at some angle $\theta$ to the stream direction, while it crosses from one bank to the other. Find the heading angle $\theta$ that minimizes the boat’s downstream drift, the distance along the bank between the launch point and where the boat actually reaches the far side.

The velocity parallelogram at an arbitrary heading: the stream carries the boat at u, the boat adds v at angle θ to it, and the two sum to the ground velocity V. The drawn lengths hold the given ratio v=u/n. — The velocity parallelogram at an arbitrary heading: the stream carries the boat at $\vect u$ , the boat adds $\vect v$ at angle $\theta$ to it, and the two sum to the ground velocity $\vect V$ . The drawn lengths hold the given ratio $v=u/n$ .

Given

$v = u/n$ , $n = 2.0$

Find

the heading angle

\theta

that minimizes the drift

Solution

Step 1Resolve the ground velocity along and across the stream

Taking $x$ along the stream and $y$ across it, with $\vect u$ purely along $x$ and $\vect v$ at angle $\theta$ to it,

\begin{aligned} V_x &= u+v\cos\theta\\ V_y &= v\sin\theta \end{aligned} \tag{1}

Step 2Write the drift as a function of the heading

For a crossing of width $d$ , the across-stream component sets the crossing time, $t=d/V_y$ , and the along-stream component carries the boat downstream by $V_x t$ during that same time.

x(\theta) = V_x t

Substituting the components (1) gives the drift explicitly as a function of the heading.

x(\theta) = \dfrac{d\,(u+v\cos\theta)}{v\sin\theta}

(2)

Step 3Minimize the drift over the heading

Treating $d$ and $v$ as fixed and differentiating (2) by the quotient rule,

\begin{align*}\frac{dx}{d\theta} &= \frac{d}{v}\cdot\frac{(-v\sin\theta)(\sin\theta)-(u+v\cos\theta)(\cos\theta)}{\sin^2\theta}\\ &= \frac{d}{v}\cdot\frac{-v(\sin^2\theta+\cos^2\theta)-u\cos\theta}{\sin^2\theta}\\ &= \frac{d}{v}\cdot\frac{-v-u\cos\theta}{\sin^2\theta}\end{align*}

Setting $dx/d\theta=0$ leaves the numerator’s bracket, and solving it for $\cos\theta$ gives the extremum condition. Since $x(\theta)\to+\infty$ at both $\theta\to0^+$ and $\theta\to\pi^-$ (the crossing time blows up as $V_y\to0$ ) while $x$ stays finite in between, this single interior stationary point is the minimum.

-v-u\cos\theta = 0

\cos\theta = -\dfrac{v}{u}

(3)

Step 4Substitute the given speed ratio

With $v=u/n$ from the given data, $v/u=1/n$ , so (3) becomes $\cos\theta=-1/n$ . Writing $\theta=\pi/2+\varphi$ and using $\cos(\pi/2+\varphi)=-\sin\varphi$ turns this into a statement about $\varphi$ alone.

\begin{align*}\cos\theta &= -\frac1n\\ \cos\!\left(\frac{\pi}{2}+\varphi\right) &= -\frac1n\\ -\sin\varphi &= -\frac1n \;\Longrightarrow\; \varphi = \arcsin\frac1n\end{align*}

With $n=2.0$ the argument $1/n=0.50$ , and evaluating the arcsine gives the numerical heading.

\varphi = \arcsin(0.50) = 30^\circ

\theta = \dfrac{\pi}{2}+\arcsin\dfrac1n = 90^\circ+30^\circ = 120^\circ

(4)

The same parallelogram redrawn at the heading just found, θ = 120°. Posed this way the resultant V comes out perpendicular to the boat's own v, which is the geometric form of the condition checked below. — The same parallelogram redrawn at the heading just found, $\theta = 120^\circ$ . Posed this way the resultant $\vect V$ comes out perpendicular to the boat’s own $\vect v$ , which is the geometric form of the condition checked below.

Check

At the optimum, $\vect V\cdot\vect v=(\vect u+\vect v)\cdot\vect v=uv\cos\theta+v^2=uv(-v/u)+v^2=0$ : the resultant ground velocity is exactly perpendicular to the boat’s own heading, an independent geometric statement of the same condition $\cos\theta=-v/u$ found above by calculus.

Remark

The perpendicularity $\vect V\perp\vect v$ found in the check is the solution to a genuinely different optimization: if the resultant’s direction were fixed instead (the boat must land along some prescribed bearing) and only its own speed $|\vect v|$ were free to minimize, the smallest $|\vect v|$ achieving that exact resultant direction is again the one perpendicular to $\vect V$ , this time because the shortest vector from a fixed point to a line meets that line at a right angle.