Boat on fire.

Oh, come on- if you're gonna take a problem from Irodov, at least give him credit ;)! Ok, so as for solving the problem, here we go: By minimizing drifting, they mean minimize the distance the boat is carried downriver; so, we want to minimize how far in the x direction the boat moves. If we minimize it for a given distance to travel across the river, we minimize it for the whole trip.

So the boat wants to travel a distance D (in the y-direction) across the river, and move as little as possible in the x direction (the amount that it moves we will call R). We know that the velocity must be at an angle greater than 90 degrees to the stream direction, or the boat would use some of its own motor to go downstream, which we don't want happening (the less downstream, the better). However, the boat can't go completely at 180 degrees (or something near that) because its velocity is half that of the stream's. The boat would drift less, but it would never go across, so in the long run it would just be stuck, drifting endlessly. Thus the angle is somewhere in between 90 and 180 degrees. Let's make an angle x between the velocity of the boat and the vertical; this angle plus 90 will be our answer. If the boat travels for a time t, we know that v•cosx•t=D, and (u-v•sinx)•t=R, where v is the boat velocity and u is the stream velocity. D/(v•cosx)•(u-v•sinx)=R=(D•u-D•v•sinx)/(v•cosx)=(D•2v-D•v•sinx)/(v•cosx)=(2D-D•sinx)/cosx. Taking the derivative to minimize, we get that 0=d•(2sinx-1)/cos^2(x). Then sin(x)=1/2, or x=30 degrees.

Remembering that x is only the angle with respect to the vertical, we add those extra 90 degrees to find the angle with respect to the stream's velocity. We finally get: 120 degrees. Whoooooo!

$$ It's difficult for me to explain without a picture, but I'll make as simple as possible. Let $v_b$ be velocity of the boat, $v_s$ be velocity of the stream, and $\,\theta$ be the angle between boat and stream direction. If $\,d$ denotes the width of the river, then the time taken to cross the river is $$ $t=\frac{d}{v_b \sin\theta}.$ Therefore, the drifting of the boat is $$ $\begin{aligned} s&=(v_b\cos\theta + v_s)t\\ &=(v_b\cos\theta + v_s)\frac{d}{v_b \sin\theta}\\ &=(\cot\theta + 2\csc\theta)d. \end{aligned}$ $$ To minimize drifting can be obtained by setting the first derivative of $s$ with respect to $\,\theta$ equal to zero. $$ $\begin{aligned} \frac{ds}{d\theta}&=0\\ \frac{d}{d\theta}(\cot\theta + 2\csc\theta)d&=0\\ -\csc^2\theta-2\cot\theta\csc\theta&=0\\ \csc^2\theta&=-2\cot\theta\csc\theta\\ \csc\theta&=-2\cot\theta\\ \cos\theta&=-\frac{1}{2}\\ \theta&=\boxed{120^\circ} \end{aligned}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

TAKE ANY ANGLE FROM PERPENDICULAR LINE SO TIME TAKEN TO CROSS THE RIVER IS D/VCOS ANGLE. SO DRIFT IS Vr * TIME TAKEN FOR MINIMUM VALUE dx/ dof angle is zero then we find angle from perpendicular is 30 so angle from stream is 120